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A-Level Chemistry · ⁨A-Level 化学⁩

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A-Level Chemistry (9701) is really three subjects that must talk to each other: physical (energy, equilibrium, rates, electrochemistry), inorganic (periodicity, Group 2, Group 17, transition elements) and organic, which grows from alkanes into a long chain of mechanisms and ends with NMR and mass spectrometry.

Most lost marks here are not from hard chemistry. They come from missing conditions, unbalanced equations, no state symbols, and organic answers that give a product but not the mechanism arrows the scheme pays for.

Organic needs a different kind of revision: draw one reaction map by hand and keep adding to it. The exam asks you to get from one functional group to another, not to recite a single reaction.

  • 1

    Atomic structure · ⁨原子構造⁩

    Watch lesson · ⁨レッスンを視聴⁩
    1.1

    What an atom is made of

    Syllabus · ⁨シラバス⁩
    English
    1. understand that atoms are mostly empty space surrounding a very small, dense nucleus that contains protons and neutrons; electrons are found in shells in the empty space around the nucleus
    2. identify and describe protons, neutrons and electrons in terms of their relative charges and relative masses
    3. understand the terms atomic and proton number; mass and nucleon number
    4. describe the distribution of mass and charge within an atom
    5. describe the behaviour of beams of protons, neutrons and electrons moving at the same velocity in an electric field
    6. determine the numbers of protons, neutrons and electrons present in both atoms and ions given atomic or proton number, mass or nucleon number and charge
    7. state and explain qualitatively the variations in atomic radius and ionic radius across a period and down a group
    日本語
    1. 原子は非常に小さく密度の高い核を取り囲む大部分が空洞であることを理解すること;陽子と中性子が含まれる核の周囲の空洞部分に電子が電子殻で存在することを理解する
    2. 相対電荷および相対質量に基づき、陽子、中性子、電子を同定・記述する
    3. 原子番号および质子数、質量数および核子数の用語を理解する
    4. 原子内の質量と電荷の分布について記述する
    5. 電気場内を同じ速度で移動する陽子、中性子、電子のビームの挙動について記述する
    6. 原子番号または质子数、質量数または核子数、および電荷を与えられたとき、原子およびイオンに含まれる陽子、中性子、電子の数を求める
    7. 周期および族に沿った原子半径およびイオン半径の変化を記述し、定性的に説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Everything is made of atoms 原子. An atom is mostly empty space. At its centre is a tiny, heavy nucleus 原子核. The nucleus holds two kinds of particle: protons 质子 and neutrons 中子. Around the nucleus, in the empty space, move the electrons 电子. The electrons stay in shells 壳层 — layers at set distances from the nucleus.

    The nucleus is very small but holds almost all the mass. The electrons take up almost all the space but have almost no mass.

    Relative charge and relative mass

    We compare the three particles using relative charge 相对电荷 and relative mass 相对质量. These are simple numbers, not real units.

    Particle Relative charge Relative mass
    proton $+1$ $1$
    neutron $0$ $1$
    electron $-1$ $\tfrac{1}{1836}$ (about $0$)

    A proton and a neutron have almost the same mass. An electron is about 1836 times lighter. The proton is positive, the electron is negative, and the neutron has no charge — it is neutral 中性.

    Proton number and nucleon number

    Two numbers describe the nucleus:

    • the proton number 质子数 (also called the atomic number 原子序数), symbol $Z$ — the number of protons.
    • the nucleon number 核子数 (also called the mass number 质量数), symbol $A$ — the total number of protons and neutrons. Protons and neutrons are both nucleons 核子.

    So the number of neutrons is $A - Z$.

    Counting particles in an atom or ion

    For a neutral atom, the number of electrons equals the number of protons, which equals $Z$.

    An ion 离子 is an atom that has lost or gained electrons, so it has a charge:

    • a positive ion has fewer electrons than protons.
    • a negative ion has more electrons than protons.

    Example: $^{27}_{13}\text{Al}^{3+}$ has $13$ protons, $27 - 13 = 14$ neutrons, and $13 - 3 = 10$ electrons (it lost 3 electrons to become $3+$).

    How mass and charge are spread out

    Almost all the mass sits in the nucleus, because protons and neutrons are heavy and electrons are very light. All the positive charge is in the nucleus (the protons). The negative charge is spread out in the shells (the electrons).

    Beams of particles in an electric field

    Imagine beams of protons, neutrons and electrons moving at the same speed into an electric field 电场 between two charged plates:

    • the proton beam bends towards the negative plate (protons are positive).
    • the electron beam bends the other way, towards the positive plate. It bends much more, because the electron is far lighter — the same force gives a bigger deflection 偏转 to a smaller mass.
    • the neutron beam goes straight through. It has no charge, so the field gives it no force.

    Atomic radius and ionic radius

    The atomic radius 原子半径 is the size of an atom. The ionic radius 离子半径 is the size of an ion.

    Across a period 周期 (left to right), the atomic radius gets smaller. The nuclear charge 核电荷 (the pull from the protons) rises, but the electrons go into the same outer shell, so the shielding 屏蔽 by inner shells stays about the same. The stronger pull draws the outer shell inwards.

    Down a group 族 (top to bottom), the atomic radius gets larger. Each step down adds a new shell, so the outer electrons are further out and feel more shielding from the nucleus.

    For ions:

    • a positive ion (cation 阳离子) is smaller than its atom. It has lost its outer shell, and the electrons that remain feel a stronger pull each.
    • a negative ion (anion 阴离子) is larger than its atom. It has gained electrons, so there is more repulsion 排斥 between the electrons.
    • among ions that have the same number of electrons, the one with more protons is smaller.
    日本語
    A scanning tunnelling microscope image of atoms
    A scanning tunnelling microscope can image individual atoms.

    Everything is made of atoms 原子. An atom is mostly empty space. At its centre is a tiny, heavy nucleus 原子核. The nucleus holds two kinds of particle: protons 质子 and neutrons 中子. Around the nucleus, in the empty space, move the electrons 电子. The electrons stay in shells 壳层 — layers at set distances from the nucleus.

    The nucleus is very small but holds almost all the mass. The electrons take up almost all the space but have almost no mass.

    An atom with a central nucleus of protons and neutrons, surrounded by electrons in two shells
    An atom is mostly empty space: protons and neutrons sit in the tiny central nucleus, while electrons move in shells around it

    Relative charge and relative mass

    We compare the three particles using relative charge 相对电荷 and relative mass 相对质量. These are simple numbers, not real units.

    Particle Relative charge Relative mass
    proton $+1$ $1$
    neutron $0$ $1$
    electron $-1$ $\tfrac{1}{1836}$ (about $0$)

    A proton and a neutron have almost the same mass. An electron is about 1836 times lighter. The proton is positive, the electron is negative, and the neutron has no charge — it is neutral 中性.

    Proton number and nucleon number

    Two numbers describe the nucleus:

    • the proton number 质子数 (also called the atomic number 原子序数), symbol $Z$ — the number of protons.
    • the nucleon number 核子数 (also called the mass number 质量数), symbol $A$ — the total number of protons and neutrons. Protons and neutrons are both nucleons 核子.

    So the number of neutrons is $A - Z$.

    Counting particles in an atom or ion

    For a neutral atom, the number of electrons equals the number of protons, which equals $Z$.

    An ion 离子 is an atom that has lost or gained electrons, so it has a charge:

    • a positive ion has fewer electrons than protons.
    • a negative ion has more electrons than protons.

    Example: $^{27}_{13}\text{Al}^{3+}$ has $13$ protons, $27 - 13 = 14$ neutrons, and $13 - 3 = 10$ electrons (it lost 3 electrons to become $3+$).

    How mass and charge are spread out

    Almost all the mass sits in the nucleus, because protons and neutrons are heavy and electrons are very light. All the positive charge is in the nucleus (the protons). The negative charge is spread out in the shells (the electrons).

    Beams of particles in an electric field

    Imagine beams of protons, neutrons and electrons moving at the same speed into an electric field 电场 between two charged plates:

    • the proton beam bends towards the negative plate (protons are positive).
    • the electron beam bends the other way, towards the positive plate. It bends much more, because the electron is far lighter — the same force gives a bigger deflection 偏转 to a smaller mass.
    • the neutron beam goes straight through. It has no charge, so the field gives it no force.
    Three beams between charged plates: the electron bends sharply to the positive plate, the proton bends gently to the negative plate, the neutron goes straight
    In an electric field the proton bends towards the $-$ plate and the electron bends the opposite way and far more (it is much lighter); the neutron passes straight through

    Atomic radius and ionic radius

    The atomic radius 原子半径 is the size of an atom. The ionic radius 离子半径 is the size of an ion.

    Across a period 周期 (left to right), the atomic radius gets smaller. The nuclear charge 核电荷 (the pull from the protons) rises, but the electrons go into the same outer shell, so the shielding 屏蔽 by inner shells stays about the same. The stronger pull draws the outer shell inwards.

    Down a group 族 (top to bottom), the atomic radius gets larger. Each step down adds a new shell, so the outer electrons are further out and feel more shielding from the nucleus.

    A row of atoms getting smaller across a period, and a column of atoms getting larger down a group
    Across a period the atoms shrink (stronger nuclear pull on the same outer shell); down a group they grow (each step adds a shell)

    For ions:

    • a positive ion (cation 阳离子) is smaller than its atom. It has lost its outer shell, and the electrons that remain feel a stronger pull each.
    • a negative ion (anion 阴离子) is larger than its atom. It has gained electrons, so there is more repulsion 排斥 between the electrons.
    • among ions that have the same number of electrons, the one with more protons is smaller.
    Explore · ⁨探索⁩

    Explore the atom · ⁨原子を探検する⁩

    Tap each part. A tiny dense nucleus of protons and neutrons holds the mass; light electrons orbit it in shells. · ⁨各部分をクリックしてください。陽子と中性子からなる微小で高密度な原子核が質量の大部分を担っています。軽量の電子は殻状の軌道でその周囲を回っています。⁩

    Explore · ⁨探索⁩

    Atomic and ionic radius trends · ⁨原子半径およびイオン半径の傾向⁩

    Atomic radius falls across a period (rising nuclear charge pulls the same shell in) and rises down a group (an extra shell each time). Step across Period 3 to see it. · ⁨周期に沿って原子半径は減少する(核電荷が増加して同じ殻を引き寄せるため)、族に沿って増加する(每次都に新しい殻が加わる)。周期3を横断して確認してみよう。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    atom/ˈætəm/ 原子
    nucleus/ˈnjuːklɪəs/ 核
    proton/ˈprəʊtɒn/ 陽子
    neutron/ˈnjuːtrɒn/ 中性子
    electron/ɪˈlektrɒn/ 電子
    shell/ʃel/ 空壳
    relative charge/ˈrelətɪv tʃɑːdʒ/ 相対電荷
    relative mass/ˈrelətɪv mæs/ 相対質量
    neutral/ˈnjuːtrəl/ 中性
    proton number/ˈprəʊtɒn ˈnʌmbə/ 原子番号
    atomic number/əˈtɒmɪk ˈnʌmbə/ 原子番号
    nucleon number/ˈnjuːklɪən ˈnʌmbə/ 核子数
    mass number/mæs ˈnʌmbə/ 質量数
    nucleon/ˈnjuːklɪən/ 核子
    ion/ˈaɪɒn/ イオン
    electric field/ɪˈlektrɪk fiːld/ 電界
    deflection/dɪˈflekʃn/ 偏光
    atomic radius/əˈtɒmɪk ˈreɪdɪəs/ 原子半径
    ionic radius/aɪˈɒnɪk ˈreɪdɪəs/ イオン半径
    period/ˈpɪərɪəd/ 周期の変化率である
    nuclear charge/ˈnjuːklɪə tʃɑːdʒ/ 核電荷
    shielding/ˈʃiːldɪŋ/ 遮蔽効果
    group/ɡruːp/ グループ
    cation/ˈkætaɪən/ カチオン
    anion/ˈænaɪən/ アニオン
    repulsion/rɪˈpʌlʃn/ 反発
    1.2

    Isotopes

    Syllabus · ⁨シラバス⁩
    English
    1. define the term isotope in terms of numbers of protons and neutrons
    2. understand the notation $_y^x\text{A}$ for isotopes, where $x$ is the mass or nucleon number and $y$ is the atomic or proton number
    3. state that and explain why isotopes of the same element have the same chemical properties
    4. state that and explain why isotopes of the same element have different physical properties, limited to mass and density
    日本語
    1. 陽子数と中性子数の観点から同位体の定義を行う
    2. 同位体の表記 $_y^x\text{A}$ を理解する。ここで $x$ は質量数または核子数、$y$ は原子数または质子数である
    3. 同じ元素の同位体が同じ化学的性質を持つことを述べ、その理由を説明する
    4. 同じ元素の同位体が異なる物理的性質(質量と密度に限る)を持つことを述べ、その理由を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Isotopes 同位素 are atoms of the same element with the same number of protons but a different number of neutrons. So isotopes have the same proton number $Z$ but a different nucleon number $A$.

    We write an isotope as $^{A}_{Z}\text{X}$: the nucleon number $A$ on top, the proton number $Z$ below. For example, chlorine has two main isotopes, $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$.

    Same chemical properties

    Chemical properties 化学性质 depend on the electrons, especially the outer electrons. Isotopes of one element have the same number of electrons arranged in the same way. So they react in exactly the same way — they have the same chemical properties.

    Different physical properties

    Some physical properties 物理性质 depend on mass, so they differ between isotopes. A heavier isotope has more neutrons, so more mass, and therefore a higher density 密度. (The syllabus limits this difference to mass and density.)

    日本語

    Isotopes 同位素 are atoms of the same element with the same number of protons but a different number of neutrons. So isotopes have the same proton number $Z$ but a different nucleon number $A$.

    Two chlorine isotopes — both have 17 protons but one has 18 neutrons and the other 20
    Two chlorine isotopes: same protons, different neutrons

    We write an isotope as $^{A}_{Z}\text{X}$: the nucleon number $A$ on top, the proton number $Z$ below. For example, chlorine has two main isotopes, $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$.

    Same chemical properties

    Chemical properties 化学性质 depend on the electrons, especially the outer electrons. Isotopes of one element have the same number of electrons arranged in the same way. So they react in exactly the same way — they have the same chemical properties.

    Different physical properties

    Some physical properties 物理性质 depend on mass, so they differ between isotopes. A heavier isotope has more neutrons, so more mass, and therefore a higher density 密度. (The syllabus limits this difference to mass and density.)

    Explore · ⁨探索⁩

    Isotope lab · ⁨同位体の実験⁩

    Classify isotope facts by what changes and what stays the same. · ⁨変化することと不変であることを基に同位体の事実を分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    isotope/ˈaɪsətəʊp/ 同位体 (isotope)
    chemical properties/ˈkemɪkl ˈprɒpətiz/ 化学的性質
    physical properties/ˈfɪzɪkl ˈprɒpətiz/ 物理的性質
    density/ˈdensɪti/ 密度
    Watch lesson · ⁨レッスンを視聴⁩
    1.3

    Electrons, energy levels and orbitals

    Syllabus · ⁨シラバス⁩
    English
    1. understand the terms: shells, sub-shells and orbitals; principal quantum number (n); ground state, limited to electronic configuration
    2. describe the number of orbitals making up s, p and d sub-shells, and the number of electrons that can fill s, p and d sub-shells
    3. describe the order of increasing energy of the sub-shells within the first three shells and the 4s and 4p sub-shells
    4. describe the electronic configurations to include the number of electrons in each shell, sub-shell and orbital
    5. explain the electronic configurations in terms of energy of the electrons and inter-electron repulsion
    6. determine the electronic configuration of atoms and ions given the atomic or proton number and charge, using either of the following conventions: e.g. for Fe: $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2$ (full electronic configuration) or [Ar] $3\text{d}^6 4\text{s}^2$ (shorthand electronic configuration)
    7. understand and use the electrons in boxes notation
    8. describe and sketch the shapes of s and p orbitals
    9. describe a free radical as a species with one or more unpaired electrons
    日本語
    1. 用語の理解:シェル、サブシェルおよび軌道;主量子数 (n);基底状態(電子配置に限定)
    2. s, p, d サブシェルを構成する軌道の数と、s, p, d サブシェルに充填可能な電子の数を記述する
    3. 最初の3つのシェル内のサブシェルのエネルギー増加順および4sと4pサブシェルの順序を記述する
    4. 電子配置を記述し、各シェル、サブシェル、軌道に含まれる電子の数を含む
    5. 電子のエネルギーおよび電子間反発に基づいて電子配置を説明する
    6. 原子番号または陽子数および電荷を与えられたとき、以下のいずれかの表記法を用いて原子やイオンの電子配置を決定する:例:Feの場合:$1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2$(完全な電子配置)または [Ar] $3\text{d}^6 4\text{s}^2$(略記電子配置)
    7. ボックス内電子表記を理解し使用する
    8. sおよびp軌道の形状を記述し、スケッチする
    9. フリーラジカルとは、未対電子が1つ以上ある種であることを記述する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Electrons are arranged in shells, sub-shells and orbitals.

    Shells and the principal quantum number

    Each shell is labelled by the principal quantum number 主量子数 $n = 1, 2, 3, \dots$ A larger $n$ means a shell that is further from the nucleus and higher in energy.

    Sub-shells and orbitals

    Each shell is split into sub-shells 亚层, named s, p and d. Each sub-shell is built from orbitals 轨道. An orbital is a small region that can hold up to two electrons.

    Sub-shell Number of orbitals Maximum electrons
    s 1 2
    p 3 6
    d 5 10

    So an s sub-shell holds 2 electrons, a p sub-shell holds 6, and a d sub-shell holds 10.

    Order of increasing energy

    Electrons fill the lowest-energy sub-shell first. For the first three shells, plus 4s and 4p, the order of rising energy is:

    $$1\text{s} < 2\text{s} < 2\text{p} < 3\text{s} < 3\text{p} < 4\text{s} < 3\text{d} < 4\text{p}$$

    Notice the surprise: 4s is slightly lower in energy than 3d, so 4s fills first.

    Electronic configuration

    The electronic configuration 电子排布 lists how many electrons are in each sub-shell. The lowest-energy arrangement is the ground state 基态.

    For iron (Fe, $Z = 26$):

    $$1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6\,3\text{d}^6\,4\text{s}^2$$

    You can write a shorthand using the nearest noble gas 稀有气体 in square brackets:

    $$[\text{Ar}]\,3\text{d}^6\,4\text{s}^2$$

    Here $[\text{Ar}]$ stands for the full configuration of argon.

    For ions, you add or remove electrons. One key rule: when a transition metal forms a positive ion, it loses its 4s electrons before its 3d electrons. So $\text{Fe}^{3+}$ is $[\text{Ar}]\,3\text{d}^5$.

    Electrons in boxes

    The electrons in boxes notation draws each orbital as a box and each electron as an arrow. Two electrons in the same orbital must point opposite ways, because each electron has a property called spin 自旋, and a shared orbital needs opposite spins.

    Within a sub-shell, electrons fill empty orbitals one at a time, with parallel arrows, before any orbital gets a second electron. Spreading out like this keeps the electrons apart and lowers the repulsion between them.

    Why the configuration takes this shape

    Electrons fill from low energy to high energy because that gives the most stable (lowest-energy) atom. Within a sub-shell they spread out singly first to reduce the repulsion between the negative electrons.

    Shapes of s and p orbitals

    • an s orbital is a sphere 球形 centred on the nucleus.
    • a p orbital has two lobes, like a dumbbell, pointing along one axis. The three p orbitals point along three directions at right angles (the $x$, $y$ and $z$ axes).

    Free radicals

    A free radical 自由基 is a species with one or more unpaired electrons 未成对电子. Free radicals are very reactive.

    日本語
    The visible emission spectrum of hydrogen: four bright lines on black
    Hydrogen emits light only at discrete, characteristic wavelengths — its line emission spectrum.

    Electrons are arranged in shells, sub-shells and orbitals.

    Shells and the principal quantum number

    Each shell is labelled by the principal quantum number 主量子数 $n = 1, 2, 3, \dots$ A larger $n$ means a shell that is further from the nucleus and higher in energy.

    Sub-shells and orbitals

    Each shell is split into sub-shells 亚层, named s, p and d. Each sub-shell is built from orbitals 轨道. An orbital is a small region that can hold up to two electrons.

    Sub-shell Number of orbitals Maximum electrons
    s 1 2
    p 3 6
    d 5 10

    So an s sub-shell holds 2 electrons, a p sub-shell holds 6, and a d sub-shell holds 10.

    Order of increasing energy

    Electrons fill the lowest-energy sub-shell first. For the first three shells, plus 4s and 4p, the order of rising energy is:

    $$1\text{s} < 2\text{s} < 2\text{p} < 3\text{s} < 3\text{p} < 4\text{s} < 3\text{d} < 4\text{p}$$

    Notice the surprise: 4s is slightly lower in energy than 3d, so 4s fills first.

    An energy-level diagram of the sub-shells from 1s up to 4p, with 4s drawn just below 3d
    The sub-shells in order of increasing energy. 4s lies just below 3d, so 4s fills first

    Electronic configuration

    The electronic configuration 电子排布 lists how many electrons are in each sub-shell. The lowest-energy arrangement is the ground state 基态.

    For iron (Fe, $Z = 26$):

    $$1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6\,3\text{d}^6\,4\text{s}^2$$

    You can write a shorthand using the nearest noble gas 稀有气体 in square brackets:

    $$[\text{Ar}]\,3\text{d}^6\,4\text{s}^2$$

    Here $[\text{Ar}]$ stands for the full configuration of argon.

    For ions, you add or remove electrons. One key rule: when a transition metal forms a positive ion, it loses its 4s electrons before its 3d electrons. So $\text{Fe}^{3+}$ is $[\text{Ar}]\,3\text{d}^5$.

    Electrons in boxes

    The electrons in boxes notation draws each orbital as a box and each electron as an arrow. Two electrons in the same orbital must point opposite ways, because each electron has a property called spin 自旋, and a shared orbital needs opposite spins.

    Within a sub-shell, electrons fill empty orbitals one at a time, with parallel arrows, before any orbital gets a second electron. Spreading out like this keeps the electrons apart and lowers the repulsion between them.

    Box diagram for nitrogen: filled 1s and 2s boxes with paired opposite arrows, and three 2p boxes each with one upward arrow
    Electrons in boxes for nitrogen ($1\text{s}^2\,2\text{s}^2\,2\text{p}^3$): paired electrons point opposite ways, and the 2p orbitals fill singly with parallel spins

    Why the configuration takes this shape

    Electrons fill from low energy to high energy because that gives the most stable (lowest-energy) atom. Within a sub-shell they spread out singly first to reduce the repulsion between the negative electrons.

    Shapes of s and p orbitals

    • an s orbital is a sphere 球形 centred on the nucleus.
    • a p orbital has two lobes, like a dumbbell, pointing along one axis. The three p orbitals point along three directions at right angles (the $x$, $y$ and $z$ axes).
    On the left a spherical s orbital with the nucleus at its centre; on the right three dumbbell-shaped p orbitals pointing along the x, y and z axes
    An s orbital is a sphere; each p orbital is a dumbbell, and the three p orbitals point along the $x$, $y$ and $z$ axes

    Free radicals

    A free radical 自由基 is a species with one or more unpaired electrons 未成对电子. Free radicals are very reactive.

    Explore · ⁨探索⁩

    Filling the electron shells · ⁨電子殻の充填⁩

    Change the atomic number Z and watch the electrons fill the shells (2, 8, 8, …) — the pattern that builds the Periodic Table. · ⁨原子番号Zを変化させ、電子が殻(2, 8, 8, …)に充填される様子を見てください。これが周期表の構築パターンです。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    principal quantum number/ˈprɪnsɪpl ˈkwɒntəm ˈnʌmbə/ 主量子数
    sub-shell/sʌb ʃel/ 副殻
    orbital/ˈɔːbɪtl/ 軌道
    electronic configuration/ɪlekˈtrɒnɪk kənˌfɪɡjəˈreɪʃn/ 電子配置
    ground state/ɡraʊnd steɪt/ 基底状態
    noble gas/ˈnəʊbl ɡæs/ 希ガス
    spin/spɪn/ スピン
    sphere/sfɪə/ 球体
    free radical/friː ˈrædɪkl/ 自由ラジカル
    unpaired electrons/ʌnˈpeəd ɪˈlektrɒnz/ 不対電子
    Watch lesson · ⁨レッスンを視聴⁩
    1.4

    Ionisation energy

    Syllabus · ⁨シラバス⁩
    English
    1. define and use the term first ionisation energy, IE
    2. construct equations to represent first, second and subsequent ionisation energies
    3. identify and explain the trends in ionisation energies across a period and down a group of the Periodic Table
    4. identify and explain the variation in successive ionisation energies of an element
    5. understand that ionisation energies are due to the attraction between the nucleus and the outer electron
    6. explain the factors influencing the ionisation energies of elements in terms of nuclear charge, atomic/ionic radius, shielding by inner shells and sub-shells and spin-pair repulsion
    7. deduce the electronic configurations of elements using successive ionisation energy data
    8. deduce the position of an element in the Periodic Table using successive ionisation energy data
    日本語
    1. 用語第一電離エネルギー、IEを定義し使用する
    2. 第一、第二およびそれ以降の電離エネルギーを表す化学方程式を作成する
    3. 周期表の周期に沿ったおよび族に沿った電離エネルギーの傾向を特定・説明する
    4. 元素の連続した電離エネルギーの変動を特定・説明する
    5. 電離エネルギーは原子核と外側の電子間の引力によるものであることを理解する
    6. 核電荷、原子/イオン半径、内殻およびサブシェルによる遮蔽効果、およびスピン対反発の観点から、元素の電離エネルギーに影響を与える要因を説明する
    7. 連続した電離エネルギーデータを使用して元素の電子配置を導き出す
    8. 連続した電離エネルギーデータを使用して元素の周期表上の位置を導き出す

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    First ionisation energy

    The first ionisation energy 第一电离能 (IE) is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous $+1$ ions.

    We use gaseous atoms so there are no forces between the particles. The unit is $\text{kJ mol}^{-1}$. As an equation, for an element X:

    $$\text{X}(\text{g}) \rightarrow \text{X}^{+}(\text{g}) + \text{e}^{-}$$

    The $(\text{g})$ shows the species is a gas.

    Successive ionisation energies

    After removing one electron, you can remove another. The second ionisation energy removes one electron from each $+1$ ion:

    $$\text{X}^{+}(\text{g}) \rightarrow \text{X}^{2+}(\text{g}) + \text{e}^{-}$$

    You can keep going. These are the successive ionisation energies 逐级电离能. Each is larger than the one before, because every electron is pulled away from a more positive ion.

    What ionisation energy depends on

    Ionisation energy comes from the attraction between the positive nucleus and the outer electron. Three main factors set how strong that attraction is:

    • nuclear charge: more protons pull the electrons more strongly, so the ionisation energy is higher.
    • atomic radius: the further the outer electron sits from the nucleus, the weaker the pull, so the ionisation energy is lower.
    • shielding: inner shells block some of the pull on the outer electron. More inner shells mean more shielding and a lower ionisation energy.

    There is a smaller effect too — spin-pair repulsion 自旋成对排斥. When two electrons share one orbital, they push each other a little, so one is easier to remove.

    Trends in first ionisation energy

    Across a period, the first ionisation energy generally rises. The nuclear charge grows while shielding stays about the same, so the outer electrons are held more tightly.

    Down a group, the first ionisation energy falls. Lower elements have more shells, so more shielding and a larger radius, and the outer electron is easier to remove.

    The dips are evidence for sub-shells

    The rise across a period is not smooth. Two small dips appear, and you should be able to explain both:

    • Group 2 to Group 13 (for example Mg to Al): the electron removed from Al comes from a 3p sub-shell, which is higher in energy than the full 3s sub-shell in Mg. A 3p electron is easier to remove, so the value dips.
    • Group 15 to Group 16 (for example P to S): in S, one 3p orbital now holds a pair of electrons. Spin-pair repulsion makes one of them easier to remove, so the value dips.

    These dips are evidence that sub-shells exist.

    Successive ionisation energies are evidence for shells

    If you plot the successive ionisation energies of one element, the values rise, with big jumps at certain points. A big jump happens when the next electron must come from a shell closer to the nucleus.

    Count how many electrons come off easily before the first big jump — that is the number of electrons in the outer shell, which tells you the group the element is in. You can also use the pattern to work out the electronic configuration and the position of the element in the Periodic Table.

    Electrons removed before the first big jump Group
    1 Group 1
    2 Group 2
    3 Group 13

    Worked example. The first five successive ionisation energies of an element are $590$, $1150$, $4940$, $6480$ and $8120\ \text{kJ}\,\text{mol}^{-1}$. Which group is it in? Look for the big jump, not the biggest number. From the 1st to the 2nd the value roughly doubles, which is a normal rise. From the 2nd ($1150$) to the 3rd ($4940$) it more than quadruples: that is the jump. So two electrons come off easily before it, the outer shell holds 2 electrons, and the element is in Group 2. Count the electrons removed before the jump, and explain the jump properly: the next electron is being pulled from a shell closer to the nucleus, not from a different element.

    日本語

    First ionisation energy

    The first ionisation energy 第一电离能 (IE) is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous $+1$ ions.

    We use gaseous atoms so there are no forces between the particles. The unit is $\text{kJ mol}^{-1}$. As an equation, for an element X:

    $$\text{X}(\text{g}) \rightarrow \text{X}^{+}(\text{g}) + \text{e}^{-}$$

    The $(\text{g})$ shows the species is a gas.

    Successive ionisation energies

    After removing one electron, you can remove another. The second ionisation energy removes one electron from each $+1$ ion:

    $$\text{X}^{+}(\text{g}) \rightarrow \text{X}^{2+}(\text{g}) + \text{e}^{-}$$

    You can keep going. These are the successive ionisation energies 逐级电离能. Each is larger than the one before, because every electron is pulled away from a more positive ion.

    What ionisation energy depends on

    Ionisation energy comes from the attraction between the positive nucleus and the outer electron. Three main factors set how strong that attraction is:

    • nuclear charge: more protons pull the electrons more strongly, so the ionisation energy is higher.
    • atomic radius: the further the outer electron sits from the nucleus, the weaker the pull, so the ionisation energy is lower.
    • shielding: inner shells block some of the pull on the outer electron. More inner shells mean more shielding and a lower ionisation energy.

    There is a smaller effect too — spin-pair repulsion 自旋成对排斥. When two electrons share one orbital, they push each other a little, so one is easier to remove.

    Trends in first ionisation energy

    Across a period, the first ionisation energy generally rises. The nuclear charge grows while shielding stays about the same, so the outer electrons are held more tightly.

    Down a group, the first ionisation energy falls. Lower elements have more shells, so more shielding and a larger radius, and the outer electron is easier to remove.

    The dips are evidence for sub-shells

    The rise across a period is not smooth. Two small dips appear, and you should be able to explain both:

    • Group 2 to Group 13 (for example Mg to Al): the electron removed from Al comes from a 3p sub-shell, which is higher in energy than the full 3s sub-shell in Mg. A 3p electron is easier to remove, so the value dips.
    • Group 15 to Group 16 (for example P to S): in S, one 3p orbital now holds a pair of electrons. Spin-pair repulsion makes one of them easier to remove, so the value dips.

    These dips are evidence that sub-shells exist.

    A graph of first ionisation energy across Period 3 rising overall but dipping at aluminium and at sulfur
    First ionisation energy rises across Period 3 but dips at Al and at S — evidence that sub-shells exist

    Successive ionisation energies are evidence for shells

    If you plot the successive ionisation energies of one element, the values rise, with big jumps at certain points. A big jump happens when the next electron must come from a shell closer to the nucleus.

    Count how many electrons come off easily before the first big jump — that is the number of electrons in the outer shell, which tells you the group the element is in. You can also use the pattern to work out the electronic configuration and the position of the element in the Periodic Table.

    A log-scale graph of the eleven successive ionisation energies of sodium, with two big jumps splitting the points into groups of 1, 8 and 2
    Successive ionisation energies of sodium (log scale): the big jumps reveal the $2,8,1$ shell structure
    Electrons removed before the first big jump Group
    1 Group 1
    2 Group 2
    3 Group 13

    Worked example. The first five successive ionisation energies of an element are $590$, $1150$, $4940$, $6480$ and $8120\ \text{kJ}\,\text{mol}^{-1}$. Which group is it in? Look for the big jump, not the biggest number. From the 1st to the 2nd the value roughly doubles, which is a normal rise. From the 2nd ($1150$) to the 3rd ($4940$) it more than quadruples: that is the jump. So two electrons come off easily before it, the outer shell holds 2 electrons, and the element is in Group 2. Count the electrons removed before the jump, and explain the jump properly: the next electron is being pulled from a shell closer to the nucleus, not from a different element.

    Explore · ⁨探索⁩

    The ionisation-energy trend — and its dips · ⁨イオン化エネルギーの傾向と、その谷⁩

    First ionisation energy generally rises across a period, but DIPS where a new p sub-shell starts and where a p-orbital pair first forms. Step across to find the two tell-tale dips. · ⁨第一イオン化エネルギーは一般に周期全体で上昇しますが、新しい p 副殻が始まる場所や、p 軌道のペアが初めて形成される場所で谷(DIPS)ができます。2つの典型的な谷を見つけるためにステップを進めてください。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    first ionisation energy/fɜːst ˌaɪənaɪˈzeɪʃn ˈenədʒi/ 第一イオン化エネルギー
    successive ionisation energies/səkˈsesɪv ˌaɪənaɪˈzeɪʃn ˈenədʒiz/ 連続イオン化エネルギー
    spin-pair repulsion/spɪn peə rɪˈpʌlʃn/ スピン対反発
    1.4

    Exam tips

    • Define isotopes in full: atoms of the same element with the same number of protons but a different number of neutrons — the mark scheme wants both halves.
    • 4s fills before 3d, but electrons are removed from 4s first when forming ions, so $\text{Fe}^{3+}$ is $[\text{Ar}]3\text{d}^5$ (not $[\text{Ar}]3\text{d}^3 4\text{s}^2$).
    • In successive ionisation energies, a big jump marks the start of a new (inner) shell — use the jumps to place the element in its group.
    • Explain every ionisation-energy trend with the same three factors: nuclear charge, distance and shielding, plus sub-shell effects for the small dips.
    • Learn the exact reason first ionisation energy of oxygen is below nitrogen: oxygen's paired 2p electrons repel, so one is easier to remove.
  • 2

    Atoms, molecules and stoichiometry · ⁨原子、分子および化学量論⁩

    Watch lesson · ⁨レッスンを視聴⁩
    2.1

    Relative masses of atoms and molecules

    Syllabus · ⁨シラバス⁩
    1. define the unified atomic mass unit as one twelfth of the mass of a carbon-12 atom
    2. define relative atomic mass, $A_r$, relative isotopic mass, relative molecular mass, $M_r$, and relative formula mass in terms of the unified atomic mass unit

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Atoms 原子 are far too light to weigh in grams, so we compare every mass to one standard. The standard is the unified atomic mass unit 统一原子质量单位 (symbol u), defined as exactly one twelfth of the mass of one carbon-12 atom.

    Using this unit, we state masses as simple numbers:

    • the relative atomic mass 相对原子质量 $A_r$ of an element is the average mass of its atoms compared with $\tfrac{1}{12}$ of a carbon-12 atom. It is an average over all the isotopes 同位素, weighted by how common each one is.
    • the relative isotopic mass 相对同位素质量 is the mass of one atom of a single isotope, compared with $\tfrac{1}{12}$ of a carbon-12 atom.
    • the relative molecular mass 相对分子质量 $M_r$ of a molecule 分子 is the sum of the relative atomic masses of all its atoms.
    • the relative formula mass 相对式量 is the same idea for a substance that is not made of molecules (such as an ionic compound). Add up the relative atomic masses shown in the formula.

    To find $A_r$ from isotope data, multiply each isotope mass by its percentage, add these up, and divide by 100.

    Worked example. Chlorine is $75\%$ ${}^{35}\text{Cl}$ and $25\%$ ${}^{37}\text{Cl}$. Find its relative atomic mass.

    $$A_r = \frac{(75 \times 35) + (25 \times 37)}{100} = \frac{2625 + 925}{100} = 35.5.$$

    Worked example. A mass spectrometer shows copper is $69.2\%$ ${}^{63}\text{Cu}$ and $30.8\%$ ${}^{65}\text{Cu}$. Find its relative atomic mass.

    $$A_r = \frac{(69.2 \times 63) + (30.8 \times 65)}{100} = \frac{4359.6 + 2002.0}{100} = 63.6.$$
    日本語

    Atoms 原子 are far too light to weigh in grams, so we compare every mass to one standard. The standard is the unified atomic mass unit 统一原子质量单位 (symbol u), defined as exactly one twelfth of the mass of one carbon-12 atom.

    Using this unit, we state masses as simple numbers:

    • the relative atomic mass 相对原子质量 $A_r$ of an element is the average mass of its atoms compared with $\tfrac{1}{12}$ of a carbon-12 atom. It is an average over all the isotopes 同位素, weighted by how common each one is.
    • the relative isotopic mass 相对同位素质量 is the mass of one atom of a single isotope, compared with $\tfrac{1}{12}$ of a carbon-12 atom.
    • the relative molecular mass 相对分子质量 $M_r$ of a molecule 分子 is the sum of the relative atomic masses of all its atoms.
    • the relative formula mass 相对式量 is the same idea for a substance that is not made of molecules (such as an ionic compound). Add up the relative atomic masses shown in the formula.
    A bar showing chlorine is 75% chlorine-35 and 25% chlorine-37, with the weighted-average calculation giving a relative atomic mass of 35.5
    Relative atomic mass is a weighted average: chlorine's two isotopes (75% ³⁵Cl, 25% ³⁷Cl) average to Aᵣ = 35.5

    To find $A_r$ from isotope data, multiply each isotope mass by its percentage, add these up, and divide by 100.

    Worked example. Chlorine is $75\%$ ${}^{35}\text{Cl}$ and $25\%$ ${}^{37}\text{Cl}$. Find its relative atomic mass.

    $$A_r = \frac{(75 \times 35) + (25 \times 37)}{100} = \frac{2625 + 925}{100} = 35.5.$$

    Worked example. A mass spectrometer shows copper is $69.2\%$ ${}^{63}\text{Cu}$ and $30.8\%$ ${}^{65}\text{Cu}$. Find its relative atomic mass.

    $$A_r = \frac{(69.2 \times 63) + (30.8 \times 65)}{100} = \frac{4359.6 + 2002.0}{100} = 63.6.$$
    Explore · ⁨探索⁩

    Relative mass lab · ⁨相対質量の実験室⁩

    relative atomic mass = weighted mean · ⁨相対原子質量 = 加重平均⁩

    Change isotope abundance and see the weighted mean move. · ⁨同位体の存在比を変えると、加重平均がどのように動くか見ることができます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    atom/ˈætəm/ 原子
    unified atomic mass unit/ˈjuːnɪfaɪd əˈtɒmɪk mæs ˈjuːnɪt/ 統一原子質量単位
    relative atomic mass/ˈrelətɪv əˈtɒmɪk mæs/ 相対原子質量
    isotope/ˈaɪsətəʊp/ 同位体 (isotope)
    relative isotopic mass/ˈrelətɪv ˌaɪsəˈtɒpɪk mæs/ 相対同位体質量
    relative molecular mass/ˈrelətɪv məˈlekjʊlə mæs/ 相対分子量
    molecule/ˈmɒlɪkjuːl/ 分子
    relative formula mass/ˈrelətɪv ˈfɔːmjʊlə mæs/ 相対式質量
    2.2

    The mole and the Avogadro constant

    Syllabus · ⁨シラバス⁩
    English
    1. define and use the term mole in terms of the Avogadro constant
    日本語
    1. アボガドロ定数を用いてモルという用語を定義し、使用する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Chemists count particles in groups called moles, just as we count eggs in dozens.

    One mole 摩尔 (symbol mol) is the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant 阿伏伽德罗常量:

    $$N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$$

    So one mole of anything contains $6.02 \times 10^{23}$ particles. The particles may be atoms, molecules or ions 离子 — always say which.

    The mass of one mole in grams equals the relative mass ($A_r$ or $M_r$). This is the molar mass 摩尔质量, with units $\text{g mol}^{-1}$. The key equation is:

    $$n = \frac{m}{M}$$

    where $n$ is the amount in moles, $m$ is the mass in grams, and $M$ is the molar mass.

    Worked example. How many moles are in $8.0\ \text{g}$ of methane, $\text{CH}_4$? ($A_r$: C $= 12$, H $= 1$.)

    The molar mass is $M = 12 + 4(1) = 16\ \text{g mol}^{-1}$, so

    $$n = \frac{m}{M} = \frac{8.0}{16} = 0.50\ \text{mol}.$$
    日本語
    A modern electronic laboratory balance
    A modern electronic balance measures mass — the basis of mole calculations.

    Chemists count particles in groups called moles, just as we count eggs in dozens.

    One mole 摩尔 (symbol mol) is the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant 阿伏伽德罗常量:

    $$N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$$

    So one mole of anything contains $6.02 \times 10^{23}$ particles. The particles may be atoms, molecules or ions 离子 — always say which.

    The mass of one mole in grams equals the relative mass ($A_r$ or $M_r$). This is the molar mass 摩尔质量, with units $\text{g mol}^{-1}$. The key equation is:

    $$n = \frac{m}{M}$$

    where $n$ is the amount in moles, $m$ is the mass in grams, and $M$ is the molar mass.

    Worked example. How many moles are in $8.0\ \text{g}$ of methane, $\text{CH}_4$? ($A_r$: C $= 12$, H $= 1$.)

    The molar mass is $M = 12 + 4(1) = 16\ \text{g mol}^{-1}$, so

    $$n = \frac{m}{M} = \frac{8.0}{16} = 0.50\ \text{mol}.$$
    A central moles box linked by double arrows to mass, number of particles, gas volume and solution concentration, each arrow labelled with its conversion
    The mole is the hub of every amount calculation: convert to mass ($n=m/M$), particles ($\times N_A$), gas volume ($n=V/24$) or solution ($n=cV$)
    Explore · ⁨探索⁩

    Mole mass lab · ⁨モル質量実験⁩

    n = m / M

    Change mass and see moles scale through molar mass. · ⁨質量を変化させ、モール質量を通じてモル数の変化を確認する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    mole/məʊl/ モル
    Avogadro constant/ˌævəˈɡædrəʊ ˈkɒnstənt/ アボガドロ定数
    ion/ˈaɪɒn/ イオン
    molar mass/ˈməʊlə mæs/ モル質量
    2.3

    Formulas

    Syllabus · ⁨シラバス⁩
    English
    1. write formulas of ionic compounds from ionic charges and oxidation numbers (shown by a Roman numeral), including: (a) the prediction of ionic charge from the position of an element in the Periodic Table (b) recall of the names and formulas for the following ions: $\text{NO}_3^-$, $\text{CO}_3^{2-}$, $\text{SO}_4^{2-}$, $\text{OH}^-$, $\text{NH}_4^+$, $\text{Zn}^{2+}$, $\text{Ag}^+$, $\text{HCO}_3^-$, $\text{PO}_4^{3-}$
    2. (a) write and construct equations (which should be balanced), including ionic equations (which should not include spectator ions) (b) use appropriate state symbols in equations
    3. define and use the terms empirical and molecular formula
    4. understand and use the terms anhydrous, hydrated and water of crystallisation
    5. calculate empirical and molecular formulas, using given data
    日本語
    1. イオン電荷および酸化数(ローマ数字で示される)からイオン化合物の化学式を書く。以下を含む: (a) 元素が周期表に位置する位置からのイオン電荷の予測 (b) 以下のイオンの名称および化学式の暗記: $\text{NO}_3^-$, $\text{CO}_3^{2-}$, $\text{SO}_4^{2-}$, $\text{OH}^-$, $\text{NH}_4^+$, $\text{Zn}^{2+}$, $\text{Ag}^+$, $\text{HCO}_3^-$, $\text{PO}_4^{3-}$
    2. (a) 方程式(バランスさせること)およびイオン方程式(傍観イオンを含めないこと)を書く、構成する (b) 方程式に適切な状態記号を使用する
    3. 実験式および分子式という用語を定義し、使用する
    4. 無水物、水和物および結晶水という用語を理解し、使用する
    5. 与えられたデータを用いて実験式および分子式を計算する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A compound 化合物 is a substance made of two or more elements chemically joined.

    Charges and formulas of ionic compounds

    In an ionic compound 离子化合物 the total positive charge balances the total negative charge, so the compound is neutral overall.

    You can predict the charge of many ions from the element's position in the Periodic Table:

    Group 1 2 13 15 16 17
    Usual ion charge $+1$ $+2$ $+3$ $-3$ $-2$ $-1$

    Hydrogen forms $\text{H}^+$, and the Group 18 noble gases do not normally form ions.

    Some ions you must know by name and formula:

    Name Formula
    nitrate $\text{NO}_3^{-}$
    carbonate $\text{CO}_3^{2-}$
    sulfate $\text{SO}_4^{2-}$
    hydroxide $\text{OH}^{-}$
    ammonium $\text{NH}_4^{+}$
    zinc $\text{Zn}^{2+}$
    silver $\text{Ag}^{+}$
    hydrogencarbonate $\text{HCO}_3^{-}$
    phosphate $\text{PO}_4^{3-}$

    For a metal that can have more than one charge, a Roman numeral shows the oxidation number 氧化数. For example, iron(II) is $\text{Fe}^{2+}$ and iron(III) is $\text{Fe}^{3+}$. To write a formula, balance the charges: iron(III) oxide is $\text{Fe}_2\text{O}_3$, because two $\text{Fe}^{3+}$ balance three $\text{O}^{2-}$.

    Equations and state symbols

    A chemical equation must be balanced 配平 — the same number of each kind of atom on both sides. Add state symbols 状态符号 to show the state of each species: (s) solid, (l) liquid, (g) gas, and (aq) aqueous 水溶液 (dissolved in water).

    An ionic equation 离子方程式 shows only the ions and molecules that actually change. The ions that do not change are spectator ions 旁观离子, and you leave them out. For example, the reaction that forms silver chloride is:

    $$\text{Ag}^{+}(\text{aq}) + \text{Cl}^{-}(\text{aq}) \rightarrow \text{AgCl}(\text{s})$$

    Empirical and molecular formulas

    The empirical formula 实验式 is the simplest whole-number ratio of the atoms of each element in a compound. The molecular formula 分子式 shows the actual number of atoms of each element in one molecule.

    For example, ethane has empirical formula $\text{CH}_3$ but molecular formula $\text{C}_2\text{H}_6$.

    Hydrated and anhydrous solids

    Some solids hold water inside their crystals. This water is the water of crystallisation 结晶水. A solid that contains it is hydrated 水合的; the same solid with the water removed is anhydrous 无水的.

    For example, hydrated copper(II) sulfate is $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$. Heating it drives off the water to leave anhydrous $\text{CuSO}_4$.

    Calculating empirical and molecular formulas

    To find the empirical formula from masses (or percentages by mass):

    1. divide each element's mass by its $A_r$ to get the moles.
    2. divide all the mole values by the smallest one.
    3. round to the nearest whole numbers — that ratio is the empirical formula.

    To get the molecular formula, you also need $M_r$. Find how many times the empirical formula mass fits into $M_r$, then multiply the formula by that number.

    Worked example. A compound is $40.0\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen by mass. Find its empirical formula. ($A_r$: C $= 12$, H $= 1$, O $= 16$.)

    Take $100\ \text{g}$ and divide each mass by its $A_r$ to get moles: C $= 40.0/12 = 3.33$, H $= 6.7/1 = 6.7$, O $= 53.3/16 = 3.33$. Dividing through by the smallest ($3.33$) gives a ratio C : H : O $= 1 : 2 : 1$, so the empirical formula is $\text{CH}_2\text{O}$.

    日本語

    A compound 化合物 is a substance made of two or more elements chemically joined.

    Charges and formulas of ionic compounds

    In an ionic compound 离子化合物 the total positive charge balances the total negative charge, so the compound is neutral overall.

    You can predict the charge of many ions from the element's position in the Periodic Table:

    Group 1 2 13 15 16 17
    Usual ion charge $+1$ $+2$ $+3$ $-3$ $-2$ $-1$

    Hydrogen forms $\text{H}^+$, and the Group 18 noble gases do not normally form ions.

    Some ions you must know by name and formula:

    Name Formula
    nitrate $\text{NO}_3^{-}$
    carbonate $\text{CO}_3^{2-}$
    sulfate $\text{SO}_4^{2-}$
    hydroxide $\text{OH}^{-}$
    ammonium $\text{NH}_4^{+}$
    zinc $\text{Zn}^{2+}$
    silver $\text{Ag}^{+}$
    hydrogencarbonate $\text{HCO}_3^{-}$
    phosphate $\text{PO}_4^{3-}$

    For a metal that can have more than one charge, a Roman numeral shows the oxidation number 氧化数. For example, iron(II) is $\text{Fe}^{2+}$ and iron(III) is $\text{Fe}^{3+}$. To write a formula, balance the charges: iron(III) oxide is $\text{Fe}_2\text{O}_3$, because two $\text{Fe}^{3+}$ balance three $\text{O}^{2-}$.

    Equations and state symbols

    A chemical equation must be balanced 配平 — the same number of each kind of atom on both sides. Add state symbols 状态符号 to show the state of each species: (s) solid, (l) liquid, (g) gas, and (aq) aqueous 水溶液 (dissolved in water).

    An ionic equation 离子方程式 shows only the ions and molecules that actually change. The ions that do not change are spectator ions 旁观离子, and you leave them out. For example, the reaction that forms silver chloride is:

    $$\text{Ag}^{+}(\text{aq}) + \text{Cl}^{-}(\text{aq}) \rightarrow \text{AgCl}(\text{s})$$

    Empirical and molecular formulas

    The empirical formula 实验式 is the simplest whole-number ratio of the atoms of each element in a compound. The molecular formula 分子式 shows the actual number of atoms of each element in one molecule.

    For example, ethane has empirical formula $\text{CH}_3$ but molecular formula $\text{C}_2\text{H}_6$.

    Hydrated and anhydrous solids

    Some solids hold water inside their crystals. This water is the water of crystallisation 结晶水. A solid that contains it is hydrated 水合的; the same solid with the water removed is anhydrous 无水的.

    For example, hydrated copper(II) sulfate is $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$. Heating it drives off the water to leave anhydrous $\text{CuSO}_4$.

    Calculating empirical and molecular formulas

    To find the empirical formula from masses (or percentages by mass):

    1. divide each element's mass by its $A_r$ to get the moles.
    2. divide all the mole values by the smallest one.
    3. round to the nearest whole numbers — that ratio is the empirical formula.

    To get the molecular formula, you also need $M_r$. Find how many times the empirical formula mass fits into $M_r$, then multiply the formula by that number.

    Worked example. A compound is $40.0\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen by mass. Find its empirical formula. ($A_r$: C $= 12$, H $= 1$, O $= 16$.)

    Take $100\ \text{g}$ and divide each mass by its $A_r$ to get moles: C $= 40.0/12 = 3.33$, H $= 6.7/1 = 6.7$, O $= 53.3/16 = 3.33$. Dividing through by the smallest ($3.33$) gives a ratio C : H : O $= 1 : 2 : 1$, so the empirical formula is $\text{CH}_2\text{O}$.

    The worked example laid out as a table: percentage by mass divided by Ar gives moles, dividing by the smallest gives the ratio 1 to 2 to 1, so the empirical formula is CH2O
    The empirical-formula recipe: divide by Ar, divide by the smallest, read off the ratio
    Explore · ⁨探索⁩

    Equation balancing route · ⁨方程式の平衡化处理法⁩

    Follow atoms through a chemical equation so both sides match. · ⁨化学方程式を通じて原子を追跡し、両辺が一致するようにする。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    compound/ˈkɒmpaʊnd/ 化合物
    ionic compound/aɪˈɒnɪk ˈkɒmpaʊnd/ イオン化合物
    oxidation number/ˌɒksɪˈdeɪʃn ˈnʌmbə/ 酸化数
    balanced/ˈbælənst/ バランス取れた
    state symbols/steɪt ˈsɪmblz/ 状態記号
    aqueous/ˈeɪkwɪəs/ 水溶液(_aqueous_)
    ionic equation/aɪˈɒnɪk ɪˈkweɪʒn/ イオン方程式
    spectator ions/spekˈteɪtə ˈaɪɒnz/ 観測イオン
    empirical formula/emˈpɪrɪkl ˈfɔːmjʊlə/ 経験式
    molecular formula/məˈlekjʊlə ˈfɔːmjʊlə/ 分子式
    water of crystallisation/ˈwɔːtə ɒv ˌkrɪstəlaɪˈzeɪʃn/ 結晶水
    hydrated/haɪˈdreɪtɪd/ 水和物(_attrated_)
    anhydrous/ænˈhaɪdrəs/ 無水物(_anhydrous_)
    2.4

    Reacting masses and volumes

    Syllabus · ⁨シラバス⁩
    English
    1. perform calculations including use of the mole concept, involving: (a) reacting masses (from formulas and equations) including percentage yield calculations (b) volumes of gases (e.g. in the burning of hydrocarbons) (c) volumes and concentrations of solutions (d) limiting reagent and excess reagent (When performing calculations, candidates’ answers should reflect the number of significant figures given or asked for in the question. When rounding up or down, candidates should ensure that significant figures are neither lost unnecessarily nor used beyond what is justified (see also Mathematical requirements section).) (e) deduce stoichiometric relationships from calculations such as those in 2.4.1(a)–(d)
    日本語
    1. 以下の計算を行う(モル概念の使用を含む): (a) 反応物の質量(化学式および方程式より)、収率計算を含む (b) 気体の体積(例:炭化水素の燃焼時) (c) 溶液の体積および濃度 (d) 限剤および過剰剤 (計算を行う際、受験者の回答は問題で指定または要求された有効数字の数と一致すべきである。四捨五入を行う際は、受験者は有効数字が無駄に失われたり、正当な範囲を超えて使用されたりしないように注意すべきである(数学的要項セクションも参照)。) (e) 2.4.1(a)–(d)のような計算から化学量論的関係を導く

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Reacting masses and percentage yield

    The numbers in front of each species in a balanced equation give the mole ratio — this is the stoichiometry 化学计量. To find a reacting mass: change the known mass to moles, use the mole ratio to find the moles you want, then change back to mass.

    Worked example. What mass of magnesium oxide forms when $4.8\ \text{g}$ of magnesium burns completely? $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. ($A_r$: Mg $= 24$, O $= 16$.)

    Moles of Mg $= 4.8/24 = 0.20\ \text{mol}$. The ratio Mg : MgO is $1 : 1$, so $0.20\ \text{mol}$ of MgO forms. Its molar mass is $24 + 16 = 40\ \text{g mol}^{-1}$, so

    $$m = nM = 0.20 \times 40 = 8.0\ \text{g}.$$

    In real reactions you usually get less product than the maximum. The percentage yield 产率 compares the amount you actually made with the most you could make:

    $$\text{percentage yield} = \frac{\text{actual amount of product}}{\text{maximum possible amount}} \times 100\%$$

    Limiting and excess reagent

    When two reactants are mixed, one usually runs out first. The limiting reagent 限量试剂 is the one that runs out — it decides how much product forms. The other is the excess reagent 过量试剂, because there is more than enough of it. Always base the calculation on the limiting reagent.

    To find it: work out the moles of each reactant, divide each by its number in the equation, and the smallest result is the limiting reagent.

    Volumes of gases

    At the same temperature and pressure, equal volumes of any gases contain equal numbers of molecules. At room temperature and pressure (r.t.p.), one mole of any gas takes up $24.0\ \text{dm}^3$, so:

    $$n = \frac{V}{24.0}\qquad (V \text{ in } \text{dm}^3 \text{ at r.t.p.})$$

    This is used when burning hydrocarbons 碳氢化合物 (compounds of only carbon and hydrogen), where you compare gas volumes.

    Volumes and concentrations of solutions

    The concentration 浓度 of a solution is the amount of solute 溶质 in each cubic decimetre of solution 溶液, measured in $\text{mol dm}^{-3}$:

    $$n = c \times V$$

    where $c$ is the concentration and $V$ is the volume in $\text{dm}^3$. Remember that $1000\ \text{cm}^3 = 1\ \text{dm}^3$.

    Worked example. In a titration, $25.0\ \text{cm}^3$ of sodium hydroxide solution is exactly neutralised by $20.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ hydrochloric acid: $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$. Find the concentration of the sodium hydroxide.

    Moles of HCl $= cV = 0.100 \times \dfrac{20.0}{1000} = 2.00 \times 10^{-3}\ \text{mol}$. The ratio is $1 : 1$, so there are $2.00 \times 10^{-3}\ \text{mol}$ of NaOH in $25.0\ \text{cm}^3$:

    $$c = \frac{n}{V} = \frac{2.00 \times 10^{-3}}{25.0/1000} = 0.0800\ \text{mol dm}^{-3}.$$

    This is the basis of a titration 滴定, where you find an unknown concentration by reacting it with a solution whose concentration you already know.

    Significant figures

    Give your answer to a sensible number of significant figures 有效数字 — usually match the data in the question. Do not write more digits than the data supports, and do not round so early that you lose accuracy.

    日本語
    A titration being carried out
    A titration finds reacting volumes precisely.

    Reacting masses and percentage yield

    The numbers in front of each species in a balanced equation give the mole ratio — this is the stoichiometry 化学计量. To find a reacting mass: change the known mass to moles, use the mole ratio to find the moles you want, then change back to mass.

    Worked example. What mass of magnesium oxide forms when $4.8\ \text{g}$ of magnesium burns completely? $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. ($A_r$: Mg $= 24$, O $= 16$.)

    Moles of Mg $= 4.8/24 = 0.20\ \text{mol}$. The ratio Mg : MgO is $1 : 1$, so $0.20\ \text{mol}$ of MgO forms. Its molar mass is $24 + 16 = 40\ \text{g mol}^{-1}$, so

    $$m = nM = 0.20 \times 40 = 8.0\ \text{g}.$$
    The mole bridge for the magnesium example: 4.8 grams of magnesium divided by Ar gives 0.20 moles, the 1 to 1 ratio gives 0.20 moles of magnesium oxide, times Mr gives 8.0 grams
    The mole bridge: mass to moles, ratio, then back to mass

    In real reactions you usually get less product than the maximum. The percentage yield 产率 compares the amount you actually made with the most you could make:

    $$\text{percentage yield} = \frac{\text{actual amount of product}}{\text{maximum possible amount}} \times 100\%$$

    Limiting and excess reagent

    When two reactants are mixed, one usually runs out first. The limiting reagent 限量试剂 is the one that runs out — it decides how much product forms. The other is the excess reagent 过量试剂, because there is more than enough of it. Always base the calculation on the limiting reagent.

    To find it: work out the moles of each reactant, divide each by its number in the equation, and the smallest result is the limiting reagent.

    Burning 4 hydrogen molecules with 1 oxygen molecule makes 2 water molecules and leaves 2 hydrogen molecules unreacted
    The limiting reagent runs out first and decides how much product forms; the leftover reactant is in excess

    Volumes of gases

    At the same temperature and pressure, equal volumes of any gases contain equal numbers of molecules. At room temperature and pressure (r.t.p.), one mole of any gas takes up $24.0\ \text{dm}^3$, so:

    $$n = \frac{V}{24.0}\qquad (V \text{ in } \text{dm}^3 \text{ at r.t.p.})$$
    Two equal-sized boxes, one of hydrogen and one of carbon dioxide, each holding six molecules
    Equal volumes of gases at the same temperature and pressure hold equal numbers of molecules, whatever the gas

    This is used when burning hydrocarbons 碳氢化合物 (compounds of only carbon and hydrogen), where you compare gas volumes.

    Volumes and concentrations of solutions

    The concentration 浓度 of a solution is the amount of solute 溶质 in each cubic decimetre of solution 溶液, measured in $\text{mol dm}^{-3}$:

    $$n = c \times V$$

    where $c$ is the concentration and $V$ is the volume in $\text{dm}^3$. Remember that $1000\ \text{cm}^3 = 1\ \text{dm}^3$.

    Worked example. In a titration, $25.0\ \text{cm}^3$ of sodium hydroxide solution is exactly neutralised by $20.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ hydrochloric acid: $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$. Find the concentration of the sodium hydroxide.

    Moles of HCl $= cV = 0.100 \times \dfrac{20.0}{1000} = 2.00 \times 10^{-3}\ \text{mol}$. The ratio is $1 : 1$, so there are $2.00 \times 10^{-3}\ \text{mol}$ of NaOH in $25.0\ \text{cm}^3$:

    $$c = \frac{n}{V} = \frac{2.00 \times 10^{-3}}{25.0/1000} = 0.0800\ \text{mol dm}^{-3}.$$

    This is the basis of a titration 滴定, where you find an unknown concentration by reacting it with a solution whose concentration you already know.

    A burette clamped on a stand above a conical flask on a white tile
    In a titration a burette adds a solution of known concentration to the unknown in the conical flask, until the indicator changes

    Significant figures

    Give your answer to a sensible number of significant figures 有效数字 — usually match the data in the question. Do not write more digits than the data supports, and do not round so early that you lose accuracy.

    Explore · ⁨探索⁩

    Reacting mass route · ⁨反応質量ルート⁩

    Follow a balanced equation from known mass to predicted product mass. · ⁨平衡化学方程式に従って、既知の質量から予測される生成物の質量へ進める。⁩

    Explore · ⁨探索⁩

    Gas volume lab · ⁨気体体積実験⁩

    n = V / 24 dm3

    Change gas volume and see moles scale at room conditions. · ⁨室温条件下で気体体積を変化させ、物質量がどのように比例するかを確認する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    stoichiometry/ˌstəʊɪkɪˈɒmətri/ 化学量論
    percentage yield/pəˈsentɪdʒ jiːld/ 収率
    limiting reagent/ˈlɪmɪtɪŋ rɪˈeɪdʒənt/ 限界試薬
    excess reagent/ekˈses rɪˈeɪdʒənt/ 過剰試薬
    hydrocarbons/ˈhaɪdrəkɑːbənz/ 炭化水素
    concentration/ˌkɒnsənˈtreɪʃn/ 濃度
    solute/ˈsɒljuːt/ 溶質
    solution/səˈluːʃn/ 溶液
    titration/taɪˈtreɪʃn/ 滴定
    significant figures/sɪɡˈnɪfɪkənt ˈfɪɡəz/ 有効数字
    Watch lesson · ⁨レッスンを視聴⁩
    2.4

    Exam tips

    • Convert volumes to $\text{dm}^3$ before using concentration ($25.0\ \text{cm}^3 = 0.0250\ \text{dm}^3$); the missing $\div 1000$ is the most common titration error.
    • Use the balancing numbers as the mole ratio between species — never the $M_r$ values.
    • Empirical formula: divide each mass/percentage by $A_r$, then by the smallest, then scale to whole numbers; use $M_r$ to reach the molecular formula.
    • For gases at r.t.p. use $\text{volume} = \text{moles} \times 24\ \text{dm}^3$; quote the equation you use every time.
    • Give the answer to the same significant figures as the data (usually 3) and always include units.
  • 3

    Chemical bonding · ⁨化学結合⁩

    Watch lesson · ⁨レッスンを視聴⁩
    3.1

    Electronegativity

    Syllabus · ⁨シラバス⁩
    English
    1. define electronegativity as the power of an atom to attract electrons to itself
    2. explain the factors influencing the electronegativities of the elements in terms of nuclear charge, atomic radius and shielding by inner shells and sub-shells
    3. state and explain the trends in electronegativity across a period and down a group of the Periodic Table
    4. use the differences in Pauling electronegativity values to predict the formation of ionic and covalent bonds (the presence of covalent character in some ionic compounds will not be assessed) (Pauling electronegativity values will be given where necessary)
    日本語
    1. 電気陰性度を、原子が自身のほうに電子を引き寄せる力を指すと定義する
    2. 核電荷、原子半径、内殻および亜殻による遮蔽効果の観点から、元素の電気陰性度に影響を与える要因を説明する
    3. 周期表において周期方向および族方向に現れる電気陰性度の傾向を述べ、説明する
    4. ポリング電気陰性度の値の差を用いて、イオン結合および共有結合の形成を予測する(一部のイオン化合物に見られる共有結合性の存在は評価対象外とする)(必要な場合にポリング電気陰性度の値が提示される)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Electronegativity 电负性 is the power of an atom to attract the electrons 电子 in a bond towards itself.

    Three factors decide how electronegative an atom is:

    • nuclear charge 核电荷: more protons pull the bonding electrons more strongly.
    • atomic radius 原子半径: the closer the bond is to the nucleus, the stronger the pull.
    • shielding 屏蔽 by inner shells and sub-shells: more inner electrons weaken the pull on the bonding electrons.

    So electronegativity rises across a period (more nuclear charge, smaller radius) and falls down a group (larger radius, more shielding). Fluorine is the most electronegative element.

    You can use the difference in Pauling electronegativity 鲍林电负性 values to predict the bond type. A large difference gives an ionic bond; a small difference gives a covalent bond.

    日本語

    Electronegativity 电负性 is the power of an atom to attract the electrons 电子 in a bond towards itself.

    Three factors decide how electronegative an atom is:

    • nuclear charge 核电荷: more protons pull the bonding electrons more strongly.
    • atomic radius 原子半径: the closer the bond is to the nucleus, the stronger the pull.
    • shielding 屏蔽 by inner shells and sub-shells: more inner electrons weaken the pull on the bonding electrons.

    So electronegativity rises across a period (more nuclear charge, smaller radius) and falls down a group (larger radius, more shielding). Fluorine is the most electronegative element.

    A grid shaded from pale at the bottom-left to deep at the top-right, with fluorine marked in the top-right corner
    Electronegativity rises across a period and falls down a group, so fluorine is the most electronegative element

    You can use the difference in Pauling electronegativity 鲍林电负性 values to predict the bond type. A large difference gives an ionic bond; a small difference gives a covalent bond.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    electronegativity/ɪˌlektrəʊŋɡəˈtɪvɪti/ 電気陰性度
    electron/ɪˈlektrɒn/ 電子
    nuclear charge/ˈnjuːklɪə tʃɑːdʒ/ 核電荷
    atomic radius/əˈtɒmɪk ˈreɪdɪəs/ 原子半径
    shielding/ˈʃiːldɪŋ/ 遮蔽効果
    Pauling electronegativity/ˈpɔːlɪŋ ɪˌlektrəʊŋɡəˈtɪvɪti/ ポーリング電気陰性度
    3.2

    Ionic bonding

    Syllabus · ⁨シラバス⁩
    English
    1. define ionic bonding as the electrostatic attraction between oppositely charged ions (positively charged cations and negatively charged anions)
    2. describe ionic bonding including the examples of sodium chloride, magnesium oxide and calcium fluoride
    日本語
    1. イオン結合を、反対電荷を持つイオン(正電荷を持つ陽イオンおよび負電荷を持つ陰イオン)間の静電的引力として定義する
    2. イオン結合を記述する。塩化ナトリウム、酸化マグネシウム、フッ化カルシウムの例を含む

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Ionic bonding: electron transfer

    Ionic bonding 离子键 is the electrostatic attraction 静电引力 between oppositely charged ions 离子 — positive cations 阳离子 and negative anions 阴离子.

    It forms when a metal gives electrons to a non-metal. Good examples are sodium chloride ($\text{NaCl}$), magnesium oxide ($\text{MgO}$) and calcium fluoride ($\text{CaF}_2$). The ions pack into a regular giant lattice 晶格, held together by the attraction in every direction.

    日本語
    Ionic bonding: electron transfer

    Ionic bonding 离子键 is the electrostatic attraction 静电引力 between oppositely charged ions 离子 — positive cations 阳离子 and negative anions 阴离子.

    It forms when a metal gives electrons to a non-metal. Good examples are sodium chloride ($\text{NaCl}$), magnesium oxide ($\text{MgO}$) and calcium fluoride ($\text{CaF}_2$). The ions pack into a regular giant lattice 晶格, held together by the attraction in every direction.

    A dot-and-cross diagram showing a sodium atom with one cross transferring it to a chlorine atom, giving Na+ and Cl- ions in brackets
    Ionic bonding in NaCl: sodium transfers its single outer electron to chlorine, giving Na$^+$ and a full-octet Cl$^-$
    A clear-grey rock salt crystal on a black background, made of many small cube-shaped blocks with right-angled faces
    A real crystal of rock salt (halite, NaCl); the cubic shapes mirror the giant ionic lattice inside
    Explore · ⁨探索⁩

    Forming an ionic bond (NaCl) · ⁨イオン結合の形成(NaCl)⁩

    Step through it. A metal hands its outer electron to a non-metal; the oppositely charged ions then attract in a giant lattice. · ⁨形成過程:金属が最外殻電子を非金属に渡し、反対の電荷を持つイオン同士が巨大格子内で引き合います。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    ionic bonding/aɪˈɒnɪk ˈbɒndɪŋ/ イオン結合
    electrostatic attraction/ɪˌlektrəʊˈstætɪk əˈtrækʃn/ 静電気的引力
    ion/ˈaɪɒn/ イオン
    cation/ˈkætaɪən/ カチオン
    anion/ˈænaɪən/ アニオン
    lattice/ˈlætɪs/ 格子
    3.3

    Metallic bonding

    Syllabus · ⁨シラバス⁩
    English
    1. define metallic bonding as the electrostatic attraction between positive metal ions and delocalised electrons
    日本語
    1. 金属結合を、正の金属イオンと非局在化電子の間の静電的引力として定義する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Metallic bonding 金属键 is the electrostatic attraction between positive metal ions and a "sea" of delocalised electrons 离域电子.

    The outer electrons are free to move through the whole metal. This explains why metals conduct electricity and are strong.

    日本語

    Metallic bonding 金属键 is the electrostatic attraction between positive metal ions and a "sea" of delocalised electrons 离域电子.

    The outer electrons are free to move through the whole metal. This explains why metals conduct electricity and are strong.

    A regular grid of positive metal ions with small electrons scattered in the gaps between them
    Metallic bonding: positive metal ions sit in a sea of delocalised electrons that are free to move
    Explore · ⁨探索⁩

    Inside a metal — and why it behaves that way · ⁨金属内部の構造と、その物理的性質の理由⁩

    Step through it. Positive ions sit in a shared sea of delocalised electrons. That one picture explains conduction, malleability, and strength. · ⁨形成過程:正のイオンが非局在電子の海の中に配置されています。この一枚のモデルで、伝導性、延性、強度が説明できます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    metallic bonding/məˈtælɪk ˈbɒndɪŋ/ 金属結合
    delocalised electrons/dɪˈlɒkəlaɪzd ɪˈlektrɒnz/ 非局在電子
    3.4

    Covalent and coordinate bonding

    Syllabus · ⁨シラバス⁩
    English
    1. define covalent bonding as electrostatic attraction between the nuclei of two atoms and a shared pair of electrons (a) describe covalent bonding in molecules including: • hydrogen, $\text{H}_2$ • oxygen, $\text{O}_2$ • nitrogen, $\text{N}_2$ • chlorine, $\text{Cl}_2$ • hydrogen chloride, $\text{HCl}$ • carbon dioxide, $\text{CO}_2$ • ammonia, $\text{NH}_3$ • methane, $\text{CH}_4$ • ethane, $\text{C}_2\text{H}_6$ • ethene, $\text{C}_2\text{H}_4$ (b) understand that elements in period 3 can expand their octet including in the compounds sulfur dioxide, $\text{SO}_2$, phosphorus pentachloride, $\text{PCl}_5$, and sulfur hexafluoride, $\text{SF}_6$ (c) describe coordinate (dative covalent) bonding, including in the reaction between ammonia and hydrogen chloride gases to form the ammonium ion, $\text{NH}_4^+$, and in the $\text{Al}_2\text{Cl}_6$ molecule
    2. (a) describe covalent bonds in terms of orbital overlap giving $\sigma$ and $\pi$ bonds: • $\sigma$ bonds are formed by direct overlap of orbitals between the bonding atoms • $\pi$ bonds are formed by the sideways overlap of adjacent p orbitals above and below the $\sigma$ bond (b) describe how the $\sigma$ and $\pi$ bonds form in molecules including $\text{H}_2$, $\text{C}_2\text{H}_6$, $\text{C}_2\text{H}_4$, $\text{HCN}$ and $\text{N}_2$ (c) use the concept of hybridisation to describe $\text{sp}$, $\text{sp}^2$ and $\text{sp}^3$ orbitals
    3. (a) define the terms: • bond energy as the energy required to break one mole of a particular covalent bond in the gaseous state • bond length as the internuclear distance of two covalently bonded atoms (b) use bond energy values and the concept of bond length to compare the reactivity of covalent molecules
    日本語
    1. 共有結合を、2つの原子の原子核と共有電子対の間の静電的引力として定義する (a) 以下の分子における共有結合を記述する: • 水素、 $\text{H}_2$ • 酸素、 $\text{O}_2$ • 窒素、 $\text{N}_2$ • 塩素、 $\text{Cl}_2$ • 塩化水素、 $\text{HCl}$ • 二酸化炭素、 $\text{CO}_2$ • アンモニア、 $\text{NH}_3$ • メタン、 $\text{CH}_4$ • エタン、 $\text{C}_2\text{H}_6$ • エチレン、 $\text{C}_2\text{H}_4$ (b) 第3周期の元素が、二酸化硫黄 $\text{SO}_2$、五塩化リン $\text{PCl}_5$、六フッ化硫黄 $\text{SF}_6$ のような化合物においてオクテット拡張を行うことができることを理解する (c) 配位結合(供与共有結合)を記述する。アンモニアと塩化水素ガスの反応によりアンモニウムイオン $\text{NH}_4^+$ が生成される場合、ならびに $\text{Al}_2\text{Cl}_6$ 分子における case を含む
    2. (a) 軌道の重なりについて記述し、$\sigma$および$\pi$結合を説明すること: • $\sigma$結合は結合原子間の軌道の直接重なりによって形成される • $\pi$結合は、$\sigma$結合の上下面に隣接するp軌道の側方重なりにより形成される (b) $\sigma$および$\pi$結合が、$\text{H}_2$、$\text{C}_2\text{H}_6$、$\text{C}_2\text{H}_4$、$\text{HCN}$、$\text{N}_2$を含む分子内でどのように形成されるかを記述すること (c) 混成の概念を用いて、$\text{sp}$、$\text{sp}^2$、$\text{sp}^3$軌道を記述すること。
    3. (a) 以下の用語を定義せよ。 • 結合エネルギー: 気体状態で特定の共有結合1モルを切断するために必要なエネルギー • 結合距離: 共有結合した2つの原子核間の距離 (b) 結合エネルギーの値および結合距離の概念を用いて、共有結合分子の反応性を比較せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Covalent bonding: a shared pair

    Covalent bonding 共价键 is the electrostatic attraction between the nuclei of two atoms and a shared pair of electrons.

    Simple molecules with covalent bonds include $\text{H}_2$, $\text{O}_2$, $\text{N}_2$, $\text{Cl}_2$, $\text{HCl}$, $\text{CO}_2$, $\text{NH}_3$, $\text{CH}_4$, $\text{C}_2\text{H}_6$ and $\text{C}_2\text{H}_4$. A double bond shares two pairs; a triple bond (as in $\text{N}_2$) shares three pairs.

    Atoms in Period 3 and below can expand the octet 扩展八隅体 — hold more than eight electrons in their outer shell. Examples are $\text{SO}_2$, $\text{PCl}_5$ and $\text{SF}_6$.

    A coordinate bond 配位键 (also called a dative covalent bond) is a covalent bond where both shared electrons come from the same atom. For example, when ammonia and hydrogen chloride gases meet, the lone pair on the nitrogen forms a coordinate bond to $\text{H}^+$, making the ammonium ion $\text{NH}_4^+$. Coordinate bonds also join the two halves of the $\text{Al}_2\text{Cl}_6$ molecule.

    Sigma and pi bonds

    Covalent bonds form when orbitals 轨道 overlap:

    • a sigma bond σ键 forms by the direct, head-on overlap 重叠 of orbitals between the two atoms.
    • a pi bond π键 forms by the sideways overlap of two p orbitals, above and below the sigma bond.

    A single bond is one sigma bond. A double bond (as in $\text{C}_2\text{H}_4$) is one sigma plus one pi bond. A triple bond (as in $\text{N}_2$ and $\text{HCN}$) is one sigma plus two pi bonds.

    Hybridisation

    Hybridisation 杂化 mixes orbitals in the same shell to make new, equal orbitals for bonding:

    • $\text{sp}$: two equal orbitals, used in a linear molecule.
    • $\text{sp}^2$: three equal orbitals, used in a flat molecule like $\text{C}_2\text{H}_4$.
    • $\text{sp}^3$: four equal orbitals, used in $\text{CH}_4$.

    Bond energy and bond length

    • bond energy 键能 is the energy needed to break one mole of a particular covalent bond in the gas state.
    • bond length 键长 is the distance between the centres of the two bonded atoms.

    A shorter bond is usually stronger (higher bond energy). Triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds. Stronger bonds make a molecule harder to react.

    日本語
    Covalent bonding: a shared pair

    Covalent bonding 共价键 is the electrostatic attraction between the nuclei of two atoms and a shared pair of electrons.

    Simple molecules with covalent bonds include $\text{H}_2$, $\text{O}_2$, $\text{N}_2$, $\text{Cl}_2$, $\text{HCl}$, $\text{CO}_2$, $\text{NH}_3$, $\text{CH}_4$, $\text{C}_2\text{H}_6$ and $\text{C}_2\text{H}_4$. A double bond shares two pairs; a triple bond (as in $\text{N}_2$) shares three pairs.

    Atoms in Period 3 and below can expand the octet 扩展八隅体 — hold more than eight electrons in their outer shell. Examples are $\text{SO}_2$, $\text{PCl}_5$ and $\text{SF}_6$.

    A coordinate bond 配位键 (also called a dative covalent bond) is a covalent bond where both shared electrons come from the same atom. For example, when ammonia and hydrogen chloride gases meet, the lone pair on the nitrogen forms a coordinate bond to $\text{H}^+$, making the ammonium ion $\text{NH}_4^+$. Coordinate bonds also join the two halves of the $\text{Al}_2\text{Cl}_6$ molecule.

    Ammonia's lone pair forming a new bond to an incoming hydrogen ion, shown as an arrow, to make the ammonium ion in brackets with a plus charge
    A coordinate (dative) bond: nitrogen's lone pair forms the fourth N–H bond, both electrons coming from N

    Sigma and pi bonds

    Covalent bonds form when orbitals 轨道 overlap:

    • a sigma bond σ键 forms by the direct, head-on overlap 重叠 of orbitals between the two atoms.
    • a pi bond π键 forms by the sideways overlap of two p orbitals, above and below the sigma bond.

    A single bond is one sigma bond. A double bond (as in $\text{C}_2\text{H}_4$) is one sigma plus one pi bond. A triple bond (as in $\text{N}_2$ and $\text{HCN}$) is one sigma plus two pi bonds.

    On the left two orbitals overlapping head-on between the nuclei; on the right two vertical p orbitals overlapping sideways above and below the axis
    A $\sigma$ bond forms by direct head-on overlap; a $\pi$ bond forms by the sideways overlap of two p orbitals, above and below

    Hybridisation

    Hybridisation 杂化 mixes orbitals in the same shell to make new, equal orbitals for bonding:

    • $\text{sp}$: two equal orbitals, used in a linear molecule.
    • $\text{sp}^2$: three equal orbitals, used in a flat molecule like $\text{C}_2\text{H}_4$.
    • $\text{sp}^3$: four equal orbitals, used in $\text{CH}_4$.

    Bond energy and bond length

    • bond energy 键能 is the energy needed to break one mole of a particular covalent bond in the gas state.
    • bond length 键长 is the distance between the centres of the two bonded atoms.

    A shorter bond is usually stronger (higher bond energy). Triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds. Stronger bonds make a molecule harder to react.

    Explore · ⁨探索⁩

    Sharing a pair of electrons · ⁨電子対の共有⁩

    Step through a covalent bond: two atoms overlap and share a pair so each reaches a full shell — when the two pull equally the bond is non-polar. · ⁨共有結合の形成:2つの原子が重なり合い、電子対を共有してそれぞれ完全な外殻を得る。2つの原子が等しく引き合う場合、結合は非極性となる。⁩

    Explore · ⁨探索⁩

    Covalent bonding (sharing) · ⁨共有結合(共有)⁩

    Two non-metal atoms overlap and share a pair of electrons — counted for both — so each reaches a full outer shell. O₂ shares two pairs (a double bond). · ⁨2つの非金属原子が重なり合い、電子対を共有する — 両方の原子でカウントされるため、それぞれ最外殻が満たされる。O₂ は2組の電子を共有する(二重結合)。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    covalent bonding/ˈkəʊvələnt ˈbɒndɪŋ/ 共有結合
    expand the octet/ekˈspænd ðɪ ɒkˈtet/ オクテット拡張
    coordinate bond/kəʊˈɔːdɪnət bɒnd/ 配位結合
    orbital/ˈɔːbɪtl/ 軌道
    sigma bond/ˈsɪɡmə bɒnd/ シグマ結合
    overlap/ˌəʊvəˈlæp/ 重なり
    pi bond/paɪ bɒnd/ π結合
    hybridisation/ˌhaɪbrɪdaɪˈzeɪʃn/ 雑種交配
    bond energy/bɒnd ˈenədʒi/ 結合エネルギー
    bond length/bɒnd leŋθ/ 結合長
    3.5

    Shapes of molecules

    Syllabus · ⁨シラバス⁩
    English
    1. state and explain the shapes of, and bond angles in, molecules by using VSEPR theory, including as simple examples: • $\text{BF}_3$ (trigonal planar, 120°) • $\text{CO}_2$ (linear, 180°) • $\text{CH}_4$ (tetrahedral, 109.5°) • $\text{NH}_3$ (pyramidal, 107°) • $\text{H}_2\text{O}$ (non-linear, 104.5°) • $\text{SF}_6$ (octahedral, 90°) • $\text{PF}_5$ (trigonal bipyramidal, 120° and 90°)
    2. predict the shapes of, and bond angles in, molecules and ions analogous to those specified in 3.5.1
    日本語
    1. VSEPR理論を用いて、分子の形状および結合角を記述・説明せよ。簡単な例として以下を含める: • $\text{BF}_3$ (平面三角形, 120°) • $\text{CO}_2$ (直線形, 180°) • $\text{CH}_4$ (正四面体形, 109.5°) • $\text{NH}_3$ (ピラミッド形, 107°) • $\text{H}_2\text{O}$ (非直線形, 104.5°) • $\text{SF}_6$ (正八面体形, 90°) • $\text{PF}_5$ (三角双錐形, 120°および90°)
    2. 3.5.1で指定された分子やイオンに類似する分子およびイオンの形状および結合角を予測せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    To work out a shape, use VSEPR theory 价层电子对互斥理论: the pairs of electrons around the central atom push apart as far as possible, because like charges repel.

    A lone pair 孤对电子 (not in a bond) pushes more strongly than a bonding pair 成键电子对. Each lone pair squeezes the bond angle 键角 by about $2.5°$.

    Molecule Shape Bond angle
    $\text{CO}_2$ linear 直线形 $180°$
    $\text{BF}_3$ trigonal planar 平面三角形 $120°$
    $\text{CH}_4$ tetrahedral 四面体形 $109.5°$
    $\text{NH}_3$ pyramidal 三角锥形 $107°$
    $\text{H}_2\text{O}$ bent 角形 $104.5°$
    $\text{PF}_5$ trigonal bipyramidal 三角双锥形 $120°$ and $90°$
    $\text{SF}_6$ octahedral 八面体形 $90°$

    $\text{NH}_3$ has one lone pair and $\text{H}_2\text{O}$ has two, which is why their angles drop below the $109.5°$ of $\text{CH}_4$. You can predict the shapes of similar molecules and ions in the same way.

    Worked example. Predict the shape and bond angle of $\text{NH}_3$ and of $\text{H}_2\text{O}$. Nitrogen has 5 outer electrons and forms 3 bonds, leaving 3 bonding pairs and 1 lone pair - four pairs in total, so they start from the tetrahedral $109.5°$. The lone pair repels more strongly and squeezes the angle by about $2.5°$, giving a pyramidal shape at about $107°$. Oxygen forms 2 bonds and keeps 2 lone pairs: still four pairs, but now two squeezes, so the shape is bent at about $104.5°$. Count all the pairs to fix the basic geometry, subtract $2.5°$ for each lone pair, and name the shape from the atoms - $\text{NH}_3$ has four pairs but is pyramidal, not tetrahedral.

    日本語
    A ball-and-stick molecular model
    A molecular model shows the three-dimensional shape of a covalent molecule.

    To work out a shape, use VSEPR theory 价层电子对互斥理论: the pairs of electrons around the central atom push apart as far as possible, because like charges repel.

    A lone pair 孤对电子 (not in a bond) pushes more strongly than a bonding pair 成键电子对. Each lone pair squeezes the bond angle 键角 by about $2.5°$.

    Molecule Shape Bond angle
    $\text{CO}_2$ linear 直线形 $180°$
    $\text{BF}_3$ trigonal planar 平面三角形 $120°$
    $\text{CH}_4$ tetrahedral 四面体形 $109.5°$
    $\text{NH}_3$ pyramidal 三角锥形 $107°$
    $\text{H}_2\text{O}$ bent 角形 $104.5°$
    $\text{PF}_5$ trigonal bipyramidal 三角双锥形 $120°$ and $90°$
    $\text{SF}_6$ octahedral 八面体形 $90°$

    $\text{NH}_3$ has one lone pair and $\text{H}_2\text{O}$ has two, which is why their angles drop below the $109.5°$ of $\text{CH}_4$. You can predict the shapes of similar molecules and ions in the same way.

    Seven molecular shapes drawn with a central atom and bonded atoms: linear, trigonal planar, tetrahedral, pyramidal, bent, trigonal bipyramidal and octahedral, each with its bond angle
    The seven shapes from VSEPR theory; the lone pairs on NH$_3$ and H$_2$O push harder, squeezing the bond angle below $109.5°$

    Worked example. Predict the shape and bond angle of $\text{NH}_3$ and of $\text{H}_2\text{O}$. Nitrogen has 5 outer electrons and forms 3 bonds, leaving 3 bonding pairs and 1 lone pair - four pairs in total, so they start from the tetrahedral $109.5°$. The lone pair repels more strongly and squeezes the angle by about $2.5°$, giving a pyramidal shape at about $107°$. Oxygen forms 2 bonds and keeps 2 lone pairs: still four pairs, but now two squeezes, so the shape is bent at about $104.5°$. Count all the pairs to fix the basic geometry, subtract $2.5°$ for each lone pair, and name the shape from the atoms - $\text{NH}_3$ has four pairs but is pyramidal, not tetrahedral.

    Explore · ⁨探索⁩

    Shape from bonding and lone pairs · ⁨結合電子対と非共有電子対からの形状⁩

    Count the bonding pairs and lone pairs around the central atom; they repel into the shape with least strain. Three bonds and one lone pair give a pyramid, like ammonia (NH3). · ⁨中心原子周围的の結合電子対と非共有電子対を数え、これらは最小の歪みを持つ形状に反発する。3つの結合と1つの非共有電子対はピラミッド形を生じ、アンモニア(NH3)のように振る舞う。⁩

    Explore · ⁨探索⁩

    Predicting molecular shape · ⁨分子形状の予測⁩

    Set the bonding and lone pairs. Electron pairs repel and spread out as far apart as possible — that fixes the shape and bond angle. · ⁨結合対と非共有電子対を設定してください。電子対は互いに反発し、可能な限り遠く離れるように配置されるため、形状と結合角が決まります。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    VSEPR theory/ˈvespə ˈθɪəri/ VSEPR理論
    lone pair/ləʊn peə/ 非共有電子対
    bonding pair/ˈbɒndɪŋ peə/ 結合電子対
    bond angle/bɒnd ˈæŋɡl/ 結合角
    linear/ˈlɪnɪə/ 直線的である
    trigonal planar/ˈtrɪɡənl ˈpleɪnə/ 平面三角形
    tetrahedral/ˌtetrəˈhiːdrəl/ 四面体形
    pyramidal/pɪˈræmɪdl/ 三角錐
    bent/bent/ 湾曲型
    trigonal bipyramidal/ˈtrɪɡənl baɪˈpɪrəmɪdl/ 三角双錐形
    octahedral/ˌɒktəˈhiːdrəl/ 八面体
    3.6

    Intermolecular forces

    Syllabus · ⁨シラバス⁩
    English
    1. (a) describe hydrogen bonding, limited to molecules containing N–H and O–H groups, including ammonia and water as simple examples (b) use the concept of hydrogen bonding to explain the anomalous properties of $\text{H}_2\text{O}$ (ice and water): • its relatively high melting and boiling points • its relatively high surface tension • the density of the solid ice compared with the liquid water
    2. use the concept of electronegativity to explain bond polarity and dipole moments of molecules
    3. (a) describe van der Waals’ forces as the intermolecular forces between molecular entities other than those due to bond formation, and use the term van der Waals’ forces as a generic term to describe all intermolecular forces (b) describe the types of van der Waals’ forces: • instantaneous dipole–induced dipole (id-id) forces, also called London dispersion forces • permanent dipole–permanent dipole (pd-pd) forces, including hydrogen bonding (c) describe hydrogen bonding and understand that hydrogen bonding is a special case of permanent dipole–permanent dipole forces between molecules where hydrogen is bonded to a highly electronegative atom
    4. state that, in general, ionic, covalent and metallic bonding are stronger than intermolecular forces
    日本語
    1. (a) 水素結合を、N–HおよびO–H基を含む分子に限って説明せよ。アンモニアと水を単純な例として含める (b) 水素結合の概念を用いて、$\text{H}_2\text{O}$(氷と水)の異常な性質を説明せよ。 • 比較的高い融点および沸点 • 比較的高い表面張力 • 固体の氷の密度と液体の水との比較
    2. 電気陰性度の概念を用いて、結合の極性および分子の双極子モーメントを説明せよ
    3. (a) ファンデルワールス力を、結合形成によるもの以外の分子間力として記述し、ファンデルワールス力をすべての分子間力を記述するための汎用的な用語として用いる (b) ファンデルワールス力の種類を記述せよ。 • 瞬間双極子-誘起双極子 (id-id) 力、またはロンドン分散力とも呼ばれる • 永久双極子-永久双極子 (pd-pd) 力、水素結合を含む (c) 水素結合を記述し、水素結合が、水素が高電気陰性な原子に結合している分子間の永久双極子-永久双極子力という特殊なケースであることを理解せよ
    4. 一般的に、イオン結合、共有結合および金属結合は分子間力よりも強いことを述べよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Intermolecular forces 分子间作用力 are the forces between molecules 分子. They are much weaker than the ionic, covalent and metallic bonding inside substances.

    Bond polarity and dipoles

    When two atoms with different electronegativity share a bond, the electrons sit closer to the more electronegative atom. The bond then has a polarity 极性: one end is slightly negative ($\delta-$) and the other slightly positive ($\delta+$). This separation of charge is a dipole 偶极.

    If the dipoles in a molecule do not cancel, the whole molecule has a dipole moment 偶极矩 and is polar. If they cancel by symmetry (as in $\text{CO}_2$), the molecule is non-polar.

    Van der Waals' forces

    Van der Waals' forces 范德华力 is the general name for all intermolecular forces. There are two main types.

    The first type is the instantaneous dipole–induced dipole force, also called the London dispersion force 伦敦色散力. Moving electrons make a brief instantaneous dipole 瞬时偶极, which then creates a matching induced dipole 诱导偶极 in a nearby molecule. These forces act between all molecules and get stronger when there are more electrons.

    The second type is the permanent dipole–permanent dipole force. It acts between molecules that are always polar, because each one has a permanent dipole 永久偶极.

    Hydrogen bonding

    Hydrogen bonding 氢键 is a strong, special case of permanent dipole forces. It forms when hydrogen is bonded to a very electronegative atom — nitrogen, oxygen or fluorine — and is attracted to a lone pair on an N, O or F atom in a neighbour. Look for N–H and O–H groups, as in ammonia and water.

    Hydrogen bonding explains the strange behaviour of water:

    • its high melting and boiling point 沸点, because many hydrogen bonds must be broken.
    • its high surface tension 表面张力.
    • ice is less dense than liquid water, because hydrogen bonds hold the molecules in an open, spread-out structure, so ice floats.
    日本語

    Intermolecular forces 分子间作用力 are the forces between molecules 分子. They are much weaker than the ionic, covalent and metallic bonding inside substances.

    Bond polarity and dipoles

    When two atoms with different electronegativity share a bond, the electrons sit closer to the more electronegative atom. The bond then has a polarity 极性: one end is slightly negative ($\delta-$) and the other slightly positive ($\delta+$). This separation of charge is a dipole 偶极.

    If the dipoles in a molecule do not cancel, the whole molecule has a dipole moment 偶极矩 and is polar. If they cancel by symmetry (as in $\text{CO}_2$), the molecule is non-polar.

    Van der Waals' forces

    Van der Waals' forces 范德华力 is the general name for all intermolecular forces. There are two main types.

    The first type is the instantaneous dipole–induced dipole force, also called the London dispersion force 伦敦色散力. Moving electrons make a brief instantaneous dipole 瞬时偶极, which then creates a matching induced dipole 诱导偶极 in a nearby molecule. These forces act between all molecules and get stronger when there are more electrons.

    Two molecules side by side: a brief instantaneous dipole in one induces a matching dipole in its neighbour, so the two attract
    A London force: a momentary dipole in one molecule induces a dipole in its neighbour, so they attract — this acts between all molecules

    The second type is the permanent dipole–permanent dipole force. It acts between molecules that are always polar, because each one has a permanent dipole 永久偶极.

    Hydrogen bonding

    Hydrogen bonding 氢键 is a strong, special case of permanent dipole forces. It forms when hydrogen is bonded to a very electronegative atom — nitrogen, oxygen or fluorine — and is attracted to a lone pair on an N, O or F atom in a neighbour. Look for N–H and O–H groups, as in ammonia and water.

    Three water molecules linked by dashed hydrogen bonds, each running from a slightly positive hydrogen to the slightly negative oxygen of a neighbour
    Hydrogen bonding in water: a $\delta+$ hydrogen is attracted to a lone pair on the $\delta-$ oxygen of a neighbouring molecule

    Hydrogen bonding explains the strange behaviour of water:

    • its high melting and boiling point 沸点, because many hydrogen bonds must be broken.
    • its high surface tension 表面张力.
    • ice is less dense than liquid water, because hydrogen bonds hold the molecules in an open, spread-out structure, so ice floats.
    Explore · ⁨探索⁩

    Polarity and intermolecular forces lab · ⁨極性と分子間力の実験⁩

    Classify molecules by the feature that controls attractions. · ⁨吸引力を支配する特徴によって分子を分類せよ。⁩

    Explore · ⁨探索⁩

    Why hydrogen bonds make water special · ⁨水素結合が水を特別にする理由⁩

    Step through it. One weak-but-strong force — the hydrogen bond — explains water's high boiling point, why ice floats, and why it dissolves so much. · ⁨段階的に確認しよう。 weak but strong force — the hydrogen bond — explains water's high boiling point, why ice floats, and why it dissolves so much.⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    intermolecular forces/ˌɪntəməˈlekjʊlə ˈfɔːsɪz/ 分子間力
    molecule/ˈmɒlɪkjuːl/ 分子
    polarity/pəʊˈlærɪti/ 極性
    dipole/ˈdaɪpəʊl/ 双極子
    dipole moment/ˈdaɪpəʊl ˈməʊmənt/ 双極子モーメント
    van der Waals' forces/ˈvændɜː wɑːlz ˈfɔːsɪz/ ファンデルワールス力
    London dispersion forces/ˈlʌndn dɪˈspɜːʃn ˈfɔːsɪz/ ロンダン分散力
    instantaneous dipole/ˌɪnstənˈteɪnɪəs ˈdaɪpəʊl/ 瞬間双極子
    induced dipole/ɪnˈdjuːst ˈdaɪpəʊl/ 誘起双極子
    permanent dipole/ˈpɜːmənənt ˈdaɪpəʊl/ 常時双極子
    hydrogen bonding/ˈhaɪdrədʒn ˈbɒndɪŋ/ 水素結合
    boiling point/ˈbɔɪlɪŋ pɔɪnt/ 沸点
    surface tension/ˈsɜːfɪs ˈtenʃn/ 表面張力
    3.7

    Dot-and-cross diagrams

    Syllabus · ⁨シラバス⁩
    English
    1. use dot-and-cross diagrams to illustrate ionic, covalent and coordinate bonding including the representation of any compounds stated in 3.4 and 3.5 (dot-and-cross diagrams may include species with atoms which have an expanded octet or species with an odd number of electrons)
    日本語
    1. ドットとクロス図を用いて、イオン結合、共有結合および配位結合を説明せよ。3.4および3.5で述べられた化合物の表現を含める(ドットとクロス図には、拡張オクテットを持つ種や電子数が奇数の種が含まれ得る)。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A dot-and-cross diagram 点叉图 shows the outer electrons of each atom, using dots for one atom and crosses for the other. This makes it clear where each bonding electron came from. You can draw them for ionic, covalent and coordinate bonding, including molecules with an expanded octet or an odd number of electrons.

    日本語

    A dot-and-cross diagram 点叉图 shows the outer electrons of each atom, using dots for one atom and crosses for the other. This makes it clear where each bonding electron came from. You can draw them for ionic, covalent and coordinate bonding, including molecules with an expanded octet or an odd number of electrons.

    Dot-and-cross diagrams for water and nitrogen: water shows two bonding pairs and two lone pairs on oxygen, nitrogen shows three shared pairs and a lone pair on each atom
    Covalent dot-and-cross: each shared pair is one electron from each atom. Water has two bonding pairs and two lone pairs; nitrogen shares three pairs (a triple bond)
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    dot-and-cross diagram/dɒt ænd krɒs ˈdaɪəɡræm/ 点線と×印の図
    3.7

    Exam tips

    • For shapes, count bonding pairs and lone pairs, name the shape, then give the exact bond angle (e.g. $\text{NH}_3$: pyramidal, $107^\circ$) — each lone pair lowers the angle by about $2.5^\circ$.
    • A dative (coordinate) bond has both electrons from one atom (e.g. $\text{NH}_4^+$, $\text{H}_3\text{O}^+$); draw the arrow from the lone pair.
    • Name the intermolecular force precisely: hydrogen bonding needs H bonded to N, O or F; otherwise it is permanent-dipole or induced-dipole (van der Waals). Never call van der Waals forces "bonds".
    • Explain a physical property by stating which forces are broken, not just "strong bonds".
  • 4

    States of matter · ⁨物質の状態⁩

    Watch lesson · ⁨レッスンを視聴⁩
    4.1

    The gaseous state · ⁨気体状態⁩

    Syllabus · ⁨シラバス⁩
    English
    1. explain the origin of pressure in a gas in terms of collisions between gas molecules and the wall of the container
    2. understand that ideal gases have zero particle volume and no intermolecular forces of attraction
    3. state and use the ideal gas equation $pV = nRT$ in calculations, including in the determination of $M_r$
    日本語
    1. 気体分子と容器壁の衝突によって生じる圧力の起源を説明する
    2. 理想気体が粒子体積ゼロであり、分子間引力がないことを理解する
    3. 理想気体方程式 $pV = nRT$ を用いて計算を行い、$M_r$ の決定を含む: 式を述べて適用する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Where gas pressure comes from

    Gas molecules move fast in all directions. They keep hitting — colliding 碰撞 with — the walls of their container. Each hit gives the wall a tiny push. The pressure 压强 of the gas is the overall result of these many collisions on the walls.

    Ideal gases

    An ideal gas 理想气体 is a simple model. We assume two things:

    • the particles themselves take up zero volume.
    • there are no intermolecular forces 分子间作用力 of attraction between the particles.

    A real gas 实际气体 follows this model closely at low pressure and high temperature. It behaves least like an ideal gas at high pressure and low temperature, when the particles are squeezed close together and the forces between them start to matter.

    The ideal gas equation

    The ideal gas equation 理想气体方程 links pressure, volume, amount and temperature:

    $$pV = nRT$$

    where $p$ is the pressure in Pa, $V$ is the volume in $\text{m}^3$, $n$ is the amount in moles, $T$ is the temperature in kelvin 开尔文 (K), and $R$ is the gas constant 气体常量 ($8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$).

    Always change the units first: °C to K (add 273), and $\text{cm}^3$ or $\text{dm}^3$ to $\text{m}^3$.

    Worked example. Find the volume of $0.50\ \text{mol}$ of an ideal gas at $27\ ^{\circ}\text{C}$ and $100\ \text{kPa}$. ($R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$.)

    Convert first: $T = 300\ \text{K}$, $p = 1.00 \times 10^{5}\ \text{Pa}$. Then

    $$V = \frac{nRT}{p} = \frac{0.50 \times 8.31 \times 300}{1.00 \times 10^{5}} = 0.0125\ \text{m}^3\ (= 12.5\ \text{dm}^3).$$

    You can also use the equation to find a molar mass 摩尔质量. Since $n = m/M$:

    $$pV = \frac{m}{M}RT \qquad\Rightarrow\qquad M = \frac{mRT}{pV}$$

    This lets you work out $M_r$ from the mass (or the density) of a gas.

    Worked example. A flask holds $0.96\ \text{g}$ of a gas in $600\ \text{cm}^3$ at $100\ \text{kPa}$ and $27\ ^{\circ}\text{C}$. Find the molar mass of the gas. ($R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$.)

    Convert: $V = 6.00 \times 10^{-4}\ \text{m}^3$, $T = 300\ \text{K}$. Then

    $$M = \frac{mRT}{pV} = \frac{0.96 \times 8.31 \times 300}{(1.00 \times 10^{5})(6.00 \times 10^{-4})} \approx 40\ \text{g mol}^{-1}.$$
    日本語
    沸騰する水から立ち上る蒸気
    沸騰は液体水を蒸気に変える — 物質の状態変化である。

    気体圧力の発生原因

    気体分子はあらゆる方向に高速で運動しています。容器の壁に繰り返し衝突 — 衝突 — を繰り返します。各衝突が壁に微小な押す力を加えます。気体の圧力とは、壁に対するこれらの多数の衝突による全体的な結果です。

    箱の中の高速粒子:各粒子に速度ベクトルがあり、一部が壁に衝突して衝撃星印が付いている
    気体圧力:高速分子はあらゆる方向に動き、壁に衝突する;多数の微小な押し合わせが加算されて圧力となる

    理想気体

    理想気体は単純なモデルです。次の2点を仮定します:

    • 粒子自体の体積はゼロである。
    • 粒子の間に分子間引力はない。

    実在気体は低圧・高温の条件でこのモデルにほぼ従います。高圧・低温では粒子が密に詰まり、粒子間の力が無視できなくなるため、理想気体としての振る舞いは最も少なくなります。

    2つの箱:左は速度ベクトルと相互作用のない微小な点状粒子の理想気体、右は高圧・低温の実在気体で、大きな詰まった粒子が破線の引力線 연결されている
    理想気体はモデル:粒子間に力が存在しない点状粒子。実在気体は高圧・低温でこのモデルからの逸脱が最も大きく、粒子の詰まりや実際のサイズ、引力が無視できなくなる

    理想気体方程式

    理想気体方程式は、圧力、体積、物質量、温度を結びつけます:

    $$pV = nRT$$

    ここで $p$ は圧力(Pa)、 $V$ は体積($\text{m}^3$ )、 $n$ は物質量(mol)、 $T$ は温度(ケルビン(K))、 $R$ は気体定数($8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$ )である。

    必ず最初に単位を変換する:°C を K に変換するには 273 を加算し、 $\text{cm}^3$ または $\text{dm}^3$ を $\text{m}^3$ に変換する。

    ** worked example **。$0.50\ \text{mol}$ の理想気体の体積を、$27\ ^{\circ}\text{C}$ と $100\ \text{kPa}$ の条件下で求めよ。($R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$。)

    まず変換する:$T = 300\ \text{K}$, $p = 1.00 \times 10^{5}\ \text{Pa}$。次に

    $$V = \frac{nRT}{p} = \frac{0.50 \times 8.31 \times 300}{1.00 \times 10^{5}} = 0.0125\ \text{m}^3\ (= 12.5\ \text{dm}^3).$$

    この式を用いてモル質量を求めることもできます。$n = m/M$ が成り立つため:

    $$pV = \frac{m}{M}RT \qquad\Rightarrow\qquad M = \frac{mRT}{pV}$$

    これにより、気体の質量(または密度)から$M_r$ を求めることができます。

    ** worked example **。フラスコに$0.96\ \text{g}$ の気体が$600\ \text{cm}^3$ 、$100\ \text{kPa}$ 、$27\ ^{\circ}\text{C}$ で入っています。この気体のモル質量を求めよ。($R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$。)

    変換する:$V = 6.00 \times 10^{-4}\ \text{m}^3$, $T = 300\ \text{K}$。次に

    $$M = \frac{mRT}{pV} = \frac{0.96 \times 8.31 \times 300}{(1.00 \times 10^{5})(6.00 \times 10^{-4})} \approx 40\ \text{g mol}^{-1}.$$
    Explore · ⁨探索⁩

    The gaseous state · ⁨気体状態⁩

    p = k / V

    Boyle's law: at constant temperature pressure ∝ 1/volume. · ⁨ボーイの法則:一定温度では圧力∝1/体積。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    collide/kəˈlaɪd/ 衝突
    pressure/ˈpreʃə/ 圧力
    ideal gas/aɪˈdɪəl ɡæs/ 理想ガス
    intermolecular forces/ˌɪntəməˈlekjʊlə ˈfɔːsɪz/ 分子間力
    real gas/rɪəl ɡæs/ 実在ガス
    ideal gas equation/aɪˈdɪəl ɡæs ɪˈkweɪʒn/ 理想気体方程式
    kelvin/ˈkelvɪn/ ケルビン
    gas constant/ɡæs ˈkɒnstənt/ 気体定数
    molar mass/ˈməʊlə mæs/ モル質量
    crystalline solid/ˈkrɪstəlaɪn ˈsɒlɪd/ 結晶固体
    giant ionic/ˈdʒaɪənt aɪˈɒnɪk/ 巨大イオン
    lattice/ˈlætɪs/ 格子
    ion/ˈaɪɒn/ イオン
    simple molecular/ˈsɪmpl məˈlekjʊlə/ 简单分子
    molecule/ˈmɒlɪkjuːl/ 分子
    fullerene/ˈfʊləren/ フルレレン
    giant molecular/ˈdʒaɪənt məˈlekjʊlə/ 巨大分子
    covalent bonds/ˈkəʊvələnt bɒndz/ 共有結合
    silicon(IV) oxide/ˈsɪlɪkən ˈɒksaɪd/ 四酸化ケイ素
    4.2

    Bonding and structure · ⁨結合と構造⁩

    Syllabus · ⁨シラバス⁩
    English
    1. describe, in simple terms, the lattice structure of a crystalline solid which is: (a) giant ionic, including sodium chloride and magnesium oxide (b) simple molecular, including iodine, buckminsterfullerene $\text{C}_{60}$ and ice (c) giant molecular, including silicon(IV) oxide, graphite and diamond (d) giant metallic, including copper
    2. describe, interpret and predict the effect of different types of structure and bonding on the physical properties of substances, including melting point, boiling point, electrical conductivity and solubility
    3. deduce the type of structure and bonding present in a substance from given information
    日本語
    1. 以下の結晶性固体の格子構造を簡潔に説明する: (a) 巨大イオン性: 塩化ナトリウムおよび酸化マグネシウムを含む、(b) 単一分子性: ヨウ素、バッキンガーフルレレン $\text{C}_{60}$ および氷を含む、(c) 巨大分子性: 四酸化ケイ素、グラファイトおよびダイヤモンドを含む、(d) 巨大金属性: 銅を含む
    2. 異なる種類の構造および結合が物質の物理的性質に与える影響を説明、解釈、予測する。これには融点、沸点、電気伝導度および溶解度が含まれる
    3. 与えられた情報から、物質に含まれる構造および結合の種類を導き出す

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    How a substance behaves depends on how its particles are joined. There are four main structures of a crystalline solid 晶体.

    Giant ionic

    A giant ionic 离子晶体 structure is a huge regular lattice 晶格 of positive and negative ions 离子, held together by strong attraction in every direction. Examples are sodium chloride and magnesium oxide.

    Simple molecular

    A simple molecular 分子晶体 structure is made of small molecules 分子. The bonds inside each molecule are strong, but the intermolecular forces between the molecules are weak. Examples are iodine ($\text{I}_2$), fullerene 富勒烯 ($\text{C}_{60}$) and ice.

    Giant molecular

    A giant molecular 原子晶体 structure (also called giant covalent) is a huge network of atoms joined by strong covalent bonds 共价键. Examples are silicon(IV) oxide 二氧化硅, graphite 石墨 and diamond 金刚石.

    Giant metallic

    A giant metallic 金属晶体 structure is a lattice of positive metal ions in a "sea" of delocalised electrons 离域电子. An example is copper.

    Physical properties

    The structure decides the physical properties:

    Structure Melting/boiling point Conducts electricity? Solubility in water
    giant ionic high only when molten or dissolved usually soluble
    simple molecular low no usually low
    giant molecular very high no (except graphite) insoluble
    giant metallic high yes (solid and molten) insoluble
    • melting point 熔点 and boiling point 沸点 are high when strong forces (ionic, covalent or metallic) must be broken, and low when only weak intermolecular forces break.
    • electrical conductivity 导电性 needs charged particles that can move — ions that are free (when molten or dissolved) or delocalised electrons. Graphite conducts because some of its electrons are delocalised.
    • solubility 溶解度 in water is usually high for ionic solids and low for molecular and giant covalent solids.

    You can work backwards too: from the melting point, conductivity and solubility of an unknown substance, deduce the type of structure and bonding it has.

    日本語
    石英結晶のクラスター
    石英は珪素と酸素からなる巨大共有結合構造です。

    物質の性質は粒子がどのように結びついているかによって決まります。結晶性固体には4つの主要な構造があります。

    4つのパネル:正負イオンが交互に並ぶ巨大イオン格子、分離した小さな分子、結合した共有ネットワーク、電子海内の金属イオン
    結晶性固体の4つの構造 — 構造が融点、電気伝導性、溶解性を決定します

    巨大イオン

    巨大イオン構造とは、正イオンと負イオンの大きな規則正しい格子であり、あらゆる方向からの強い引力によって保持されています。例として、塩化ナトリウムや酸化マグネシウムがあります。

    単純分子

    単純分子構造は小さな分子から構成されます。各分子内の結合は強いため、分子間の力は弱いです。例として、ヨウ素 ($\text{I}_2$ )、フルレネ ($\text{C}_{60}$ )、氷があります。

    巨大分子

    巨大分子構造(巨大共有結合とも呼ばれる)は、強い共有結合で結びついた原子の巨大なネットワークです。例として、四酸化ケイ素、グラファイト、ダイヤモンドがあります。

    左側は各炭素が他の4つの炭素に結合したダイヤモンドノード、右側は弱い力で保持された六角形の層が積まれたグラファイト
    炭素の2つの巨大共有結合形態:ダイヤモンドは剛直な3Dネットワーク(非常に硬い)、グラファイトは滑りやすい層と伝導性の自由電子を持つ
    不規則な八面体のダイヤモンド結晶
    岩石中の天然ダイヤモンド。その結晶全体が一つの巨大分子であり、各炭素が他の4つの炭素に結合した単一の途切れないネットワークです。これを壊すには無数の強い共有結合を切断する必要がありますが、そのためダイヤモンドは最も硬度の高い天然物質です

    巨大金属

    巨大金属構造とは、「離域電子の海」内の正の金属イオンの格子です。例として銅があります。

    輝く赤茶色の金属光沢と小さな緑色の変質斑点を持つ丸みを帯びた銅の塊
    純粋な銅の金属片。光沢、金属特有の伝導性と可塑性はすべて、離域電子の海に座る巨大な銅イオンの格子によるものです

    物理的性質

    構造が物理的性質を決定します:

    構造 融点・沸点 電気伝導性あり? 水への溶解性
    巨大イオン 高い 溶融時または水溶液のみ 通常可溶
    単純分子 低い なし 通常低い
    巨大分子 非常に高い なし(グラファイトを除く) 不溶
    巨大金属 高い はい(固形および溶融時) 不溶
    • 融点と沸点は、強い力(イオン性、共有結合、金属間結合)を切断する必要がある場合が高く、弱い分子間力のみが切断される場合は低いです。
    • 電気伝導性には、移動可能な荷電粒子が必要です。これは、自由なイオン(溶融時または水溶液中)または離域電子です。グラファイトは一部の電子が離域しているため伝導します。
    • 水に対する溶解性は、一般的にイオン性固体では高く、分子性および巨大共有結合固体では低いです。

    逆にも計算できます:未知物質の融点、伝導性、溶解性から、その構造および結合の種類を推論できます。

    Explore · ⁨探索⁩

    Giant structure lab · ⁨巨大構造実験⁩

    Compare giant structures by particles and bonding. · ⁨粒子と結合に基づき巨大構造を比較せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    graphite/ˈɡræfaɪt/ グラファイト
    diamond/ˈdaɪəmənd/ ダイヤモンド
    giant metallic/ˈdʒaɪənt məˈtælɪk/ 巨大金属
    delocalised electrons/dɪˈlɒkəlaɪzd ɪˈlektrɒnz/ 非局在電子
    melting point/ˈmeltɪŋ pɔɪnt/ 融点
    boiling point/ˈbɔɪlɪŋ pɔɪnt/ 沸点
    electrical conductivity/ɪˈlektrɪkl kɒndəkˈtɪvɪti/ 電気伝導性
    solubility/ˌsɒljuːˈbɪlɪti/ 溶解度
    4.2

    Exam tips · ⁨試験対策⁩

    English
    • Use SI units in $pV = nRT$: pressure in Pa, volume in $\text{m}^3$ ($\text{cm}^3 \times 10^{-6}$), temperature in K ($^\circ\text{C} + 273$).
    • State the ideal-gas assumptions (negligible molecular volume, no intermolecular forces) and when real gases deviate (high pressure, low temperature).
    • Link each property to structure: giant ionic (high m.p., conducts molten), giant covalent (very high m.p.), simple molecular (low m.p.), giant metallic (conducts, malleable).
    • Graphite conducts because each carbon has a delocalised electron; diamond does not — a favourite comparison.
    日本語
    • $pV = nRT$ ではSI単位を使用する:圧力はPa、体積は$\text{m}^3$ ($\text{cm}^3 \times 10^{-6}$ )、温度はK ($^\circ\text{C} + 273$ )。
    • 理想気体の仮定(分子体積の無視、分子間力の欠如)と、実在ガスが逸脱する条件(高圧、低温)を述べる。
    • 各性質を構造に結びつける:巨大イオン(高い融点、溶融時の伝導)、巨大共有結合(非常に高い融点)、単純分子(低い融点)、巨大金属(伝導性、延性)。
    • グラファイトは各炭素が離域電子を持つため伝導しますが、ダイヤモンドは持たない — 比較対象としてよく出題されます。
  • 5

    Chemical energetics · ⁨化学熱力学⁩

    Watch lesson · ⁨レッスンを視聴⁩
    5.1

    Enthalpy change, ΔH

    Syllabus · ⁨シラバス⁩
    English
    1. understand that chemical reactions are accompanied by enthalpy changes and these changes can be exothermic ($\Delta H$ is negative) or endothermic ($\Delta H$ is positive)
    2. construct and interpret a reaction pathway diagram, in terms of the enthalpy change of the reaction and of the activation energy
    3. define and use the terms: (a) standard conditions (this syllabus assumes that these are $298\text{ K}$ and $101\text{ kPa}$) shown by $^{\ominus}$. (b) enthalpy change with particular reference to: reaction, $\Delta H_r$, formation, $\Delta H_f$, combustion, $\Delta H_c$, neutralisation, $\Delta H_{\text{neut}}$
    4. understand that energy transfers occur during chemical reactions because of the breaking and making of chemical bonds
    5. use bond energies ($\Delta H$ positive, i.e. bond breaking) to calculate enthalpy change of reaction, $\Delta H_r$
    6. understand that some bond energies are exact and some bond energies are averages
    7. calculate enthalpy changes from appropriate experimental results, including the use of the relationships $q = mc\Delta T$ and $\Delta H = -mc\Delta T/n$
    日本語
    1. 化学反応に伴いエンタルピー変化が生じ、これらの変化が発熱反応($\Delta H$は負)または吸熱反応($\Delta H$は正)であることを理解する
    2. 反応のエンタルピー変化および活性化エネルギーの観点から、反応経路図を構築・解釈する
    3. 以下の用語を定義して使用する: (a) 標準状態: このカリキュラムではこれらが $298\text{ K}$ および $101\text{ kPa}$ であると仮定しており、$^{\ominus}$ で示される。(b) 特定の文脈におけるエンタルピー変化: 反応、$\Delta H_r$、生成、$\Delta H_f$、燃焼、$\Delta H_c$、中和、$\Delta H_{\text{neut}}$
    4. 化学結合の切断および形成によりエネルギー移動が起こることを理解する
    5. 結合エネルギー($\Delta H$は正、つまり結合の切断)を用いて、反応エンタルピー変化 $\Delta H_r$ を計算する
    6. 一部の結合エネルギーは正確な値であり、他のものは平均値であることを理解する
    7. 適切な実験結果からエンタルピー変化を計算する。これには関係式 $q = mc\Delta T$ および $\Delta H = -mc\Delta T/n$ の使用が含まれる

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Reaction profile: activation energy and ΔH

    Every chemical reaction takes in or gives out energy. This energy change, measured at constant pressure, is the enthalpy change 焓变, with symbol $\Delta H$.

    • in an exothermic 放热 reaction the system gives out heat, so the products have less energy than the reactants and $\Delta H$ is negative.
    • in an endothermic 吸热 reaction the system takes in heat, so the products have more energy than the reactants and $\Delta H$ is positive.

    Reaction pathway diagrams

    A reaction pathway diagram 反应路径图 shows the energy of the reactants and products, and the energy "hill" between them. The height of the hill is the activation energy 活化能 — the least energy the particles need before they can react.

    • exothermic: products sit lower than reactants ($\Delta H < 0$).
    • endothermic: products sit higher than reactants ($\Delta H > 0$).

    Standard conditions and types of enthalpy change

    Energy values are compared under standard conditions 标准条件: $298\ \text{K}$ and $101\ \text{kPa}$, shown by the symbol $^{\ominus}$. Each substance is in its normal physical state at those conditions.

    Symbol Name Definition (per mole, under standard conditions)
    $\Delta H_r^{\ominus}$ enthalpy change of reaction 反应焓变 for the amounts shown in the equation
    $\Delta H_f^{\ominus}$ enthalpy change of formation 生成焓变 one mole of a compound forms from its elements
    $\Delta H_c^{\ominus}$ enthalpy change of combustion 燃烧焓变 one mole of a substance burns completely in oxygen
    $\Delta H_{\text{neut}}^{\ominus}$ enthalpy change of neutralisation 中和焓变 one mole of water forms from an acid and an alkali

    Energy from breaking and making bonds

    During a reaction, old bonds break and new bonds form. Breaking a bond needs energy (endothermic); making a bond releases energy (exothermic). The enthalpy change of the reaction is the difference between the two:

    $$\Delta H_r = \sum (\text{bond energies broken}) - \sum (\text{bond energies made})$$

    The bond energy 键能 is the energy needed to break one mole of a particular bond in the gas state, so it is always positive. Some bond energies are exact (for one specific molecule); others are averages taken over many different molecules, so calculations using them are only approximate.

    Worked example. Use bond energies to find $\Delta H$ for $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$. Bond energies (kJ mol⁻¹): H–H $= 436$, Cl–Cl $= 242$, H–Cl $= 431$.

    $$\Delta H = \sum(\text{broken}) - \sum(\text{made}) = (436 + 242) - (2 \times 431) = 678 - 862 = -184\ \text{kJ mol}^{-1}.$$

    Measuring enthalpy change in the lab

    When a reaction heats up (or cools down) a known mass of water or solution, the heat transferred is:

    $$q = mc\Delta T$$

    where $m$ is the mass, $c$ is the specific heat capacity 比热容 (how much energy raises 1 g by 1 K), and $\Delta T$ is the temperature change. The enthalpy change per mole is then:

    $$\Delta H = -\frac{mc\Delta T}{n}$$

    The minus sign makes $\Delta H$ negative when the temperature rises (an exothermic reaction).

    Worked example. Burning $0.50\ \text{g}$ of methanol ($M_r = 32$) raises the temperature of $100\ \text{g}$ of water by $18\ ^{\circ}\text{C}$. Find the enthalpy change of combustion per mole. ($c = 4.18\ \text{J g}^{-1}\,\text{K}^{-1}$.)

    The heat released is $q = mc\Delta T = 100 \times 4.18 \times 18 = 7520\ \text{J}$. The amount burnt is $n = 0.50/32 = 0.0156\ \text{mol}$, so

    $$\Delta H_c = -\frac{q}{n} = -\frac{7520}{0.0156} \approx -480\ \text{kJ mol}^{-1}.$$
    日本語
    Reaction profile: activation energy and ΔH
    A blue Bunsen burner flame
    Burning fuel is exothermic, releasing energy to the surroundings.
    An instant cold pack
    An instant cold pack uses an endothermic reaction that takes in heat.

    Every chemical reaction takes in or gives out energy. This energy change, measured at constant pressure, is the enthalpy change 焓变, with symbol $\Delta H$.

    • in an exothermic 放热 reaction the system gives out heat, so the products have less energy than the reactants and $\Delta H$ is negative.
    • in an endothermic 吸热 reaction the system takes in heat, so the products have more energy than the reactants and $\Delta H$ is positive.

    Reaction pathway diagrams

    A reaction pathway diagram 反应路径图 shows the energy of the reactants and products, and the energy "hill" between them. The height of the hill is the activation energy 活化能 — the least energy the particles need before they can react.

    • exothermic: products sit lower than reactants ($\Delta H < 0$).
    • endothermic: products sit higher than reactants ($\Delta H > 0$).
    Two reaction pathway diagrams side by side: an exothermic one with products below the reactants, and an endothermic one with products above
    Exothermic reactions end lower than they start ($\Delta H<0$); endothermic reactions end higher ($\Delta H>0$)

    Standard conditions and types of enthalpy change

    Energy values are compared under standard conditions 标准条件: $298\ \text{K}$ and $101\ \text{kPa}$, shown by the symbol $^{\ominus}$. Each substance is in its normal physical state at those conditions.

    Symbol Name Definition (per mole, under standard conditions)
    $\Delta H_r^{\ominus}$ enthalpy change of reaction 反应焓变 for the amounts shown in the equation
    $\Delta H_f^{\ominus}$ enthalpy change of formation 生成焓变 one mole of a compound forms from its elements
    $\Delta H_c^{\ominus}$ enthalpy change of combustion 燃烧焓变 one mole of a substance burns completely in oxygen
    $\Delta H_{\text{neut}}^{\ominus}$ enthalpy change of neutralisation 中和焓变 one mole of water forms from an acid and an alkali
    Two processes: formation goes from elements to one mole of a compound; combustion goes from one mole of a substance plus oxygen to carbon dioxide and water
    Don't confuse them: formation builds 1 mol of a compound from its elements; combustion burns 1 mol of a substance in oxygen

    Energy from breaking and making bonds

    During a reaction, old bonds break and new bonds form. Breaking a bond needs energy (endothermic); making a bond releases energy (exothermic). The enthalpy change of the reaction is the difference between the two:

    $$\Delta H_r = \sum (\text{bond energies broken}) - \sum (\text{bond energies made})$$

    The bond energy 键能 is the energy needed to break one mole of a particular bond in the gas state, so it is always positive. Some bond energies are exact (for one specific molecule); others are averages taken over many different molecules, so calculations using them are only approximate.

    Worked example. Use bond energies to find $\Delta H$ for $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$. Bond energies (kJ mol⁻¹): H–H $= 436$, Cl–Cl $= 242$, H–Cl $= 431$.

    $$\Delta H = \sum(\text{broken}) - \sum(\text{made}) = (436 + 242) - (2 \times 431) = 678 - 862 = -184\ \text{kJ mol}^{-1}.$$
    A stepped energy diagram: reactants rise to separate gaseous atoms as bonds break, then fall to products as bonds form
    Breaking bonds takes energy in; making bonds gives energy out. $\Delta H$ is the difference between the two

    Measuring enthalpy change in the lab

    When a reaction heats up (or cools down) a known mass of water or solution, the heat transferred is:

    $$q = mc\Delta T$$

    where $m$ is the mass, $c$ is the specific heat capacity 比热容 (how much energy raises 1 g by 1 K), and $\Delta T$ is the temperature change. The enthalpy change per mole is then:

    $$\Delta H = -\frac{mc\Delta T}{n}$$

    The minus sign makes $\Delta H$ negative when the temperature rises (an exothermic reaction).

    Worked example. Burning $0.50\ \text{g}$ of methanol ($M_r = 32$) raises the temperature of $100\ \text{g}$ of water by $18\ ^{\circ}\text{C}$. Find the enthalpy change of combustion per mole. ($c = 4.18\ \text{J g}^{-1}\,\text{K}^{-1}$.)

    The heat released is $q = mc\Delta T = 100 \times 4.18 \times 18 = 7520\ \text{J}$. The amount burnt is $n = 0.50/32 = 0.0156\ \text{mol}$, so

    $$\Delta H_c = -\frac{q}{n} = -\frac{7520}{0.0156} \approx -480\ \text{kJ mol}^{-1}.$$
    An insulated cup of solution with a lid and a thermometer dipping in, used to measure the temperature change
    An insulated cup and a thermometer measure the temperature change of a known mass of solution
    Explore · ⁨探索⁩

    Exothermic and endothermic reactions · ⁨発熱反応と吸熱反応⁩

    Drag ΔH. An exothermic reaction releases energy (products lower); an endothermic one takes it in (products higher). · ⁨ΔHをドラッグしてください。発熱反応はエネルギーを放出し(生成物側が低い)、吸熱反応はエネルギーを取り込みます(生成物側が高い)。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    enthalpy change/enˈθælpi tʃeɪndʒ/ エンタルピー変化
    exothermic/eɡzəˈðɜːmɪk/ 発熱反応
    endothermic/ˌendəʊˈθɜːmɪk/ 吸熱反応
    reaction pathway diagram/rɪˈækʃn ˈpæθweɪ ˈdaɪəɡræm/ 反応経路図
    activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ 活性化エネルギー
    standard conditions/ˈstændəd kənˈdɪʃnz/ 標準状態
    enthalpy change of reaction/enˈθælpi tʃeɪndʒ ɒv rɪˈækʃn/ 反応エンタルピー変化
    enthalpy change of formation/enˈθælpi tʃeɪndʒ ɒv fɔːˈmeɪʃn/ 生成エンタルピー変化
    enthalpy change of combustion/enˈθælpi tʃeɪndʒ ɒv kəmˈbʌstʃn/ 燃焼エンタルピー変化
    enthalpy change of neutralisation/enˈθælpi tʃeɪndʒ ɒv ˌnjuːtrəlaɪˈzeɪʃn/ 中和エンタルピー変化
    bond energy/bɒnd ˈenədʒi/ 結合エネルギー
    specific heat capacity/spəˈsɪfɪk hiːt kəˈpæsɪti/ 比熱容
    5.2

    Hess's law

    Syllabus · ⁨シラバス⁩
    English
    1. apply Hess’s law to construct simple energy cycles
    2. carry out calculations using cycles and relevant energy terms, including: (a) determining enthalpy changes that cannot be found by direct experiment (b) use of bond energy data
    日本語
    1. ヘスの法則を用いて単純なエネルギーサイクルを構築する
    2. サイクルおよび関連するエネルギー項を用いて計算を行う。これには: (a) 直接実験で得られないエンタルピー変化の決定、(b) 結合エネルギーデータの使用 が含まれる

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Hess's law 盖斯定律 says that the total enthalpy change for a reaction is the same, no matter which route you take from reactants to products. This is because energy is conserved.

    This lets you draw an energy cycle 能量循环: link the reactants and products by a direct step and by an indirect route, then add the steps so that both routes give the same total.

    Hess's law is useful in two ways:

    • it lets you find an enthalpy change that you cannot measure directly (for example, the formation of a compound that forms slowly or with side reactions).
    • it lets you calculate $\Delta H_r$ from bond energy data, or from formation or combustion data given in the question.
    日本語

    Hess's law 盖斯定律 says that the total enthalpy change for a reaction is the same, no matter which route you take from reactants to products. This is because energy is conserved.

    This lets you draw an energy cycle 能量循环: link the reactants and products by a direct step and by an indirect route, then add the steps so that both routes give the same total.

    A triangular Hess cycle linking reactants and products directly and through an intermediate, with the three enthalpy changes labelled
    The direct route equals the indirect route, so $\Delta H_r = \Delta H_1 + \Delta H_2$

    Hess's law is useful in two ways:

    • it lets you find an enthalpy change that you cannot measure directly (for example, the formation of a compound that forms slowly or with side reactions).
    • it lets you calculate $\Delta H_r$ from bond energy data, or from formation or combustion data given in the question.
    Explore · ⁨探索⁩

    Hess's law cycle · ⁨ヘスの法則サイクル⁩

    Enthalpy change is the same whichever route you take — so an unknown ΔH can be found by an alternative path. · ⁨どの経路を取ってもエンタルピー変化は同じである。したがって、未知のΔHは代替の経路を通じて求めることができる。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Hess's law/ˈhesɪz lɔː/ ヘスの法則
    energy cycle/ˈenədʒi ˈsaɪkl/ エネルギーサイクル
    5.2

    Exam tips

    • Define each enthalpy change with its exact standard conditions (e.g. combustion = one mole burned completely in excess oxygen).
    • Use $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$; breaking is endothermic ($+$), making is exothermic ($-$) — getting the sign the wrong way round is the classic error.
    • In $q = mc\Delta T$ use the mass of the water/solution, then divide by moles and add the minus sign for an exothermic reaction.
    • Draw Hess cycles with arrows the same way round, follow the alternative route, and always give $\Delta H$ a sign and units ($\text{kJ mol}^{-1}$).
  • 6

    Electrochemistry · ⁨電気化学⁩

    Watch lesson · ⁨レッスンを視聴⁩
    6.1

    Oxidation number · ⁨酸化数⁩

    Syllabus · ⁨シラバス⁩
    English
    1. calculate oxidation numbers of elements in compounds and ions
    2. use changes in oxidation numbers to help balance chemical equations
    3. explain and use the terms redox, oxidation, reduction and disproportionation in terms of electron transfer and changes in oxidation number
    4. explain and use the terms oxidising agent and reducing agent
    5. use a Roman numeral to indicate the magnitude of the oxidation number of an element
    日本語
    1. 化合物およびイオン中の元素の酸化数を計算する
    2. 酸化数の変化を利用して化学方程式の係数を合わせるのを助ける
    3. 電子移動および酸化数の変化の観点から、酸化還元反応(redox)、酸化、還元、自己酸化還元(disproportionation) という用語を説明して使用する
    4. 酸化剤および還元剤という用語を説明して使用する
    5. 元素の酸化数の大きさを示すためにローマ数字を使用する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The oxidation number 氧化数 (also called the oxidation state) shows how many electrons 电子 an atom has lost or gained compared with the free element. You work it out using simple rules:

    • an uncombined element has an oxidation number of $0$.
    • a simple ion has an oxidation number equal to its charge (so $\text{Mg}^{2+}$ is $+2$).
    • Group 1 is always $+1$, Group 2 is always $+2$.
    • hydrogen is $+1$ (but $-1$ in metal hydrides).
    • oxygen is $-2$ (but $-1$ in peroxides).
    • fluorine is always $-1$.
    • the oxidation numbers in a neutral compound add up to $0$; in an ion they add up to the charge.

    A Roman numeral shows the size of the oxidation number of an element, for example iron(II) means $+2$ and manganese(VII) in $\text{KMnO}_4$ means $+7$.

    Worked example. Find the oxidation number of sulfur in the sulfate ion, $\text{SO}_4^{2-}$.

    Each oxygen is $-2$, so the four oxygens give $4 \times (-2) = -8$. In an ion the numbers add up to the charge, here $-2$. If sulfur is $x$:

    $$x + (-8) = -2 \quad\Rightarrow\quad x = +6.$$

    Worked example. Find the oxidation number of nitrogen in the nitrate ion, $\text{NO}_3^-$.

    $$x + 3 \times (-2) = -1 \quad\Rightarrow\quad x - 6 = -1 \quad\Rightarrow\quad x = +5.$$
    日本語

    酸化数(酸化状態とも呼ばれる)は、自由元素と比較して、ある原子が何個の電子を失ったかまたは得たかを示す値です。以下の単純なルールを使って求めます:

    • 単体(未結合の元素)の酸化数は $0$ です。
    • 単イオンの酸化数はその電荷と等しい(つまり $\text{Mg}^{2+}$ は $+2$ です)。
    • Group 1 は常に $+1$、Group 2 は常に $+2$ です。
    • water は $+1$ (ただし金属水化物内では $-1$)。
    • oxygen は $-2$ (ただしペルキシド内では $-1$)。
    • フッ素は常に $-1$ です。
    • 中性化合物内の酸化数の総和は $0$ となり、イオン内の酸化数の総和は電荷と等しくなります。

    ローマ数字は元素の酸化数の大きさを示し、例えば鉄(II)は $+2$ 、マンガン(VII)は $\text{KMnO}_4$ において $+7$ を意味します。

    KMnO4中のマンガンの酸化数を計算:カリウムは+1、4つの酸素は-8、合計がゼロであるため、マンガンは+7でなければならない
    酸化数の割り当て:既知の原子(K = +1, O = 各−2)を固定し、「合計がゼロになる」を使って未知のもの(Mn = +7)を見つけます

    ** worked example.** スルホ酸イオン $\text{SO}_4^{2-}$ 中の硫黄の酸化数を見つけます。

    各酸素は $-2$ なので、4つの酸素で $4 \times (-2) = -8$ となります。イオンでは数が電荷に等しくなるため、ここでは $-2$ です。硫黄が $x$ ならば:

    $$x + (-8) = -2 \quad\Rightarrow\quad x = +6.$$

    ** worked example.** ニトレートイオン $\text{NO}_3^-$ 中の窒素の酸化数を見つけます。

    $$x + 3 \times (-2) = -1 \quad\Rightarrow\quad x - 6 = -1 \quad\Rightarrow\quad x = +5.$$
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    oxidation number/ˌɒksɪˈdeɪʃn ˈnʌmbə/ 酸化数
    electron/ɪˈlektrɒn/ 電子
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    6.1

    Redox in terms of electrons · ⁨電子に関する酸化還元反応⁩

    English

    A redox 氧化还原 reaction is one where electrons move from one species to another.

    • oxidation 氧化 is the loss of electrons. The oxidation number goes up.
    • reduction 还原 is the gain of electrons. The oxidation number goes down.

    A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain.

    Oxidation and reduction always happen together, because the electrons lost by one species are gained by another. This electron transfer 电子转移 is why we call it a redox reaction.

    日本語
    複数のAA形乾電池
    電池は回路内を電子を動かすために酸化還元反応を使います。

    酸化還元反応とは、電子が一方の種から他方へ移動する反応です。

    • 酸化とは電子の喪失です。酸化数は上がります。
    • 還元とは電子の獲得です。酸化数は下がります。

    useful memory aid は OIL RIG: Oxidation Is Loss, Reduction Is Gain です。

    酸化と還元の両方が必ず同時に起こるのは、ある種が喪失した電子を別の種が得るからです。この 電子移動 が私たちがこれを酸化還元反応と呼ぶ理由です。

    亜鉛が電子2つを失ってZn2+となり、銅イオンがそれを受け取って銅になる様子。電子移動を示す矢印付き
    酸化還元反応とは電子移動です: 還元剤は電子を喪失(酸化され、数上昇)、酸化剤は電子を得る(還元され、数下降)
    Explore · ⁨探索⁩

    Redox is electron transfer · ⁨酸化還元反応とは電子の移動である⁩

    Step through it: the metal loses (is oxidised) and the non-metal gains (is reduced) — oxidation number rises for one, falls for the other. · ⁨手順を追って考える:金属は電子を失う(酸化される)、非金属は電子を得る(還元される)—oneの酸化数は上昇し、もう一方のそれは低下する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    redox/rɪˈdɒks/ 酸化還元
    reduction/rɪˈdʌkʃn/ 還元
    electron transfer/ɪˈlektrɒn ˈtrænsfɜː/ 電子の移動
    6.1

    Using oxidation numbers to balance equations · ⁨酸化数を用いた化学方程式の釣り合い⁩

    English

    Changes in oxidation number help you balance a redox equation:

    1. find the element whose oxidation number rises (it is oxidised) and the one whose number falls (it is reduced).
    2. the total rise must equal the total fall, because every electron lost is gained somewhere.
    3. choose the ratio of the two species so the rise and fall match, then balance the rest of the equation.

    Worked example. Balance the reaction of manganate(VII) with iron(II) in acid: $\text{MnO}_4^- + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_2\text{O}$.

    Manganese falls from $+7$ to $+2$ (a fall of $5$); iron rises from $+2$ to $+3$ (a rise of $1$). To make the total fall equal the total rise, take $5$ iron ions for every $1$ manganate ion. Then balance oxygen with water ($4\,\text{H}_2\text{O}$) and hydrogen with $\text{H}^+$ ($8\,\text{H}^+$):

    $$\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}.$$
    日本語

    酸化数の変化は酸化還元方程式を釣り合わせるのに役立ちます:

    1. 酸化数が上がる元素(酸化されるもの)と、数が下がるものを見つけます。
    2. 喪失されたすべての電子どこかで得られるため、合計の上昇値は合計の下降値と等しくなります。
    3. 上昇と下降が一致するように2つの種の比率を選び、方程式の残りを釣り合わせます。
    鉄が +2 から +3 に1つずつ上昇し、マンガンが +7 から +2 に5つずつ下降するため、5個の鉄が1個のパーマングネートで釣り合う様子
    酸化還元方程式の釣り合い: 酸化数の合計上昇値は合計下降値と等しくなり、これが比率を決定します

    ** worked example.** 酸性中でのパーマンガン酸(II)と鉄(II)の反応を釣り合わせます: $\text{MnO}_4^- + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_2\text{O}$。

    マンガンは $+7$ から $+2$ に下がる(下降幅 $5$)、鉄は $+2$ から $+3$ に上がる(上昇幅 $1$)。合計の下降と上昇を等しくするために、$5$ 個の鉄イオンに対して $1$ 個のマンガネートイオンをとる。次に酸素を水 ($4\,\text{H}_2\text{O}$) で、水素を $\text{H}^+$ ($8\,\text{H}^+$) でバランスさせる:

    $$\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}.$$
    6.1

    Disproportionation · ⁨自己酸化還元反応⁩

    English

    Disproportionation 歧化 is a special redox reaction in which the same element is both oxidised and reduced at the same time. For example, when chlorine reacts with cold water, some chlorine atoms are reduced (to $\text{Cl}^-$) and others are oxidised (to $\text{ClO}^-$).

    日本語

    自己酸化還元反応とは、同じ元素が同時に酸化も還元も受ける特別な酸化還元反応です。例えば、塩素が冷水と反応すると、一部の塩素原子は還元され($\text{Cl}^-$ へ)、他の一部は酸化されます($\text{ClO}^-$ へ)。

    酸化数の線に示された塩素が 0 から ClO- では +1 まで上がり、Cl- では -1 まで下がる様子
    自己酸化還元反応: 塩素($0$)は同時にClO$^-$ ($+1$) へ酸化され、Cl$^-$ ($-1$) へ還元されます
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    disproportionation/ˌdɪsprəˈpɔːʃəneɪʃn/ 歧化反応
    oxidising agent/ˈɒksɪdaɪzɪŋ ˈeɪdʒənt/ 酸化剤
    reducing agent/rɪˈdjuːsɪŋ ˈeɪdʒənt/ 還元剤
    6.1

    Oxidising and reducing agents · ⁨酸化剤と還元剤⁩

    English
    • an oxidising agent 氧化剂 takes electrons away from another species. In doing so, it is itself reduced.
    • a reducing agent 还原剂 gives electrons to another species. In doing so, it is itself oxidised.

    So in any redox reaction, the oxidising agent causes the oxidation of the other species, while the reducing agent causes the reduction.

    日本語
    鉄の表面の錆
    錆びつきは酸素による鉄の酸化です。
    • 酸化剤は他の種から電子を取り除きます。その過程で自身は還元されます。
    • 還元剤は他の種に電子を与えます。その過程で自身は酸化されます。

    したがって、いかなる酸化還元反応においても、酸化剤は他の種の酸化を引き起こし、還元剤は還元を引き起こします。

    6.1

    Exam tips · ⁨試験対策⁩

    English
    • Apply the oxidation-number rules in order: O is $-2$ except in peroxides ($-1$); H is $+1$ except in metal hydrides ($-1$).
    • In a redox equation the total increase equals the total decrease in oxidation number — use this to fix the ratio, then balance O with $\text{H}_2\text{O}$ and H with $\text{H}^+$.
    • An oxidising agent is itself reduced; name both what is oxidised/reduced and the agent for full marks.
    • Disproportionation is the same element both oxidised and reduced — show both oxidation-number changes.
    日本語
    • 酸化数のルールを順に適用します: O は $-2$ ですが、パ過酸化物($-1$)を除きます;H は $+1$ ですが、金属水素化物($-1$)を除きます。
    • 酸化還元方程式では酸化数の合計増加値は合計減少値と等しくなります——これを使って比率を固定し、次で$\text{H}_2\text{O}$ でO、$\text{H}^+$ でHを釣り合わせます。
    • 酸化剤は自身は還元されます;酸化・還元されるものと、それぞれの剤の名前を両方示さないと満点にはなりません。
    • 自己酸化還元反応とは同じ元素が酸化も還元も受けること——両方の酸化数の変化を示してください。
  • 7

    Equilibria · ⁨平衡⁩

    Watch lesson · ⁨レッスンを視聴⁩
    7.1

    Reversible reactions and dynamic equilibrium

    Syllabus · ⁨シラバス⁩
    English
    1. (a) understand what is meant by a reversible reaction (b) understand what is meant by dynamic equilibrium in terms of the rate of forward and reverse reactions being equal and the concentration of reactants and products remaining constant (c) understand the need for a closed system in order to establish dynamic equilibrium
    2. define Le Chatelier’s principle as: if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change
    3. use Le Chatelier’s principle to deduce qualitatively (from appropriate information) the effects of changes in temperature, concentration, pressure or presence of a catalyst on a system at equilibrium
    4. deduce expressions for equilibrium constants in terms of concentrations, $K_c$
    5. use the terms mole fraction and partial pressure
    6. deduce expressions for equilibrium constants in terms of partial pressures, $K_p$ (use of the relationship between $K_p$ and $K_c$ is not required)
    7. use the $K_c$ and $K_p$ expressions to carry out calculations (such calculations will not require the solving of quadratic equations)
    8. calculate the quantities present at equilibrium, given appropriate data
    9. state whether changes in temperature, concentration or pressure or the presence of a catalyst affect the value of the equilibrium constant for a reaction
    10. describe and explain the conditions used in the Haber process and the Contact process, as examples of the importance of an understanding of dynamic equilibrium in the chemical industry and the application of Le Chatelier’s principle
    日本語
    1. (a) 可逆反応の意味を理解する (b) 正方向および逆方向の反応速度が等しく、反応物および生成物の濃度が一定に保たれるという点から、動態平衡の意味を理解する (c) 動態平衡を確立するためには閉鎖系が必要であることを理解する
    2. ルシャトリエの原理を以下のように定義する:動態平衡にある系に対して変化を加えると、その変化を最小限にするために平衡位置は移動する
    3. ルシャトリエの原理を用いて、温度、濃度、圧力または触媒の有無の変化が平衡状態にある系に与える影響を、適切な情報に基づいて定性的に導き出す
    4. 平衡定数の式を濃度を用いて導き出し、$K_c$
    5. モル分率および分圧の用語を使用する
    6. 分圧を用いた平衡定数の式を導き出し、$K_p$($K_p$と$K_c$の関係式の使用は不要)
    7. $K_c$および$K_p$の式を用いて計算を行う(これらの計算には二次方程式を解く必要はない)
    8. 適切なデータを与えられたとき、平衡時における各物質の量を計算する
    9. 温度、濃度、圧力の変化や触媒の有無が、ある反応の平衡定数の値に影響を与えるかどうかを述べる
    10. 化学工業における動態平衡の理解の重要性およびルシャトリエの原理の応用例として、ハーバー法およびコンタクト法で用いられる条件を説明・解説する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Le Chatelier's principle
    Dynamic equilibrium: the rates converge

    A reversible reaction 可逆反应 can go both ways. The forward reaction 正反应 makes products; the reverse reaction 逆反应 turns products back into reactants. We show this with the sign $\rightleftharpoons$.

    In a closed system 封闭系统 (nothing enters or leaves), the reaction reaches a dynamic equilibrium 动态平衡 when:

    • the rate of the forward reaction equals the rate of the reverse reaction.
    • the concentrations of reactants and products stay constant.

    It is called dynamic because both reactions are still happening — they just cancel out. A closed system is needed, or products would escape and equilibrium could never be reached.

    日本語
    Le Chatelier's principle
    Dynamic equilibrium: the rates converge
    Cobalt(II) chloride solutions of different colours
    Cobalt chloride changes colour as its equilibrium shifts.

    A reversible reaction 可逆反应 can go both ways. The forward reaction 正反应 makes products; the reverse reaction 逆反应 turns products back into reactants. We show this with the sign $\rightleftharpoons$.

    In a closed system 封闭系统 (nothing enters or leaves), the reaction reaches a dynamic equilibrium 动态平衡 when:

    • the rate of the forward reaction equals the rate of the reverse reaction.
    • the concentrations of reactants and products stay constant.

    It is called dynamic because both reactions are still happening — they just cancel out. A closed system is needed, or products would escape and equilibrium could never be reached.

    A graph where the forward rate falls and the reverse rate rises until the two meet and stay equal
    At dynamic equilibrium the forward and reverse rates have become equal, so the concentrations stay constant
    Explore · ⁨探索⁩

    Dynamic equilibrium

    forward rate = reverse rate

    At equilibrium the position can sit anywhere — change a condition and watch it shift.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    reversible reaction/rɪˈvɜːsɪbl rɪˈækʃn/ 可逆反応
    forward reaction/ˈfɔːwəd rɪˈækʃn/ 順反応
    reverse reaction/rɪˈvɜːs rɪˈækʃn/ 逆反応
    closed system/kləʊzd ˈsɪstəm/ 閉じた系
    dynamic equilibrium/daɪˈnæmɪk ˌiːkwɪˈlɪbrɪəm/ 動態平衡
    Watch lesson · ⁨レッスンを視聴⁩
    7.1

    Le Chatelier's principle

    English

    Le Chatelier's principle 勒夏特列原理 says: if you change a system at equilibrium, the position of equilibrium 平衡 moves to oppose (reduce) that change.

    Change you make Which way the equilibrium moves
    increase concentration of a reactant towards the products
    increase pressure towards the side with fewer gas molecules
    increase temperature towards the endothermic direction
    add a catalyst 催化剂 no shift (it speeds up both ways equally)

    A catalyst lets equilibrium be reached faster, but it does not change the position of equilibrium.

    日本語

    Le Chatelier's principle 勒夏特列原理 says: if you change a system at equilibrium, the position of equilibrium 平衡 moves to oppose (reduce) that change.

    Change you make Which way the equilibrium moves
    increase concentration of a reactant towards the products
    increase pressure towards the side with fewer gas molecules
    increase temperature towards the endothermic direction
    add a catalyst 催化剂 no shift (it speeds up both ways equally)

    A catalyst lets equilibrium be reached faster, but it does not change the position of equilibrium.

    Three changes each with an arrow showing which way the equilibrium moves: add reactant towards products, raise pressure to the fewer-molecules side, raise temperature in the endothermic direction
    Le Chatelier's principle: the equilibrium always shifts to oppose the change you make
    Explore · ⁨探索⁩

    Shifting an equilibrium

    Change the temperature, pressure or concentration and watch the equilibrium shift to oppose your change — Le Chatelier's principle in action.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Le Chatelier's principle/lə ˈtʃeɪtlɪəz ˈprɪnsɪpl/ ル・シャトリエの原理
    equilibrium/ˌiːkwɪˈlɪbrɪəm/ 平衡
    catalyst/ˈkætəlɪst/ 触媒
    7.1

    Equilibrium constants

    English

    For a reaction at equilibrium, the equilibrium constant 平衡常数 links the amounts of products and reactants. For the reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$:

    $$K_c = \frac{[\text{C}]^c\,[\text{D}]^d}{[\text{A}]^a\,[\text{B}]^b}$$

    where the square brackets mean concentration in $\text{mol dm}^{-3}$.

    Worked example. At equilibrium the mixture $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ has $[\text{H}_2] = 0.20$, $[\text{I}_2] = 0.20$ and $[\text{HI}] = 1.6\ \text{mol dm}^{-3}$. Find $K_c$.

    $$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{1.6^2}{0.20 \times 0.20} = 64$$

    (no units here, because the concentration powers cancel top and bottom).

    Worked example (ICE table). $1.00\ \text{mol}$ each of $\text{H}_2$ and $\text{I}_2$ are sealed in a $1.00\ \text{dm}^3$ flask; at equilibrium $0.20\ \text{mol}$ of $\text{H}_2$ remains. Set out an ICE table (Initial, Change, Equilibrium):

    $\text{H}_2$ $\text{I}_2$ $2\text{HI}$
    Initial 1.00 1.00 0
    Change $-0.80$ $-0.80$ $+1.60$
    Equilibrium 0.20 0.20 1.60

    $0.80\ \text{mol}$ of $\text{H}_2$ reacted, so $2\times0.80 = 1.60\ \text{mol}$ HI formed. In the $1.00\ \text{dm}^3$ flask the concentrations equal the moles, so $K_c = \dfrac{1.60^2}{0.20\times0.20} = 64$.

    For reactions of gases, we use partial pressure 分压 instead of concentration. The partial pressure of a gas is the share of the total pressure that it provides. It is found from the mole fraction 摩尔分数 (the fraction of all the moles that are that gas):

    $$\text{partial pressure} = \text{mole fraction} \times \text{total pressure}$$

    The constant written with partial pressures is $K_p$.

    Worked example. A gas mixture holds $2.0\ \text{mol}$ of $\text{N}_2$ and $6.0\ \text{mol}$ of $\text{H}_2$ at a total pressure of $200\ \text{kPa}$. Find the partial pressure of each gas.

    There are $8.0\ \text{mol}$ in total, so

    $$p(\text{N}_2) = \frac{2.0}{8.0} \times 200 = 50\ \text{kPa}, \qquad p(\text{H}_2) = \frac{6.0}{8.0} \times 200 = 150\ \text{kPa}.$$

    Worked example (a numeric $K_p$). For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ the equilibrium partial pressures are $p(\text{N}_2)=20$, $p(\text{H}_2)=40$ and $p(\text{NH}_3)=10\ \text{kPa}$. Then

    $$K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2)\,p(\text{H}_2)^3} = \frac{10^2}{20\times40^3} = 7.8\times10^{-5}\ \text{kPa}^{-2},$$
    the units coming from $\dfrac{\text{kPa}^2}{\text{kPa}\times\text{kPa}^3}=\text{kPa}^{-2}$.

    Only temperature changes the value of $K_c$ or $K_p$. Changing concentration or pressure shifts the position of equilibrium but leaves the constant unchanged; adding a catalyst changes neither the constant nor the position — it only makes equilibrium arrive faster.

    日本語

    For a reaction at equilibrium, the equilibrium constant 平衡常数 links the amounts of products and reactants. For the reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$:

    $$K_c = \frac{[\text{C}]^c\,[\text{D}]^d}{[\text{A}]^a\,[\text{B}]^b}$$

    where the square brackets mean concentration in $\text{mol dm}^{-3}$.

    Worked example. At equilibrium the mixture $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ has $[\text{H}_2] = 0.20$, $[\text{I}_2] = 0.20$ and $[\text{HI}] = 1.6\ \text{mol dm}^{-3}$. Find $K_c$.

    $$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{1.6^2}{0.20 \times 0.20} = 64$$

    (no units here, because the concentration powers cancel top and bottom).

    Worked example (ICE table). $1.00\ \text{mol}$ each of $\text{H}_2$ and $\text{I}_2$ are sealed in a $1.00\ \text{dm}^3$ flask; at equilibrium $0.20\ \text{mol}$ of $\text{H}_2$ remains. Set out an ICE table (Initial, Change, Equilibrium):

    $\text{H}_2$ $\text{I}_2$ $2\text{HI}$
    Initial 1.00 1.00 0
    Change $-0.80$ $-0.80$ $+1.60$
    Equilibrium 0.20 0.20 1.60

    $0.80\ \text{mol}$ of $\text{H}_2$ reacted, so $2\times0.80 = 1.60\ \text{mol}$ HI formed. In the $1.00\ \text{dm}^3$ flask the concentrations equal the moles, so $K_c = \dfrac{1.60^2}{0.20\times0.20} = 64$.

    For reactions of gases, we use partial pressure 分压 instead of concentration. The partial pressure of a gas is the share of the total pressure that it provides. It is found from the mole fraction 摩尔分数 (the fraction of all the moles that are that gas):

    $$\text{partial pressure} = \text{mole fraction} \times \text{total pressure}$$

    The constant written with partial pressures is $K_p$.

    Worked example. A gas mixture holds $2.0\ \text{mol}$ of $\text{N}_2$ and $6.0\ \text{mol}$ of $\text{H}_2$ at a total pressure of $200\ \text{kPa}$. Find the partial pressure of each gas.

    There are $8.0\ \text{mol}$ in total, so

    $$p(\text{N}_2) = \frac{2.0}{8.0} \times 200 = 50\ \text{kPa}, \qquad p(\text{H}_2) = \frac{6.0}{8.0} \times 200 = 150\ \text{kPa}.$$

    Worked example (a numeric $K_p$). For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ the equilibrium partial pressures are $p(\text{N}_2)=20$, $p(\text{H}_2)=40$ and $p(\text{NH}_3)=10\ \text{kPa}$. Then

    $$K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2)\,p(\text{H}_2)^3} = \frac{10^2}{20\times40^3} = 7.8\times10^{-5}\ \text{kPa}^{-2},$$
    the units coming from $\dfrac{\text{kPa}^2}{\text{kPa}\times\text{kPa}^3}=\text{kPa}^{-2}$.

    Only temperature changes the value of $K_c$ or $K_p$. Changing concentration or pressure shifts the position of equilibrium but leaves the constant unchanged; adding a catalyst changes neither the constant nor the position — it only makes equilibrium arrive faster.

    Explore · ⁨探索⁩

    Le Chatelier and Kc

    Change temperature, pressure or concentration and watch the equilibrium shift to oppose it — Kc itself only changes with temperature.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    equilibrium constant/ˌiːkwɪˈlɪbrɪəm ˈkɒnstənt/ 平衡定数
    partial pressure/ˈpɑːʃl ˈpreʃə/ 分圧
    mole fraction/məʊl ˈfrækʃn/ 物質量分数
    7.1

    The Haber and Contact processes

    English

    These two industrial processes are chosen by balancing yield, rate and cost using Le Chatelier's principle.

    • the Haber process 哈伯法 makes ammonia: $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ (exothermic). It uses about $450\,°\text{C}$, $200\ \text{atm}$ and an iron catalyst. A low temperature would give more ammonia but too slowly, so a moderate temperature is a compromise.
    • the Contact process 接触法 makes sulfur trioxide: $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ (exothermic). It uses about $450\,°\text{C}$, near $1$–$2\ \text{atm}$ and a vanadium(V) oxide catalyst.
    日本語
    A large industrial chemical complex
    Ammonia for fertiliser is made by the Haber process in large industrial plants like this one.

    These two industrial processes are chosen by balancing yield, rate and cost using Le Chatelier's principle.

    • the Haber process 哈伯法 makes ammonia: $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ (exothermic). It uses about $450\,°\text{C}$, $200\ \text{atm}$ and an iron catalyst. A low temperature would give more ammonia but too slowly, so a moderate temperature is a compromise.
    • the Contact process 接触法 makes sulfur trioxide: $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ (exothermic). It uses about $450\,°\text{C}$, near $1$–$2\ \text{atm}$ and a vanadium(V) oxide catalyst.
    A graph of ammonia yield against temperature for three pressures, yield falling with temperature and rising with pressure, with the operating point marked
    The Haber equilibrium gives more ammonia at lower temperature and higher pressure; about $450\,°$C and $200$ atm is the working compromise
    Explore · ⁨探索⁩

    The Haber & Contact processes

    compromise: yield vs rate

    High pressure boosts yield; high temperature speeds it up but lowers yield — a compromise.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Haber process/ˈheɪbə ˈprəʊses/ ハーバー法
    Contact process/ˈkɒntækt ˈprəʊses/ コンタクト法
    7.2

    Acids and bases: the Brønsted–Lowry theory

    Syllabus · ⁨シラバス⁩
    English
    1. state the names and formulas of the common acids, limited to hydrochloric acid, $\text{HCl}$, sulfuric acid, $\text{H}_2\text{SO}_4$, nitric acid, $\text{HNO}_3$, ethanoic acid, $\text{CH}_3\text{COOH}$
    2. state the names and formulas of the common alkalis, limited to sodium hydroxide, $\text{NaOH}$, potassium hydroxide, $\text{KOH}$, ammonia, $\text{NH}_3$
    3. describe the Brønsted–Lowry theory of acids and bases
    4. describe strong acids and strong bases as fully dissociated in aqueous solution and weak acids and weak bases as partially dissociated in aqueous solution
    5. appreciate that water has pH of 7, acid solutions pH of below 7 and alkaline solutions pH of above 7
    6. explain qualitatively the differences in behaviour between strong and weak acids including the reaction with a reactive metal and difference in pH values by use of a pH meter, universal indicator or conductivity
    7. understand that neutralisation reactions occur when $\text{H}^+(\text{aq})$ and $\text{OH}^-(\text{aq})$ form $\text{H}_2\text{O}(\text{l})$
    8. understand that salts are formed in neutralisation reactions
    9. sketch the pH titration curves of titrations using combinations of strong and weak acids with strong and weak alkalis
    10. select suitable indicators for acid-alkali titrations, given appropriate data ($\text{p}K_a$ values will not be used)
    日本語
    1. 一般的な酸の名前と化学式を述べる。ただし、塩酸 $\text{HCl}$、硫酸 $\text{H}_2\text{SO}_4$、硝酸 $\text{HNO}_3$、酢酸 $\text{CH}_3\text{COOH}$に限定する
    2. 一般的なアルカリ(塩基)の名前と化学式を述べる。ただし、水酸化ナトリウム $\text{NaOH}$、水酸化カリウム $\text{KOH}$、アンモニア $\text{NH}_3$に限定する
    3. ブレンステッド-ローリーの酸塩基理論を説明する
    4. 強酸および強塩基は水溶液中で完全に解離し、弱酸および弱塩基は水溶液中で部分的に解離することを述べる
    5. 水のpHは7であり、酸性溶液のpHは7未満、アルカリ性溶液のpHは7以上であることを理解する
    6. pHメーター、万能指示薬または電気伝導度の測定を用いて、強酸と弱酸の挙動の違い(活性金属との反応、pH値の違いなど)を定性的に説明する
    7. $\text{H}^+(\text{aq})$と$\text{OH}^-(\text{aq})$が結合して$\text{H}_2\text{O}(\text{l})$を形成したときに中和反応が起こることを理解する
    8. 中和反応によって塩が生成されることを理解する
    9. 強酸・弱酸と強塩基・弱塩基の組み合わせによる滴定のpH滴定曲線を描く
    10. 適切なデータに基づき、酸塩基滴定に適した指示薬を選択する($\text{p}K_a$の値は使用しない)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A proton 质子 is simply an $\text{H}^+$ ion. The Brønsted–Lowry theory defines acids and bases by what they do with protons:

    • an acid 酸 is a proton donor 质子供体 (it gives away $\text{H}^+$).
    • a base 碱 is a proton acceptor 质子受体 (it takes $\text{H}^+$). A base that dissolves in water is called an alkali.

    Common acids you must know: hydrochloric acid ($\text{HCl}$), sulfuric acid ($\text{H}_2\text{SO}_4$), nitric acid ($\text{HNO}_3$) and ethanoic acid ($\text{CH}_3\text{COOH}$). Common alkalis: sodium hydroxide ($\text{NaOH}$), potassium hydroxide ($\text{KOH}$) and ammonia ($\text{NH}_3$).

    日本語

    A proton 质子 is simply an $\text{H}^+$ ion. The Brønsted–Lowry theory defines acids and bases by what they do with protons:

    • an acid 酸 is a proton donor 质子供体 (it gives away $\text{H}^+$).
    • a base 碱 is a proton acceptor 质子受体 (it takes $\text{H}^+$). A base that dissolves in water is called an alkali.
    Hydrogen chloride donating a proton to a water molecule, forming a hydroxonium ion and a chloride ion, with the two conjugate pairs marked
    Brønsted–Lowry: the acid donates a proton ($\text{H}^+$) to the base, making two conjugate acid–base pairs

    Common acids you must know: hydrochloric acid ($\text{HCl}$), sulfuric acid ($\text{H}_2\text{SO}_4$), nitric acid ($\text{HNO}_3$) and ethanoic acid ($\text{CH}_3\text{COOH}$). Common alkalis: sodium hydroxide ($\text{NaOH}$), potassium hydroxide ($\text{KOH}$) and ammonia ($\text{NH}_3$).

    Explore · ⁨探索⁩

    The pH scale

    Slide the pH or tap a substance — each step down in pH means ten times more H⁺ ions; acids are below 7, alkalis above.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    proton/ˈprəʊtɒn/ 陽子
    acid/ˈæsɪd/ 酸
    proton donor/ˈprəʊtɒn ˈdəʊnə/ プロトン供与体
    base/beɪs/ 底
    proton acceptor/ˈprəʊtɒn əkˈseptə/ プロトン受容体
    Watch lesson · ⁨レッスンを視聴⁩
    7.2

    Strong and weak acids and bases

    English

    This is about how fully an acid or base splits up in water — not how concentrated it is.

    • a strong acid 强酸 or strong base 强碱 is fully dissociated 解离 in water (almost every molecule splits into ions).
    • a weak acid 弱酸 or weak base 弱碱 is only partly dissociated (most molecules stay whole).

    The pH scale measures how acidic a solution is: pure water is pH 7, acids are below 7, and alkalis are above 7. A strong acid has a lower pH than a weak acid of the same concentration.

    You can tell strong and weak acids apart by:

    • reaction with a reactive metal: a strong acid fizzes faster.
    • pH: measured with a pH meter or universal indicator 通用指示剂.
    • electrical conductivity: a strong acid conducts better, because it has more ions.
    日本語

    This is about how fully an acid or base splits up in water — not how concentrated it is.

    • a strong acid 强酸 or strong base 强碱 is fully dissociated 解离 in water (almost every molecule splits into ions).
    • a weak acid 弱酸 or weak base 弱碱 is only partly dissociated (most molecules stay whole).
    Two beakers at the same concentration: the strong acid full of separate ions, the weak acid full of whole molecules with only a few ions
    Same concentration, different ionisation: a strong acid is fully dissociated into ions; a weak acid stays mostly as whole molecules

    The pH scale measures how acidic a solution is: pure water is pH 7, acids are below 7, and alkalis are above 7. A strong acid has a lower pH than a weak acid of the same concentration.

    You can tell strong and weak acids apart by:

    • reaction with a reactive metal: a strong acid fizzes faster.
    • pH: measured with a pH meter or universal indicator 通用指示剂.
    • electrical conductivity: a strong acid conducts better, because it has more ions.
    Explore · ⁨探索⁩

    Strong vs weak acids

    A strong acid fully ionises (low pH); a weak acid only partly ionises, so at the same concentration its pH is higher. Slide to compare.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    strong acid/strɒŋ ˈæsɪd/ 強酸
    strong base/strɒŋ beɪs/ 強塩基
    dissociated/dɪˈsəʊsɪeɪtɪd/ 解離
    weak acid/wiːk ˈæsɪd/ 弱酸
    weak base/wiːk beɪs/ 弱塩基
    universal indicator/ˌjuːnɪˈvɜːsl ˈɪndɪkeɪtə/ 万能指示薬
    7.2

    Neutralisation, salts and titration curves

    English

    Neutralisation 中和 happens when the hydrogen ions from an acid react with the hydroxide ions from an alkali:

    $$\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$$

    A salt 盐 is also formed, from the rest of the acid and base.

    In a titration 滴定 you add one solution to another and follow the pH. The titration curve 滴定曲线 has a steep, almost vertical jump near the end point. The exact shape depends on whether each reactant is strong or weak.

    To pick an indicator 指示剂, choose one whose colour change falls inside that steep jump. For a strong acid with a strong base most indicators work; for a weak acid with a strong base you need one that changes in the higher pH range.

    日本語

    Neutralisation 中和 happens when the hydrogen ions from an acid react with the hydroxide ions from an alkali:

    $$\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$$

    A salt 盐 is also formed, from the rest of the acid and base.

    In a titration 滴定 you add one solution to another and follow the pH. The titration curve 滴定曲线 has a steep, almost vertical jump near the end point. The exact shape depends on whether each reactant is strong or weak.

    To pick an indicator 指示剂, choose one whose colour change falls inside that steep jump. For a strong acid with a strong base most indicators work; for a weak acid with a strong base you need one that changes in the higher pH range.

    Two titration curves of pH against base added, each with a steep jump at the end point; the weak-acid curve jumps over a higher pH range, with an indicator band shaded
    Titration curves each have a steep pH jump at the end point. The weak-acid jump sits higher, so the indicator must change colour in that range
    Explore · ⁨探索⁩

    A titration curve

    Add alkali to acid and watch the pH climb. The steep jump is the equivalence point, where the acid is just neutralised.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Neutralisation/ˌnjuːtrəlaɪˈzeɪʃn/ 中和反応
    salt/sɒlt/ 塩
    titration/taɪˈtreɪʃn/ 滴定
    titration curve/taɪˈtreɪʃn kɜːv/ 滴定曲線
    indicator/ˈɪndɪkeɪtə/ 指示薬
    7.2

    Exam tips

    • Define dynamic equilibrium with all three points: forward and reverse rates equal, concentrations constant, closed system.
    • Answer Le Chatelier questions by stating the change, the direction of shift, and the reason; a catalyst does not shift the position (it speeds both rates).
    • Write $K_c$ as products over reactants, each raised to its balancing number; include state symbols to decide what appears.
    • Justify Haber/Contact conditions as a compromise between yield, rate and cost — do not just quote them.
    • Brønsted: acid = proton donor, base = proton acceptor; conjugate pairs differ by one $\text{H}^+$.
  • 8

    Reaction kinetics · ⁨反応速度論⁩

    Watch lesson · ⁨レッスンを視聴⁩
    8.1

    Rate of reaction

    Syllabus · ⁨シラバス⁩
    English
    1. explain and use the term rate of reaction, frequency of collisions, effective collisions and non-effective collisions
    2. explain qualitatively, in terms of frequency of effective collisions, the effect of concentration and pressure changes on the rate of a reaction
    3. use experimental data to calculate the rate of a reaction
    日本語
    1. 反応速度、衝突頻度、有効衝突および非有効衝突の用語を説明・使用する
    2. 有効衝突の頻度を terms に用いて、濃度および圧力の変化が反応速度に与える影響を定性的に説明する
    3. 実験データを用いて反応速度を計算する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Collision theory: energy and orientation

    The rate of reaction 反应速率 is how fast reactants turn into products. We measure it as the change in concentration (or amount) in each unit of time.

    To react, particles must collide 碰撞. The collision frequency 碰撞频率 is how often the particles hit each other. But not every collision leads to a reaction:

    • an effective collision 有效碰撞 has enough energy and the correct direction, so a reaction happens.
    • a non-effective collision 无效碰撞 does not have enough energy, or the particles hit at the wrong angle, so nothing happens.

    So the rate depends on the frequency of effective collisions — how many useful collisions happen each second.

    Concentration and pressure

    If you increase the concentration of a solution (or the pressure of a gas), the particles are packed closer together. They collide more often, so there are more effective collisions each second, and the rate goes up.

    You can calculate a rate from experimental data — for example, the volume of gas made divided by the time taken.

    Worked example. A reaction gives off carbon dioxide. In the first $30\ \text{s}$, $48\ \text{cm}^3$ of gas is collected. Find the average rate of reaction over this time.

    $$\text{average rate} = \frac{\text{volume of gas}}{\text{time}} = \frac{48}{30} = 1.6\ \text{cm}^3\,\text{s}^{-1}.$$

    The rate is fastest at the start (the graph is steepest there) because the reactants are most concentrated, then it slows as they are used up.

    日本語
    Collision theory: energy and orientation

    The rate of reaction 反应速率 is how fast reactants turn into products. We measure it as the change in concentration (or amount) in each unit of time.

    To react, particles must collide 碰撞. The collision frequency 碰撞频率 is how often the particles hit each other. But not every collision leads to a reaction:

    • an effective collision 有效碰撞 has enough energy and the correct direction, so a reaction happens.
    • a non-effective collision 无效碰撞 does not have enough energy, or the particles hit at the wrong angle, so nothing happens.

    So the rate depends on the frequency of effective collisions — how many useful collisions happen each second.

    Two molecules colliding: when their reactive ends line up a reaction happens, but with the wrong orientation there is no reaction
    A collision only reacts with the right orientation and enough energy (coloured ends = the reactive part)

    Concentration and pressure

    If you increase the concentration of a solution (or the pressure of a gas), the particles are packed closer together. They collide more often, so there are more effective collisions each second, and the rate goes up.

    Two boxes of the same size, the second holding more particles, which therefore collide more often
    More particles in the same volume collide more often, so the rate rises

    You can calculate a rate from experimental data — for example, the volume of gas made divided by the time taken.

    Worked example. A reaction gives off carbon dioxide. In the first $30\ \text{s}$, $48\ \text{cm}^3$ of gas is collected. Find the average rate of reaction over this time.

    $$\text{average rate} = \frac{\text{volume of gas}}{\text{time}} = \frac{48}{30} = 1.6\ \text{cm}^3\,\text{s}^{-1}.$$

    The rate is fastest at the start (the graph is steepest there) because the reactants are most concentrated, then it slows as they are used up.

    Explore · ⁨探索⁩

    Rate of reaction · ⁨反応速度⁩

    [A] = [A]₀·bᵗ

    Reactant concentration falls over time as it's used up. · ⁨試薬が消費されるにつれて、時間とともに反応物濃度が低下する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    rate of reaction/reɪt ɒv rɪˈækʃn/ 反応速度
    collide/kəˈlaɪd/ 衝突
    collision frequency/kəˈlɪʒn ˈfriːkwənsi/ 衝突頻度
    effective collision/ɪˈfektɪv kəˈlɪʒn/ 有効衝突
    non-effective collision/nɒn ɪˈfektɪv kəˈlɪʒn/ 無効衝突
    Watch lesson · ⁨レッスンを視聴⁩
    8.2

    Temperature and activation energy

    Syllabus · ⁨シラバス⁩
    English
    1. define activation energy, $E_A$, as the minimum energy required for a collision to be effective
    2. sketch and use the Boltzmann distribution to explain the significance of activation energy
    3. explain qualitatively, in terms both of the Boltzmann distribution and of frequency of effective collisions, the effect of temperature change on the rate of a reaction
    日本語
    1. 活性化エネルギー $E_A$を、有効な衝突に必要な最小エネルギーとして定義する
    2. ボルツマン分布を描き、それを用いて活性化エネルギーの意義を説明する
    3. ボルツマン分布および有効衝突の頻度を terms に用いて、温度変化が反応速度に与える影響を定性的に説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Maxwell-Boltzmann distribution

    The activation energy 活化能 ($E_A$) is the minimum energy a collision needs in order to be effective.

    The Boltzmann distribution 玻尔兹曼分布 is a graph showing how the energies of the molecules are spread out at one temperature. The curve starts at the origin, rises to a peak, then falls away in a long tail. The total area under the curve is the total number of molecules. Only the molecules to the right of $E_A$ have enough energy to react.

    When you raise the temperature:

    • the curve flattens and spreads to the right, so a much larger fraction of molecules now have energy greater than $E_A$.
    • the molecules also move faster and collide more often.

    The first effect is the bigger one. This is why a small rise in temperature gives a large rise in rate.

    日本語
    Maxwell-Boltzmann distribution

    The activation energy 活化能 ($E_A$) is the minimum energy a collision needs in order to be effective.

    The Boltzmann distribution 玻尔兹曼分布 is a graph showing how the energies of the molecules are spread out at one temperature. The curve starts at the origin, rises to a peak, then falls away in a long tail. The total area under the curve is the total number of molecules. Only the molecules to the right of $E_A$ have enough energy to react.

    When you raise the temperature:

    • the curve flattens and spreads to the right, so a much larger fraction of molecules now have energy greater than $E_A$.
    • the molecules also move faster and collide more often.

    The first effect is the bigger one. This is why a small rise in temperature gives a large rise in rate.

    Boltzmann distribution curves at a lower and a higher temperature, with the activation energy marked and the reacting fraction shaded
    At higher temperature the curve spreads to the right, so a larger fraction of molecules can react
    Explore · ⁨探索⁩

    Why heat speeds up reactions · ⁨熱が反応速度を上げる理由⁩

    Raise the temperature: the curve spreads right, so more molecules have at least the activation energy and can react. · ⁨温度を上げると曲線が右に広がり、より多くの分子が活性化エネルギー以上を持ち、反応できるようになります。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ 活性化エネルギー
    Boltzmann distribution/ˈbɒltsmən ˌdɪstrɪˈbjuːʃn/ ボルツマン分布
    Watch lesson · ⁨レッスンを視聴⁩
    8.3

    Catalysts

    Syllabus · ⁨シラバス⁩
    English
    1. explain and use the terms catalyst and catalysis: (a) explain that, in the presence of a catalyst, a reaction has a different mechanism, i.e. one of lower activation energy (b) explain this catalytic effect in terms of the Boltzmann distribution (c) construct and interpret a reaction pathway diagram, for a reaction in the presence and absence of an effective catalyst
    日本語
    1. 触媒および触媒作用の用語を説明・使用する: (a) 触媒が存在する場合、反応は異なる反応経路(すなわち低い活性化エネルギーを持つもの)をとることを説明 (b) ボルツマン分布を terms に触媒効果について説明 (c) 有効な触媒の有無における反応の反応経路図を描き、解釈する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Reaction profile: a catalyst's lower-energy route

    A catalyst 催化剂 speeds up a reaction but is not used up itself. Catalysis 催化作用 is the name for this action.

    A catalyst works by giving the reaction a different reaction mechanism 反应机理 — a new route with a lower activation energy. On the Boltzmann distribution, lowering $E_A$ moves the line to the left, so more molecules now have enough energy. This means more effective collisions each second, and a faster rate.

    On a reaction pathway diagram 反应路径图, the catalysed route has a lower energy "hill". The enthalpy change of the reaction, $\Delta H$, is not changed by the catalyst.

    There are two types:

    • a homogeneous catalyst 均相催化剂 is in the same physical state as the reactants — for example, an acid catalyst dissolved in a solution of liquids.
    • a heterogeneous catalyst 多相催化剂 is in a different state from the reactants — for example, solid iron speeding up the reaction of gases in the Haber process.
    日本語
    Reaction profile: a catalyst's lower-energy route

    A catalyst 催化剂 speeds up a reaction but is not used up itself. Catalysis 催化作用 is the name for this action.

    A catalyst works by giving the reaction a different reaction mechanism 反应机理 — a new route with a lower activation energy. On the Boltzmann distribution, lowering $E_A$ moves the line to the left, so more molecules now have enough energy. This means more effective collisions each second, and a faster rate.

    On a reaction pathway diagram 反应路径图, the catalysed route has a lower energy "hill". The enthalpy change of the reaction, $\Delta H$, is not changed by the catalyst.

    Reaction pathway diagram comparing the catalysed and uncatalysed routes, with a lower activation energy hill for the catalysed route and the same enthalpy change
    A catalyst gives a route with lower activation energy; the enthalpy change is unchanged

    There are two types:

    • a homogeneous catalyst 均相催化剂 is in the same physical state as the reactants — for example, an acid catalyst dissolved in a solution of liquids.
    • a heterogeneous catalyst 多相催化剂 is in a different state from the reactants — for example, solid iron speeding up the reaction of gases in the Haber process.
    Two panels: a homogeneous catalyst mixed in with the reactant particles in the same state, and a heterogeneous catalyst as a separate solid surface with reactants above it
    A homogeneous catalyst is mixed in with the reactants (same state); a heterogeneous catalyst is a separate surface (different state), where reactants meet and react
    A cylindrical catalytic converter block with a fine honeycomb surface — a real heterogeneous catalyst
    A car's catalytic converter is a heterogeneous catalyst; its honeycomb gives a huge surface area
    Explore · ⁨探索⁩

    Catalysts · ⁨触媒⁩

    a catalyst lowers Ea · ⁨触媒はEₐを下げる⁩

    A catalyst lowers Ea, so a bigger fraction of molecules can react — without heating. · ⁨触媒はEₐを下げることで、より多くの分子が加熱せずに反応できるようになります。⁩

    Explore · ⁨探索⁩

    How a catalyst works · ⁨触媒の働き⁩

    Add a catalyst and watch the activation-energy barrier drop — it gives an easier route, without changing ΔH. · ⁨触媒を加えると活性化エネルギーの障壁が下がります。ΔHは変化せず、より易しい経路を提供します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    catalyst/ˈkætəlɪst/ 触媒
    catalysis/kæˈtæləsɪs/ 触媒作用
    reaction mechanism/rɪˈækʃn ˈmekənɪzəm/ 反応機構
    reaction pathway diagram/rɪˈækʃn ˈpæθweɪ ˈdaɪəɡræm/ 反応経路図
    homogeneous catalyst/həˈməʊdʒnɪəs ˈkætəlɪst/ 均一触媒
    heterogeneous catalyst/ˌhetrəˈdʒiːnɪəs ˈkætəlɪst/ 不均一触媒
    Watch lesson · ⁨レッスンを視聴⁩
    8.3

    Exam tips

    • Explain rate changes with collision theory — more frequent and/or more energetic effective collisions.
    • On a Boltzmann distribution mark $E_A$, shade to its right, and show the curve flatten and shift right at higher temperature; it starts at the origin and never touches the axis.
    • A catalyst gives an alternative route with lower $E_A$ and does not change $\Delta H$.
    • Say why a small temperature rise gives a large rate rise: a much greater proportion of molecules now exceed $E_A$ (the main effect).
  • 9

    The Periodic Table: chemical periodicity · ⁨周期表:化学的周期性⁩

    Watch lesson · ⁨レッスンを視聴⁩
    9.1

    Physical properties across Period 3

    Syllabus · ⁨シラバス⁩
    English
    1. describe qualitatively (and indicate the periodicity in) the variations in atomic radius, ionic radius, melting point and electrical conductivity of the elements
    2. explain the variation in melting point and electrical conductivity in terms of the structure and bonding of the elements
    日本語
    1. 原子半径、イオン半径、融点および電気伝導度の変化を定性的に説明し、その周期性を示す
    2. 元素の構造および結合を terms に、融点および電気伝導度の変化を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Periodicity 周期性 means that properties repeat in a regular pattern as you go across each period 周期 of the Periodic Table. Period 3 (Na to Ar) is the standard example.

    Property Trend across Period 3
    atomic radius 原子半径 gets smaller (more nuclear charge pulls the same shell in)
    ionic radius 离子半径 positive ions are small; from $\text{P}^{3-}$ onwards the negative ions are larger
    melting point 熔点 rises to a peak at silicon, then falls sharply
    electrical conductivity 导电性 high for Na, Mg, Al; almost zero from Si onwards

    The melting point and conductivity follow from the structure and bonding:

    • Na, Mg, Al are giant metallic 金属晶体. Melting points rise (Na → Al) because each atom gives more delocalised electrons and the ions get smaller, so the bonding is stronger. They conduct well.
    • Si is giant molecular 原子晶体 (giant covalent). It has the highest melting point, because strong covalent bonds must be broken. It barely conducts.
    • P, S, Cl, Ar are simple molecular 分子晶体 (or single atoms). Their melting points are low, because the only forces that break are the weak van der Waals' forces 范德华力 between the molecules - the strong covalent bonds inside each molecule are never broken at all. They do not conduct.

    Name van der Waals' forces when you explain that drop: it is the marking point, and "the covalent bonds are weak" is the error that loses it. Melting $\text{S}_8$ pulls whole molecules apart from each other; it does not touch a single $\text{S–S}$ bond. Their sizes still follow the molecule: $\text{S}_8$ melts higher than $\text{P}_4$ because it is bigger, with more electrons and so stronger van der Waals' forces, while $\text{Ar}$ (a single atom) is lowest of all.

    日本語
    A periodic table display
    Properties repeat in a regular pattern across each period of the table.

    Periodicity 周期性 means that properties repeat in a regular pattern as you go across each period 周期 of the Periodic Table. Period 3 (Na to Ar) is the standard example.

    A row of atoms from sodium to chlorine shrinking from left to right, with nuclear charge rising from +11 to +17
    Atomic radius decreases across Period 3: the rising nuclear charge pulls the same outer shell inwards
    Property Trend across Period 3
    atomic radius 原子半径 gets smaller (more nuclear charge pulls the same shell in)
    ionic radius 离子半径 positive ions are small; from $\text{P}^{3-}$ onwards the negative ions are larger
    melting point 熔点 rises to a peak at silicon, then falls sharply
    electrical conductivity 导电性 high for Na, Mg, Al; almost zero from Si onwards

    The melting point and conductivity follow from the structure and bonding:

    • Na, Mg, Al are giant metallic 金属晶体. Melting points rise (Na → Al) because each atom gives more delocalised electrons and the ions get smaller, so the bonding is stronger. They conduct well.
    • Si is giant molecular 原子晶体 (giant covalent). It has the highest melting point, because strong covalent bonds must be broken. It barely conducts.
    • P, S, Cl, Ar are simple molecular 分子晶体 (or single atoms). Their melting points are low, because the only forces that break are the weak van der Waals' forces 范德华力 between the molecules - the strong covalent bonds inside each molecule are never broken at all. They do not conduct.

    Name van der Waals' forces when you explain that drop: it is the marking point, and "the covalent bonds are weak" is the error that loses it. Melting $\text{S}_8$ pulls whole molecules apart from each other; it does not touch a single $\text{S–S}$ bond. Their sizes still follow the molecule: $\text{S}_8$ melts higher than $\text{P}_4$ because it is bigger, with more electrons and so stronger van der Waals' forces, while $\text{Ar}$ (a single atom) is lowest of all.

    A graph of melting point across Period 3 rising through the metals to a sharp peak at silicon, then dropping to the simple molecular elements
    Melting point across Period 3 peaks at silicon (giant covalent); it is high for the metals and low for the simple molecular elements
    Explore · ⁨探索⁩

    Trends across Period 3 · ⁨第3周期の傾向⁩

    Switch between atomic radius, ionisation energy and melting point, and step across Period 3 to see each periodic trend. · ⁨原子半径、第一イオン化エネルギー、融点の切り替えを行い、第3周期を横断して各周期の傾向を確認します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    periodicity/ˌpɪərɪˌɒˈdɪsɪti/ 周期性
    period/ˈpɪərɪəd/ 周期の変化率である
    atomic radius/əˈtɒmɪk ˈreɪdɪəs/ 原子半径
    ionic radius/aɪˈɒnɪk ˈreɪdɪəs/ イオン半径
    melting point/ˈmeltɪŋ pɔɪnt/ 融点
    electrical conductivity/ɪˈlektrɪkl kɒndəkˈtɪvɪti/ 電気伝導性
    giant metallic/ˈdʒaɪənt məˈtælɪk/ 巨大金属
    giant molecular/ˈdʒaɪənt məˈlekjʊlə/ 巨大分子
    simple molecular/ˈsɪmpl məˈlekjʊlə/ 简单分子
    van der Waals' forces/ˈvændɜː wɑːlz ˈfɔːsɪz/ ファンデルワールス力
    9.2

    Chemical properties across Period 3

    Syllabus · ⁨シラバス⁩
    English
    1. describe, and write equations for, the reactions of the elements with oxygen (to give $\text{Na}_2\text{O}$, $\text{MgO}$, $\text{Al}_2\text{O}_3$, $\text{P}_4\text{O}_{10}$, $\text{SO}_2$), chlorine (to give $\text{NaCl}$, $\text{MgCl}_2$, $\text{AlCl}_3$, $\text{SiCl}_4$, $\text{PCl}_5$) and water ($\text{Na}$ and $\text{Mg}$ only)
    2. state and explain the variation in the oxidation number of the oxides ($\text{Na}_2\text{O}$, $\text{MgO}$, $\text{Al}_2\text{O}_3$, $\text{P}_4\text{O}_{10}$, $\text{SO}_2$ and $\text{SO}_3$ only) and chlorides ($\text{NaCl}$, $\text{MgCl}_2$, $\text{AlCl}_3$, $\text{SiCl}_4$, $\text{PCl}_5$ only) in terms of their outer shell (valence shell) electrons
    3. describe, and write equations for, the reactions, if any, of the oxides $\text{Na}_2\text{O}$, $\text{MgO}$, $\text{Al}_2\text{O}_3$, $\text{SiO}_2$, $\text{P}_4\text{O}_{10}$, $\text{SO}_2$ and $\text{SO}_3$ with water including the likely pHs of the solutions obtained
    4. describe, explain, and write equations for, the acid/base behaviour of the oxides $\text{Na}_2\text{O}$, $\text{MgO}$, $\text{Al}_2\text{O}_3$, $\text{P}_4\text{O}_{10}$, $\text{SO}_2$ and $\text{SO}_3$ and the hydroxides $\text{NaOH}$, $\text{Mg(OH)}_2$ and $\text{Al(OH)}_3$ including, where relevant, amphoteric behaviour in reactions with acids and bases (sodium hydroxide only)
    5. describe, explain, and write equations for, the reactions of the chlorides $\text{NaCl}$, $\text{MgCl}_2$, $\text{AlCl}_3$, $\text{SiCl}_4$, $\text{PCl}_5$ with water including the likely pHs of the solutions obtained
    6. explain the variations and trends in 9.2.2, 9.2.3, 9.2.4 and 9.2.5 in terms of bonding and electronegativity
    7. suggest the types of chemical bonding present in the chlorides and oxides from observations of their chemical and physical properties
    日本語
    1. 元素の酸素($\text{Na}_2\text{O}$、$\text{MgO}$、$\text{Al}_2\text{O}_3$、$\text{P}_4\text{O}_{10}$、$\text{SO}_2$を与える)、塩素($\text{NaCl}$、$\text{MgCl}_2$、$\text{AlCl}_3$、$\text{SiCl}_4$、$\text{PCl}_5$を与える)および水($\text{Na}$および$\text{Mg}$のみ)との反応を説明し、化学方程式を書く
    2. 酸化物($\text{Na}_2\text{O}$、$\text{MgO}$、$\text{Al}_2\text{O}_3$、$\text{P}_4\text{O}_{10}$、$\text{SO}_2$、$\text{SO}_3$のみ)および塩化物($\text{NaCl}$、$\text{MgCl}_2$、$\text{AlCl}_3$、$\text{SiCl}_4$、$\text{PCl}_5$のみ)の酸化数の変化を、最外殻電子(価電子)を terms に述べて説明する
    3. 酸化物$\text{Na}_2\text{O}$、$\text{MgO}$、$\text{Al}_2\text{O}_3$、$\text{SiO}_2$、$\text{P}_4\text{O}_{10}$、$\text{SO}_2$、$\text{SO}_3$の水との反応(あれば)を説明し、化学方程式を書き、得られる溶液の想定されるpHを述べる
    4. 酸化物$\text{Na}_2\text{O}$、$\text{MgO}$、$\text{Al}_2\text{O}_3$、$\text{P}_4\text{O}_{10}$、$\text{SO}_2$、$\text{SO}_3$および水酸化物$\text{NaOH}$、$\text{Mg(OH)}_2$、$\text{Al(OH)}_3$の酸塩基挙動を説明・解説し、化学方程式を書く。関連する場合は、酸および塩基(水酸化ナトリウムのみ)との反応における両性挙動を含む
    5. 塩化物 $\text{NaCl}$, $\text{MgCl}_2$, $\text{AlCl}_3$, $\text{SiCl}_4$, $\text{PCl}_5$ の水との反応を説明・記述し、化学方程式を書き、得られる溶液のpHの推定値を示す
    6. 結合と電気陰性度を用いて、9.2.2, 9.2.3, 9.2.4 および 9.2.5 の変化と傾向を説明する
    7. 化学的および物理的性質の観察に基づき、塩化物および酸化物に見られる化学結合の種類を提案する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Reactions with oxygen, chlorine and water

    With oxygen:

    $$4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} \qquad 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \qquad 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$$
    $$\text{P}_4 + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10} \qquad \text{S} + \text{O}_2 \rightarrow \text{SO}_2$$

    With chlorine:

    $$2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl} \qquad \text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2 \qquad 2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$$
    $$\text{Si} + 2\text{Cl}_2 \rightarrow \text{SiCl}_4 \qquad 2\text{P} + 5\text{Cl}_2 \rightarrow 2\text{PCl}_5$$

    With water (only Na and Mg react):

    $$2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \qquad \text{Mg} + 2\text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 + \text{H}_2$$

    Sodium reacts fast; magnesium reacts only very slowly with cold water.

    Oxidation number of the oxides and chlorides

    The oxidation number 氧化数 of the Period 3 element in its oxide or chloride rises across the period, because it equals the number of outer-shell (valence shell 价层) electrons 电子 the atom uses in bonding:

    Oxides $\text{Na}_2\text{O}$ $\text{MgO}$ $\text{Al}_2\text{O}_3$ $\text{P}_4\text{O}_{10}$ $\text{SO}_2$ / $\text{SO}_3$
    oxidation number $+1$ $+2$ $+3$ $+5$ $+4$ / $+6$

    The chlorides $\text{NaCl}$, $\text{MgCl}_2$, $\text{AlCl}_3$, $\text{SiCl}_4$, $\text{PCl}_5$ show oxidation numbers $+1$ to $+5$ in the same way.

    Oxides with water, and acid–base behaviour

    Across the period the oxides 氧化物 change from basic to acidic:

    Oxide With water Acid–base nature Approximate pH
    $\text{Na}_2\text{O}$ $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$ basic 13–14
    $\text{MgO}$ $\text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2$ basic 9–10
    $\text{Al}_2\text{O}_3$ insoluble amphoteric 两性 7
    $\text{SiO}_2$ insoluble weakly acidic 7
    $\text{P}_4\text{O}_{10}$ $\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4$ acidic 1–2
    $\text{SO}_3$ $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4$ strongly acidic 0–1

    Metal oxides (left) are basic; non-metal oxides (right) are acidic. $\text{Al}_2\text{O}_3$ and its hydroxide 氢氧化物 $\text{Al(OH)}_3$ are amphoteric — they react with both acids and bases:

    $$\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}$$
    $$\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{NaAl(OH)}_4$$

    Chlorides with water

    • $\text{NaCl}$ and $\text{MgCl}_2$ are ionic. They simply dissolve, giving a near-neutral solution.
    • $\text{SiCl}_4$ and $\text{PCl}_5$ are covalent. They undergo hydrolysis 水解 (react with water) to make acidic solutions and fumes of $\text{HCl}$:
    $$\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl} \qquad \text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}$$

    Explaining the trends

    These trends follow from the change in bonding and electronegativity 电负性. On the left, the elements are metals with low electronegativity, so their oxides and chlorides are ionic and basic (or neutral). On the right, the elements are non-metals with high electronegativity, so their oxides and chlorides are covalent and acidic. You can use a chloride's or oxide's properties (melting point, conductivity, effect on water) to suggest whether its bonding is ionic or covalent.

    Worked example. Predict the pH when $\text{Na}_2\text{O}$, $\text{Al}_2\text{O}_3$ and $\text{SO}_3$ are each added to water. Acid-base character follows the metal to non-metal change across Period 3. $\text{Na}_2\text{O}$ is ionic and basic: it dissolves to give $\text{NaOH}$, so the pH is about 13. $\text{SO}_3$ is covalent and acidic: it reacts to give $\text{H}_2\text{SO}_4$, so the pH is about 1. $\text{Al}_2\text{O}_3$ sits at the changeover - it is amphoteric and essentially insoluble in water, so the pH stays about 7. The trap is that last one: amphoteric does not mean neutral, it means the oxide reacts with acids and with alkalis - it simply does not react with water.

    日本語

    Reactions with oxygen, chlorine and water

    With oxygen:

    $$4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} \qquad 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \qquad 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$$
    $$\text{P}_4 + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10} \qquad \text{S} + \text{O}_2 \rightarrow \text{SO}_2$$

    With chlorine:

    $$2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl} \qquad \text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2 \qquad 2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$$
    $$\text{Si} + 2\text{Cl}_2 \rightarrow \text{SiCl}_4 \qquad 2\text{P} + 5\text{Cl}_2 \rightarrow 2\text{PCl}_5$$

    With water (only Na and Mg react):

    $$2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \qquad \text{Mg} + 2\text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 + \text{H}_2$$

    Sodium reacts fast; magnesium reacts only very slowly with cold water.

    Oxidation number of the oxides and chlorides

    The oxidation number 氧化数 of the Period 3 element in its oxide or chloride rises across the period, because it equals the number of outer-shell (valence shell 价层) electrons 电子 the atom uses in bonding:

    Oxides $\text{Na}_2\text{O}$ $\text{MgO}$ $\text{Al}_2\text{O}_3$ $\text{P}_4\text{O}_{10}$ $\text{SO}_2$ / $\text{SO}_3$
    oxidation number $+1$ $+2$ $+3$ $+5$ $+4$ / $+6$

    The chlorides $\text{NaCl}$, $\text{MgCl}_2$, $\text{AlCl}_3$, $\text{SiCl}_4$, $\text{PCl}_5$ show oxidation numbers $+1$ to $+5$ in the same way.

    Oxides with water, and acid–base behaviour

    Across the period the oxides 氧化物 change from basic to acidic:

    A coloured bar split into basic, amphoteric and acidic regions, with the Period 3 oxides placed along it from Na2O to SO3 and approximate pH values
    The Period 3 oxides change from basic (the metals) through amphoteric (Al$_2$O$_3$) to acidic (the non-metals)
    Oxide With water Acid–base nature Approximate pH
    $\text{Na}_2\text{O}$ $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$ basic 13–14
    $\text{MgO}$ $\text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2$ basic 9–10
    $\text{Al}_2\text{O}_3$ insoluble amphoteric 两性 7
    $\text{SiO}_2$ insoluble weakly acidic 7
    $\text{P}_4\text{O}_{10}$ $\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4$ acidic 1–2
    $\text{SO}_3$ $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4$ strongly acidic 0–1

    Metal oxides (left) are basic; non-metal oxides (right) are acidic. $\text{Al}_2\text{O}_3$ and its hydroxide 氢氧化物 $\text{Al(OH)}_3$ are amphoteric — they react with both acids and bases:

    $$\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}$$
    $$\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{NaAl(OH)}_4$$
    Aluminium oxide in the centre reacting to the left with acid and to the right with base
    Aluminium oxide is amphoteric — it reacts with acids (behaving as a base) and with bases (behaving as an acid)

    Chlorides with water

    • $\text{NaCl}$ and $\text{MgCl}_2$ are ionic. They simply dissolve, giving a near-neutral solution.
    • $\text{SiCl}_4$ and $\text{PCl}_5$ are covalent. They undergo hydrolysis 水解 (react with water) to make acidic solutions and fumes of $\text{HCl}$:
    $$\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl} \qquad \text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}$$
    Two routes: the ionic chlorides dissolve to a neutral solution, while the covalent chlorides hydrolyse to an acidic solution and HCl fumes
    The ionic chlorides just dissolve (neutral); the covalent chlorides react with water (hydrolyse) to give an acidic solution and HCl fumes

    Explaining the trends

    These trends follow from the change in bonding and electronegativity 电负性. On the left, the elements are metals with low electronegativity, so their oxides and chlorides are ionic and basic (or neutral). On the right, the elements are non-metals with high electronegativity, so their oxides and chlorides are covalent and acidic. You can use a chloride's or oxide's properties (melting point, conductivity, effect on water) to suggest whether its bonding is ionic or covalent.

    Worked example. Predict the pH when $\text{Na}_2\text{O}$, $\text{Al}_2\text{O}_3$ and $\text{SO}_3$ are each added to water. Acid-base character follows the metal to non-metal change across Period 3. $\text{Na}_2\text{O}$ is ionic and basic: it dissolves to give $\text{NaOH}$, so the pH is about 13. $\text{SO}_3$ is covalent and acidic: it reacts to give $\text{H}_2\text{SO}_4$, so the pH is about 1. $\text{Al}_2\text{O}_3$ sits at the changeover - it is amphoteric and essentially insoluble in water, so the pH stays about 7. The trap is that last one: amphoteric does not mean neutral, it means the oxide reacts with acids and with alkalis - it simply does not react with water.

    Explore · ⁨探索⁩

    Period 3 reaction lab · ⁨第3周期反応実験室⁩

    Classify Period 3 reactions by what they reveal about bonding and acidity. · ⁨結合および酸性について示す内容に基づき、第3周期の反応を分類せよ。⁩

    Explore · ⁨探索⁩

    Oxide and chloride hydrolysis route · ⁨酸化物および塩化物の加水分解ルート⁩

    Follow how Period 3 oxides and chlorides change pH in water. · ⁨第3周期の酸化物および塩化物が水中でpHを変化させる様子を追跡せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    oxidation number/ˌɒksɪˈdeɪʃn ˈnʌmbə/ 酸化数
    valence shell/ˈveɪləns ʃel/ 価電子殻
    electron/ɪˈlektrɒn/ 電子
    oxide/ˈɒksaɪd/ 酸化物
    amphoteric/ˌæmfəʊˈterɪk/ 両性である
    hydroxide/haɪˈdrɒksaɪd/ 水酸化物
    hydrolysis/haɪˈdrɒləsɪs/ 加水分解反応
    electronegativity/ɪˌlektrəʊŋɡəˈtɪvɪti/ 電気陰性度
    9.3

    Periodicity of other elements

    Syllabus · ⁨シラバス⁩
    English
    1. predict the characteristic properties of an element in a given group by using knowledge of chemical periodicity
    2. deduce the nature, possible position in the Periodic Table and identity of unknown elements from given information about physical and chemical properties
    日本語
    1. 化学的周期律に関する知識を用い、特定の族に属する元素の特徴的な性質を予測する
    2. 与えられた物理的・化学的性質に関する情報から、未知の元素の性質、周期表上の位置および正体を読み取る

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    The same idea works for any group. If you know the pattern down a group and across a period, you can:

    • predict the properties of an element from its position (for example, a Group 1 element will be a reactive metal forming a $+1$ ion).
    • deduce the likely position and identity of an unknown element from its physical and chemical properties.
    Explore · ⁨探索⁩

    Periodic trends · ⁨周期的傾向⁩

    Step across Period 3 to see a property rise or fall, then repeat in the next period — that recurring pattern is periodicity. · ⁨第3周期を横切ってある性質が上昇するか下降するかを見る、そして次の周期でそれを繰り返す — その繰り返しのパターンは周期性である。⁩

    9.3

    Exam tips

    • Always pair the trend with its reason: atomic radius falls across Period 3 (rising nuclear charge, similar shielding).
    • Explain the melting-point pattern by structure: giant metallic (Na→Al), giant covalent (Si, highest), then simple molecular ($\text{P}_4$, $\text{S}_8$, $\text{Cl}_2$) and monatomic (Ar).
    • Oxides go basic → amphoteric ($\text{Al}_2\text{O}_3$) → acidic across the period; link to ionic-to-covalent bonding.
    • For reactions of oxides/chlorides with water, give the products and the resulting pH.
  • 10

    Group 2 · ⁨第2族⁩

    Watch lesson · ⁨レッスンを視聴⁩
    10.1

    The Group 2 metals · ⁨第2族金属⁩

    Syllabus · ⁨シラバス⁩
    English
    1. describe, and write equations for, the reactions of the elements with oxygen, water and dilute hydrochloric and sulfuric acids
    2. describe, and write equations for, the reactions of the oxides, hydroxides and carbonates with water and dilute hydrochloric and sulfuric acids
    3. describe, and write equations for, the thermal decomposition of the nitrates and carbonates, to include the trend in thermal stabilities
    4. describe, and make predictions from, the trends in physical and chemical properties of the elements involved in the reactions in 10.1.1 and the compounds involved in 10.1.2, 10.1.3 and 10.1.5
    5. state the variation in the solubilities of the hydroxides and sulfates
    日本語
    1. 元素の酸素、水、希塩酸および希硫酸との反応を記述し、化学方程式を立てる
    2. 酸化物、水酸化物および炭酸塩の水、希塩酸および希硫酸との反応を記述し、化学方程式を立てる
    3. 熱分解について、窒素酸塩および炭酸塩の熱分解を記述し、熱安定性の傾向を含む
    4. 10.1.1の反応に関わる元素および10.1.2、10.1.3、10.1.5の反応に関わる化合物の物理的および化学的性質の傾向を記述し、予測を行う
    5. 水酸化物および硫酸塩の溶解度の変異を述べる

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Group 族 2 holds the metals magnesium, calcium, strontium and barium. They all have two outer electrons, which they lose to form $2+$ ions. Going down the group, the atoms get larger and the outer electrons are easier to lose, so the metals get more reactive — their reactivity 反应活性 increases down the group.

    Reactions of the elements

    With oxygen, they burn to form an oxide:

    $$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$

    With water, they form a hydroxide and hydrogen. The reaction gets faster down the group:

    $$\text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2$$

    Magnesium is slow with cold water but reacts fast with steam, giving $\text{MgO}$ and $\text{H}_2$.

    With dilute hydrochloric or sulfuric acid, they form a salt and hydrogen:

    $$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \qquad \text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2$$

    With sulfuric acid the reaction slows down the group, because the sulfates 硫酸盐 formed (such as $\text{BaSO}_4$) are insoluble and coat the metal.

    Reactions of the compounds

    The oxides 氧化物 and hydroxides 氢氧化物 are basic. They react with water and with dilute acids:

    $$\text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 \qquad \text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}$$

    The carbonates 碳酸盐 react with dilute acids to give a salt, water and carbon dioxide:

    $$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$$

    Thermal decomposition

    Thermal decomposition 热分解 means breaking a compound apart by heating it.

    • the carbonates break into the oxide and carbon dioxide:
    $$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$$
    • the nitrates 硝酸盐 break into the oxide, brown nitrogen dioxide gas and oxygen:
    $$2\text{Ca(NO}_3)_2 \rightarrow 2\text{CaO} + 4\text{NO}_2 + \text{O}_2$$

    The thermal stability 热稳定性 of both the carbonates and the nitrates increases down the group. A larger metal ion pulls less on the carbonate or nitrate ion, so the compound is harder to break apart — it needs a higher temperature. So magnesium carbonate decomposes most easily, and barium carbonate is the hardest.

    Trends in solubility

    Compound Trend in solubility 溶解度 down the group
    hydroxides increase (Mg(OH)$_2$ almost insoluble; Ba(OH)$_2$ soluble)
    sulfates decrease (MgSO$_4$ soluble; BaSO$_4$ insoluble)

    From these trends you can predict the properties of the next element down, radium, and of its compounds.

    Worked example. $\text{MgCO}_3$ decomposes at about $540\ °\text{C}$ and $\text{BaCO}_3$ at about $1360\ °\text{C}$. Explain the trend and predict where $\text{CaCO}_3$ sits. Going down Group 2 the cation gets larger, so its charge is spread over a bigger surface and its polarising power falls. A less polarising cation distorts the carbonate ion less, so the $\text{C-O}$ bond is weakened less and more heat is needed to break it. Thermal stability therefore increases down the group, and $\text{CaCO}_3$, lying between Mg and Ba, decomposes at an intermediate temperature of about $900\ °\text{C}$. Argue the whole chain - cation size, then polarising power, then distortion of the anion; "it is more reactive" explains nothing here.

    日本語
    カルシウム炭酸塩の海岸石灰岩崖
    石灰岩はカルシウム炭酸塩である——第2族金属カルシウムの化合物。

    第2族には、マグネシウム、カルシウム、ストロンチウム、バリウムという金属が含まれる。これらすべては2つの最外殻電子を持ち、失って $2+$ イオンを形成する。族を下 going するにつれて、原子は大きくなり、最外殻電子は失われやすくなるため、金属はより活性になる——その反応性は族を下 going して増加する。

    第2族原子がマグネシウムからバリウムまで大きくなる様子と、反応性が増すことを示す下向きの矢印
    第2族で反応性が増加する:大きな原子は2つの最外殻電子をより緩やかに保持するため、失われやすい

    元素の反応

    酸素に対して、燃焼して酸化物を形成する:

    $$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$
    ペンチで持ったマグネシウムリボンが眩しい白光を放ちながら燃え、酸化マグネシウムの白色煙を出す
    マグネシウムは空気中で鮮やかな白色の炎を放ちながら燃え、白色の酸化マグネシウムを形成する

    水に対して、水酸化物と水素ガスを形成する。反応は族を下 going して速くなる:

    $$\text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2$$

    マグネシウムは冷水に対して遅いが、蒸気に対して速く反応して $\text{MgO}$ と $\text{H}_2$ を生じる。

    希薄塩酸または希薄硫酸に対して、塩と水素ガスを形成する:

    $$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \qquad \text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2$$

    硫酸との反応は族を下 going して遅くなる。なぜなら、生成される硫酸塩(例えば $\text{BaSO}_4$ など)は難溶であり、金属を覆うからである。

    化合物の反応

    酸化物と水酸化物は塩基性である。水および希薄酸と反応する:

    $$\text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 \qquad \text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}$$

    炭酸塩は希薄酸と反応して塩、水、二酸化炭素を生じる:

    $$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$$

    熱分解

    熱分解とは、加熱によって化合物を崩壊させることである。

    • 炭酸塩は酸化物と二酸化炭素に分解する:
    $$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$$
    • 硝酸塩は酸化物、茶色の窒素酸化物ガス、酸素に分解する:
    $$2\text{Ca(NO}_3)_2 \rightarrow 2\text{CaO} + 4\text{NO}_2 + \text{O}_2$$

    炭酸塩と硝酸塩の熱安定性は族を下 going して増加する。大きな金属イオンは炭酸塩イオンや硝酸塩イオンをより弱く引き寄せるため、化合物は崩壊しにくく、高い温度が必要となる。したがって、炭酸マグネシウムは最も容易に分解し、炭酸バリウムは最も分解されにくい。

    小さなマグネシウムイオンが炭酸塩イオンを強く歪ませる様子、隣にはわずかにしか歪ませない大きなバリウムイオン
    熱安定性は族を下 going して上昇する:小さなカチオンは炭酸塩イオンをより強く分極(歪ませ)して弱め、分解されやすくする

    溶解度の傾向

    化合物 族を下 going する溶解度の傾向
    水酸化物 増加 (Mg(OH)$_2$ はほとんど難溶;Ba(OH)$_2$ は可溶)
    硫酸塩 減少 (MgSO$_4$ は可溶;BaSO$_4$ は難溶)
    対数スケールの2本の線が交差:水酸化物の溶解度はMgからBaへ上昇し、硫酸塩の溶解度は低下
    第2族を下 going するにつれ、水酸化物はより可溶になり、硫酸塩はより難溶になる(対数スケールに注意)

    これらの傾向から、族を下 going した次の元素であるラジウムおよびその化合物の性質を予測できる。

    計算例. $\text{MgCO}_3$ は約 $540\ °\text{C}$ で分解し、$\text{BaCO}_3$ は約 $1360\ °\text{C}$ で分解する。この傾向を説明し、$\text{CaCO}_3$ がどの位置にあるかを予測せよ。第2族を下るにつれてカチオンは大きくなるため、電荷がより広い表面に分散され、分極力が低下する。分極力が小さいカチオンは炭酸イオンをより少しくずすため、$\text{C-O}$ 結合は弱くならず、分解するために必要な熱も増える。したがって、熱安定性は族を下るにつれて増加し、MgとBaの間に位置する$\text{CaCO}_3$ は中間温度である約 $900\ °\text{C}$ で分解する。一連の流れを論じよ:カチオンのサイズ → 分極力 → アニオンの歪み。単に「反応性が高いから」と説明してはならない。

    Explore · ⁨探索⁩

    Group 2 reactivity ladder · ⁨第2族の反応性階段⁩

    Move down Group 2 and see why reactions become more vigorous. · ⁨第2族を下に進んで、なぜ反応が激しくなるかを確認せよ。⁩

    Explore · ⁨探索⁩

    Group 2 trend lab · ⁨第2族の傾向実験室⁩

    Compare thermal stability and solubility trends down Group 2. · ⁨第2族を下に進む際の熱的安定性と溶解度の傾向を比較せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    group/ɡruːp/ グループ
    reactivity/rɪəkˈtɪvɪti/ 反応性
    sulfate/ˈsʌlfeɪt/ 硫酸塩
    oxide/ˈɒksaɪd/ 酸化物
    hydroxide/haɪˈdrɒksaɪd/ 水酸化物
    carbonate/ˈkɑːbəneɪt/ 炭酸塩
    thermal decomposition/ˈθɜːml ˌdiːkɒmpəˈzɪʃn/ 熱分解
    nitrate/ˈnaɪtreɪt/ 窒素酸
    thermal stability/ˈθɜːml stəˈbɪlɪti/ 熱安定性
    solubility/ˌsɒljuːˈbɪlɪti/ 溶解度
    10.1

    Exam tips · ⁨試験対策⁩

    English
    • Reactivity increases down Group 2: ionisation energy falls as atoms get larger, so electrons are lost more easily.
    • Reactions with water become more vigorous down the group; Mg reacts slowly with cold water but fast with steam (→ MgO).
    • Hydroxides get more soluble down the group; sulfates get less soluble ($\text{BaSO}_4$ insoluble — the basis of the sulfate test).
    • Write equations with the correct $+2$ ion and state the observations (fizzing, dissolving).
    日本語
    • 第2族を下るにつれて反応性は増加する:原子が大きくなるにつれて第一イオン化エネルギーが低下するため、電子が失われやすくなる。
    • 水との反応は族を下るにつれて激しくなる;Mgは冷水とはゆっくり反応するが、蒸気とは速く反応する(→ MgO)。
    • 水酸化物は族を下るにつれて溶解度が増加し、硫酸塩は溶解度が減少する($\text{BaSO}_4$ は不溶 — 硫酸塩の検出法に基づく)。
    • 正しい$+2$ イオンを用いて化学方程式を書き、観察現象(泡立ち、溶解)を記述せよ。
  • 11

    Group 17 · ⁨第17族⁩

    Watch lesson · ⁨レッスンを視聴⁩
    11.1

    Physical properties of the halogens

    Syllabus · ⁨シラバス⁩
    English
    1. describe the colours and the trend in volatility of chlorine, bromine and iodine
    2. describe and explain the trend in the bond strength of the halogen molecules
    3. interpret the volatility of the elements in terms of instantaneous dipole–induced dipole forces
    日本語
    1. クロリル、臭素、ヨウ素の色および揮発性の傾向を記述する
    2. ハロゲン分子の結合能の傾向を記述・説明する
    3. 一時的双極子-誘起双極子力に基づいて元素の揮発性を解釈する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The halogens 卤素 are the Group 族 17 elements. They exist as diatomic molecules ($\text{Cl}_2$, $\text{Br}_2$, $\text{I}_2$). Going down the group:

    Element Colour and state at room temperature
    chlorine pale green gas
    bromine red-brown liquid
    iodine grey-black solid (purple vapour)

    The volatility 挥发性 (how easily a substance turns to vapour) decreases down the group: chlorine is a gas, but iodine is a solid. This is because the molecules get larger and have more electrons, so the instantaneous dipole 瞬时偶极 and induced dipole 诱导偶极 forces between them get stronger. Stronger forces are harder to break, so the boiling point rises and volatility falls.

    The bond energy 键能 (bond strength) of the $\text{X}\text{–}\text{X}$ molecules generally falls from $\text{Cl}_2$ to $\text{I}_2$, because the shared electrons are further from the nuclei in the larger atoms.

    日本語
    Bromine: a dark liquid giving off an orange vapour in a sealed ampoule
    Bromine is a Group 17 halogen — a dark liquid that gives off an orange vapour.

    The halogens 卤素 are the Group 族 17 elements. They exist as diatomic molecules ($\text{Cl}_2$, $\text{Br}_2$, $\text{I}_2$). Going down the group:

    Element Colour and state at room temperature
    chlorine pale green gas
    bromine red-brown liquid
    iodine grey-black solid (purple vapour)

    The volatility 挥发性 (how easily a substance turns to vapour) decreases down the group: chlorine is a gas, but iodine is a solid. This is because the molecules get larger and have more electrons, so the instantaneous dipole 瞬时偶极 and induced dipole 诱导偶极 forces between them get stronger. Stronger forces are harder to break, so the boiling point rises and volatility falls.

    Three gas jars: chlorine as a pale green gas, bromine as a red-brown liquid, iodine as a grey-black solid
    Down Group 17 the halogens change from a pale green gas to a red-brown liquid to a grey-black solid as volatility falls

    The bond energy 键能 (bond strength) of the $\text{X}\text{–}\text{X}$ molecules generally falls from $\text{Cl}_2$ to $\text{I}_2$, because the shared electrons are further from the nuclei in the larger atoms.

    Explore · ⁨探索⁩

    Halogen physical trend lab · ⁨ハロゲン物理的傾向実験⁩

    Follow halogens down the group and link state, colour and volatility. · ⁨族を下っていき、状態、色、揮発性を関連付けよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    halogen/ˈhælədʒn/ ハロゲン
    group/ɡruːp/ グループ
    volatility/ˌvɒləˈtɪlɪti/ 揮発性
    instantaneous dipole/ˌɪnstənˈteɪnɪəs ˈdaɪpəʊl/ 瞬間双極子
    induced dipole/ɪnˈdjuːst ˈdaɪpəʊl/ 誘起双極子
    bond energy/bɒnd ˈenədʒi/ 結合エネルギー
    11.2

    Chemical properties of the halogens and hydrogen halides

    Syllabus · ⁨シラバス⁩
    English
    1. describe the relative reactivity of the elements as oxidising agents
    2. describe the reactions of the elements with hydrogen and explain their relative reactivity in these reactions
    3. describe the relative thermal stabilities of the hydrogen halides and explain these in terms of bond strengths
    日本語
    1. 元素の酸化剤としての相対的な反応性を記述する
    2. 元素の水素との反応を記述し、これらの反応における相対的な反応性を説明する
    3. ハロゲン化水素の相対的な熱安定性を記述し、結合能に基づいて説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Halogens as oxidising agents

    Each halogen reacts by gaining one electron to form a $1-$ ion, so it acts as an oxidising agent 氧化剂. This power decreases down the group, because the larger atoms attract an extra electron less strongly. A more reactive halogen can push out a less reactive one from its salt:

    $$\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2$$

    Reactions with hydrogen

    Each halogen reacts with hydrogen to form a hydrogen halide 卤化氢:

    $$\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$$

    The reaction gets less vigorous down the group: chlorine reacts explosively in light, bromine needs heat, and iodine reacts slowly and only partly.

    Thermal stability of the hydrogen halides

    The thermal stability 热稳定性 of the hydrogen halides decreases down the group. The H–X bond gets weaker as the halogen atom gets larger, so HI breaks apart on gentle heating while HCl is very stable.

    日本語

    Halogens as oxidising agents

    Each halogen reacts by gaining one electron to form a $1-$ ion, so it acts as an oxidising agent 氧化剂. This power decreases down the group, because the larger atoms attract an extra electron less strongly. A more reactive halogen can push out a less reactive one from its salt:

    Chlorine added to colourless potassium bromide turns the solution orange as bromine is displaced
    Chlorine displaces bromine: the solution turns orange
    $$\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2$$

    Reactions with hydrogen

    Each halogen reacts with hydrogen to form a hydrogen halide 卤化氢:

    $$\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$$

    The reaction gets less vigorous down the group: chlorine reacts explosively in light, bromine needs heat, and iodine reacts slowly and only partly.

    Thermal stability of the hydrogen halides

    The thermal stability 热稳定性 of the hydrogen halides decreases down the group. The H–X bond gets weaker as the halogen atom gets larger, so HI breaks apart on gentle heating while HCl is very stable.

    Explore · ⁨探索⁩

    Halogen and hydrogen halide lab · ⁨ハロゲンと水酸ハロゲン実験⁩

    Classify halogen chemistry by oxidising and reducing strength. · ⁨ハロゲン化学を酸化力と還元力の強さによって分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    oxidising agent/ˈɒksɪdaɪzɪŋ ˈeɪdʒənt/ 酸化剤
    hydrogen halide/ˈhaɪdrədʒn ˈhælaɪd/ 水素ハライド
    thermal stability/ˈθɜːml stəˈbɪlɪti/ 熱安定性
    11.3

    Reactions of the halide ions

    Syllabus · ⁨シラバス⁩
    English
    1. describe the relative reactivity of halide ions as reducing agents
    2. describe and explain the reactions of halide ions with: (a) aqueous silver ions followed by aqueous ammonia (the formation and formula of the $[\text{Ag}(\text{NH}_3)_2]^+$ complex is not required) (b) concentrated sulfuric acid, to include balanced chemical equations
    日本語
    1. ハロゲン化物イオンの還元剤としての相対的な反応性を記述する
    2. ハロゲン化物イオンの次のような反応を記述・説明する:(a) 水溶液銀イオン followed by 水溶液アンモニア($[\text{Ag}(\text{NH}_3)_2]^+$錯体形成および式は不要)(b) 濃硫酸、配位数化学方程式を含む

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Halide ions as reducing agents

    A halide ion 卤离子 (such as $\text{Cl}^-$) can give away an electron, acting as a reducing agent 还原剂. This power increases down the group, because a larger ion holds its outer electron less tightly.

    Reaction with aqueous silver ions

    Add aqueous silver nitrate, then aqueous ammonia 氨, to identify the halide from the colour of the silver halide precipitate 沉淀:

    Halide Precipitate with $\text{Ag}^+$ Solubility in ammonia
    $\text{Cl}^-$ white dissolves in dilute ammonia
    $\text{Br}^-$ cream dissolves only in concentrated ammonia
    $\text{I}^-$ yellow insoluble in ammonia

    Reaction with concentrated sulfuric acid

    All the halides first give the hydrogen halide. The lower halides are then oxidised by the acid, because they are stronger reducing agents:

    $$\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}$$
    • chloride gives only $\text{HCl}$ (no redox).
    • bromide also gives some brown $\text{Br}_2$ and $\text{SO}_2$.
    • iodide gives $\text{I}_2$ and the smelly gases $\text{H}_2\text{S}$ and $\text{SO}_2$, because $\text{I}^-$ is the strongest reducing agent.
    日本語

    Halide ions as reducing agents

    A halide ion 卤离子 (such as $\text{Cl}^-$) can give away an electron, acting as a reducing agent 还原剂. This power increases down the group, because a larger ion holds its outer electron less tightly.

    Two columns down the group: the halogens with a downward arrow for falling oxidising power, the halide ions with an upward arrow for rising reducing power
    Two opposite trends: the oxidising power of the halogens falls down the group, while the reducing power of the halide ions rises

    Reaction with aqueous silver ions

    Add aqueous silver nitrate, then aqueous ammonia 氨, to identify the halide from the colour of the silver halide precipitate 沉淀:

    Halide Precipitate with $\text{Ag}^+$ Solubility in ammonia
    $\text{Cl}^-$ white dissolves in dilute ammonia
    $\text{Br}^-$ cream dissolves only in concentrated ammonia
    $\text{I}^-$ yellow insoluble in ammonia
    Three test tubes with silver halide precipitates: white silver chloride, cream silver bromide and yellow silver iodide
    Silver halide precipitates: AgCl is white, AgBr cream and AgI yellow — and their solubility in ammonia confirms which halide is present
    Three real test tubes in a wooden rack, labelled AgCl, AgBr and AgI, holding a white, a pale cream and a pale yellow precipitate from left to right
    The silver halide test: AgCl is white, AgBr cream and AgI yellow

    Reaction with concentrated sulfuric acid

    All the halides first give the hydrogen halide. The lower halides are then oxidised by the acid, because they are stronger reducing agents:

    $$\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}$$
    • chloride gives only $\text{HCl}$ (no redox).
    • bromide also gives some brown $\text{Br}_2$ and $\text{SO}_2$.
    • iodide gives $\text{I}_2$ and the smelly gases $\text{H}_2\text{S}$ and $\text{SO}_2$, because $\text{I}^-$ is the strongest reducing agent.
    Three rows: chloride gives only HCl, bromide also gives bromine and sulfur dioxide, iodide gives iodine, hydrogen sulfide and sulfur dioxide, with an arrow showing reducing power increases down the group
    With concentrated sulfuric acid, chloride gives only HCl, but bromide and iodide (stronger reducing agents) are also oxidised to the halogen
    Explore · ⁨探索⁩

    Halide ion test lab · ⁨ハロゲン化物イオン検出実験⁩

    Match halide ion evidence to the ion present. · ⁨ハロゲン化物イオンの証拠と存在するイオンを対応させよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    halide ion/ˈhælaɪd ˈaɪɒn/ ハロゲン化物イオン
    reducing agent/rɪˈdjuːsɪŋ ˈeɪdʒənt/ 還元剤
    ammonia/æˈməʊnɪə/ アンモニア
    precipitate/prɪˈsɪpɪteɪt/ 沈殿物
    11.4

    Reactions of chlorine

    Syllabus · ⁨シラバス⁩
    English
    1. describe and interpret, in terms of changes in oxidation number, the reaction of chlorine with cold and with hot aqueous sodium hydroxide and recognise these as disproportionation reactions
    2. explain, including by use of an equation, the use of chlorine in water purification to include the production of the active species $\text{HOCl}$ and $\text{ClO}^-$ which kill bacteria
    日本語
    1. 酸化数の変化に基づいて、クロリンが冷水および熱水水酸化ナトリウム溶液と反応することを記述・解釈し、これらが歧化反応であることを認識する
    2. クロリンの水道水浄化での使用を説明し、化学式を用い、細菌を殺害する活性種である$\text{HOCl}$および$\text{ClO}^-$の生成を含む

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    With sodium hydroxide

    With cold, dilute sodium hydroxide, chlorine reacts to form chloride and chlorate(I):

    $$\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}$$

    With hot, concentrated sodium hydroxide, it forms chloride and chlorate(V):

    $$3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}$$

    In both, the oxidation number 氧化数 of chlorine goes both up and down (from $0$), so both are disproportionation 歧化 reactions.

    Chlorine in water purification

    A little chlorine is added to water for water purification 水净化. It reacts with water:

    $$\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HOCl} + \text{HCl}$$

    The active species $\text{HOCl}$ and $\text{ClO}^-$ kill bacteria 细菌, making the water safe to drink.

    Worked example. Solid $\text{NaCl}$, $\text{NaBr}$ and $\text{NaI}$ are each warmed with concentrated $\text{H}_2\text{SO}_4$. Predict the products. Reducing power increases down the group, so how far each halide reduces the sulfuric acid differs. $\text{Cl}^{-}$ is too weak to reduce it at all, so you get only steamy $\text{HCl}$ - an acid-base reaction. $\text{Br}^{-}$ reduces it a little: $\text{HBr}$ plus brown $\text{Br}_2$ and $\text{SO}_2$. $\text{I}^{-}$ is the strongest reducing agent: $\text{HI}$ plus $\text{I}_2$, and it drives the sulfur all the way down to $\text{H}_2\text{S}$, with its bad-egg smell. Every halide gives the hydrogen halide first; the extra products appear only where the halide is a strong enough reducing agent to attack the sulfur.

    日本語
    Containers of liquid pool chlorine
    Chlorine is added to pool water to kill microbes.

    With sodium hydroxide

    With cold, dilute sodium hydroxide, chlorine reacts to form chloride and chlorate(I):

    $$\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}$$

    With hot, concentrated sodium hydroxide, it forms chloride and chlorate(V):

    $$3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}$$

    In both, the oxidation number 氧化数 of chlorine goes both up and down (from $0$), so both are disproportionation 歧化 reactions.

    Chlorine in water purification

    A little chlorine is added to water for water purification 水净化. It reacts with water:

    $$\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HOCl} + \text{HCl}$$

    The active species $\text{HOCl}$ and $\text{ClO}^-$ kill bacteria 细菌, making the water safe to drink.

    Worked example. Solid $\text{NaCl}$, $\text{NaBr}$ and $\text{NaI}$ are each warmed with concentrated $\text{H}_2\text{SO}_4$. Predict the products. Reducing power increases down the group, so how far each halide reduces the sulfuric acid differs. $\text{Cl}^{-}$ is too weak to reduce it at all, so you get only steamy $\text{HCl}$ - an acid-base reaction. $\text{Br}^{-}$ reduces it a little: $\text{HBr}$ plus brown $\text{Br}_2$ and $\text{SO}_2$. $\text{I}^{-}$ is the strongest reducing agent: $\text{HI}$ plus $\text{I}_2$, and it drives the sulfur all the way down to $\text{H}_2\text{S}$, with its bad-egg smell. Every halide gives the hydrogen halide first; the extra products appear only where the halide is a strong enough reducing agent to attack the sulfur.

    Explore · ⁨探索⁩

    Chlorine reaction route · ⁨塩素反応ルート⁩

    Follow chlorine from water treatment to redox reactions. · ⁨水道処理から酸化還元反応まで塩素を追跡せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    oxidation number/ˌɒksɪˈdeɪʃn ˈnʌmbə/ 酸化数
    disproportionation/ˌdɪsprəˈpɔːʃəneɪʃn/ 歧化反応
    water purification/ˈwɔːtə ˌpjʊərɪfɪˈkeɪʃn/ 水道水浄化
    bacteria/bækˈtɪərɪə/ 細菌
    11.4

    Exam tips

    • Halogens get less reactive down the group (harder to gain an electron); oxidising power decreases.
    • Displacement: a more reactive halogen displaces a less reactive halide — state the colour change.
    • Test halide ions with $\text{AgNO}_3$: white ($\text{Cl}^-$), cream ($\text{Br}^-$), yellow ($\text{I}^-$), then confirm with dilute/concentrated ammonia.
    • Chlorine with water and with cold $\text{NaOH}$ are disproportionation — show the oxidation-number changes.
  • 12

    Nitrogen and sulfur · ⁨窒素と硫黄⁩

    Watch lesson · ⁨レッスンを視聴⁩
    12.1

    Why nitrogen is unreactive · ⁨ニトロンが不活性である理由⁩

    Syllabus · ⁨シラバス⁩
    English
    1. explain the lack of reactivity of nitrogen, with reference to triple bond strength and lack of polarity
    2. describe and explain: (a) the basicity of ammonia, using the Brønsted–Lowry theory (b) the structure of the ammonium ion and its formation by an acid–base reaction (c) the displacement of ammonia from ammonium salts by an acid–base reaction
    3. state and explain the natural and man-made occurrences of oxides of nitrogen and their catalytic removal from the exhaust gases of internal combustion engines
    4. understand that atmospheric oxides of nitrogen ($\text{NO}$ and $\text{NO}_2$) can react with unburned hydrocarbons to form peroxyacetyl nitrate, PAN, which is a component of photochemical smog
    5. describe the role of $\text{NO}$ and $\text{NO}_2$ in the formation of acid rain both directly and in their catalytic role in the oxidation of atmospheric sulfur dioxide
    日本語
    1. 三重結合の強さと非極性の欠如に言及して、ニトロンの反応性の欠如を説明する
    2. 記述・説明する:(a) Brønsted–Lowry理論を用いたアンモニアの塩基性(b) アンモニウムイオンの構造および酸塩基反応による形成(c) 酸塩基反応によるアンモニウム塩からのアンモニア置換
    3. 窒素酸化物の自然および人為的な発生を述べ、内燃機関排ガス中での触媒除去について説明する
    4. 大気中の窒素酸化物($\text{NO}$および$\text{NO}_2$)が燃焼不完全炭化水素と反応してペルシアシル硝酸、PANを形成すること、これが光化学スモッグの構成要素であることを理解する
    5. $\text{NO}$および$\text{NO}_2$が大気中二酸化硫黄の酸化における触媒作用を含め、酸雨形成において直接的および触媒的役割を果たすことを記述する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Nitrogen gas, $\text{N}_2$, makes up most of the air but reacts with very little. There are two reasons:

    • the two nitrogen atoms are joined by a triple bond 三键, which has a very high bond energy 键能. A lot of energy is needed to break it.
    • the molecule has no polarity 极性 — it is perfectly symmetrical, so nothing pulls other molecules towards it.
    日本語

    窒素ガス$\text{N}_2$は空気の大部分を占めていますが、ほとんど反応しません。その理由は2つあります:

    • 2つの窒素原子は三重結合で結びついており、非常に高い結合エネルギーを持っています。これを切断するには多くのエネルギーが必要です。
    • この分子は分極を持っていません — 完全に対称的であるため、他の分子を引き寄せる力は働かない。
    三重結合で結ばれた2つの窒素原子、それぞれに非共有電子対を持ち、強い結合および対称的で非極性分子としてラベル付けされている
    窒素が不活性である理由は2つ:極めて強い三重結合、および対称的で非極性の分子
    Explore · ⁨探索⁩

    The shape of ammonia · ⁨アンモニアの形状⁩

    Ammonia has three bonding pairs and one lone pair — the lone pair pushes the bonds down into a pyramidal shape (about 107°). · ⁨アンモニアには結合電子対が3つ、非共有電子対が1つある——非共有電子対が結合を下へ押し下げ、三角錐形(約107°)を形成する。⁩

    12.1

    Ammonia and the ammonium ion · ⁨アンモニアとアンモニウムイオン⁩

    English

    Basicity of ammonia

    The basicity 碱性 of ammonia (its ability to act as a base) comes from the lone pair 孤对电子 of electrons on the nitrogen atom. Using the Brønsted–Lowry theory, ammonia is a base because this lone pair can accept a proton 质子 ($\text{H}^+$):

    $$\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+$$

    The ammonium ion

    When the lone pair forms a bond to $\text{H}^+$, it makes the ammonium ion 铵离子, $\text{NH}_4^+$. Because both shared electrons came from the nitrogen, this new bond is a coordinate bond 配位键. The ion has four identical N–H bonds and a tetrahedral shape.

    Displacement of ammonia from its salts

    If you warm an ammonium salt with a base (such as sodium hydroxide), you push out ammonia gas. This is an acid–base displacement 置换:

    $$\text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NaCl} + \text{NH}_3 + \text{H}_2\text{O}$$

    The sharp smell of ammonia, and damp red litmus turning blue, is a test for an ammonium salt.

    日本語
    大規模な工業用化学プラント
    アンモニアは工業施設で大規模に窒素肥料へ変換されます。

    アンモニアの塩基性

    アンモニアの塩基性(塩基としての機能)は、窒素原子上の非共有電子対に由来します。ブレンステッド-ローリー説によると、この非共有電子対がプロトン($\text{H}^+$)を受け取れるため、アンモニアは塩基です:

    $$\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+$$
    窒素に非共有電子対を持つアンモニアが水素イオンを受け取り、四面体構造のアンモニウムイオンを形成する
    アンモニアは塩基として振る舞い、窒素上の非共有電子対がプロトンを受け取ることでアンモニウムイオンを生成する

    アンモニウムイオン

    非共有電子対が$\text{H}^+$に結合すると、アンモニウムイオン$\text{NH}_4^+$が形成されます。この新結合の両方の共有電子対は窒素由来であるため、これは配位結合です。このイオンには4つの等しいN–H結合があり、四面体構造をとります。

    塩からのアンモニアの遊離

    アンモニウム塩を塩基(例:水酸化ナトリウム)と温めると、アンモニアガスが遊離します。これは酸塩基置換反応です:

    $$\text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NaCl} + \text{NH}_3 + \text{H}_2\text{O}$$

    アンモニア特有の刺激臭と、湿った赤リトマス紙の青色転化は、アンモニウム塩のテスト法です。

    12.1

    Oxides of nitrogen and air pollution · ⁨窒素酸化物と大気汚染⁩

    English

    Where they come from

    Oxides of nitrogen ($\text{NO}$ and $\text{NO}_2$, together called $\text{NO}_x$) come from two sources:

    • natural: lightning gives enough energy for nitrogen and oxygen in the air to combine.
    • man-made: the high temperature inside an internal combustion engine 内燃机 makes nitrogen and oxygen react:
    $$\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}$$

    Removing them from car exhaust

    A catalytic converter 催化转化器 cleans the exhaust gases 尾气. It lets the harmful gases react together to form harmless ones:

    $$2\text{NO} + 2\text{CO} \rightarrow \text{N}_2 + 2\text{CO}_2$$

    Photochemical smog

    In sunlight, $\text{NO}$ and $\text{NO}_2$ react with unburned hydrocarbons to form peroxyacetyl nitrate (PAN). PAN is a harmful part of photochemical smog 光化学烟雾, the brown haze seen over busy cities.

    Acid rain

    The oxides of nitrogen also help make acid rain 酸雨 in two ways:

    • directly: $\text{NO}_2$ dissolves in rain to form nitric acid.
    • as a catalyst: $\text{NO}_2$ speeds up the oxidation of atmospheric sulfur dioxide 二氧化硫 ($\text{SO}_2$) into $\text{SO}_3$, which then forms sulfuric acid in the rain.

    Worked example. A white solid is warmed with aqueous $\text{NaOH}$, and a gas is released that turns damp red litmus blue. Identify the gas and the ion in the solid, and explain the reaction. The only common gas that turns damp red litmus blue is ammonia, $\text{NH}_3$, so the solid contains the ammonium ion, $\text{NH}_4^{+}$. The hydroxide ion is the stronger base, so it takes the proton back from the ammonium ion:

    $$\text{NH}_4^{+} + \text{OH}^{-} \rightarrow \text{NH}_3 + \text{H}_2\text{O}$$

    This is the standard test for $\text{NH}_4^{+}$. Always say the litmus is damp: the ammonia must dissolve in the water before it can show its basicity, so dry litmus would give no colour change at all.

    日本語

    発生源

    窒素酸化物($\text{NO}$と$\text{NO}_2$、统称$\text{NO}_x$)は2つの起源から生じます:

    • 自然由来: 雷により、空気中の窒素と酸素が結合する十分なエネルギーが与えられます。
    • 人為由来: 内燃機関内部の高温により、窒素と酸素が反応します:
    $$\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}$$

    排ガスからの除去

    触媒コンバーターは排気ガスを浄化します。有害ガスを無害なものに変えるために互いに反応させます:

    $$2\text{NO} + 2\text{CO} \rightarrow \text{N}_2 + 2\text{CO}_2$$
    有害ガスNOとCOが触媒ブロックを通過し、無害な窒素と二酸化炭素として排出される様子
    触媒コンバーター内では、有害ガスNOとCOが触媒上で反応し、無害なN$_2$とCO$_2$を生成する

    光化学スモッグ

    日光下で、$\text{NO}$と$\text{NO}_2$が未燃焼の炭化水素と反応し、過アセチル硝酸(PAN)を生成します。PANは都会の上空に見られる茶色の霞である光化学スモッグの有害な成分です。

    青空の下、建物の上に茶灰色の霾層が宿る大都市の航空写真
    大都市の上空の光化学スモッグ。茶色の層は交通排気ガスの位置に近い低層に閉じ込められており、日光によって窒素酸化物と未燃焼炭化水素が視覚的に確認できる霾へと変換される

    酸性雨

    窒素酸化物は次の2つの方法で酸性雨の形成に関与します:

    • 直接:$\text{NO}_2$が雨水に溶解して硝酸を生成する。
    • 触媒として:$\text{NO}_2$が大気中の二酸化硫黄($\text{SO}_2$)の酸化を促進し、$\text{SO}_3$を生成することで、これが雨水の中で硫酸となる。
    フロー図:NO2が硝酸を生成し、SO2が硫酸を生成する。NO2は硫黄経路も触媒化し、両者が酸性雨につながる
    窒素と硫黄の酸化物が酸性雨を形成する:NO$_2$は直接硝酸を生成し、SO$_2$の酸化を触媒化して硫酸を生成する

    ** worked example.** 白色固体を $\text{NaOH}$ とともに温めると、湿った赤リトマス紙を青色に変える気体が発生した。この気体と固体中のイオンを特定し、反応式を説明せよ。湿った赤リトマス紙を青色に変える一般的な気体はアンモニアのみであり、 $\text{NH}_3$ therefore 固体にはアンモニウムイオン $\text{NH}_4^{+}$ が含まれている。水酸化物イオンの方が塩基性が強いため、プロトンをアンモニウムイオンから取り戻す:

    $$\text{NH}_4^{+} + \text{OH}^{-} \rightarrow \text{NH}_3 + \text{H}_2\text{O}$$

    これは $\text{NH}_4^{+}$ の標準的な検出法である。必ずリトマス紙が湿っていることを述べるべきである。アンモニアは水に溶解して初めて塩基性を示すため、乾燥したリトマス紙では色変化は全く生じない。

    Explore · ⁨探索⁩

    NOx pollution route · ⁨NOx汚染ルート⁩

    Trace nitrogen oxides from hot engines to environmental harm. · ⁨高温エンジンからの微量窒素酸化物から環境への害まで。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    triple bond/ˈtrɪpl bɒnd/ 三重結合
    bond energy/bɒnd ˈenədʒi/ 結合エネルギー
    polarity/pəʊˈlærɪti/ 極性
    basicity/beɪˈsɪsɪti/ 塩基性
    lone pair/ləʊn peə/ 非共有電子対
    proton/ˈprəʊtɒn/ 陽子
    ammonium ion/æˈməʊnɪəm ˈaɪɒn/ アンモニウムイオン
    coordinate bond/kəʊˈɔːdɪnət bɒnd/ 配位結合
    displacement/dɪˈspleɪsmənt/ 変位
    internal combustion engine/ɪnˈtɜːnl kəmˈbʌstʃn ˈendʒɪn/ 内燃機関
    catalytic converter/ˌkætəˈlɪtɪk kənˈvɜːtə/ 触媒変換装置
    exhaust gases/eɡˈzɔːst ˈɡæsɪz/ 排気ガス
    photochemical smog/ˌfəʊtəʊˈkemɪkl smɒɡ/ 光化学スモッグ
    acid rain/ˈæsɪd reɪn/ 酸性雨
    sulfur dioxide/ˈsʌlfɜː daɪˈɒksaɪd/ 二酸化硫黄
    12.1

    Exam tips · ⁨試験対策⁩

    English
    • $\text{N}_2$ is unreactive because of its strong triple bond (very high bond energy) — state this exactly.
    • Ammonia is a base and a ligand because of its lone pair; the ammonium ion forms by a dative bond.
    • Explain how oxides of nitrogen form (high temperature in engines) and their link to acid rain and photochemical smog.
    • Give balanced equations and correct observations for reactions of ammonia and the ammonium ion.
    日本語
    • $\text{N}_2$ は強い三重結合(非常に高い結合エネルギー)のため不活性である — この点を正確に述べよ。
    • アンモニアは非共有電子対を持つため塩基および配位子であり、アンモニウムイオンは配位結合によって形成される。
    • 窒素酸化物の生成過程(エンジン内の高温条件下)と、酸性雨および光化学スモッグとの関連を説明せよ。
    • アンモニアおよびアンモニウムイオンの反応について、平衡方程式および正しい観察結果を記せ。
  • 13

    An introduction to AS Level organic chemistry · ⁨AS Level有機化学への入門⁩

    Watch lesson · ⁨レッスンを視聴⁩
    13.1

    Formulas, functional groups and naming

    Syllabus · ⁨シラバス⁩
    English
    1. define the term hydrocarbon as a compound made up of C and H atoms only
    2. understand that alkanes are simple hydrocarbons with no functional group
    3. understand that the compounds in the table on pages 29 and 30 contain a functional group which dictates their physical and chemical properties
    4. interpret and use the general, structural, displayed and skeletal formulas of the classes of compound stated in the table on pages 29 and 30
    5. understand and use systematic nomenclature of simple aliphatic organic molecules with functional groups detailed in the table on pages 29 and 30, up to six carbon atoms (six plus six for esters, straight chains only for esters and nitriles)
    6. deduce the molecular and/or empirical formula of a compound, given its structural, displayed or skeletal formula
    日本語
    1. 用語炭化水素をCとH原子のみで構成される化合物として定義する
    2. アルカンが官能基を持たない単純な炭化水素であることを理解する
    3. 29ページおよび30ページの表にある化合物が物理的および化学的性質を決定する官能基を含むことを理解する
    4. 表29ページおよび30ページに示された化合物のクラスに関する、一般式、構造式、表示式、骨格式の解釈と使用
    5. 表29ページおよび30ページに記載されている官能基を有する炭化水素族有機分子の系名法を理解し、最大6個の炭素原子(エステルは6+6、直鎖のみ、ニトリルも直鎖のみ)まで適用する
    6. 構造式、表示式、または骨格式が与えられたとき、その化合物の分子式および/または実験式を導き出す

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A hydrocarbon 碳氢化合物 is a compound of only carbon and hydrogen. Alkanes 烷烃 are the simplest hydrocarbons and have no functional group.

    A functional group 官能团 is the reactive part of a molecule. It decides the physical and chemical properties of the compound, so molecules with the same functional group behave alike (for example the $\text{–OH}$ group in alcohols).

    Types of formula

    Formula What it shows
    general formula 通式 the pattern for a whole family, e.g. alkanes are $\text{C}_n\text{H}_{2n+2}$
    molecular formula 分子式 the actual number of each atom, e.g. $\text{C}_4\text{H}_{10}$
    structural formula 结构式 the groups in order, e.g. $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$
    displayed formula 展开式 every atom and every bond drawn out
    skeletal formula 骨架式 lines for bonds; carbons at corners, hydrogens on carbon not shown

    You can read off the empirical formula 实验式 (simplest ratio) from any of these.

    Naming

    Use systematic nomenclature 命名法: a stem for the number of carbons (meth-, eth-, prop-, but-, pent-, hex- for 1 to 6), an ending for the functional group, and numbers to show where groups are.

    日本語
    A sample of dark crude oil
    Crude oil is a complex mixture of hydrocarbons — the feedstock for organic chemistry.

    A hydrocarbon 碳氢化合物 is a compound of only carbon and hydrogen. Alkanes 烷烃 are the simplest hydrocarbons and have no functional group.

    A functional group 官能团 is the reactive part of a molecule. It decides the physical and chemical properties of the compound, so molecules with the same functional group behave alike (for example the $\text{–OH}$ group in alcohols).

    Types of formula

    Formula What it shows
    general formula 通式 the pattern for a whole family, e.g. alkanes are $\text{C}_n\text{H}_{2n+2}$
    molecular formula 分子式 the actual number of each atom, e.g. $\text{C}_4\text{H}_{10}$
    structural formula 结构式 the groups in order, e.g. $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$
    displayed formula 展开式 every atom and every bond drawn out
    skeletal formula 骨架式 lines for bonds; carbons at corners, hydrogens on carbon not shown
    Butane drawn four ways: the molecular formula, the structural formula, the displayed formula with every atom and bond, and the skeletal zigzag
    The same molecule (butane) shown four ways — molecular, structural, displayed and skeletal — each hiding more detail than the last

    You can read off the empirical formula 实验式 (simplest ratio) from any of these.

    Naming

    Use systematic nomenclature 命名法: a stem for the number of carbons (meth-, eth-, prop-, but-, pent-, hex- for 1 to 6), an ending for the functional group, and numbers to show where groups are.

    Explore · ⁨探索⁩

    Functional group lab · ⁨官能基実験室⁩

    Sort organic molecules by the group that controls their reactions. · ⁨反応を制御する官能基に基づいて有機分子を分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    hydrocarbon/ˈhaɪdrəkɑːbən/ 炭化水素
    alkane/ˈælkeɪn/ アルカン
    functional group/ˈfʌŋkʃənl ɡruːp/ 官能基
    general formula/ˈdʒenərəl ˈfɔːmjʊlə/ 一般式
    molecular formula/məˈlekjʊlə ˈfɔːmjʊlə/ 分子式
    structural formula/ˈstrʌktʃərəl ˈfɔːmjʊlə/ 構造式
    displayed formula/dɪˈspleɪd ˈfɔːmjʊlə/ 展開式
    skeletal formula/ˈskelɪtl ˈfɔːmjʊlə/ 骨格式
    empirical formula/emˈpɪrɪkl ˈfɔːmjʊlə/ 経験式
    nomenclature/nəˈmeŋklətʃə/ 命名法
    13.2

    Characteristic organic reactions

    Syllabus · ⁨シラバス⁩
    English
    1. interpret and use the following terminology associated with types of organic compounds and reactions: (a) homologous series (b) saturated and unsaturated (c) homolytic and heterolytic fission (d) free radical, initiation, propagation, termination (e) nucleophile, electrophile, nucleophilic, electrophilic (f) addition, substitution, elimination, hydrolysis, condensation (g) oxidation and reduction (in equations for organic redox reactions, the symbol [O] can be used to represent one atom of oxygen from an oxidising agent and the symbol [H] to represent one atom of hydrogen from a reducing agent)
    2. understand and use the following terminology associated with types of organic mechanisms: (a) free-radical substitution (b) electrophilic addition (c) nucleophilic substitution (d) nucleophilic addition (in organic reaction mechanisms, the use of curly arrows to represent movement of electron pairs is expected; the arrow should begin at a bond or a lone pair of electrons)
    日本語
    1. 有機化合物および反応の種類に関連する以下の用語を解釈して使用する:(a) ホモロゲス系列 (b) 飽和・不飽和 (c) 均裂・不均裂 (d) 自由ラジカル、開始、伝播、終端 (e) 求核剤、求電子剤、求核的、求電子的 (f) 付加、置換、脱離、加水分解、縮合 (g) 酸化と還元(有機酸化還元反応の式では、[O] を酸化剤由来の酸素原子1個を表す記号として、[H] を還元剤由来の水素原子1個を表す記号として用いることができる)
    2. 有機反応機構の種類に関連する以下の用語を理解して使用する:(a) 自由ラジカル置換 (b) 求電子付加 (c) 求核置換 (d) 求核付加(有機反応機構では、電子対の移動を表すためにカールド矢印を用いることが期待される;矢印は結合または非共有電子対から始まるべきである)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Some key terms

    • a homologous series 同系列 is a family of compounds with the same functional group and general formula, where each member differs by $\text{CH}_2$.
    • a saturated 饱和 compound has only single C–C bonds; an unsaturated 不饱和 compound has a C=C double bond (or a triple bond).

    Breaking bonds

    A covalent bond can break in two ways:

    • homolytic fission 均裂: the bond splits evenly, one electron to each atom. This makes two free radicals 自由基 (species with an unpaired electron).
    • heterolytic fission 异裂: the bond splits unevenly, both electrons going to one atom. This makes two ions.

    Attacking species

    • a nucleophile 亲核试剂 is a species with a lone pair that is attracted to a positive (electron-poor) centre.
    • an electrophile 亲电试剂 is a species attracted to a negative (electron-rich) centre, such as a C=C double bond.

    Types of reaction

    Reaction What happens
    addition 加成 two molecules join to make one
    substitution 取代 one atom or group is swapped for another
    elimination 消去 a small molecule is removed, making a double bond
    hydrolysis 水解 a molecule is split apart by water
    condensation 缩合 two molecules join and a small molecule (such as water) is lost

    For organic redox, the symbol $[\text{O}]$ stands for one oxygen atom from an oxidising agent, and $[\text{H}]$ for one hydrogen atom from a reducing agent.

    Types of mechanism

    A free-radical reaction happens in three steps: initiation 引发 (radicals are made), propagation 增长 (radicals react and make new radicals), and termination 终止 (two radicals join and stop the chain).

    The main mechanisms you meet are free-radical substitution 自由基取代 (alkanes), electrophilic addition 亲电加成 (alkenes), nucleophilic substitution 亲核取代 (halogenoalkanes) and nucleophilic addition 亲核加成 (carbonyls). In mechanisms, a curly arrow 弯箭头 shows a pair of electrons moving; it starts at a bond or a lone pair 孤对电子.

    日本語

    Some key terms

    • a homologous series 同系列 is a family of compounds with the same functional group and general formula, where each member differs by $\text{CH}_2$.
    • a saturated 饱和 compound has only single C–C bonds; an unsaturated 不饱和 compound has a C=C double bond (or a triple bond).

    Breaking bonds

    A covalent bond can break in two ways:

    • homolytic fission 均裂: the bond splits evenly, one electron to each atom. This makes two free radicals 自由基 (species with an unpaired electron).
    • heterolytic fission 异裂: the bond splits unevenly, both electrons going to one atom. This makes two ions.
    Homolytic fission splitting a Cl–Cl bond into two chlorine radicals; heterolytic fission sending both bonding electrons to bromine to give two ions
    Homolytic fission gives one electron to each atom (two radicals); heterolytic fission gives both electrons to one atom (two ions)

    Attacking species

    • a nucleophile 亲核试剂 is a species with a lone pair that is attracted to a positive (electron-poor) centre.
    • an electrophile 亲电试剂 is a species attracted to a negative (electron-rich) centre, such as a C=C double bond.
    A nucleophile using its lone pair to attack a slightly positive carbon; an electrophile being drawn to a C=C double bond
    A nucleophile uses its lone pair to attack an electron-poor centre; an electrophile is drawn to an electron-rich one such as a C=C bond

    Types of reaction

    Reaction What happens
    addition 加成 two molecules join to make one
    substitution 取代 one atom or group is swapped for another
    elimination 消去 a small molecule is removed, making a double bond
    hydrolysis 水解 a molecule is split apart by water
    condensation 缩合 two molecules join and a small molecule (such as water) is lost

    For organic redox, the symbol $[\text{O}]$ stands for one oxygen atom from an oxidising agent, and $[\text{H}]$ for one hydrogen atom from a reducing agent.

    Types of mechanism

    A free-radical reaction happens in three steps: initiation 引发 (radicals are made), propagation 增长 (radicals react and make new radicals), and termination 终止 (two radicals join and stop the chain).

    The main mechanisms you meet are free-radical substitution 自由基取代 (alkanes), electrophilic addition 亲电加成 (alkenes), nucleophilic substitution 亲核取代 (halogenoalkanes) and nucleophilic addition 亲核加成 (carbonyls). In mechanisms, a curly arrow 弯箭头 shows a pair of electrons moving; it starts at a bond or a lone pair 孤对电子.

    Explore · ⁨探索⁩

    Bond fission and attack route · ⁨結合開裂と攻撃経路⁩

    Watch a polar bond lead to electrophiles, nucleophiles and radicals. · ⁨極性結合が求電子剤、求核剤、ラジカルを生み出す様子を見よ。⁩

    Explore · ⁨探索⁩

    Organic reaction type lab · ⁨有機反応タイプ実験室⁩

    Classify reaction examples by the pattern of bonds changing. · ⁨結合の変化パターンに基づいて反応例を分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    homologous series/həˈmɒləɡəs ˈsɪəriːz/ writings系列
    saturated/ˈsætʃəreɪtɪd/ 飽和溶液
    unsaturated/ʌnˈsætʃəreɪtɪd/ 不飽和溶液
    homolytic fission/ˌhɒməˈlɪtɪk ˈfɪʃn/ 均等開裂
    free radical/friː ˈrædɪkl/ 自由ラジカル
    heterolytic fission/ˌhetrəˈlɪtɪk ˈfɪʃn/ 不均等開裂
    nucleophile/ˈnjuːklɪɒfaɪl/ 求核剤
    electrophile/ɪˌlektrəʊˈfaɪl/ 求電子体
    addition/əˈdɪʃn/ 付加反応
    substitution/ˌsʌbstɪˈtjuːʃn/ 置換反応
    elimination/ɪˌlɪmɪˈneɪʃn/ 脱離反応
    hydrolysis/haɪˈdrɒləsɪs/ 加水分解反応
    condensation/kɒndenˈseɪʃn/ 縮合
    initiation/ɪˌnɪʃɪˈeɪʃn/ 開始
    propagation/ˌprɒpəˈɡeɪʃn/ 増幅
    termination/ˌtɜːmɪˈneɪʃn/ 終端
    free-radical substitution/friː ˈrædɪkl ˌsʌbstɪˈtjuːʃn/ 自由ラジカル置換
    electrophilic addition/ɪˌlektrəʊˈfɪlɪk əˈdɪʃn/ 求電付加
    nucleophilic substitution/ˌnjuːklɪəˈfɪlɪk ˌsʌbstɪˈtjuːʃn/ 求核置換
    nucleophilic addition/ˌnjuːklɪəˈfɪlɪk əˈdɪʃn/ 求核付加
    curly arrow/ˈkɜːli ˈærəʊ/ カーリー矢印
    lone pair/ləʊn peə/ 非共有電子対
    13.3

    Shapes of organic molecules

    Syllabus · ⁨シラバス⁩
    1. describe organic molecules as either straight-chained, branched or cyclic
    2. describe and explain the shape of, and bond angles in, molecules containing $\text{sp}$, $\text{sp}^2$ and $\text{sp}^3$ hybridised atoms
    3. describe the arrangement of $\sigma$ and $\pi$ bonds in molecules containing $\text{sp}$, $\text{sp}^2$ and $\text{sp}^3$ hybridised atoms
    4. understand and use the term planar when describing the arrangement of atoms in organic molecules, for example ethene

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Organic molecules can be straight-chained, branched or cyclic 环状 (in a ring).

    The shape around a carbon depends on its hybridisation 杂化:

    Hybridisation Bonds Shape Angle
    $\text{sp}^3$ 4 single tetrahedral $109.5°$
    $\text{sp}^2$ 1 double + 2 single planar 平面 (flat) $120°$
    $\text{sp}$ 1 triple (or 2 doubles) linear $180°$

    Every single bond is a sigma bond σ键, made by direct overlap. A double bond is one sigma bond plus one pi bond π键, made by sideways overlap of p orbitals. Ethene is planar because of its $\text{sp}^2$ carbons.

    日本語
    A ball-and-stick molecular model
    Organic molecules have definite three-dimensional shapes.

    Organic molecules can be straight-chained, branched or cyclic 环状 (in a ring).

    The shape around a carbon depends on its hybridisation 杂化:

    Hybridisation Bonds Shape Angle
    $\text{sp}^3$ 4 single tetrahedral $109.5°$
    $\text{sp}^2$ 1 double + 2 single planar 平面 (flat) $120°$
    $\text{sp}$ 1 triple (or 2 doubles) linear $180°$

    Every single bond is a sigma bond σ键, made by direct overlap. A double bond is one sigma bond plus one pi bond π键, made by sideways overlap of p orbitals. Ethene is planar because of its $\text{sp}^2$ carbons.

    Three carbons: an sp3 carbon in tetrahedral methane, an sp2 carbon in planar ethene, and an sp carbon in linear ethyne, each with its bond angle
    The shape at a carbon follows from its hybridisation: sp$^3$ is tetrahedral ($109.5°$), sp$^2$ is planar ($120°$), sp is linear ($180°$)
    Explore · ⁨探索⁩

    Shapes of organic molecules · ⁨有機分子の形状⁩

    VSEPR around each carbon · ⁨各炭素におけるVSEPR⁩

    Around a single-bonded carbon the four pairs are tetrahedral (109.5°). · ⁨単結合炭素の周囲で4組の電子は四面体型(109.5°)をとる。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    cyclic/ˈsaɪklɪk/ 循環型
    hybridisation/ˌhaɪbrɪdaɪˈzeɪʃn/ 雑種交配
    planar/ˈpleɪnə/ 平面構造
    sigma bond/ˈsɪɡmə bɒnd/ シグマ結合
    pi bond/paɪ bɒnd/ π結合
    13.4

    Isomerism

    Syllabus · ⁨シラバス⁩
    English
    1. describe structural isomerism and its division into chain, positional and functional group isomerism
    2. describe stereoisomerism and its division into geometrical (cis/trans) and optical isomerism (use of E/Z nomenclature is acceptable but is not required)
    3. describe geometrical (cis/trans) isomerism in alkenes, and explain its origin in terms of restricted rotation due to the presence of $\pi$ bonds
    4. explain what is meant by a chiral centre and that such a centre gives rise to two optical isomers (enantiomers) (Candidates should appreciate that compounds can contain more than one chiral centre, but knowledge of meso compounds, or nomenclature such as diastereoisomers is not required.)
    5. identify chiral centres and geometrical (cis/trans) isomerism in a molecule of given structural formula including cyclic compounds
    6. deduce the possible isomers for an organic molecule of known molecular formula
    日本語
    1. 構造異性体を説明し、鎖状異性体、位置異性体、官能基異性体への分類を述べる
    2. 立体異性体を説明し、幾何異性体(シス/トランス)および光学異性体への分類を述べる(E/Z名法の使用は許容されるが必須ではない)
    3. アルケンの幾何異性体(シス/トランス)を説明し、その起源が $\pi$ 結合の存在による回転制限にあることを説明する
    4. キラル中心の定義を述べ、そのような中心が2つの光学異性体(エナンチオマー)を生じる理由を説明する(受験者は、化合物が複数のキラル中心を持つ可能性があることを理解すべきだが、メソ化合物やジアステレオマーなどの名法に関する知識は不要である)
    5. 与えられた構造式を持つ分子(環状化合物を含む)において、キラル中心および幾何異性体(シス/トランス)を同定する
    6. 既知の分子式を持つ有機分子 possible 可能な異性体を導き出す

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Isomers are compounds with the same molecular formula but a different arrangement of atoms. This is called isomerism 异构.

    Structural isomerism

    In structural isomerism 结构异构 the atoms are joined in a different order. There are three kinds:

    • chain isomerism 链异构: the carbon chain is branched in different ways.
    • positional isomerism 位置异构: the functional group is on a different carbon.
    • functional group isomerism 官能团异构: the atoms form a different functional group (for example an alcohol and an ether).

    Stereoisomerism

    In stereoisomerism 立体异构 the atoms are joined in the same order but point in different directions in space.

    • geometrical isomerism 几何异构 (cis/trans) happens at a C=C double bond. The pi bond stops the two carbons rotating (restricted rotation 受限旋转), so groups are fixed on the same side (cis 顺式) or opposite sides (trans 反式).
    • optical isomerism 旋光异构 happens at a chiral 手性 carbon — a chiral centre 手性中心 is a carbon with four different groups attached. Such a carbon gives two mirror-image forms called enantiomers 对映体. A molecule may have more than one chiral centre.

    From a molecular formula you can deduce the possible isomers by trying different chains, positions and functional groups.

    Worked example. Explain why but-2-ene shows cis-trans (E/Z) isomerism but but-1-ene does not. Stereoisomerism at a C=C needs two things: restricted rotation about the double bond (both alkenes have that), and two different groups on each of the two double-bond carbons. In but-2-ene, $\text{CH}_3\text{CH=CHCH}_3$, each double-bond carbon carries an $\text{H}$ and a $\text{CH}_3$ - different, so the methyl groups can sit on the same side (cis / Z) or on opposite sides (trans / E). In but-1-ene, $\text{CH}_2\text{=CHCH}_2\text{CH}_3$, the first carbon carries two hydrogens - identical, so swapping them changes nothing and only one form exists. Test each double-bond carbon separately: two identical groups on either carbon kills the isomerism, however different the other carbon may be.

    日本語

    Isomers are compounds with the same molecular formula but a different arrangement of atoms. This is called isomerism 异构.

    Structural isomerism

    In structural isomerism 结构异构 the atoms are joined in a different order. There are three kinds:

    • chain isomerism 链异构: the carbon chain is branched in different ways.
    • positional isomerism 位置异构: the functional group is on a different carbon.
    • functional group isomerism 官能团异构: the atoms form a different functional group (for example an alcohol and an ether).
    Three pairs of skeletal structures: butane vs 2-methylpropane, propan-1-ol vs propan-2-ol, and ethanol vs methoxymethane
    The three kinds of structural isomerism — chain, positional and functional-group — each pair sharing the same molecular formula

    Stereoisomerism

    In stereoisomerism 立体异构 the atoms are joined in the same order but point in different directions in space.

    • geometrical isomerism 几何异构 (cis/trans) happens at a C=C double bond. The pi bond stops the two carbons rotating (restricted rotation 受限旋转), so groups are fixed on the same side (cis 顺式) or opposite sides (trans 反式).
    But-2-ene drawn twice: cis with both methyl groups on the same side of the double bond, trans with them on opposite sides
    Cis–trans isomerism at a C=C bond: the methyl groups are fixed on the same side (cis) or opposite sides (trans) because the bond cannot rotate
    • optical isomerism 旋光异构 happens at a chiral 手性 carbon — a chiral centre 手性中心 is a carbon with four different groups attached. Such a carbon gives two mirror-image forms called enantiomers 对映体. A molecule may have more than one chiral centre.
    A carbon bonded to four different atoms (F, H, Cl, Br) drawn next to its mirror image, the two not superimposable
    Optical isomerism: a carbon with four different groups gives two mirror-image forms (enantiomers) that cannot be superimposed

    From a molecular formula you can deduce the possible isomers by trying different chains, positions and functional groups.

    Worked example. Explain why but-2-ene shows cis-trans (E/Z) isomerism but but-1-ene does not. Stereoisomerism at a C=C needs two things: restricted rotation about the double bond (both alkenes have that), and two different groups on each of the two double-bond carbons. In but-2-ene, $\text{CH}_3\text{CH=CHCH}_3$, each double-bond carbon carries an $\text{H}$ and a $\text{CH}_3$ - different, so the methyl groups can sit on the same side (cis / Z) or on opposite sides (trans / E). In but-1-ene, $\text{CH}_2\text{=CHCH}_2\text{CH}_3$, the first carbon carries two hydrogens - identical, so swapping them changes nothing and only one form exists. Test each double-bond carbon separately: two identical groups on either carbon kills the isomerism, however different the other carbon may be.

    Explore · ⁨探索⁩

    Structural isomerism lab · ⁨構造異性体の実験⁩

    Compare molecules with the same formula but different structures. · ⁨同じ化学式だが異なる構造を持つ分子を比較せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    isomerism/ˈaɪsəmərɪzəm/ 異性体
    structural isomerism/ˈstrʌktʃərəl ˈaɪsəmərɪzəm/ 構造異性
    chain isomerism/tʃeɪn ˈaɪsəmərɪzəm/ 鎖状異性体
    positional isomerism/pəˈzɪʃənl ˈaɪsəmərɪzəm/ 位置異性体
    functional group isomerism/ˈfʌŋkʃənl ɡruːp ˈaɪsəmərɪzəm/ 官能基異性体
    stereoisomerism/ˈsterɪəʊɪsəmərɪzəm/ 立体異性
    geometrical isomerism/ˌdʒiːəʊˈmetrɪkl ˈaɪsəmərɪzəm/ 立体的異性体(幾何異性体)
    restricted rotation/rɪˈstrɪktɪd rəʊˈteɪʃn/ 回転制限
    cis/sɪs/ シス
    trans/trænz/ トランス
    optical isomerism/ˈɒptɪkl ˈaɪsəmərɪzəm/ 光学異性
    chiral/ˈkaɪrəl/ キラル
    chiral centre/ˈkaɪrəl ˈsentə/ キラル中心
    enantiomers/eˈnæntɪəməz/ 鏡像異性体
    Watch lesson · ⁨レッスンを視聴⁩
    13.4

    Exam tips

    • Know the difference between general, molecular, structural, displayed and skeletal formulas — questions ask for a specific one.
    • Name systematically: longest chain, lowest locants, substituents alphabetical; a wrong locant loses the mark.
    • Classify each reaction by type and reagent (addition, substitution, elimination, oxidation).
    • E/Z isomerism needs restricted rotation about a $\text{C}=\text{C}$ and two different groups on each carbon.
  • 14

    Hydrocarbons · ⁨炭化水素⁩

    Watch lesson · ⁨レッスンを視聴⁩
    14.1

    Alkanes

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions (reagents and conditions) by which alkanes can be produced: (a) addition of hydrogen to an alkene in a hydrogenation reaction, $\text{H}_2\text{(g)}$ and $\text{Pt/Ni}$ catalyst and heat (b) cracking of a longer chain alkane, heat with $\text{Al}_2\text{O}_3$
    2. describe: (a) the complete and incomplete combustion of alkanes (b) the free-radical substitution of alkanes by $\text{Cl}_2$ or $\text{Br}_2$ in the presence of ultraviolet light, as exemplified by the reactions of ethane
    3. describe the mechanism of free-radical substitution with reference to the initiation, propagation and termination steps
    4. suggest how cracking can be used to obtain more useful alkanes and alkenes of lower $M_r$ from heavier crude oil fractions
    5. understand the general unreactivity of alkanes, including towards polar reagents in terms of the strength of the $\text{C–H}$ bonds and their relative lack of polarity
    6. recognise the environmental consequences of carbon monoxide, oxides of nitrogen and unburnt hydrocarbons arising from the combustion of alkanes in the internal combustion engine and of their catalytic removal
    日本語
    1. アルカンを合成するための反応(試薬および条件)を思い出す:(a) 水素化反応におけるアルケンへの水素付加、$\text{H}_2\text{(g)}$ および $\text{Pt/Ni}$触媒、加熱 (b) 長い鎖のアルカンのクラッキング、$\text{Al}_2\text{O}_3$との加熱
    2. 以下を説明する:(a) アルカンの完全燃焼および不完全燃焼 (b) 紫外線下でのアルカンの自由ラジカル置換、$\text{Cl}_2$ または $\text{Br}_2$ によるもの(エタンの反応を例とする)
    3. 開始、伝播、終端の各ステップに言及して、自由ラジカル置換の反応機構を説明する
    4. クラッキングにより、より重質の原油画分から有用な低$M_r$ のアルカンおよびアルケンを得る方法を提案する
    5. $\text{C–H}$結合の強さとその相対的な非極性という観点から、アルカンの一般的な反応性の低さ、特に極性試薬に対するものを含めて理解する
    6. 内燃機関におけるアルカンの燃焼によって生じる一酸化炭素、窒素酸化物、未燃焼炭化水素の環境影響、およびそれらの触媒除去を理解する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Alkanes 烷烃 are saturated hydrocarbons (general formula $\text{C}_n\text{H}_{2n+2}$).

    Making alkanes

    • hydrogenation 氢化: add hydrogen to an alkene 烯烃, using a $\text{Pt}$ or $\text{Ni}$ catalyst and heat.
    • cracking 裂化: break a long-chain alkane into shorter ones by heating with $\text{Al}_2\text{O}_3$.

    Combustion

    In combustion 燃烧 an alkane burns in oxygen:

    • complete combustion 完全燃烧 (plenty of oxygen) gives carbon dioxide and water.
    • incomplete combustion 不完全燃烧 (not enough oxygen) gives water plus toxic carbon monoxide 一氧化碳 ($\text{CO}$) and soot (carbon).

    You are often asked to write the balanced equation. Balance it in a fixed order - carbon first, then hydrogen, and oxygen last - because oxygen is the only element left on just one side:

    $$\text{C}_6\text{H}_{14} + 9\tfrac{1}{2}\,\text{O}_2 \rightarrow 6\,\text{CO}_2 + 7\,\text{H}_2\text{O}$$

    The 6 carbons fix $6\,\text{CO}_2$; the 14 hydrogens fix $7\,\text{H}_2\text{O}$; counting the oxygens on the right gives $12 + 7 = 19$, so the left needs $19 \div 2 = 9\tfrac{1}{2}$. A half of $\text{O}_2$ is perfectly acceptable in this equation - and if the question asks for whole numbers, simply double everything ($2\,\text{C}_6\text{H}_{14} + 19\,\text{O}_2 \rightarrow 12\,\text{CO}_2 + 14\,\text{H}_2\text{O}$).

    Free-radical substitution

    Alkanes react with chlorine or bromine by free-radical substitution 自由基取代, in ultraviolet light 紫外线. For ethane and chlorine the mechanism has three steps:

    • initiation 引发 — UV light splits the halogen into two free radicals 自由基: $\;\text{Cl}_2 \rightarrow 2\,\text{Cl}\cdot$ This kind of break is homolytic fission 均裂: the bond splits evenly, one electron going to each atom, which is what makes two radicals. (The opposite, heterolytic fission 异裂, sends both electrons to one atom and makes a pair of ions - that is what happens in the polar mechanisms such as electrophilic addition.) Examiners ask for this word by name, so use it.
    • propagation 增长 — radicals react and make new radicals:
      $$\text{Cl}\cdot + \text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_5\cdot + \text{HCl} \qquad \text{C}_2\text{H}_5\cdot + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}\cdot$$
    • termination 终止 — two radicals join, ending the chain: $\;\text{Cl}\cdot + \text{C}_2\text{H}_5\cdot \rightarrow \text{C}_2\text{H}_5\text{Cl}$

    Why cracking is useful, and why alkanes are unreactive

    Cracking turns heavy fractions of crude oil 原油 into more useful, lower-$M_r$ alkanes and alkenes. (A fraction 馏分 is a group of molecules with a similar boiling-point range.)

    Alkanes are generally unreactive, especially towards polar reagents. This is because the C–H and C–C bonds are strong and have little polarity 极性, so there is no charge to attract an attacking species.

    Environmental effects

    Burning alkanes in an internal combustion engine gives off carbon monoxide, oxides of nitrogen and unburnt hydrocarbons. A catalytic converter removes these by turning them into harmless gases.

    日本語
    Blue flames on a gas stove
    Natural gas — mainly methane, the simplest alkane — burns on a stove.

    Alkanes 烷烃 are saturated hydrocarbons (general formula $\text{C}_n\text{H}_{2n+2}$).

    Making alkanes

    • hydrogenation 氢化: add hydrogen to an alkene 烯烃, using a $\text{Pt}$ or $\text{Ni}$ catalyst and heat.
    • cracking 裂化: break a long-chain alkane into shorter ones by heating with $\text{Al}_2\text{O}_3$.

    Combustion

    In combustion 燃烧 an alkane burns in oxygen:

    Complete combustion gives CO2 and water; incomplete combustion also gives CO and soot
    Complete combustion gives CO2 and water; incomplete also gives CO and soot
    • complete combustion 完全燃烧 (plenty of oxygen) gives carbon dioxide and water.
    • incomplete combustion 不完全燃烧 (not enough oxygen) gives water plus toxic carbon monoxide 一氧化碳 ($\text{CO}$) and soot (carbon).

    You are often asked to write the balanced equation. Balance it in a fixed order - carbon first, then hydrogen, and oxygen last - because oxygen is the only element left on just one side:

    $$\text{C}_6\text{H}_{14} + 9\tfrac{1}{2}\,\text{O}_2 \rightarrow 6\,\text{CO}_2 + 7\,\text{H}_2\text{O}$$

    The 6 carbons fix $6\,\text{CO}_2$; the 14 hydrogens fix $7\,\text{H}_2\text{O}$; counting the oxygens on the right gives $12 + 7 = 19$, so the left needs $19 \div 2 = 9\tfrac{1}{2}$. A half of $\text{O}_2$ is perfectly acceptable in this equation - and if the question asks for whole numbers, simply double everything ($2\,\text{C}_6\text{H}_{14} + 19\,\text{O}_2 \rightarrow 12\,\text{CO}_2 + 14\,\text{H}_2\text{O}$).

    Free-radical substitution

    Alkanes react with chlorine or bromine by free-radical substitution 自由基取代, in ultraviolet light 紫外线. For ethane and chlorine the mechanism has three steps:

    • initiation 引发 — UV light splits the halogen into two free radicals 自由基: $\;\text{Cl}_2 \rightarrow 2\,\text{Cl}\cdot$ This kind of break is homolytic fission 均裂: the bond splits evenly, one electron going to each atom, which is what makes two radicals. (The opposite, heterolytic fission 异裂, sends both electrons to one atom and makes a pair of ions - that is what happens in the polar mechanisms such as electrophilic addition.) Examiners ask for this word by name, so use it.
    • propagation 增长 — radicals react and make new radicals:
      $$\text{Cl}\cdot + \text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_5\cdot + \text{HCl} \qquad \text{C}_2\text{H}_5\cdot + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}\cdot$$
    • termination 终止 — two radicals join, ending the chain: $\;\text{Cl}\cdot + \text{C}_2\text{H}_5\cdot \rightarrow \text{C}_2\text{H}_5\text{Cl}$
    The three steps of free-radical substitution: initiation splitting chlorine into radicals, propagation carrying the chain, and termination joining two radicals
    Free-radical substitution in three steps: initiation makes radicals, propagation carries the chain, and termination ends it

    Why cracking is useful, and why alkanes are unreactive

    Cracking turns heavy fractions of crude oil 原油 into more useful, lower-$M_r$ alkanes and alkenes. (A fraction 馏分 is a group of molecules with a similar boiling-point range.)

    Alkanes are generally unreactive, especially towards polar reagents. This is because the C–H and C–C bonds are strong and have little polarity 极性, so there is no charge to attract an attacking species.

    Environmental effects

    Burning alkanes in an internal combustion engine gives off carbon monoxide, oxides of nitrogen and unburnt hydrocarbons. A catalytic converter removes these by turning them into harmless gases.

    Explore · ⁨探索⁩

    The tetrahedral carbon · ⁨四面体型炭素⁩

    Each carbon in an alkane has four bonding pairs and no lone pairs — they spread as far apart as possible into a tetrahedron (109.5°). · ⁨アルカンの各炭素は4つの結合電子対を持ち、非共有電子対はない。これらは可能な限り離れて四面体(109.5°)を形成する。⁩

    Explore · ⁨探索⁩

    Free-radical substitution · ⁨自由ラジカル置換反応⁩

    Methane reacts with chlorine in UV light by a chain of radical steps. · ⁨メタンは紫外線下で塩素と反応し、ラジカル段階の連鎖反応を起こす。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    alkane/ˈælkeɪn/ アルカン
    hydrogenation/haɪˈdrɒdʒəneɪʃn/ 水素添加
    alkene/ˈælkiːn/ アルケン
    cracking/ˈkrækɪŋ/ クラッキング
    combustion/kəmˈbʌstʃn/ 燃焼反応
    complete combustion/kəmˈpliːt kəmˈbʌstʃn/ 完全燃焼
    incomplete combustion/ɪŋkəmˈpliːt kəmˈbʌstʃn/ 不完全燃焼
    carbon monoxide/ˈkɑːbən mʌˈnɒksaɪd/ 一酸化炭素
    free-radical substitution/friː ˈrædɪkl ˌsʌbstɪˈtjuːʃn/ 自由ラジカル置換
    ultraviolet light/ˌʊltrəˈvaɪəlɪt laɪt/ 紫外線
    initiation/ɪˌnɪʃɪˈeɪʃn/ 開始
    free radical/friː ˈrædɪkl/ 自由ラジカル
    propagation/ˌprɒpəˈɡeɪʃn/ 増幅
    termination/ˌtɜːmɪˈneɪʃn/ 終端
    crude oil/kruːd ɔɪl/ 原油
    fraction/ˈfrækʃn/ 蒸留
    polarity/pəʊˈlærɪti/ 極性
    homolytic fission/ˌhɒməˈlɪtɪk ˈfɪʃn/ 均等開裂
    heterolytic fission/ˌhetrəˈlɪtɪk ˈfɪʃn/ 不均等開裂
    14.2

    Alkenes

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions (including reagents and conditions) by which alkenes can be produced: (a) elimination of $\text{HX}$ from a halogenoalkane by ethanolic $\text{NaOH}$ and heat (b) dehydration of an alcohol, by using a heated catalyst (e.g. $\text{Al}_2\text{O}_3$) or a concentrated acid (e.g. concentrated $\text{H}_2\text{SO}_4$) (c) cracking of a longer chain alkane
    2. describe the following reactions of alkenes: (a) the electrophilic addition of (i) hydrogen in a hydrogenation reaction, $\text{H}_2\text{(g)}$ and $\text{Pt/Ni}$ catalyst and heat (ii) steam, $\text{H}_2\text{O(g)}$ and $\text{H}_3\text{PO}_4$ catalyst (iii) a hydrogen halide, $\text{HX(g)}$, at room temperature (iv) a halogen, $\text{X}_2$ (b) the oxidation by cold dilute acidified $\text{KMnO}_4$ to form the diol (c) the oxidation by hot concentrated acidified $\text{KMnO}_4$ leading to the rupture of the carbon–carbon double bond and the identities of the subsequent products to determine the position of alkene linkages in larger molecules (d) addition polymerisation exemplified by the reactions of ethene and propene
    3. describe the use of aqueous bromine to show the presence of a C=C bond
    4. describe the mechanism of electrophilic addition in alkenes, using bromine/ethene and hydrogen bromide/propene as examples
    5. describe and explain the inductive effects of alkyl groups on the stability of primary, secondary and tertiary cations formed during electrophilic addition (this should be used to explain Markovnikov addition)
    日本語
    1. アルケンを合成するための反応(試薬および条件を含む)を思い出す:(a) 塩化アルキルからの$\text{HX}$ の脱離、エタノール性$\text{NaOH}$ と加熱 (b) アルコールの脱水、加熱した触媒(例:$\text{Al}_2\text{O}_3$)または濃酸(例:濃$\text{H}_2\text{SO}_4$)による (c) 長い鎖のアルカンのクラッキング
    2. アルケンの以下の反応を説明する:(a) 求電子付加による (i) 水素化反応における水素付加、$\text{H}_2\text{(g)}$ および $\text{Pt/Ni}$触媒、加熱 (ii) スチーム、$\text{H}_2\text{O(g)}$ および $\text{H}_3\text{PO}_4$触媒 (iii) ハロゲン化水素、$\text{HX(g)}$、室温 (iv) ハロゲン、$\text{X}_2$ (b) 冷たい希薄酸性$\text{KMnO}_4$による酸化によるジオールの生成 (c) 熱い濃酸性$\text{KMnO}_4$による酸化による炭素-炭素二重結合の切断および後続生成物の同定によるアルケン結合位置の決定 (d) エチレンおよびプロピレンの反応を例とした付加重合法
    3. 水溶液ブROMINEを用いてC=C結合の存在を示す方法について説明する
    4. ブロミン/エチレンおよび臭化水素/プロピレンを例として、アルケンにおける求電子付加の反応機構を説明する
    5. 求電子付加中に生成される一次、二次、三次 カチオンの安定性に対するアルキル基の誘起効果について説明し、これを用いてマルコフニコフ付加を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Alkenes have a C=C double bond (general formula $\text{C}_n\text{H}_{2n}$). The double bond is the functional group, so alkenes are reactive.

    Making alkenes

    • elimination 消去 of $\text{HX}$ from a halogenoalkane 卤代烷, using $\text{NaOH}$ dissolved in ethanol, with heat.
    • dehydration 脱水 of an alcohol 醇, using a hot $\text{Al}_2\text{O}_3$ catalyst or concentrated sulfuric acid.
    • cracking of a longer-chain alkane.

    Reactions of alkenes

    Most reactions are electrophilic addition 亲电加成 across the double bond:

    Reagent and conditions Product
    $\text{H}_2$, $\text{Pt/Ni}$ catalyst, heat alkane
    steam 水蒸气 ($\text{H}_2\text{O}$), $\text{H}_3\text{PO}_4$ catalyst alcohol
    a hydrogen halide 卤化氢 ($\text{HX}$), room temperature halogenoalkane
    a halogen $\text{X}_2$ a di-substituted alkane

    There are also two oxidation reactions with acidified $\text{KMnO}_4$:

    • cold, dilute $\text{KMnO}_4$ adds two $\text{–OH}$ groups to give a diol 二醇.
    • hot, concentrated $\text{KMnO}_4$ breaks the C=C bond right apart, and what each half turns into tells you exactly where the double bond used to be.

    That second reaction is worth learning properly, because it is how the exam asks you to locate a double bond in a large molecule. Oxidation is written with $[\text{O}]$, meaning "an oxygen atom from the oxidising agent". Cut the molecule at the C=C, then look at what each of the two carbons was carrying:

    The C=C carbon carries It becomes
    two alkyl groups ($\text{=CR}_2$) a ketone 酮, $\text{R}_2\text{C=O}$
    one alkyl and one $\text{H}$ ($\text{=CHR}$) a carboxylic acid 羧酸, $\text{RCOOH}$
    two hydrogens ($\text{=CH}_2$) $\text{CO}_2$ and water

    The pattern is simply how many hydrogens that carbon had: none stops at a ketone, one is pushed on to an acid, and two are oxidised all the way to $\text{CO}_2$.

    Worked example. Give the products when $(\text{CH}_3)_2\text{C=CHCH}_3$ is heated with hot concentrated acidified $\text{KMnO}_4$. Cut at the C=C and take each carbon separately. The left carbon carries two methyl groups and no hydrogen, so it stops at a ketone: $(\text{CH}_3)_2\text{C=O}$, which is propanone. The right carbon carries one methyl and one H, so it goes on to a carboxylic acid: $\text{CH}_3\text{COOH}$, ethanoic acid. So:

    $$(\text{CH}_3)_2\text{C=CHCH}_3 + 3[\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{COOH}$$

    Now run it backwards, which is the way the question is usually set. Given the products, rebuild the alkene by putting the two carbonyl carbons back together as a C=C: a ketone means that carbon had two alkyl groups, an acid means one alkyl and one H, and $\text{CO}_2$ means the chain ended in $\text{=CH}_2$. Getting $\text{CO}_2$ is the strongest clue of all - it can only come from a terminal double bond.

    Test for a C=C bond

    Shake the compound with orange bromine water 溴水. An alkene decolourises it (turns it colourless) by electrophilic addition. An alkane does not.

    Addition polymerisation

    In addition polymerisation 加成聚合, many alkene molecules join into one long chain, with no other product. Ethene gives poly(ethene). The long-chain product is a polymer 聚合物.

    The mechanism and Markovnikov's rule

    In electrophilic addition (for example bromine with ethene), the electron-rich C=C attracts the electrophile. This forms a positive intermediate called a carbocation 碳正离子, which the negative part then attacks.

    Alkyl groups push electrons towards the positive carbon — this is the inductive effect 诱导效应. So a carbocation with more alkyl groups is more stable: tertiary is more stable than secondary, which is more stable than primary. When $\text{HBr}$ adds to propene, the more stable carbocation forms, so hydrogen adds to the carbon that already has more hydrogens. This pattern is Markovnikov's rule 马氏规则.

    Worked example. Predict the major product when $\text{HBr}$ adds to propene, $\text{CH}_3\text{CH=CH}_2$. The $\text{H}^{+}$ adds first, and it adds in whichever way makes the more stable carbocation. Adding the $\text{H}$ to the end carbon puts the positive charge on the middle carbon, giving a secondary carbocation, which is stabilised by electron-releasing alkyl groups on two sides. Adding it to the middle carbon would leave a less stable primary carbocation. The bromide ion then attacks the secondary carbocation, so the major product is 2-bromopropane. That is Markovnikov's rule - but quote the reason (carbocation stability: tertiary > secondary > primary), because the rule on its own is not the explanation.

    日本語
    An oil refinery
    Cracking heavy fractions at a refinery produces alkenes for plastics and fuels.

    Alkenes have a C=C double bond (general formula $\text{C}_n\text{H}_{2n}$). The double bond is the functional group, so alkenes are reactive.

    Making alkenes

    • elimination 消去 of $\text{HX}$ from a halogenoalkane 卤代烷, using $\text{NaOH}$ dissolved in ethanol, with heat.
    • dehydration 脱水 of an alcohol 醇, using a hot $\text{Al}_2\text{O}_3$ catalyst or concentrated sulfuric acid.
    • cracking of a longer-chain alkane.

    Reactions of alkenes

    Most reactions are electrophilic addition 亲电加成 across the double bond:

    Reagent and conditions Product
    $\text{H}_2$, $\text{Pt/Ni}$ catalyst, heat alkane
    steam 水蒸气 ($\text{H}_2\text{O}$), $\text{H}_3\text{PO}_4$ catalyst alcohol
    a hydrogen halide 卤化氢 ($\text{HX}$), room temperature halogenoalkane
    a halogen $\text{X}_2$ a di-substituted alkane

    There are also two oxidation reactions with acidified $\text{KMnO}_4$:

    • cold, dilute $\text{KMnO}_4$ adds two $\text{–OH}$ groups to give a diol 二醇.
    • hot, concentrated $\text{KMnO}_4$ breaks the C=C bond right apart, and what each half turns into tells you exactly where the double bond used to be.

    That second reaction is worth learning properly, because it is how the exam asks you to locate a double bond in a large molecule. Oxidation is written with $[\text{O}]$, meaning "an oxygen atom from the oxidising agent". Cut the molecule at the C=C, then look at what each of the two carbons was carrying:

    The C=C carbon carries It becomes
    two alkyl groups ($\text{=CR}_2$) a ketone 酮, $\text{R}_2\text{C=O}$
    one alkyl and one $\text{H}$ ($\text{=CHR}$) a carboxylic acid 羧酸, $\text{RCOOH}$
    two hydrogens ($\text{=CH}_2$) $\text{CO}_2$ and water

    The pattern is simply how many hydrogens that carbon had: none stops at a ketone, one is pushed on to an acid, and two are oxidised all the way to $\text{CO}_2$.

    Worked example. Give the products when $(\text{CH}_3)_2\text{C=CHCH}_3$ is heated with hot concentrated acidified $\text{KMnO}_4$. Cut at the C=C and take each carbon separately. The left carbon carries two methyl groups and no hydrogen, so it stops at a ketone: $(\text{CH}_3)_2\text{C=O}$, which is propanone. The right carbon carries one methyl and one H, so it goes on to a carboxylic acid: $\text{CH}_3\text{COOH}$, ethanoic acid. So:

    $$(\text{CH}_3)_2\text{C=CHCH}_3 + 3[\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{COOH}$$

    Now run it backwards, which is the way the question is usually set. Given the products, rebuild the alkene by putting the two carbonyl carbons back together as a C=C: a ketone means that carbon had two alkyl groups, an acid means one alkyl and one H, and $\text{CO}_2$ means the chain ended in $\text{=CH}_2$. Getting $\text{CO}_2$ is the strongest clue of all - it can only come from a terminal double bond.

    Test for a C=C bond

    Shake the compound with orange bromine water 溴水. An alkene decolourises it (turns it colourless) by electrophilic addition. An alkane does not.

    Two test tubes of orange bromine water: the one with an alkene has turned colourless, the one with an alkane is still orange
    The bromine-water test: an alkene decolourises the orange bromine water, while an alkane leaves it orange

    Addition polymerisation

    In addition polymerisation 加成聚合, many alkene molecules join into one long chain, with no other product. Ethene gives poly(ethene). The long-chain product is a polymer 聚合物.

    n ethene molecules with double bonds on the left becoming the repeating unit of poly(ethene) in brackets on the right
    Addition polymerisation: many ethene molecules open their double bonds and join into the long chain of poly(ethene)

    The mechanism and Markovnikov's rule

    In electrophilic addition (for example bromine with ethene), the electron-rich C=C attracts the electrophile. This forms a positive intermediate called a carbocation 碳正离子, which the negative part then attacks.

    The mechanism of bromine adding to ethene: curly arrows show the C=C attacking one bromine, a carbocation forming, then bromide attacking it to give 1,2-dibromoethane
    Electrophilic addition of bromine to ethene: the C=C attacks Br$^{\delta+}$, a carbocation forms, then Br$^-$ attacks it

    Alkyl groups push electrons towards the positive carbon — this is the inductive effect 诱导效应. So a carbocation with more alkyl groups is more stable: tertiary is more stable than secondary, which is more stable than primary. When $\text{HBr}$ adds to propene, the more stable carbocation forms, so hydrogen adds to the carbon that already has more hydrogens. This pattern is Markovnikov's rule 马氏规则.

    Primary, secondary and tertiary carbocations with arrows showing alkyl groups pushing electrons towards the positive carbon, stability rising left to right
    Carbocation stability rises from primary to tertiary as more alkyl groups push electrons in — the basis of Markovnikov's rule

    Worked example. Predict the major product when $\text{HBr}$ adds to propene, $\text{CH}_3\text{CH=CH}_2$. The $\text{H}^{+}$ adds first, and it adds in whichever way makes the more stable carbocation. Adding the $\text{H}$ to the end carbon puts the positive charge on the middle carbon, giving a secondary carbocation, which is stabilised by electron-releasing alkyl groups on two sides. Adding it to the middle carbon would leave a less stable primary carbocation. The bromide ion then attacks the secondary carbocation, so the major product is 2-bromopropane. That is Markovnikov's rule - but quote the reason (carbocation stability: tertiary > secondary > primary), because the rule on its own is not the explanation.

    Explore · ⁨探索⁩

    Alkene addition route · ⁨アルケンの付加反応経路⁩

    Follow how the C=C bond opens and new atoms add. · ⁨C=C結合が開いて新たな原子が付加する様子を追跡する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    elimination/ɪˌlɪmɪˈneɪʃn/ 脱離反応
    halogenoalkane/ˈheɪləʊdʒnəʊlkeɪn/ ハロゲン化アルキル
    dehydration/ˌdiːhaɪˈdreɪʃn/ 脱水
    alcohol/ˈælkəhɒl/ アルコール
    electrophilic addition/ɪˌlektrəʊˈfɪlɪk əˈdɪʃn/ 求電付加
    steam/stiːm/ 蒸気
    hydrogen halide/ˈhaɪdrədʒn ˈhælaɪd/ 水素ハライド
    diol/dɪˈɒl/ ジオール
    bromine water/ˈbrəʊmaɪn ˈwɔːtə/ 臭素水
    addition polymerisation/əˈdɪʃn ˌpɒlɪməraɪˈzeɪʃn/ 付加重合
    polymer/ˈpɒlɪmə/ ポリマー
    carbocation/ˌkɑːbəˈkeɪʃn/ カルボカチオン
    inductive effect/ɪnˈdʌktɪv ɪˈfekt/ 誘導効果
    Markovnikov's rule/ˈmɑːkəvnɪkɒvz ruːl/ マルコフニコフの法則
    ketone/ˈketəʊn/ ケトン
    carboxylic acid/ˌkɑːbəkˈsɪlɪk ˈæsɪd/ カルボン酸
    Watch lesson · ⁨レッスンを視聴⁩
    14.2

    Exam tips

    • Alkanes: free-radical substitution needs UV light — show initiation, propagation and termination with the radical dots.
    • Alkenes: electrophilic addition via a carbocation; draw curly arrows from the $\text{C}=\text{C}$.
    • Markovnikov: H adds to the carbon with more H's, via the more stable carbocation (alkyl groups are electron-releasing).
    • Test for $\text{C}=\text{C}$: bromine water decolourises (orange → colourless) — state the observation.
  • 15

    Halogen compounds · ⁨ハロゲン化合物⁩

    Watch lesson · ⁨レッスンを視聴⁩
    15.1

    Halogenoalkanes · ⁨ハロゲンアルカン⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions (reagents and conditions) by which halogenoalkanes can be produced: (a) the free-radical substitution of alkanes by $\text{Cl}_2$ or $\text{Br}_2$ in the presence of ultraviolet light, as exemplified by the reactions of ethane (b) electrophilic addition of an alkene with a halogen, $\text{X}_2$, or hydrogen halide, $\text{HX(g)}$, at room temperature (c) substitution of an alcohol, e.g. by reaction with $\text{HX(g)}$; or with $\text{KCl}$ and concentrated $\text{H}_2\text{SO}_4$ or concentrated $\text{H}_3\text{PO}_4$; or with $\text{PCl}_3$ and heat; or with $\text{PCl}_5$; or with $\text{SOCl}_2$
    2. classify halogenoalkanes into primary, secondary and tertiary
    3. describe the following nucleophilic substitution reactions: (a) the reaction with $\text{NaOH(aq)}$ and heat to produce an alcohol (b) the reaction with $\text{KCN}$ in ethanol and heat to produce a nitrile (c) the reaction with $\text{NH}_3$ in ethanol heated under pressure to produce an amine (d) the reaction with aqueous silver nitrate in ethanol as a method of identifying the halogen present as exemplified by bromoethane
    4. describe the elimination reaction with $\text{NaOH}$ in ethanol and heat to produce an alkene as exemplified by bromoethane
    5. describe the $\text{S}_\text{N}1$ and $\text{S}_\text{N}2$ mechanisms of nucleophilic substitution in halogenoalkanes including the inductive effects of alkyl groups
    6. recall that primary halogenoalkanes tend to react via the $\text{S}_\text{N}2$ mechanism; tertiary halogenoalkanes via the $\text{S}_\text{N}1$ mechanism; and secondary halogenoalkanes by a mixture of the two, depending on structure
    7. describe and explain the different reactivities of halogenoalkanes (with particular reference to the relative strengths of the C–X bonds as exemplified by the reactions of halogenoalkanes with aqueous silver nitrates)
    日本語
    1. ハロゲン化アルキルが生成される反応(試薬および条件)を思い出す:(a) 紫外線存在下でのアルカンの自由ラジカル置換反応である$\text{Cl}_2$または$\text{Br}_2$によるもの。エタンの反応を例に示す。(b) 常温におけるアルケンのハロゲン、$\text{X}_2$、または水素ハライド、$\text{HX(g)}$との求電子付加反応。(c) アルコールの置換反応。例えば、$\text{HX(g)}$との反応、または$\text{KCl}$と濃硫酸、$\text{H}_2\text{SO}_4$、あるいは濃硝酸、$\text{H}_3\text{PO}_4$との反応、$\text{PCl}_3$と加熱、$\text{PCl}_5$、または$\text{SOCl}_2$との反応によるもの
    2. ハロゲン化アルキルを一次、二次、三次に分類する
    3. 以下の求核置換反応を説明する:(a) $\text{NaOH(aq)}$ と加熱による反応でアルコールを生成する (b) エタノール中、$\text{KCN}$ と加熱による反応でシアノ化合物(ニトリル)を生成する (c) 加圧下で加熱したエタノール中、$\text{NH}_3$ との反応でアミンを生成する (d) 水溶液硝酸銀とエタノールを用いる反応は、ハロゲン化アルカンの一種であるブロモエタンの例のように、存在するハロゲンの同定法として用いられる
    4. 脱離反応を説明する。$\text{NaOH}$ をエタノール中で加熱すると、ブロモエタンの例のようにアルケンを生成する
    5. ハロゲン化アルカンにおける求核置換反応の$\text{S}_\text{N}1$ および$\text{S}_\text{N}2$ 機構を説明し、アルキル基の誘起効果を含む
    6. 第一級ハロゲン化アルカンは主に$\text{S}_\text{N}2$ 機構で反応し、第三級ハロゲン化アルカンは$\text{S}_\text{N}1$ 機構で反応し、第二級ハロゲン化アルカンは構造に応じて両方の機構の混合物で反応することを思い出す
    7. ハロゲン化アルカンの異なる反応性を説明・解説する(C–X結合の相対的な強さに特に言及し、ハロゲン化アルカンの水溶液硝酸銀との反応がその例となる)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A halogenoalkane 卤代烷 is an alkane with one or more halogen atoms in place of hydrogen (a C–X bond, where X is a halogen).

    Making halogenoalkanes

    • free-radical substitution 自由基取代 of an alkane with $\text{Cl}_2$ or $\text{Br}_2$ in ultraviolet light.
    • electrophilic addition 亲电加成 of an alkene with a halogen $\text{X}_2$ or a hydrogen halide $\text{HX}$.
    • substitution of an alcohol 醇, for example by $\text{HX}$, by $\text{PCl}_5$, by $\text{PCl}_3$ with heat, or by $\text{SOCl}_2$.

    Three classes

    A halogenoalkane is primary 伯, secondary 仲 or tertiary 叔, depending on how many carbon atoms are joined to the carbon that holds the halogen (one, two or three).

    日本語

    ハロゲンアルカンとは、アルカンの水素原子の一つまたは複数ハロゲン原子に置き換わった化合物である(C–X 結合、X はハロゲン)。

    フリーザースプレー缶から冷たいハロゲンアルカンの蒸気が出ている様子
    揮発性のハロゲンアルカンはかつて冷媒やエアゾール噴射剤として広く用いられていた

    ハロゲンアルカンの合成

    • 自由ラジカル置換反応: アルカンを紫外線下で $\text{Cl}_2$ または $\text{Br}_2$ と反応させる。
    • 求電子付加反応: アルケンをハロゲン $\text{X}_2$ または水素ハロゲン化物 $\text{HX}$ と反応させる。
    • アルコールの置換反応: 例えば $\text{HX}$ で、$\text{PCl}_5$ で、$\text{PCl}_3$ で加熱、あるいは $\text{SOCl}_2$ で反応させる。

    3つの分類

    ハロゲンアルカンは、ハロゲンを保持する炭素に結合している炭素原子の数(1個、2個、3個)に応じて、一次、二次、または三次と分類される。

    臭素を含む炭素を強調した3つの構造:一次(炭素1つ)、二次(炭素2つ)、三次(炭素3つ)
    ハロゲンを保持する炭素に結合している炭素の数によって区別される、一次・二次・三次ハロゲンアルカン
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    halogenoalkane/ˈheɪləʊdʒnəʊlkeɪn/ ハロゲン化アルキル
    free-radical substitution/friː ˈrædɪkl ˌsʌbstɪˈtjuːʃn/ 自由ラジカル置換
    electrophilic addition/ɪˌlektrəʊˈfɪlɪk əˈdɪʃn/ 求電付加
    alcohol/ˈælkəhɒl/ アルコール
    primary/ˈpraɪməri/ 第一級
    secondary/ˈsekəndəri/ 第二級
    tertiary/ˈtɜːʃjəri/ 第三級
    nucleophilic substitution/ˌnjuːklɪəˈfɪlɪk ˌsʌbstɪˈtjuːʃn/ 求核置換
    nucleophile/ˈnjuːklɪɒfaɪl/ 求核剤
    nitrile/ˈnaɪtraɪl/ ニトリル
    amine/ˈæmaɪn/ アミン
    silver nitrate/ˈsɪlvə ˈnaɪtreɪt/ 硝酸銀
    precipitate/prɪˈsɪpɪteɪt/ 沈殿物
    15.1

    Nucleophilic substitution · ⁨求核置換反応⁩

    English

    The C–X bond is polar, so the carbon is slightly positive. A nucleophilic substitution 亲核取代 happens when a nucleophile 亲核试剂 (a lone-pair species) attacks that carbon and replaces the halogen.

    Reagent and conditions Product
    $\text{NaOH(aq)}$, heat an alcohol
    $\text{KCN}$ in ethanol, heat a nitrile 腈 (adds one carbon to the chain)
    $\text{NH}_3$ in ethanol, heated under pressure an amine 胺

    To identify the halogen, warm the halogenoalkane with silver nitrate 硝酸银 in ethanol. A silver halide precipitate 沉淀 forms, and its colour shows which halogen is present (white $\text{AgCl}$, cream $\text{AgBr}$, yellow $\text{AgI}$).

    日本語

    C–X 結合は極性を持つため、炭素はわずかに正電荷を帯びる。求核置換反応では、求核試薬(非共有電子対を持つ種)がその炭素を攻撃し、ハロゲンに置き換わる。

    沸騰液が入った丸底フラスコに垂直コンデンサーを取り付け、蒸気を還流させている様子
    多くのハロゲンアルカンの反応は、還流条件下で混合物を加熱して行われる
    試薬および条件 生成物
    $\text{NaOH(aq)}$、加熱 アルコール
    エタノール中の $\text{KCN}$、加熱 ニトリル(鎖に炭素が1つ増える)
    加圧下でエタノール中の $\text{NH}_3$、加熱 アミン
    ロータリーエバポレーター:温水浴の中で回転する溶液入りフラスコ、溶媒蒸気が別の受容フラスコで凝縮されている様子
    ロータリーエバポレーターは減圧下で溶媒を除去する装置であり、反応後の工程である

    ハロゲンの同定のため、ハロゲンアルカンをエタノール中の硝酸銀と温める。ハロゲン化銀の沈殿ができ、その色が存在するハロゲンを示す(白色 $\text{AgCl}$、クリーム色 $\text{AgBr}$、黄色 $\text{AgI}$)。

    Explore · ⁨探索⁩

    Nucleophilic substitution · ⁨求核置換反応⁩

    A nucleophile swaps in for the leaving group on a halogenoalkane. · ⁨求核試薬がハロゲン化アルキンの脱離基に替わります。⁩

    15.1

    Elimination · ⁨脱離反応⁩

    English

    The same halogenoalkane can instead undergo elimination 消去 to form an alkene 烯烃. The conditions decide which reaction wins:

    • $\text{NaOH}$ in water → nucleophilic substitution → an alcohol.
    • $\text{NaOH}$ in ethanol, heated → elimination → an alkene.
    $$\text{C}_2\text{H}_5\text{Br} + \text{NaOH} \rightarrow \text{C}_2\text{H}_4 + \text{NaBr} + \text{H}_2\text{O}$$
    日本語

    同じハロゲンアルカンは代わりに脱離反応を起こしてアルケンを生成する場合もある。反応条件によりどちらの反応が優先されるかが決まる:

    • $\text{NaOH}$ を水中では → 求核置換反応 → アルコール。
    • $\text{NaOH}$ をエタノール中で加熱すると → 脱離反応 → アルケン。
    $$\text{C}_2\text{H}_5\text{Br} + \text{NaOH} \rightarrow \text{C}_2\text{H}_4 + \text{NaBr} + \text{H}_2\text{O}$$
    枝分かれ図:臭素エタンに水酸化ナトリウム水溶液を使うと置換反応でアルコールになるが、エタノール中での水酸化ナトリウム(加熱)では脱離反応でアルケンになる
    同じハロゲンアルカン:水中の NaOH では置換反応でアルコールとなり、エタノール中の NaOH(加熱)では脱離反応でアルケンとなる
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    elimination/ɪˌlɪmɪˈneɪʃn/ 脱離反応
    alkene/ˈælkiːn/ アルケン
    transition state/trænˈsɪʃn steɪt/ 制限因子
    carbocation/ˌkɑːbəˈkeɪʃn/ カルボカチオン
    inductive effect/ɪnˈdʌktɪv ɪˈfekt/ 誘導効果
    bond energy/bɒnd ˈenədʒi/ 結合エネルギー
    15.1

    The S$_\text{N}$1 and S$_\text{N}$2 mechanisms · ⁨S$_\text{N}$1およびS$_\text{N}$2機構⁩

    English

    Nucleophilic substitution can follow two routes:

    • S$_\text{N}$2: one step. The nucleophile attacks at the same time as the halogen leaves, passing through a crowded transition state 过渡态 where both are half-bonded. The rate depends on both the halogenoalkane and the nucleophile.
    • S$_\text{N}$1: two steps. First the C–X bond breaks to give a carbocation 碳正离子; then the nucleophile attacks it. The rate depends only on the halogenoalkane.

    Alkyl groups push electrons towards the positive carbon (the inductive effect 诱导效应), so they stabilise the carbocation. This is why:

    • primary halogenoalkanes mostly react by S$_\text{N}$2.
    • tertiary halogenoalkanes mostly react by S$_\text{N}$1 (their carbocation is well stabilised).
    • secondary halogenoalkanes use a mixture of the two.
    日本語

    求核置換反応には2つの経路がある:

    • S$_\text{N}$2:1段階。求核試薬の攻撃とハロゲン离去が同時に起こり、両方が部分的に結合している混雑した遷移状態を通過します。速度はハロゲン化アルキルと求核試薬の両方に依存します。
    • S$_\text{N}$1: 二歩反応。まず C–X 結合が切断されて炭素陽イオンが生じ、次に求核試薬がそれを攻撃する。速度はハロゲンアルカンのみに依存する。
    S N 2 が一歩反応で求核試薬が攻撃しつつハロゲンが離脱する遷移状態を示し、S N 1 が二歩反応で炭素陽イオンを介する様子
    S$_\text{N}$2 は混雑した遷移状態を通る単一歩反応;S$_\text{N}$1 は炭素陽イオンを介する二歩反応

    アルキル基は正電荷を持つ炭素へ電子を押し込む(誘起効果)ため、炭素陽イオンを安定化する。そのため:

    • 一次ハロゲンアルカンは主に S$_\text{N}$2 で反応する。
    • 三次ハロゲンアルカンは主に S$_\text{N}$1 で反応する(炭素陽イオンが十分に安定化されているため)。
    • 二次ハロゲンアルカンは両方の経路の混合を用いる。
    Explore · ⁨探索⁩

    The SN2 mechanism, step by step · ⁨SN2机理,逐步解析⁩

    Step through nucleophilic substitution. The nucleophile attacks from behind as the halide leaves — all in one smooth step, flipping the molecule inside-out. · ⁨通过求核置换反应进行解析。亲核试剂从背面进攻,同时卤离子离去——整个过程一步完成,分子构型发生翻转。⁩

    15.1

    Different reactivities · ⁨異なる反応性⁩

    English

    How fast a halogenoalkane reacts depends on the strength of the C–X bond, measured by its bond energy 键能. The C–I bond is the weakest, so iodoalkanes react fastest; the C–Cl bond is the strongest of the three, so chloroalkanes react slowest. So when tested with silver nitrate, an iodoalkane gives its precipitate first — its higher reactivity 反应活性 comes from the weaker C–X bond.

    Worked example. Predict the mechanism for the hydrolysis of 1-bromobutane and of 2-bromo-2-methylpropane. Classify the halogenoalkane first. 1-bromobutane is primary: the carbon carrying the $\text{Br}$ is barely shielded, so the nucleophile can attack the back of it and the mechanism is $\text{S}_\text{N}2$ - one step, with the rate depending on both the halogenoalkane and the nucleophile. 2-bromo-2-methylpropane is tertiary: three bulky methyl groups block that attack, but they also stabilise the carbocation formed once the $\text{Br}$ leaves, so it goes $\text{S}_\text{N}1$ - two steps, with the rate depending on the halogenoalkane only. Decide from the class (primary → $\text{S}_\text{N}2$, tertiary → $\text{S}_\text{N}1$), and note that the tertiary one hydrolyses faster despite being the more crowded.

    日本語

    ハロゲンアルカンの反応速度は C–X 結合の強さ、すなわち結合エネルギーに依存する。C–I 結合が最も弱いため、ヨードアルカンの反応が最も速い。C–Cl 結合は3つの中で最も強いため、クロロアルカンの反応が最も遅い。したがって硝酸銀による試験では、ヨードアルカンは最初に沈殿を生じる——これはより弱い C–X 結合による高い反応性の結果である。

    C-FからC-Iへ向かってC-X結合エネルギーが低下する棒グラフ。弱い結合ほど反応速度が速いことを示す矢印が付いている
    C–X結合が弱いほどハロゲン化アルカンは早く反応し、ヨード化アルカンが最も早く、クロロ化アルカンが最も遅い

    ** worked example.** 1-ブロモブタンおよび2-ブロモ-2-メチルプロパンの加水分解の反応機構を予測せよ。まずハロゲン化アルカンの分類を行え。1-ブロモブタンは第一級: $\text{Br}$を持つ炭素はほとんど遮蔽されていないため、求核剤は背面から攻撃でき、反応機構は$\text{S}_\text{N}2$である: 一歩で完了し、反応速度はハロゲン化アルカンと求核剤の両方に依存する。2-ブロモ-2-メチルプロパンは第三級: 3つの大きなメチル基がその攻撃を阻害するが、それらは$\text{Br}$が脱離した後に生成するカルボニウムイオンも安定化させるため、この反応は$\text{S}_\text{N}1$に従う: 二つのステップであり、反応速度はハロゲン化アルカンのみに依存する。クラス(第一級 → $\text{S}_\text{N}2$、第三級 → $\text{S}_\text{N}1$)に基づいて決定せよ。なお、第三級化合物の方が混雑しているにもかかわらず、加水分解は速いことに注意せよ。

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    reactivity/rɪəkˈtɪvɪti/ 反応性
    15.1

    Exam tips · ⁨試験対策⁩

    English
    • Draw nucleophilic substitution with a curly arrow from the nucleophile's lone pair to the $\delta+$ carbon and one from the $\text{C-X}$ bond.
    • Match mechanism to class: $S_\text{N}1$ (tertiary, carbocation, two steps) vs $S_\text{N}2$ (primary, one step, transition state).
    • Reactivity is set by bond enthalpy: $\text{C-I}$ is weakest, so iodoalkanes react fastest (not electronegativity).
    • Elimination (hot, ethanolic KOH) vs substitution (warm, aqueous KOH) — the conditions decide the product; state them.
    日本語
    • 求核置換反応を描く: 求核剤の非共有電子対から$\delta+$炭素へ、そして$\text{C-X}$結合からそれぞれカールアローを引く。
    • 反応機構とクラスの対応: $S_\text{N}1$(第三級、カルボニウムイオン、二つのステップ)対 $S_\text{N}2$(第一級、一歩、遷移状態)。
    • 反応性は結合エンタルピーによって決まる: $\text{C-I}$が最も弱いため、ヨード化アルカンが最も早く反応する(電気陰性度ではない)。
    • 脱水素反応(熱いエタノール性KOH)対 置換反応(温かい水溶液性KOH): 条件が生成物を決定するため、条件を明記せよ。
  • 16

    Hydroxy compounds · ⁨ヒドロキシ化合物⁩

    Watch lesson · ⁨レッスンを視聴⁩
    16.1

    Alcohols · ⁨アルコール類⁩

    Syllabus · ⁨シラバス⁩
    1. recall the reactions (reagents and conditions) by which alcohols can be produced: (a) electrophilic addition of steam to an alkene, $\text{H}_2\text{O(g)}$ and $\text{H}_3\text{PO}_4$ catalyst (b) reaction of alkenes with cold dilute acidified potassium manganate(VII) to form a diol (c) substitution of a halogenoalkane using $\text{NaOH(aq)}$ and heat (d) reduction of an aldehyde or ketone using $\text{NaBH}_4$ or $\text{LiAlH}_4$ (e) reduction of a carboxylic acid using $\text{LiAlH}_4$ (f) hydrolysis of an ester using dilute acid or dilute alkali and heat
    2. describe: (a) the reaction with oxygen (combustion) (b) substitution to form halogenoalkanes, e.g. by reaction with $\text{HX(g)}$; or with $\text{KCl}$ and concentrated $\text{H}_2\text{SO}_4$ or concentrated $\text{H}_3\text{PO}_4$; or with $\text{PCl}_3$ and heat; or with $\text{PCl}_5$; or with $\text{SOCl}_2$ (c) the reaction with $\text{Na(s)}$ (d) oxidation with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or acidified $\text{KMnO}_4$ to: (i) carbonyl compounds by distillation (ii) carboxylic acids by refluxing (primary alcohols give aldehydes which can be further oxidised to carboxylic acids, secondary alcohols give ketones, tertiary alcohols cannot be oxidised) (e) dehydration to an alkene, by using a heated catalyst, e.g. $\text{Al}_2\text{O}_3$ or a concentrated acid (f) formation of esters by reaction with carboxylic acids and concentrated $\text{H}_2\text{SO}_4$ as catalyst as exemplified by ethanol
    3. (a) classify alcohols as primary, secondary and tertiary alcohols, to include examples with more than one alcohol group (b) state characteristic distinguishing reactions, e.g. mild oxidation with acidified $\text{K}_2\text{Cr}_2\text{O}_7$, colour change from orange to green
    4. deduce the presence of a $\text{CH}_3\text{CH(OH)}-$ group in an alcohol, $\text{CH}_3\text{CH(OH)}-\text{R}$, from its reaction with alkaline $\text{I}_2\text{(aq)}$ to form a yellow precipitate of tri-iodomethane and an ion, $\text{RCO}_2^-$
    5. explain the acidity of alcohols compared with water

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    An alcohol 醇 has the $\text{–OH}$ (hydroxyl) functional group.

    Making alcohols

    Method Reagents and conditions
    addition of steam to an alkene $\text{H}_2\text{O(g)}$, $\text{H}_3\text{PO}_4$ catalyst (electrophilic addition 亲电加成)
    an alkene 烯烃 with cold dilute $\text{KMnO}_4$ gives a diol 二醇 (two $\text{–OH}$ groups)
    substitution of a halogenoalkane 卤代烷 $\text{NaOH(aq)}$, heat
    reduction 还原 of an aldehyde 醛 or ketone 酮 $\text{NaBH}_4$ or $\text{LiAlH}_4$
    reduction of a carboxylic acid 羧酸 $\text{LiAlH}_4$
    hydrolysis 水解 of an ester 酯 dilute acid or alkali, heat

    Reactions of alcohols

    • combustion: alcohols burn in oxygen to give carbon dioxide and water.
    • substitution to a halogenoalkane, for example with $\text{HX}$, $\text{PCl}_5$, $\text{PCl}_3$ and heat, or $\text{SOCl}_2$.
    • with sodium: alcohols react with sodium metal to give hydrogen and a sodium alkoxide — like water, but more slowly.
    • oxidation 氧化 with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (or $\text{KMnO}_4$). The product depends on the class of alcohol (see below).
    • dehydration 脱水 to an alkene, using a hot $\text{Al}_2\text{O}_3$ catalyst or concentrated acid.
    • ester formation: an alcohol reacts with a carboxylic acid (with concentrated $\text{H}_2\text{SO}_4$ catalyst) to make an ester.

    Three classes and how oxidation tells them apart

    An alcohol is primary 伯, secondary 仲 or tertiary 叔, depending on how many carbons are joined to the carbon holding the $\text{–OH}$. Some molecules have more than one $\text{–OH}$ group.

    Class Oxidation product
    primary aldehyde (by distillation 蒸馏), then carboxylic acid (by reflux 回流)
    secondary a ketone
    tertiary not oxidised

    In a quick test, acidified $\text{K}_2\text{Cr}_2\text{O}_7$ turns from orange to green with a primary or secondary alcohol, but stays orange with a tertiary alcohol.

    • distillation removes the aldehyde as it forms, before it can be oxidised further.
    • reflux keeps boiling the mixture and returning the vapour, so the alcohol is fully oxidised to the carboxylic acid.

    The iodoform test

    If you warm an alcohol that contains the $\text{CH}_3\text{CH(OH)}-$ group with alkaline aqueous iodine, you get a pale yellow precipitate of tri-iodomethane 三碘甲烷 ($\text{CHI}_3$) and the ion $\text{RCO}_2^-$. This is a useful test for that group.

    Acidity of alcohols

    The $\text{–OH}$ group makes alcohols very weakly acidic: they can lose the $\text{H}^+$ to form an $\text{RO}^-$ ion. But their acidity 酸性 is lower than that of water. This is because the alkyl group pushes electron density onto the oxygen, which makes the $\text{RO}^-$ ion less stable, so the alcohol holds onto its $\text{H}^+$ more tightly.

    Worked example. Three unlabelled bottles hold butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. How does warming each with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ tell them apart? What matters is how many hydrogens sit on the carbon carrying the $\text{OH}$. Butan-1-ol is primary (two such hydrogens): the orange dichromate turns green, and by distilling you collect an aldehyde (butanal), or by refluxing you get the carboxylic acid (butanoic acid). Butan-2-ol is secondary (one such hydrogen): also green, but the product is a ketone (butanone), which will not oxidise further. 2-methylpropan-2-ol is tertiary (no such hydrogen): there is nothing to remove, so the dichromate stays orange. The colour only separates the tertiary from the other two - to split primary from secondary you must identify the product (an aldehyde gives a silver mirror with Tollens', a ketone does not).

    日本語

    アルコールには$\text{–OH}$(ヒドロキシ基)官能基が存在する。

    ハンドサンタイザーのボトル
    ハンドサンタイザーに含まれるアルコールのエタノールは微生物を殺菌する

    アルコールの合成

    方法 試薬および条件
    アルケンへの蒸気の付加 $\text{H}_2\text{O(g)}$、$\text{H}_3\text{PO}_4$触媒(求電付加反応)
    冷たい希$\text{KMnO}_4$とのアルケン ジオール(2つの$\text{–OH}$基)を与える
    ハロゲン化アルカンの置換 $\text{NaOH(aq)}$、加熱
    アルデヒドまたはケトンの還元 $\text{NaBH}_4$または$\text{LiAlH}_4$
    カルボン酸の還元 $\text{LiAlH}_4$
    エステルの加水分解 希酸または塩基、加熱
    箔で栓をしたコップに濁った液体が入っており、実験室で発酵している様子
    産業的にエタノールは糖の発酵によって製造され、分留によって水から分離される

    アルコールの反応

    • 燃焼: アルコールは酸素中で燃焼して二酸化炭素と水を生じる。
    • ハロゲン化アルカンへの置換: 例えば$\text{HX}$、$\text{PCl}_5$、$\text{PCl}_3$および加熱、あるいは$\text{SOCl}_2$を用いる場合。
    • ナトリウムとの反応: アルコールは金属ナトリウムと反応して水素とナトリウムアルコキシドを生じ、水と同様だがよりゆっくりとする。
    • 酸性$\text{K}_2\text{Cr}_2\text{O}_7$(または$\text{KMnO}_4$)による酸化: 生成物はアルコールの種類(以下参照)に依存する。
    • アルケンへの脱水: 熱い$\text{Al}_2\text{O}_3$触媒または濃酸を用いる。
    • エステル形成: アルコールは濃$\text{H}_2\text{SO}_4$触媒存在下でカルボン酸と反応してエステルを作る。
    アルコールが四つの異なる反応(燃焼、アルケンへの脱水、エステル化、酸化)を行う様子
    アルコールは燃焼、脱水、エステル化、酸化できる

    三つのクラスと酸化による識別法

    アルコールは、$\text{–OH}$を持つ炭素に結合している炭素の数によって第一級、第二級、第三級に分類される。ある分子には複数の$\text{–OH}$基が含まれることもある。

    中央の炭素にそれぞれOH基を持つ三つの構造: 第一級は1つのR基と2つの水素、第二級は2つのR基と1つの水素、第三級は3つのR基
    クラスは-OHを持つ炭素によって決まる: R基が1つなら第一級、2つなら第二級、3つなら第三級であり、これが酸化挙動を決定する
    クラス 酸化産物
    第一級 アルデヒド(分留により)、次にカルボン酸(還流により)
    第二級 ケトン
    第三級 酸化されない
    図式: 第一級アルコールがアルデヒドを経てカルボン酸へ、第二級アルコールがケトンへ、第三級アルコールは酸化されない様子
    クラス別の酸化: 第一級アルコールはアルデヒドを経てカルボン酸となり、第二級はケトンを生成し、第三級は酸化されない

    簡易テストにおいて、酸性$\text{K}_2\text{Cr}_2\text{O}_7$は第一級または第二級アルコールと反応すると橙色から緑色に変化するが、第三級アルコールでは橙色のままとなる。

    二つの試験管に入った橙色のジクロム酸: 第一級または第二級アルコールを含むものは緑色になり、第三級アルコールを含むものはまだ橙色のまま
    酸性ジクロム酸は第一級または第二級アルコールと反応すると橙色から緑色になるが、第三級アルコールでは橙色のままとなる
    • 分留は、アルデヒドがさらに酸化される前に生成されたまま除去する。
    • 還流は混合物を沸騰させ続け、蒸気を還流させることで、アルコールをカルボン酸まで完全に酸化させる。
    二組の実験装置: 垂直凝縮管で蒸気をフラスコに戻す還流装置、および傾斜凝縮管で回収フラスコへ導く分留装置
    分留は生成されたアルデヒドを取り除き、還流は蒸気を還流させながら沸騰を続け、酸まで完全に酸化させる
    温度計と水冷却器付きの実験室用分留カラム
    実際の分留セットアップ: 蒸気が沸き上がり、凝縮管で冷却され、回収される。この原理は発酵液からのエタノール分離にも用いられる

    イオドーホルム試験

    $\text{CH}_3\text{CH(OH)}-$基を含むアルコールをアルカリ性水溶液のヨウ素とともに温めると、淡黄色の沈殿物であるヨードホルム($\text{CHI}_3$)とイオン$\text{RCO}_2^-$が得られる。これはその基に対する有用な試験である

    アルコールの酸性

    $\text{–OH}$ 基はアルコールを非常に弱い酸とします:それらは $\text{H}^+$ を失って $\text{RO}^-$ イオンを形成できます。しかし、それらの酸性は水より低いです。これはアルキル基が電子密度を酸素に押し寄せるため、$\text{RO}^-$ イオンが不安定になり、アルコールがその $\text{H}^+$ をより強く保持するためです。

    ** worked example.** ラベルのない3つのボトルには、ブタン-1-オール、ブタン-2-オール、および2-メチルプロパン-2-オールが入っています。これらに酸性の $\text{K}_2\text{Cr}_2\text{O}_7$ を加えて温めることで、どのように区別できますか?重要なのは、$\text{OH}$ を持つ炭素に結合している水素原子の数です。ブタン-1-オールは第一級(この水素が2つあります)です:オレンジ色のジクロマートは緑色に変わり、蒸留によってアルデヒド(ブタナル)を回収するか、還流によってカルボン酸(ブタン酸)を得ます。ブタン-2-オールは第二級(この水素が1つあります):同様に緑色になりますが、生成物はケトン(ブタノン)であり、さらに酸化されません。2-メチルプロパン-2-オールは第三級(この水素はnone)です:取り除くものがありませんので、ジクロマートはオレンジ色のままです。色の変化で第三級のみを他の2つと区別できますが、第一級と第二級を分けるには生成物を同定する必要があります(アルデヒドはテロアス试剂で銀鏡反応を示しますが、ケトンは示しません)。

    Explore · ⁨探索⁩

    Alcohol reaction map · ⁨アルコール反応マップ⁩

    Choose what happens to alcohols under different reagents. · ⁨異なる試薬条件下でのアルコールの変化を選択せよ。⁩

    Explore · ⁨探索⁩

    Alcohol oxidation test lab · ⁨アルコール酸化試験実験室⁩

    Classify alcohols by oxidation product and observation. · ⁨酸化生成物および観察結果に基づいてアルコールを分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    alcohol/ˈælkəhɒl/ アルコール
    electrophilic addition/ɪˌlektrəʊˈfɪlɪk əˈdɪʃn/ 求電付加
    alkene/ˈælkiːn/ アルケン
    diol/dɪˈɒl/ ジオール
    halogenoalkane/ˈheɪləʊdʒnəʊlkeɪn/ ハロゲン化アルキル
    reduction/rɪˈdʌkʃn/ 還元
    aldehyde/ˈældɪhaɪd/ アルデヒドが得られる
    ketone/ˈketəʊn/ ケトン
    carboxylic acid/ˌkɑːbəkˈsɪlɪk ˈæsɪd/ カルボン酸
    hydrolysis/haɪˈdrɒləsɪs/ 加水分解反応
    ester/ˈestə/ エステル
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    dehydration/ˌdiːhaɪˈdreɪʃn/ 脱水
    primary/ˈpraɪməri/ 第一級
    secondary/ˈsekəndəri/ 第二級
    tertiary/ˈtɜːʃjəri/ 第三級
    distillation/dɪstɪˈleɪʃn/ 蒸留
    reflux/ˈriːflʌks/ 還流
    tri-iodomethane/traɪ ˈaɪədəʊmθeɪn/ トリヨードメタン
    acidity/æˈsɪdɪti/ 酸性度
    16.1

    Exam tips · ⁨試験対策⁩

    English
    • Classify the alcohol as primary, secondary or tertiary first — it decides the oxidation product.
    • Oxidation with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (orange → green): primary → aldehyde (distil) → acid (reflux); secondary → ketone; tertiary → no reaction.
    • Distinguish the conditions for aldehyde vs acid from a primary alcohol (distillation vs reflux).
    • Name esters correctly (the acid part comes second, ending "-oate").
    日本語
    • アルコールをまず第一級、第二級、または第三級に分類してください——これが酸化生成物を決定します。
    • 酸性の $\text{K}_2\text{Cr}_2\text{O}_7$ (オレンジ色 → 緑色)による酸化:第一級 → アルデヒド(蒸留)→ 酸(還流);第二級 → ケトン;第三級 → 反応なし。
    • 第一級アルコールからのアルデヒド vs 酸の条件を区別してください(蒸留 vs 還流)。
    • エーテルを正しく命名してください(酸の部分の方が後に来て「-oate」で終わります)。
  • 17

    Carbonyl compounds · ⁨カルボニル化合物⁩

    Watch lesson · ⁨レッスンを視聴⁩
    17.1

    Aldehydes and ketones · ⁨アルデヒドとケトン⁩

    Syllabus · ⁨シラバス⁩
    1. recall the reactions (reagents and conditions) by which aldehydes and ketones can be produced: (a) the oxidation of primary alcohols using acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or acidified $\text{KMnO}_4$ and distillation to produce aldehydes (b) the oxidation of secondary alcohols using acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or acidified $\text{KMnO}_4$ and distillation to produce ketones
    2. describe: (a) the reduction of aldehydes and ketones using $\text{NaBH}_4$ or $\text{LiAlH}_4$ to produce alcohols (b) the reaction of aldehydes and ketones with $\text{HCN}$, $\text{KCN}$ as catalyst, and heat to produce hydroxynitriles as exemplified by ethanal and propanone
    3. describe the mechanism of the nucleophilic addition reactions of hydrogen cyanide with aldehydes and ketones in 17.1.2(b)
    4. describe the use of 2,4-dinitrophenylhydrazine (2,4-DNPH reagent) to detect the presence of carbonyl compounds
    5. deduce the nature (aldehyde or ketone) of an unknown carbonyl compound from the results of simple tests (Fehling's and Tollens' reagents; ease of oxidation)
    6. deduce the presence of a $\text{CH}_3\text{CO}-$ group in an aldehyde or ketone, $\text{CH}_3\text{CO}-\text{R}$, from its reaction with alkaline $\text{I}_2(\text{aq})$ to form a yellow precipitate of tri-iodomethane and an ion, $\text{RCO}_2^-$

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Aldehydes and ketones are carbonyl compounds 羰基化合物 — they contain the C=O carbonyl 羰基 group.

    • in an aldehyde 醛 the carbonyl carbon is on the end of the chain (it also carries an H), written $\text{–CHO}$.
    • in a ketone 酮 the carbonyl carbon is in the middle, between two other carbons.

    Making aldehydes and ketones

    Both are made by the oxidation 氧化 of an alcohol 醇 with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or $\text{KMnO}_4$:

    • a primary alcohol, with distillation 蒸馏, gives an aldehyde.
    • a secondary alcohol gives a ketone.

    Reactions

    • reduction 还原 with $\text{NaBH}_4$ or $\text{LiAlH}_4$ turns a carbonyl compound back into an alcohol (an aldehyde gives a primary alcohol; a ketone gives a secondary alcohol).
    • reaction with hydrogen cyanide 氰化氢 ($\text{HCN}$), with $\text{KCN}$ as catalyst and heat, adds $\text{H}$ and $\text{CN}$ across the C=O to make a hydroxynitrile 羟基腈. This adds one carbon to the chain.

    The mechanism: nucleophilic addition

    The reaction with $\text{HCN}$ is a nucleophilic addition 亲核加成. The carbonyl carbon is slightly positive (oxygen pulls the electrons away). So:

    1. the $\text{CN}^-$ ion (a nucleophile) attacks the slightly positive carbon.
    2. this breaks the C=O double bond, leaving a negative oxygen ($\text{O}^-$).
    3. the $\text{O}^-$ takes an $\text{H}^+$ (from $\text{HCN}$) to finish the hydroxynitrile.
    日本語

    アルデヒドとケトンはカルボニル化合物です——それらは C=O カルボニル基を含んでいます。

    • アルデヒドでは、カルボニル炭素が鎖の端にあります(Hも持っています)、$\text{–CHO}$ と書きます。
    • ケトンでは、カルボニル炭素が真ん中にあり、他の2つの炭素の間にあります。
    爪切り液のボトルと綿パッド
    プロパノン(アセトン)、最も単純なケトンは、爪切り液の溶媒です
    C=O炭素が端にHを持つエタノールと、C=O炭素が2つの炭素の間にいるプロパノン
    両方とも C=O カルボニル基を持っています:アルデヒドでは鎖の端にあり(–CHO)、ケトンでは真ん中にあります

    アルデヒドとケトンの合成

    どちらも、酸性の $\text{K}_2\text{Cr}_2\text{O}_7$ または $\text{KMnO}_4$ によるアルコールの酸化によって作られます:

    • 第一級アルコールは、蒸留によりアルデヒドを与えます。
    • 第二級アルコールはケトンを生成します。

    反応

    • $\text{NaBH}_4$ または $\text{LiAlH}_4$ による還元は、カルボニル化合物をアルコールに戻します(アルデヒドは第一級アルコールを、ケトンは第二級アルコールを与えます)。
    • シアン化水素($\text{HCN}$)との反応で、$\text{KCN}$ を触媒として加熱すると、C=O に $\text{H}$ と $\text{CN}$ が付加してヒドロキシニトリルになります。これにより鎖に炭素が1つ増えます。

    反応機構:求核付加

    $\text{HCN}$ との反応は求核付加です。カルボニル炭素はわずかに正電荷を帯びています(酸素が電子を引き寄せます)。したがって:

    1. $\text{CN}^-$ イオン(求核剤)がわずかに正電荷の炭素を攻撃します。
    2. これにより C=O 二重結合が切断され、負の酸素($\text{O}^-$)が残ります。
    3. $\text{O}^-$ は $\text{H}^+$ ($\text{HCN}$ から)を取り込んでヒドロキシニトリルを完成させます。
    HCN付加の反応機構:シアニドがわずかに正電荷のカルボニル炭素を攻撃し、C=O が切断されて負の酸素となり、その後酸素が陽イオンを取る
    HCNの求核付加:CN$^-$ が $\delta+$ カルボニル炭素を攻撃し、C=O が O$^-$ に分解し、その後 O$^-$ が H$^+$ を取り込む
    Explore · ⁨探索⁩

    Carbonyl compound lab · ⁨カルボニル化合物実験室⁩

    Sort carbonyl reactions by what they reveal. · ⁨何を示すかによってカルボニル反応を分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    aldehyde/ˈældɪhaɪd/ アルデヒドが得られる
    ketone/ˈketəʊn/ ケトン
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    alcohol/ˈælkəhɒl/ アルコール
    distillation/dɪstɪˈleɪʃn/ 蒸留
    reduction/rɪˈdʌkʃn/ 還元
    hydrogen cyanide/ˈhaɪdrədʒn ˈsaɪənaɪd/ シアン水素酸
    hydroxynitrile/haɪˈdrɒksɪnaɪtraɪl/ ヒドロキシニトリル
    nucleophilic addition/ˌnjuːklɪəˈfɪlɪk əˈdɪʃn/ 求核付加
    Fehling's reagent/ˈfeɪlɪŋz rɪˈeɪdʒənt/ フェーリング試薬
    Tollens' reagent/ˈtəʊlənz rɪˈeɪdʒənt/ ト伦ス試薬
    tri-iodomethane/traɪ ˈaɪədəʊmθeɪn/ トリヨードメタン
    17.1

    Tests for carbonyl compounds · ⁨カルボニル化合物のテスト⁩

    English

    Detecting any carbonyl

    Add 2,4-DNPH reagent (2,4-dinitrophenylhydrazine). An orange precipitate confirms that the compound is an aldehyde or a ketone.

    Telling an aldehyde from a ketone

    Aldehydes are easily oxidised to carboxylic acids, but ketones are not. Two tests use this difference:

    Test Aldehyde Ketone
    Fehling's reagent 斐林试剂 (blue solution) turns to a brick-red precipitate no change
    Tollens' reagent 托伦试剂 (colourless) gives a silver mirror no change

    The iodoform test

    If the compound has the $\text{CH}_3\text{CO}-$ group, warming it with alkaline aqueous iodine gives a pale yellow precipitate of tri-iodomethane 三碘甲烷 ($\text{CHI}_3$) and the ion $\text{RCO}_2^-$.

    Worked example. A liquid gives an orange precipitate with 2,4-DNPH, gives no silver mirror with Tollens' reagent, and gives a yellow precipitate with alkaline aqueous iodine. Identify it. Take the tests one at a time, using each for exactly what it proves. The orange precipitate with 2,4-DNPH proves a carbonyl group is present - an aldehyde or a ketone, nothing else. No silver mirror with Tollens' rules out an aldehyde, so it must be a ketone. The yellow precipitate in the iodoform test proves a $\text{CH}_3\text{CO}-$ group next to the carbonyl. The simplest compound satisfying all three is propanone, $\text{CH}_3\text{COCH}_3$. Keep the roles straight: 2,4-DNPH finds any carbonyl, Tollens' separates aldehyde from ketone, and iodoform detects the methyl group beside the C=O.

    日本語

    任意のカルボニルの検出

    2,4-DNPH試薬(2,4-ジニトロフェニルヒドラジン)を加えます。オレンジ色の沈殿は、その化合物がアルデヒドまたはケトンであることを確認します。

    アルデヒドとケトンの識別

    アルデヒドはカルボン酸へ容易に酸化されますが、ケトンはそうではありません。この違いを用いた2つのテストがあります:

    試験 アルデヒド ケトン
    フェリング試薬(青色溶液) レンガ赤色の沈殿になる 変化なし
    テロアス試薬(無色) 銀鏡を作る 変化なし
    2つの試験管:アルデヒドに対してレンガ赤色の沈殿を作るフェリング試液と、銀鏡を作るテロアス試薬
    アルデヒドとケトンの識別:アルデヒドはフェリングでレンガ赤色の沈殿を、テロアスで銀鏡を作ります;ケトンは変化なし
    内壁が明るく光る銀層でコーティングされた試験管、鏡のように見える
    テロアステストの陽性反応:アルデヒドが試験管を光る銀鏡でコーティングする

    イオドーホルム試験

    化合物が $\text{CH}_3\text{CO}-$ 基を持っている場合、塩基性の水溶性ヨウ素と温めると、淡黄色のヨードホルム($\text{CHI}_3$)沈殿とイオン $\text{RCO}_2^-$ が生成します。

    ** worked example.** 液体が2,4-DNPHと反応してオレンジ色の沈殿を生じ、Tollens試薬では銀鏡を生じず、アルカリ性ヨウ素水溶液で黄色の沈殿を生じた。これを同定せよ。各試験を順に行い、それぞれの試験が証明する内容进行きonlyに用いること。2,4-DNPHとの橙色沈殿はカルボニル基の存在を示す——アルデヒドまたはケトン、それ以外ではない。Tollens試薬による銀鏡の生成なしはアルデヒドを除外するため、ケトンであることがわかる。ヨウ素ホルム試験における黄色沈殿は、カルボニルの隣に$\text{CH}_3\text{CO}-$基があることを示す。これら3つの条件を満たす最も単純な化合物はプロパノンであり、$\text{CH}_3\text{COCH}_3$である。役割を正しく理解しておく:2,4-DNPHはあらゆるカルボニルを検出し、Tollensはアルデヒドとケトンを区別し、ヨウ素ホルムはC=Oの隣のメチル基を検出する。

    Explore · ⁨探索⁩

    Carbonyl test lab · ⁨カルボニル試験実験室⁩

    Match observations to aldehydes and ketones. · ⁨観測結果をアルデヒドとケトンに対応させなさい。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    carbonyl compound/ˈkɑːbənaɪl ˈkɒmpaʊnd/ カルボニル化合物
    carbonyl/ˈkɑːbənaɪl/ カルボニル
    17.1

    Exam tips · ⁨試験対策⁩

    English
    • 2,4-DNPH gives an orange precipitate with any carbonyl — it tests for $\text{C}=\text{O}$.
    • Tollens' (silver mirror) and Fehling's (brick-red) are positive for aldehydes only — this is how you tell aldehydes from ketones.
    • Nucleophilic addition of HCN adds one carbon (→ hydroxynitrile); show the curly arrows.
    • The iodoform (tri-iodomethane) test is positive for $\text{CH}_3\text{CO}-$ groups.
    日本語
    • 2,4-DNPHはあらゆるカルボニルと橙色沈殿を生じる——$\text{C}=\text{O}$を検出するための試験である。
    • Tollens(銀鏡)およびFehling(レンガ赤)はアルデヒドのみで陽性となる——これがアルデヒドとケトンを区別する方法である。
    • HCNの求核付加により炭素が1つ増える(→ハイドロキシアシルトリン)、その際カール矢印を示すこと。
    • **ヨウ素ホルム(トリヨードメタン)**試験は$\text{CH}_3\text{CO}-$基に対して陽性となる。
  • 18

    Carboxylic acids and derivatives · ⁨カルボン酸及其び誘導体⁩

    Watch lesson · ⁨レッスンを視聴⁩
    18.1

    Carboxylic acids · ⁨カルボン酸⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions by which carboxylic acids can be produced: (a) oxidation of primary alcohols and aldehydes with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or acidified $\text{KMnO}_4$ and refluxing (b) hydrolysis of nitriles with dilute acid or dilute alkali followed by acidification (c) hydrolysis of esters with dilute acid or dilute alkali and heat followed by acidification
    2. describe: (a) the redox reaction with reactive metals to produce a salt and $\text{H}_2(\text{g})$ (b) the neutralisation reaction with alkalis to produce a salt and $\text{H}_2\text{O}(\text{l})$ (c) the acid–base reaction with carbonates to produce a salt and $\text{H}_2\text{O}(\text{l})$ and $\text{CO}_2(\text{g})$ (d) esterification with alcohols with concentrated $\text{H}_2\text{SO}_4$ as catalyst (e) reduction by $\text{LiAlH}_4$ to form a primary alcohol
    日本語
    1. カルボン酸の生成反応を思い出す:(a) 第一級アルコールおよびアルデヒドの酸性$\text{K}_2\text{Cr}_2\text{O}_7$ または酸性$\text{KMnO}_4$ による酸化と還流 (b) シアノ化合物の希酸または希塩基による加水分解後、酸化处理 (c) エステルの希酸または希塩基による加水分解と加熱後、酸化处理
    2. 以下を説明する:(a) 活性金属との酸化還元反応により塩と$\text{H}_2(\text{g})$ を生成する (b) 塩基との中和反応により塩と$\text{H}_2\text{O}(\text{l})$ を生成する (c) 炭酸塩との酸塩基反応により塩と$\text{H}_2\text{O}(\text{l})$ および$\text{CO}_2(\text{g})$ を生成する (d) エタノールなどのアルコールとのエステル化反応で、濃$\text{H}_2\text{SO}_4$ を触媒とする (e) $\text{LiAlH}_4$ による還元で第一級アルコールを生成する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A carboxylic acid 羧酸 has the $\text{–COOH}$ (carboxyl) functional group. It is a weak acid.

    Making carboxylic acids

    • oxidation 氧化 of a primary alcohol 醇 or an aldehyde 醛 with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or $\text{KMnO}_4$, with reflux 回流 (so it is fully oxidised).
    • hydrolysis 水解 of a nitrile 腈 with dilute acid or alkali, then acidifying.
    • hydrolysis of an ester 酯 with dilute acid or alkali and heat, then acidifying.

    Reactions of carboxylic acids

    These reactions all show that carboxylic acids are acids:

    • with a reactive metal (a redox 氧化还原 reaction): gives a salt 盐 and hydrogen.
    $$2\text{CH}_3\text{COOH} + \text{Mg} \rightarrow (\text{CH}_3\text{COO})_2\text{Mg} + \text{H}_2$$
    • with an alkali (a neutralisation 中和): gives a salt and water.
    $$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$
    • with a carbonate 碳酸盐: gives a salt, water and carbon dioxide. The fizzing of $\text{CO}_2$ is a test for a carboxylic acid.
    $$2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2$$

    Two reactions change the functional group:

    • esterification 酯化 with an alcohol (concentrated $\text{H}_2\text{SO}_4$ catalyst) gives an ester.
    • reduction 还原 by $\text{LiAlH}_4$ gives a primary alcohol.
    日本語

    カルボン酸は$\text{–COOH}$(カルボキシル)官能基を持つ。弱酸である。

    棚に並ぶ食酢のボトル
    食酢はエタノール酸というカルボン酸の希薄溶液である

    カルボン酸の合成

    • 酸性化された$\text{K}_2\text{Cr}_2\text{O}_7$または$\text{KMnO}_4$を用いて、第一アルコールまたはアルデヒドを酸化する(還流によって完全に酸化される)。
    • ニトリルの希酸またはアルカリによる加水分解後、酸処理を行う。
    • エステルの希酸またはアルカリおよび加熱による加水分解後、酸処理を行う。

    カルボン酸の反応

    これらの反応はすべてカルボン酸が酸であることを示している:

    • 活性金属との反応(酸化還元反応):塩と水素を生じる。
    $$2\text{CH}_3\text{COOH} + \text{Mg} \rightarrow (\text{CH}_3\text{COO})_2\text{Mg} + \text{H}_2$$
    • アルカリとの反応(中和反応):塩と水を生じる。
    $$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$
    • 炭酸塩との反応:塩、水、二酸化炭素を生じる。$\text{CO}_2$の泡立ちがカルボン酸の試験となる。
    $$2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2$$
    カルボン酸が活性金属、アルカリ、炭酸塩と反応して指定された生成物を生じる枝分かれ図
    カルボン酸は酸として振る舞う:金属と反応して塩(+ $_2$H)、アルカリと反応して塩(+ 水)、または炭酸塩と反応して塩(+ 水 + $_2$CO₂ — 泡立ちは検出試験となる)

    2つの反応が官能基を変化させる:

    • アルコールとのエステル化(濃$\text{H}_2\text{SO}_4$触媒)によりエステルが生じる。
    • $\text{LiAlH}_4$による還元により第一アルコールが生じる。
    Explore · ⁨探索⁩

    Carboxylic acid reaction map · ⁨カルボン酸反応マップ⁩

    Follow carboxylic acids through neutralisation, esterification and reduction. · ⁨中和、エステル化、還元の経路をカルボン酸を通じて追跡せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    carboxylic acid/ˌkɑːbəkˈsɪlɪk ˈæsɪd/ カルボン酸
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    alcohol/ˈælkəhɒl/ アルコール
    aldehyde/ˈældɪhaɪd/ アルデヒドが得られる
    reflux/ˈriːflʌks/ 還流
    18.2

    Esters · ⁨エステル⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reaction (reagents and conditions) by which esters can be produced: (a) the condensation reaction between an alcohol and a carboxylic acid with concentrated $\text{H}_2\text{SO}_4$ as catalyst
    2. describe the hydrolysis of esters by dilute acid and by dilute alkali and heat
    日本語
    1. エステルの生成反応(試薬および条件)を思い出す:(a) アルコールとカルボン酸の濃$\text{H}_2\text{SO}_4$ を触媒とする縮合反応
    2. 加水分解を希酸および希アルカリと加熱を用いて記述する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    An ester has the $\text{–COO–}$ group. It often smells sweet or fruity.

    Making esters

    An ester forms in a condensation 缩合 reaction between an alcohol and a carboxylic acid, with concentrated $\text{H}_2\text{SO}_4$ as catalyst. A water molecule is lost, and the reaction is reversible:

    $$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

    Hydrolysis of esters

    Hydrolysis splits the ester back apart. The conditions change the products:

    • dilute acid and heat: reversible. Gives back the carboxylic acid and the alcohol.
    • dilute alkali and heat: not reversible. Gives the alcohol and the salt of the carboxylic acid (the carboxylate ion).

    Worked example. Name the ester made from ethanol and propanoic acid, and say which part comes from which. In an ester name the alcohol gives the first word (the alkyl part) and the acid gives the second (the -oate part). Ethanol supplies $\text{C}_2\text{H}_5-$ and propanoic acid supplies $\text{CH}_3\text{CH}_2\text{COO}-$, so the ester is ethyl propanoate, $\text{CH}_3\text{CH}_2\text{COOC}_2\text{H}_5$. It is made by refluxing the two with a concentrated $\text{H}_2\text{SO}_4$ catalyst, and the reaction is reversible, so the yield is never complete. The trap is naming it backwards: propyl ethanoate is a completely different ester (made from propan-1-ol and ethanoic acid). Alcohol first, acid second.

    日本語

    エステルは$\text{–COO–}$基を持つ。甘いまたは果実のような匂いがすることが多い。

    装飾的なカットグラスの香水ボトル
    エステルは多くの果実や香りの甘い匂いの原因となる

    エステルの合成

    エステルはアルコールとカルボン酸との縮合反応によって形成され、濃$\text{H}_2\text{SO}_4$を触媒とする。水分子が脱離し、反応は可逆的である:

    $$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$
    エステル化反応式:酸由来のOHとアルコール由来のHが水として脱離する原子として強調され、新たに形成されるエステル酸素結合が強調されている
    エステル化は縮合反応である:酸由来の–OHとアルコール由来の–Hが水として脱離し、C–O–Cエステル結合を形成する

    エステルの加水分解

    加水分解によりエステルが再び分解される。条件によって生成物が異なる:

    • 希酸および加熱:可逆的。カルボン酸とアルコールが再生される。
    • 希アルカリおよび加熱:不可逆的。アルコールとカルボン酸の塩(カルボキシレートイオン)が生じる。
    枝分かれ図:エステルが希酸によって加水分解されるとカルボン酸とアルコールが、希アルカリによって加水分解されるとカルボキシレート塩とアルコールが生じる
    エステルの加水分解:希酸(可逆)はカルボン酸とアルコールを、希アルカリ(不可逆)はカルボキシレート塩とアルコールを生じる

    ** worked example.** エタノールとプロパノ酸から作られるエステル名を述べ、それぞれの由来となる部分を示せ。エステル名の付け方として、アルコールが最初の単語(アルキル基部分)、酸が2番目の単語(-oate部分)となる。エタノールは$\text{C}_2\text{H}_5-$を、プロパノ酸は$\text{CH}_3\text{CH}_2\text{COO}-$を提供するため、エステル名はプロピルプロパノエート、$\text{CH}_3\text{CH}_2\text{COOC}_2\text{H}_5$となる。これは濃$\text{H}_2\text{SO}_4$触媒を用いて還流して合成され、反応は可逆であるため収率は完全ではない。よくある間違いは逆顺で命名することである:エチルプロピルエートは全く異なるエステル(プロパン-1-オールと酢酸から作る)である。アルコールを先、酸を後にする。

    Explore · ⁨探索⁩

    Ester route lab · ⁨エステル合成実験室⁩

    Follow ester formation and hydrolysis as reversible paths. · ⁨可逆反応としてのエステル形成および加水分解を追跡せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    hydrolysis/haɪˈdrɒləsɪs/ 加水分解反応
    nitrile/ˈnaɪtraɪl/ ニトリル
    ester/ˈestə/ エステル
    redox/rɪˈdɒks/ 酸化還元
    salt/sɒlt/ 塩
    neutralisation/ˌnjuːtrəlaɪˈzeɪʃn/ 中和反応
    carbonate/ˈkɑːbəneɪt/ 炭酸塩
    esterification/ˌestəˌrɪfɪˈkeɪʃn/ エステル化
    reduction/rɪˈdʌkʃn/ 還元
    condensation/kɒndenˈseɪʃn/ 縮合
    18.2

    Exam tips · ⁨試験対策⁩

    English
    • Carboxylic acids react with carbonates to give $\text{CO}_2$ (fizzing) — this distinguishes them from phenols.
    • Explain acid strength via the delocalised carboxylate ion; electron-withdrawing groups (e.g. Cl) increase strength.
    • Esterification (with concentrated $\text{H}_2\text{SO}_4$) is reversible; name esters as alkyl alkanoate.
    • Know the products of acid vs alkaline hydrolysis of an ester (they differ).
    日本語
    • カルボン酸は炭酸塩と反応して**$\text{CO}_2$**(泡立ち)を生じる——これがフェノールとの区別方法となる。
    • 非局在化したカルボキシレートイオンにより酸の強さを説明する。電子吸引基(例:Cl)则有强酸性。
    • エステル化(濃$\text{H}_2\text{SO}_4$を用いる)は可逆的であり、エステル名はアルキルアルカノエートと称する。
    • エステルの酸性加水分解とアルカリ性加水分解の生成物を知っておく(両者は異なる)。
  • 19

    Nitrogen compounds · ⁨窒素化合物⁩

    Watch lesson · ⁨レッスンを視聴⁩
    19.1

    Primary amines · ⁨第一アミン⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions by which amines can be produced: (a) reaction of a halogenoalkane with $\text{NH}_3$ in ethanol heated under pressure Classification of amines will not be tested at AS Level.
    日本語
    1. アミンが生成される反応を思い出す: (a) 塩素化アルキルの$\text{NH}_3$とのエタノール中での反応、加圧下で加熱 ASレベルではアミンの分類は試験されない。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    An amine 胺 has an $\text{–NH}_2$ group (the nitrogen replaces a hydrogen of ammonia).

    Making a primary amine

    Heat a halogenoalkane 卤代烷 with ammonia dissolved in ethanol, under pressure:

    $$\text{C}_2\text{H}_5\text{Br} + \text{NH}_3 \rightarrow \text{C}_2\text{H}_5\text{NH}_2 + \text{HBr}$$

    This is a nucleophilic substitution 亲核取代: the lone pair on the nitrogen of ammonia attacks the slightly positive carbon and pushes out the halogen. You use an excess of ammonia, or the amine made can react again.

    日本語
    木製レールに干してある、赤、青、黄、クリーム色に染められた羊毛の紐
    染色:アゾおよびアミン化学こそが、繊維に速乾性で鮮やかな色を与えるものである

    アミンは$\text{–NH}_2$基を持つ(窒素がアンモニアの水素を置換したもの)。

    第一アミンの合成

    圧力下で、エタノールに溶解させたアンモニアとハロゲンアルカンを加熱する:

    $$\text{C}_2\text{H}_5\text{Br} + \text{NH}_3 \rightarrow \text{C}_2\text{H}_5\text{NH}_2 + \text{HBr}$$

    これは求核置換反応です。アンモニアの窒素原子に存在する非共有電子対が、わずかに正電荷を帯びた炭素原子に攻撃し、ハロゲン原子を押し出します。過剰量のアンモニアを使用するか、生成したアミンが再度反応することを防ぎます。

    ハロゲン化アルキルとアンモニアの反応:窒素の非共有電子対がわずかに正電荷を帯びた炭素に攻撃し、臭素が脱離して一次アミンとHBrが生成する
    一次アミンの合成:アンモニアの非共有電子対がδ+炭素に攻撃し、ハロゲンを押し出す——これが求核置換反応
    Explore · ⁨探索⁩

    Amine reaction lab · ⁨アミン反応実験室⁩

    Classify amine examples by basicity and nucleophilic behaviour. · ⁨アミンの例を、塩基性と求核性に基づいて分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    amine/ˈæmaɪn/ アミン
    halogenoalkane/ˈheɪləʊdʒnəʊlkeɪn/ ハロゲン化アルキル
    nucleophilic substitution/ˌnjuːklɪəˈfɪlɪk ˌsʌbstɪˈtjuːʃn/ 求核置換
    19.2

    Nitriles and hydroxynitriles · ⁨ニトリル類とヒドロキシニトリル類⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions by which nitriles can be produced: (a) reaction of a halogenoalkane with $\text{KCN}$ in ethanol and heat
    2. recall the reactions by which hydroxynitriles can be produced: (a) the reaction of aldehydes and ketones with $\text{HCN}$, $\text{KCN}$ as catalyst, and heat
    3. describe the hydrolysis of nitriles with dilute acid or dilute alkali followed by acidification to produce a carboxylic acid
    日本語
    1. ニトリルが生成される反応を思い出す: (a) 塩素化アルキルの$\text{KCN}$とのエタノール中での反応、加熱
    2. ヒドロキシニトリルが生成される反応を思い出す: (a) アルデヒドおよびケトンと$\text{HCN}$、触媒としての$\text{KCN}$、および加熱による反応
    3. ニトリルの希酸または希アルカリによる加水分解、および酸化处理してカルボン酸を得るプロセスを記述する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Making a nitrile

    Heat a halogenoalkane with potassium cyanide ($\text{KCN}$) in ethanol:

    $$\text{C}_2\text{H}_5\text{Br} + \text{KCN} \rightarrow \text{C}_2\text{H}_5\text{CN} + \text{KBr}$$

    This is also a nucleophilic substitution, with the $\text{CN}^-$ ion as the nucleophile. It is useful because it adds one carbon to the chain. The product is a nitrile 腈.

    Making a hydroxynitrile

    Add $\text{HCN}$ (with $\text{KCN}$ as catalyst, and heat) to an aldehyde 醛 or ketone 酮. The $\text{H}$ and $\text{CN}$ add across the C=O bond to give a hydroxynitrile 羟基腈. The reagent is hydrogen cyanide 氰化氢, and the mechanism is nucleophilic addition 亲核加成.

    Hydrolysis of nitriles

    Warm a nitrile with dilute acid (or dilute alkali, then acidify). This hydrolysis 水解 turns the $\text{–CN}$ group into a $\text{–COOH}$ group, giving a carboxylic acid 羧酸:

    $$\text{CH}_3\text{CN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow \text{CH}_3\text{COOH} + \text{NH}_4\text{Cl}$$

    A nitrile can also be reduced by hydrogen and a catalyst to form an amine, which is the reduction 还原 route to a longer-chain amine.

    Worked example. Starting from bromoethane, make propanoic acid. Compare the carbons first: bromoethane has 2, propanoic acid has 3, so a carbon must be added - and the $\text{KCN}$ step is the reaction that does it. Step 1: warm bromoethane with ethanolic $\text{KCN}$; nucleophilic substitution gives propanenitrile, $\text{CH}_3\text{CH}_2\text{CN}$, which now has 3 carbons because the $\text{CN}$ carbon joins the chain. Step 2: reflux the nitrile with dilute $\text{HCl}$; hydrolysis gives propanoic acid. Count the carbons before you plan: whenever the target has exactly one more than the starting material, the nitrile route is almost always the intended answer, and remember the $\text{CN}$ carbon is part of the new chain.

    日本語

    ニトリルの合成

    ハロゲン化アルキルにエタノール中シアン化カリウム($\text{KCN}$)を加えて加熱する:

    $$\text{C}_2\text{H}_5\text{Br} + \text{KCN} \rightarrow \text{C}_2\text{H}_5\text{CN} + \text{KBr}$$

    これも求核置換反応であり、$\text{CN}^-$イオンが求核剤として働きます。この反応は有用であるのは、炭素鎖に炭素を1つ追加できるからです。生成物はニトリルとなります。

    ヒドロキシニトリルの合成

    アルデヒドまたはケトンに$\text{HCN}$(触媒として$\text{KCN}$を用い、加熱下で)を加える。$\text{H}$と$\text{CN}$がC=O結合に付加してヒドロキシニトリルが生成する。試薬はシアン化水素であり、反応メカニズムは求核付加反応である。

    2つの反応:シアン化物イオンによるハロゲン化アルキルのニトリルへの置換反応と、シアン化水素によるカルボニル化合物への付加反応によるヒドロキシニトリルの生成
    シアン化物が炭素を追加する2つの方法:ハロゲン化アルキルとKCNによる置換でニトリルを、カルボニル化合物とHCNによる付加でヒドロキシニトリルを得る

    ニトリルの加水分解

    ニトリルに希酸(または希塩基 subsequently 酸処理)を加えて温める。この加水分解により$\text{–CN}$基が$\text{–COOH}$基に変化し、カルボン酸が生成する:

    $$\text{CH}_3\text{CN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow \text{CH}_3\text{COOH} + \text{NH}_4\text{Cl}$$

    ニトリルは水素と触媒によって還元され、アミンが生成することがあります。これは炭素数が多いアミンへ向かう還元法です。

    図式:ハロゲン化アルキルがKCNでニトリル(炭素1個追加)になり、その後ニトリルが加水分解されてカルボン酸になるか、あるいは還元されてアミンになる行程
    ニトリルは有用なハブとなる:KCNで炭素を追加してニトリルを作り、そこから加水分解でカルボン酸を得るか、還元でアミンを得る
    ナイロンストッキングのパッケージ
    ニトリルの還元でアミンが得られ、アミンとジカルボン酸が連結してナイロンなどのポリアミドが生成する——ストッキングで初めて有名になった繊維

    ** worked example. ** ブromoethaneからプロパノ酸を合成する。まず炭素数を比較する:bromoethaneは炭素2個、プロパノ酸は炭素3個なので、炭素を追加する必要がある——その$\text{KCN}$工程がそれに相当する。ステップ1:bromoethaneをエタノール中$\text{KCN}$と温める。求核置換反応によりプロパネニトリル$\text{CH}_3\text{CH}_2\text{CN}$が生成し、$\text{CN}$炭素が鎖に結合するため炭素数が3になる。ステップ2:ニトリルを希$\text{HCl}$中で還流する。加水分解によりプロパノ酸が生成する。計画を立てる前に炭素数を数えよ:標的物質が開始物質よりちょうど1つ多い場合、ニトリル経由法がほぼ必ず意図された解答であり、$\text{CN}$炭素が新しい鎖の一部であることを忘れるな。

    Explore · ⁨探索⁩

    Nitrile synthesis route · ⁨ニトリルの合成経路⁩

    Follow nitriles and hydroxynitriles as carbon-chain extension tools. · ⁨ニトリルおよびヒドロキシニトリルを炭素鎖伸長ツールとして学ぶ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    nitrile/ˈnaɪtraɪl/ ニトリル
    aldehyde/ˈældɪhaɪd/ アルデヒドが得られる
    ketone/ˈketəʊn/ ケトン
    hydroxynitrile/haɪˈdrɒksɪnaɪtraɪl/ ヒドロキシニトリル
    hydrogen cyanide/ˈhaɪdrədʒn ˈsaɪənaɪd/ シアン水素酸
    nucleophilic addition/ˌnjuːklɪəˈfɪlɪk əˈdɪʃn/ 求核付加
    hydrolysis/haɪˈdrɒləsɪs/ 加水分解反応
    carboxylic acid/ˌkɑːbəkˈsɪlɪk ˈæsɪd/ カルボン酸
    reduction/rɪˈdʌkʃn/ 還元
    19.2

    Exam tips · ⁨試験対策⁩

    English
    • Amines are bases (the N lone pair accepts $\text{H}^+$); aliphatic amines are stronger bases than ammonia.
    • Making amines: reduce a nitrile, or react a halogenoalkane with excess ammonia.
    • KCN adds one carbon (halogenoalkane → nitrile) — track the carbon count carefully in synthesis routes.
    • State reagents and conditions precisely for each conversion.
    日本語
    • アミンは塩基性である(Nの非共有電子対が$\text{H}^+$を受け取る);脂肪族アミンはアンモニアよりも塩基性が強い。
    • アミンの合成:ニトリルを還元するか、ハロゲン化アルキルと過剰量のアンモニアを反応させる。
    • KCNは炭素を1つ追加する(ハロゲン化アルキル → ニトリル)——合成経路における炭素数の追跡には注意が必要。
    • 各変換に対して試薬と条件を正確に記述する。
  • 20

    Polymerisation · ⁨重合反応⁩

    Watch lesson · ⁨レッスンを視聴⁩
    20.1

    Addition polymerisation · ⁨付加重合⁩

    Syllabus · ⁨シラバス⁩
    English
    1. describe addition polymerisation as exemplified by poly(ethene) and poly(chloroethene), PVC
    2. deduce the repeat unit of an addition polymer obtained from a given monomer
    3. identify the monomer(s) present in a given section of an addition polymer molecule
    4. recognise the difficulty of the disposal of poly(alkene)s, i.e. non-biodegradability and harmful combustion products
    日本語
    1. ポリ(エチレン)およびポリ(クロロエチレン)、PVCを例として付加重合を記述する
    2. 与えられたモノマーから得られる付加重合体の反復単位を導く
    3. 与えられた付加重合体分子の一部に含まれるモノマーを特定する
    4. ポリ(アルケン)、すなわち生分解性なしおよび有害な燃焼生成物のために処理が困難であることを認識する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    In addition polymerisation 加成聚合, many small molecules join into one very long chain, with no other product made.

    Each small molecule is a monomer 单体. It must be unsaturated — it has a C=C double bond. The double bond opens up so that the monomers can link together. The long chain that forms is the polymer 聚合物.

    Repeat units

    The repeat unit 重复单元 is the small part that is copied again and again along the chain. To find it, take the monomer, change the C=C to a single C–C, and draw bonds going out at each end.

    • poly(ethene) 聚乙烯 is made from ethene:
    $$n\,\text{CH}_2{=}\text{CH}_2 \rightarrow -(\text{CH}_2{-}\text{CH}_2)_n-$$
    • poly(chloroethene) 聚氯乙烯 (PVC) is made from chloroethene:
    $$n\,\text{CH}_2{=}\text{CHCl} \rightarrow -(\text{CH}_2{-}\text{CHCl})_n-$$

    Finding the monomer

    To go the other way, look at one repeat unit of the polymer, and put the C=C double bond back in. That gives you the monomer.

    Worked example. A polymer has the repeat unit $-(\text{CH}_2{-}\text{CH}(\text{CH}_3))_n-$. Identify the monomer and name the polymer. Reverse the rule you used to build it: rub out the bonds sticking out at each end, and turn the single C-C in the backbone back into a C=C. That gives $\text{CH}_2{=}\text{CH}(\text{CH}_3)$, which is propene - so the polymer is poly(propene). Two checks catch most errors: the monomer must have the same molecular formula as the repeat unit (addition polymerisation adds nothing and loses nothing), and the double bond goes back into the backbone, never into the side group.

    日本語

    付加重合では、多くの小さな分子が他の生成物を伴わずに一つの非常に長い鎖に結合する。

    各小さな分子をモノマーという。不飽和である必要があり、C=C二重結合を持つ。二重結合が開くことでモノマー同士が連結できる。形成される長い鎖がポリマーである。

    複数のエチレンモノマーがそれぞれC=C二重結合を持ち、二重結合を開いて端同士をつなぎ、単結合のみからなる長いポリマー鎖を形成する様子
    各モノマーの二重結合が開き、他の生成物なしで端同士がつながって長い鎖を形成する
    定規と共に置かれた透明なポリエチレンペレットの山(直径数ミリメートル程度)
    一般的な付加重合体であるポリエチレンは、直径数ミリメートルの小さなペレット状で供給され、これらを後に溶融・成形してボトルや袋などの製品とする

    繰り返し単位

    繰り返し単位とは、鎖に沿って反復される小さな部分である。見つけるには、モノマーを取り出し、C=Cを単一C–Cに変換し、両端から結合線を出して描く。

    • ポリエチレンはエチレンから作られる:
    $$n\,\text{CH}_2{=}\text{CH}_2 \rightarrow -(\text{CH}_2{-}\text{CH}_2)_n-$$
    • ポリ塩化ビニル(PVC)は塩化ビニルから作られる:
    $$n\,\text{CH}_2{=}\text{CHCl} \rightarrow -(\text{CH}_2{-}\text{CHCl})_n-$$
    左に二重結合を持つ塩化ビニル、右に括弧書きのPVCの繰り返し単位、C=Cを開いて戻すための矢印が描かれている
    繰り返し単位の探し方:モノマーのC=Cを単結合に変えて両端から結合を出し、逆にすることでモノマーを見つける

    モノマーの見つけ方

    逆に行うには、ポリマーの1つの繰り返し単位を見て、C=C二重結合を元に戻す。これがモノマーとなる。

    ** worked example. ** ポリマーの繰り返し単位が$-(\text{CH}_2{-}\text{CH}(\text{CH}_3))_n-$である。モノマーを同定し、ポリマーの名前を述べよ。構築時に用いたルールを逆にする:両端から出ている結合を消去し、骨格内の単一C-CをC=Cに戻す。これにより$\text{CH}_2{=}\text{CH}(\text{CH}_3)$が得られ、これはプロピレンである——したがってポリマーはポリプロピレンである。誤りを多く検出するための2つのチェック:モノマーは繰り返し単位と同じ分子式を持つべきである(付加重合では何も加えず何も失わない)、および二重結合は側鎖ではなく骨格に戻らなければならない。

    Explore · ⁨探索⁩

    Addition polymerisation route · ⁨付加重合反応の経路⁩

    Watch alkene monomers join by opening their double bonds. · ⁨アルケン単量体が二重結合を開いて結合する様子を見る。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    addition polymerisation/əˈdɪʃn ˌpɒlɪməraɪˈzeɪʃn/ 付加重合
    monomer/ˈmɒnəʊmə/ モノマー
    polymer/ˈpɒlɪmə/ ポリマー
    repeat unit/rɪˈpiːt ˈjuːnɪt/ 繰り返し単位
    poly(ethene)/ˈpɒlɪ/ ポリエチレン
    poly(chloroethene)/ˈpɒlɪ/ ポリ塩化ビニル
    non-biodegradable/nɒn ˌbaɪəʊdɪˈɡreɪdəbl/ 生分解不可能
    combustion/kəmˈbʌstʃn/ 燃焼反応
    20.1

    The problem of disposal · ⁨廃棄の問題⁩

    English

    Poly(alkene)s are very hard to get rid of:

    • they are non-biodegradable 不可生物降解 — microbes cannot break them down, so they stay in the ground for a very long time.
    • their combustion 燃烧 (burning) can release harmful gases. For example, burning PVC gives off toxic hydrogen chloride.
    日本語

    ポリ(アルケン)類は処分が非常に難しい:

    • それらは生分解不可能である——微生物がそれらを分解できないため、非常に長い間土中に残る。
    • それらの燃焼(燃やすこと)は有害なガスを放出することがある。例えば、PVCを燃やすと有毒の塩化水素が出される。
    ビーチに打ち上げられたプラスチック廃棄物
    ほとんどの付加重合ポリマーは生分解不可能であるため、環境中での廃棄物として蓄積する
    ポリアルケン廃棄物が引き起こす2つの問題を示す図:埋立地での生分解不可能な廃棄物の残留と、PVCの燃焼による有毒HClの排出
    ポリアルケン廃棄物は処理が困難である:生分解不可能であり、PVCを燃やすと有毒の塩化水素が出される
    20.1

    Exam tips · ⁨試験対策⁩

    English
    • Draw the repeat unit in brackets with the two bonds crossing them and $n$ outside; keep the backbone carbons.
    • Deduce the monomer from a polymer (and vice versa) — a very common question.
    • Addition polymers are inert and non-biodegradable — link to landfill, recycling and incineration issues.
    • Do not confuse with condensation (no small molecule is lost in addition polymerisation).
    日本語
    • 繰り返し単位を角括弧で囲み、2本の結合が括弧を横切るように描き、その外に $n$ を配置します。骨格炭素は保持してください。
    • ポリマーから単量体を導く(逆もまた然り)——非常に頻出する問題です。
    • 付加重合ポリマーは不活性かつ生分解不可能である——埋立地、 recycling、焼却との関連性について。
    • 縮合重合とは混同しないこと(付加重合では小分子は失われない)。
  • 21

    Organic synthesis · ⁨有機合成⁩

    Watch lesson · ⁨レッスンを視聴⁩
    21.1

    Organic synthesis · ⁨有機合成⁩

    Syllabus · ⁨シラバス⁩
    English
    1. for an organic molecule containing several functional groups: (a) identify organic functional groups using the reactions in the syllabus (b) predict properties and reactions
    2. devise multi-step synthetic routes for preparing organic molecules using the reactions in the syllabus
    3. analyse a given synthetic route in terms of type of reaction and reagents used for each step of it, and possible by-products
    日本語
    1. 複数の官能基を含む有機分子について: (a) 履修内容の反応を用いて有機官能基を特定 (b) 性質および反応を予測する
    2. 履修内容の反応を用いて有機分子を調製するための多段階合成ルートを考案する
    3. 与えられた合成ルートを反応の種類、各工程で使用される試薬、および副生成物について分析する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    This topic does not add new reactions. Instead it asks you to join up the reactions you already know, so you can build a target molecule in several steps.

    Identifying functional groups

    A molecule may have more than one functional group 官能团. Use the test reactions from the syllabus to identify each one, and then predict how the molecule will behave. For example:

    • decolourises bromine water → a C=C double bond (an alkene 烯烃).
    • gives a precipitate with silver nitrate → a halogenoalkane 卤代烷.
    • orange $\text{K}_2\text{Cr}_2\text{O}_7$ turns green → a primary or secondary alcohol 醇.
    • orange precipitate with 2,4-DNPH → an aldehyde 醛 or ketone 酮.
    • fizzes with a carbonate → a carboxylic acid 羧酸.

    A map of the AS reactions

    Each row turns one functional group into another. Learn it as a map you can travel around:

    Start Reagent and conditions Product
    alkene $\text{H}_2$, Ni alkane
    alkene $\text{HX}$, or $\text{X}_2$ halogenoalkane
    alkene steam, $\text{H}_3\text{PO}_4$ alcohol
    halogenoalkane $\text{NaOH(aq)}$, heat alcohol
    halogenoalkane $\text{KCN}$ in ethanol, heat a nitrile 腈 (adds one carbon)
    halogenoalkane $\text{NH}_3$ in ethanol, pressure an amine 胺
    alcohol $\text{K}_2\text{Cr}_2\text{O}_7$, distil / reflux aldehyde / carboxylic acid
    alcohol concentrated acid, heat alkene
    aldehyde or ketone $\text{NaBH}_4$ alcohol
    aldehyde or ketone $\text{HCN}$, $\text{KCN}$ hydroxynitrile
    nitrile dilute acid, heat carboxylic acid
    carboxylic acid + alcohol concentrated $\text{H}_2\text{SO}_4$ an ester 酯

    Planning a multi-step route

    To devise a synthetic route 合成路线:

    1. compare the target with the starting material — what has changed (the functional group, the number of carbons)?
    2. work backwards from the target: which single reaction could make it, and from what?
    3. repeat until you reach the starting material.
    4. write each step with its reagent 试剂 and conditions.

    If you need to add a carbon, the $\text{KCN}$ step is the key — it is the only AS reaction that lengthens the chain.

    Analysing a route

    When you are given a route, for each step state the type of reaction (such as oxidation 氧化, reduction 还原, substitution, addition or elimination) and the reagent used. Also think about possible by-products 副产物 — for example, making an amine from a halogenoalkane also gives a mixture of further-substituted amines, so the yield of the simple amine is low.

    Worked example. Devise a route from propene to propanone, $\text{CH}_3\text{COCH}_3$. Work backwards from the target. A ketone comes from oxidising a secondary alcohol, so the step before propanone is propan-2-ol with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ under reflux. Propan-2-ol comes from propene by adding steam over an $\text{H}_3\text{PO}_4$ catalyst - and Markovnikov's rule conveniently puts the $\text{OH}$ on the middle carbon, which is exactly the secondary alcohol needed. So the route is: propene, then steam with $\text{H}_3\text{PO}_4$, giving propan-2-ol; then acidified $\text{K}_2\text{Cr}_2\text{O}_7$ under reflux, giving propanone. Give a reagent and its conditions on every arrow: a route with the right intermediates but no reagents scores very little.

    日本語

    このトピックは新しい反応を追加するものではありません。代わりに、既知の反応をつなぐことで、複数の工程を経てターゲット分子を構築することを求めます。

    アルミ箔ブリスター包装された白色錠剤
    医薬品は、多段階の有機合成によって単純な出発物質から製造される

    官能基の特定

    1つの分子に複数の官能基がある場合があります。シラバスにある検出反応を用いてそれぞれを特定し、その分子の挙動を予測してください。例:

    • ブロム水を脱色させる → C=C二重結合(アルケン)。
    • 硝酸銀と沈殿を生じる → ハロゲンアルカン。
    • オレンジ $\text{K}_2\text{Cr}_2\text{O}_7$ がグリーンに変色する → 一次または二次 アルコール。
    • 2,4-DNPHと橙色の沈殿 → アルデヒドまたはケトン。
    • カルボネートと泡を立てる → カルボン酸。
    各検出試薬、その観察結果、および特定される官能基をリストする3列の表
    標準的な検出試験:各試薬は特定の官能基を示唆する特徴的な観察結果を与える

    ASの反応マップ

    各行は1つの官能基を別のものに変換します。旅する地図のように memorize してください:

    出発 試薬と条件 生成物
    アルケン $\text{H}_2$, Ni アルカン
    アルケン $\text{HX}$, または $\text{X}_2$ ハロゲンアルカン
    alkene steam, $\text{H}_3\text{PO}_4$ alcohol
    ハロゲンアルカン $\text{NaOH(aq)}$, 加熱 アルコール
    ハロゲンアルカン エタノール中の $\text{KCN}$, 加熱 ニトリル(炭素が1つ増える)
    ハロゲンアルカン エタノール中の $\text{NH}_3$, 加圧 アミン
    アルコール $\text{K}_2\text{Cr}_2\text{O}_7$, 蒸留 / 還流 アルデヒド / カルボン酸
    アルコール 濃酸, 加熱 アルケン
    アルデヒドまたはケトン $\text{NaBH}_4$ アルコール
    アルデヒドまたはケトン $\text{HCN}$, $\text{KCN}$ ヒドロキシニトリル
    ニトリル 希酸, 加熱 カルボン酸
    カルボン酸 + アルコール 濃 $\text{H}_2\text{SO}_4$ エステル
    アルケン、アルカン、ハロゲンアルカン、アルコール、アルデヒド、ニトリル、アミン、カルボン酸、エステルをラベル付きの試薬矢印で結んだネットワーク図
    ASの反応マップ:各矢印は1つの官能基を別のものに変換します。ターゲットから逆算してルートを計画してください

    多段階ルートの計画

    合成ルートを考案するには:

    1. ターゲットを出発物質と比較する — 何が変わったか(官能基、炭素数)?
    2. ターゲットから逆算する:どのような単一反応でこれを作れ、その原料は何か?
    3. 出発物質に到達するまで繰り返す。
    4. 各ステップを試薬と条件と共に書き出す。

    炭素を増やす必要がある場合、$\text{KCN}$ ステップが鍵となります — これはAS反応の中で唯一鎖を長くするものです。

    エチレンからプロパノ酸への合成連鎖:各進行方向のステップに試薬が示され、大きな逆矢印でプランニングがターゲットから出発点へ向かうことを示している
    ターゲットから逆算して、1つずつ反応を遡り、出発物質に到達するまでルートを計画する

    ルートの分析

    与えられたルートに対して、各ステップで反応の種類(酸化、還元、置換、付加、脱離など)と使用された試薬を述べてください。また、可能性のある副産物についても考えましょう。例えば、ハロゲンアルカンからアミンを作る場合、さらに置換されたアミンの混合物も生成するため、単純なアミンの収率は低くなります。

    ビーカーの上に固定された分液漏斗:混ざり合わない2層の有色液体を含んでいる
    分液漏斗は互いに溶け合わない2層を分離する — 反応混合物からの有機生成物の回収方法
    与えられたルートの各ステップに対して3つの質問をする:反応の種類、試薬と条件、副産物が収率を下げるかどうか
    与えられたルートのすべてのステップに対して、反応の種類、試薬、副産物を問う
    5つの反応種類とそれぞれの作用:酸化、還元、置換、付加、脱離
    各ステップの反応種類を名指す:酸化、還元、置換、付加、または脱離

    ** worked example.** プロペンからプロパノンへ合成する経路を立案せよ、$\text{CH}_3\text{COCH}_3$。標的物質から逆算する。ケトンは二次アルコールの酸化によって生成されるため、プロパノンの直前の中間体はプロパン-2-オールであり、還流条件下で酸性化された $\text{K}_2\text{Cr}_2\text{O}_7$ で酸化される。プロパン-2-オールはプロペンを $\text{H}_3\text{PO}_4$ 触媒上にある蒸気(スチーム)と付加することにより得られる。マルコフニコフ則により、$\text{OH}$ は中央の炭素原子に導入され、これがちょうど必要な二次アルコールとなる。したがって、合成経路は以下の通りである:プロペンを $\text{H}_3\text{PO}_4$ と蒸気で処理してプロパン-2-オールを得る;次に、還流条件下で酸性化された $\text{K}_2\text{Cr}_2\text{O}_7$ で酸化してプロパノンを得る。各矢印に対しては試薬とその反応条件を必ず記載せよ。適切な中間体は示しているものの試薬が記載されていない経路では、得点は極めて低い。

    Explore · ⁨探索⁩

    Reaction map lab · ⁨反応マップ実験室⁩

    Classify clues that identify functional groups and reaction pathways. · ⁨官能基や反応経路を同定する手がかりを分類せよ。⁩

    Explore · ⁨探索⁩

    Synthetic route planning lab · ⁨合成経路設計実験室⁩

    Follow how a target molecule is planned backwards then made forwards. · ⁨目標分子の逆合成計画と、それに続く正合成の流れを追う。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    functional group/ˈfʌŋkʃənl ɡruːp/ 官能基
    alkene/ˈælkiːn/ アルケン
    halogenoalkane/ˈheɪləʊdʒnəʊlkeɪn/ ハロゲン化アルキル
    alcohol/ˈælkəhɒl/ アルコール
    aldehyde/ˈældɪhaɪd/ アルデヒドが得られる
    ketone/ˈketəʊn/ ケトン
    carboxylic acid/ˌkɑːbəkˈsɪlɪk ˈæsɪd/ カルボン酸
    nitrile/ˈnaɪtraɪl/ ニトリル
    amine/ˈæmaɪn/ アミン
    ester/ˈestə/ エステル
    synthetic route/sɪnˈθetɪk ruːt/ 合成経路
    reagent/rɪˈeɪdʒənt/ 試薬
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    reduction/rɪˈdʌkʃn/ 還元
    by-product/baɪ ˈprɒdʌkt/ 副生成物
    21.1

    Exam tips · ⁨試験対策⁩

    English
    • Learn the reagents and conditions for each conversion — that is exactly what synthesis questions test.
    • Plan multi-step routes by functional group and watch the carbon count (KCN adds one carbon).
    • Choose the shortest valid route and state every reagent and condition.
    日本語
    • 各変換における試薬と反応条件を暗記すること——これが合成問題の核心的な出題ポイントである。
    • 官能基ごとの多段階合成経路を計画し、炭素数の変化に注意すること(KCN は炭素を1つ追加する)。
    • 最短かつ有効な合成経路を選び、すべての試薬と反応条件を明記すること。
  • 22

    Analytical techniques · ⁨分析技術⁩

    Watch lesson · ⁨レッスンを視聴⁩
    22.1

    Infrared spectroscopy

    Syllabus · ⁨シラバス⁩
    English
    1. analyse an infrared spectrum of a simple molecule to identify functional groups (see the Data section for the functional groups required)
    日本語
    1. 簡単な分子の赤外スペクトルを分析して官能基を特定する(必要な官能基についてはデータセクションを参照)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Infrared spectroscopy 红外光谱 helps you find the functional group 官能团 in a molecule. Each kind of bond soaks up (absorbs) infrared radiation at its own range of frequencies. Where the bond shows strong absorption 吸收, the spectrum has a dip.

    The position is measured in wavenumber 波数 (in $\text{cm}^{-1}$). You are given a data table, so you do not memorise the numbers. You just match the dips to bonds:

    • a broad dip around $3200$–$3650\ \text{cm}^{-1}$ → an O–H bond in an alcohol.
    • a dip around $1700\ \text{cm}^{-1}$ → a C=O bond (aldehyde, ketone, acid or ester).
    • a broad dip $2500$–$3000\ \text{cm}^{-1}$ together with a C=O dip → a carboxylic acid.

    This is useful for checking a reaction. For example, if propene has been turned into propan-2-ol, the C=C dip should be gone and an O–H dip should appear.

    日本語

    Infrared spectroscopy 红外光谱 helps you find the functional group 官能团 in a molecule. Each kind of bond soaks up (absorbs) infrared radiation at its own range of frequencies. Where the bond shows strong absorption 吸收, the spectrum has a dip.

    A benchtop infrared spectrometer in a laboratory
    An infrared spectrometer shines infrared through a sample and records which wavenumbers its bonds absorb

    The position is measured in wavenumber 波数 (in $\text{cm}^{-1}$). You are given a data table, so you do not memorise the numbers. You just match the dips to bonds:

    • a broad dip around $3200$–$3650\ \text{cm}^{-1}$ → an O–H bond in an alcohol.
    • a dip around $1700\ \text{cm}^{-1}$ → a C=O bond (aldehyde, ketone, acid or ester).
    • a broad dip $2500$–$3000\ \text{cm}^{-1}$ together with a C=O dip → a carboxylic acid.

    This is useful for checking a reaction. For example, if propene has been turned into propan-2-ol, the C=C dip should be gone and an O–H dip should appear.

    An infrared spectrum with transmittance dipping at a broad O-H band and a sharp C=O band
    An infrared spectrum: each bond gives a dip at its own wavenumber. A broad O–H dip together with a C=O dip identifies a carboxylic acid
    Explore · ⁨探索⁩

    IR spectroscopy lab · ⁨赤外分光法実験室⁩

    Match an absorption to the bond or functional group it reveals. · ⁨吸収線を対応する結合または官能基に結び付けなさい。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    infrared spectroscopy/ˌɪnfrəˈred spekˈtrɒskəpi/ 赤外分光法
    functional group/ˈfʌŋkʃənl ɡruːp/ 官能基
    absorption/əbˈsɔːpʃn/ 吸収
    wavenumber/ˈweɪvnʌmbə/ 波数
    Watch lesson · ⁨レッスンを視聴⁩
    22.2

    Mass spectrometry

    Syllabus · ⁨シラバス⁩
    1. analyse mass spectra in terms of $m/e$ values and isotopic abundances (knowledge of the working of the mass spectrometer is not required)
    2. calculate the relative atomic mass of an element given the relative abundances of its isotopes, or its mass spectrum
    3. deduce the molecular mass of an organic molecule from the molecular ion peak in a mass spectrum
    4. suggest the identity of molecules formed by simple fragmentation in a given mass spectrum
    5. deduce the number of carbon atoms, $n$, in a compound using the $[M + 1]^+$ peak and the formula
      $$n = \frac{100 \times \text{abundance of } [M + 1]^+ \text{ ion}}{1.1 \times \text{abundance of } M^+ \text{ ion}}$$
    6. deduce the presence of bromine and chlorine atoms in a compound using the $[M + 2]^+$ peak

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    In mass spectrometry 质谱, a molecule is turned into ions and sorted by its mass-to-charge ratio 质荷比 ($m/e$). The spectrum is a set of peaks at different $m/e$ values.

    Relative atomic mass from isotopes

    An element's isotope 同位素 mixture gives several peaks. From the isotopic abundance 同位素丰度 (how common each isotope is) you can find the relative atomic mass 相对原子质量 — a weighted average:

    $$A_r = \frac{\sum (\text{isotope mass} \times \text{abundance})}{\sum \text{abundance}}$$

    For example, chlorine is 75% $^{35}\text{Cl}$ and 25% $^{37}\text{Cl}$, giving $A_r = \dfrac{35 \times 75 + 37 \times 25}{100} = 35.5$.

    The molecular ion and fragmentation

    The peak at the highest $m/e$ (the molecular ion 分子离子 peak, or molecular ion peak 分子离子峰, $M^+$) gives the relative molecular mass of the whole molecule.

    The molecule also breaks into smaller pieces — this is fragmentation 碎裂. The gap between two peaks tells you the mass of the lost piece, so you can suggest each fragment 碎片. For example, a loss of 15 means a $\text{CH}_3$ group was lost, and a loss of 29 means $\text{CHO}$ or $\text{C}_2\text{H}_5$.

    The [M + 1] and [M + 2] peaks

    • a small [M + 1] peak comes from the $^{13}\text{C}$ isotope. The number of carbon atoms $n$ is:
    $$n = \frac{100 \times \text{abundance of } [M + 1]^+}{1.1 \times \text{abundance of } M^+}$$
    • an [M + 2] peak shows chlorine or bromine. One chlorine gives an $[M + 2]$ peak about one third the height of $M^+$ (from $^{37}\text{Cl}$); one bromine gives an $[M + 2]$ peak about the same height as $M^+$ (from $^{81}\text{Br}$).

    Worked example. A compound shows a molecular ion at $m/e = 108$ and a peak of almost equal height at $m/e = 110$. Its infrared spectrum shows no broad absorption near $3300\ \text{cm}^{-1}$. Deduce its identity. Two peaks two units apart with roughly equal heights are the $1:1$ signature of one bromine atom (from $^{79}\text{Br}$ and $^{81}\text{Br}$); one chlorine would have given a $3:1$ ratio instead. Take the bromine away from the molecular ion: $108 - 79 = 29$, which fits a $\text{C}_2\text{H}_5$ fragment. The absence of a broad peak near $3300\ \text{cm}^{-1}$ rules out an $\text{O-H}$, so there is no alcohol group. The compound is bromoethane, $\text{C}_2\text{H}_5\text{Br}$. Read the $[M+2]$ ratio rather than merely noting the peak: $1:1$ means bromine, $3:1$ means chlorine.

    日本語

    In mass spectrometry 质谱, a molecule is turned into ions and sorted by its mass-to-charge ratio 质荷比 ($m/e$). The spectrum is a set of peaks at different $m/e$ values.

    A scientist in a laboratory loading a small sample into a large modern mass spectrometer
    A modern mass spectrometer: the sample is loaded at the front, then the machine ionises it and sorts the ions by their mass-to-charge ratio to give the spectrum

    Relative atomic mass from isotopes

    An element's isotope 同位素 mixture gives several peaks. From the isotopic abundance 同位素丰度 (how common each isotope is) you can find the relative atomic mass 相对原子质量 — a weighted average:

    $$A_r = \frac{\sum (\text{isotope mass} \times \text{abundance})}{\sum \text{abundance}}$$

    For example, chlorine is 75% $^{35}\text{Cl}$ and 25% $^{37}\text{Cl}$, giving $A_r = \dfrac{35 \times 75 + 37 \times 25}{100} = 35.5$.

    The molecular ion and fragmentation

    The peak at the highest $m/e$ (the molecular ion 分子离子 peak, or molecular ion peak 分子离子峰, $M^+$) gives the relative molecular mass of the whole molecule.

    The molecule also breaks into smaller pieces — this is fragmentation 碎裂. The gap between two peaks tells you the mass of the lost piece, so you can suggest each fragment 碎片. For example, a loss of 15 means a $\text{CH}_3$ group was lost, and a loss of 29 means $\text{CHO}$ or $\text{C}_2\text{H}_5$.

    A mass spectrum of ethanol with peaks at several m/e values, the molecular ion at 46 marked and a gap of 15 from 46 to 31 labelled as the loss of a methyl group
    A mass spectrum: the highest-$m/e$ peak is the molecular ion ($M^+$, the $M_r$); the gaps between peaks give the masses of the lost fragments

    The [M + 1] and [M + 2] peaks

    • a small [M + 1] peak comes from the $^{13}\text{C}$ isotope. The number of carbon atoms $n$ is:
    $$n = \frac{100 \times \text{abundance of } [M + 1]^+}{1.1 \times \text{abundance of } M^+}$$
    • an [M + 2] peak shows chlorine or bromine. One chlorine gives an $[M + 2]$ peak about one third the height of $M^+$ (from $^{37}\text{Cl}$); one bromine gives an $[M + 2]$ peak about the same height as $M^+$ (from $^{81}\text{Br}$).
    Two pairs of mass-spectrum peaks two units apart: a chlorine pair in a 3 to 1 ratio and a bromine pair in a 1 to 1 ratio
    An $[M+2]$ peak two mass units above $M^+$ shows a halogen: one chlorine gives a $3:1$ ratio, one bromine a $1:1$ ratio

    Worked example. A compound shows a molecular ion at $m/e = 108$ and a peak of almost equal height at $m/e = 110$. Its infrared spectrum shows no broad absorption near $3300\ \text{cm}^{-1}$. Deduce its identity. Two peaks two units apart with roughly equal heights are the $1:1$ signature of one bromine atom (from $^{79}\text{Br}$ and $^{81}\text{Br}$); one chlorine would have given a $3:1$ ratio instead. Take the bromine away from the molecular ion: $108 - 79 = 29$, which fits a $\text{C}_2\text{H}_5$ fragment. The absence of a broad peak near $3300\ \text{cm}^{-1}$ rules out an $\text{O-H}$, so there is no alcohol group. The compound is bromoethane, $\text{C}_2\text{H}_5\text{Br}$. Read the $[M+2]$ ratio rather than merely noting the peak: $1:1$ means bromine, $3:1$ means chlorine.

    Explore · ⁨探索⁩

    Mass spectrometry route · ⁨質量分析法の経路⁩

    Follow a molecule through ionisation, separation and detection. · ⁨分子のイオン化、分離、検出の過程を追跡する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    mass spectrometry/mæs spekˈtrɒmətri/ 質量分析
    mass-to-charge ratio/mæs tə tʃɑːdʒ ˈreɪʃɪəʊ/ 質量対電荷比
    isotope/ˈaɪsətəʊp/ 同位体 (isotope)
    isotopic abundance/ˌaɪsəˈtɒpɪk əˈbʌndəns/ 同位体存在比
    relative atomic mass/ˈrelətɪv əˈtɒmɪk mæs/ 相対原子質量
    molecular ion/məˈlekjʊlə ˈaɪɒn/ 分子イオン
    molecular ion peak/məˈlekjʊlə ˈaɪɒn piːk/ 分子イオンピーク
    fragmentation/ˌfræɡmənˈteɪʃn/ 断片化
    fragment/ˈfræɡmənt/ 断片
    Watch lesson · ⁨レッスンを視聴⁩
    22.2

    Exam tips

    • On an IR spectrum quote the wavenumber and the bond (O-H broad $\sim 3200-3600$, C=O $\sim 1700$); use the data booklet.
    • The peak at the highest $m/z$ is the molecular ion $\text{M}^+$ and gives the $M_r$.
    • Explain fragmentation: the gap between peaks is the lost fragment (loss of 15 = $\text{CH}_3$, 29 = $\text{C}_2\text{H}_5$ or CHO).
    • Learn common isotope patterns (Cl gives M and M+2 in about 3:1; M+1 from $^{13}\text{C}$).
  • 23

    Chemical energetics · ⁨化学熱力学⁩

    Watch lesson · ⁨レッスンを視聴⁩
    23.1

    Lattice energy and Born–Haber cycles

    Syllabus · ⁨シラバス⁩
    English
    1. define and use the terms: (a) enthalpy change of atomisation, $\Delta H_{\text{at}}$ (b) lattice energy, $\Delta H_{\text{latt}}$ (the change from gas phase ions to solid lattice)
    2. (a) define and use the term first electron affinity, EA (b) explain the factors affecting the electron affinities of elements (c) describe and explain the trends in the electron affinities of the Group 16 and Group 17 elements
    3. construct and use Born–Haber cycles for ionic solids (limited to +1 and +2 cations, –1 and –2 anions)
    4. carry out calculations involving Born–Haber cycles
    5. explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of a lattice energy
    日本語
    1. 以下の用語を定義し、使用する: (a) 原子化エンタルピー変化、$\Delta H_{\text{at}}$ (b) ラチスエネルギー、$\Delta H_{\text{latt}}$(気相イオンから固体ラチスへの変化)
    2. (a) 第一電子親和力、EAという用語を定義し、使用する (b) 元素の電子親和力に影響を与える要因を説明する (c) Group 16およびGroup 17元素の電子親和力の傾向を記述・説明する
    3. イオン固体に対するボルン-ハーバーサイクルを構築し、使用する(+1および+2カチオン、–1および–2アニオンに限定)
    4. ボルン-ハーバーサイクルに関連する計算を実行する
    5. イオン電荷およびイオン半径がラチスエネルギーの数値的大小に与える影響を定性的に説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    These cycles use several enthalpy changes 焓变 ($\Delta H$). Two new ones are:

    • the enthalpy change of atomisation 原子化焓变, $\Delta H_{\text{at}}$ — the energy to make one mole of gaseous atoms from an element. It is always positive (bonds must break).
    • the lattice energy 晶格能, $\Delta H_{\text{latt}}$ — the energy change when one mole of a solid ionic lattice forms from its gaseous ions. It is always negative (strong bonds form).

    Electron affinity

    The first electron affinity 电子亲和能 (EA) is the energy change when one mole of gaseous atoms each gain one electron to form one mole of $1-$ ions. The first EA is usually negative.

    The same factors as ionisation energy apply (nuclear charge, atomic radius, shielding). Going down Group 16 or 17, the EA becomes less exothermic, because the atom is larger and pulls the extra electron in less strongly. (The very top element is an exception: its atom is so small that electron repulsion makes its EA less exothermic than the one below it.)

    Born–Haber cycles

    A Born–Haber cycle 玻恩哈伯循环 is an energy cycle that links the enthalpy change of formation of an ionic solid with its atomisation, ionisation energy, electron affinity and lattice energy. Using Hess's law, you go round the cycle to find any one unknown step. (You only need $+1$ and $+2$ cations and $-1$ and $-2$ anions.)

    Worked example. Find the lattice energy of sodium chloride from these data (kJ mol⁻¹): enthalpy of formation $\Delta H_f = -411$; atomisation $\Delta H_{\text{at}}(\text{Na}) = +107$ and $\Delta H_{\text{at}}(\text{Cl}) = +122$; first ionisation energy of Na $= +496$; electron affinity of Cl $= -349$.

    By Hess's law the direct formation route equals the route up and round the cycle:

    $$\Delta H_f = \Delta H_{\text{at}}(\text{Na}) + \Delta H_{\text{at}}(\text{Cl}) + \text{IE}_1 + \text{EA} + \Delta H_{\text{latt}},$$

    so $\Delta H_{\text{latt}} = -411 - (107 + 122 + 496 - 349) = -787\ \text{kJ mol}^{-1}.$

    What controls the size of a lattice energy

    The lattice energy is more negative (stronger) when:

    • the ionic charge is higher (e.g. $\text{Mg}^{2+}$ beats $\text{Na}^+$).
    • the ionic radius is smaller.

    Both make the attraction between the ions stronger.

    日本語

    These cycles use several enthalpy changes 焓变 ($\Delta H$). Two new ones are:

    • the enthalpy change of atomisation 原子化焓变, $\Delta H_{\text{at}}$ — the energy to make one mole of gaseous atoms from an element. It is always positive (bonds must break).
    • the lattice energy 晶格能, $\Delta H_{\text{latt}}$ — the energy change when one mole of a solid ionic lattice forms from its gaseous ions. It is always negative (strong bonds form).
    Clear cubic crystals of halite (rock salt)
    An ionic solid such as sodium chloride is a giant, regular lattice of ions — the lattice energy is released when it forms

    Electron affinity

    The first electron affinity 电子亲和能 (EA) is the energy change when one mole of gaseous atoms each gain one electron to form one mole of $1-$ ions. The first EA is usually negative.

    A gaseous atom gains an electron to form a 1- ion, usually releasing energy
    A gaseous atom gains an electron, usually releasing energy

    The same factors as ionisation energy apply (nuclear charge, atomic radius, shielding). Going down Group 16 or 17, the EA becomes less exothermic, because the atom is larger and pulls the extra electron in less strongly. (The very top element is an exception: its atom is so small that electron repulsion makes its EA less exothermic than the one below it.)

    Born–Haber cycles

    A Born–Haber cycle 玻恩哈伯循环 is an energy cycle that links the enthalpy change of formation of an ionic solid with its atomisation, ionisation energy, electron affinity and lattice energy. Using Hess's law, you go round the cycle to find any one unknown step. (You only need $+1$ and $+2$ cations and $-1$ and $-2$ anions.)

    An energy-level diagram for sodium chloride showing the steps up (atomisation, ionisation) and down (electron affinity, lattice energy) and the direct formation step
    A Born–Haber cycle for NaCl: going up costs energy (atomisation, ionisation); coming down releases it (electron affinity, and the large lattice energy)

    Worked example. Find the lattice energy of sodium chloride from these data (kJ mol⁻¹): enthalpy of formation $\Delta H_f = -411$; atomisation $\Delta H_{\text{at}}(\text{Na}) = +107$ and $\Delta H_{\text{at}}(\text{Cl}) = +122$; first ionisation energy of Na $= +496$; electron affinity of Cl $= -349$.

    By Hess's law the direct formation route equals the route up and round the cycle:

    $$\Delta H_f = \Delta H_{\text{at}}(\text{Na}) + \Delta H_{\text{at}}(\text{Cl}) + \text{IE}_1 + \text{EA} + \Delta H_{\text{latt}},$$

    so $\Delta H_{\text{latt}} = -411 - (107 + 122 + 496 - 349) = -787\ \text{kJ mol}^{-1}.$

    What controls the size of a lattice energy

    The lattice energy is more negative (stronger) when:

    Smaller ions and higher charges give a stronger lattice energy
    Smaller ions and higher charges give a stronger lattice energy
    • the ionic charge is higher (e.g. $\text{Mg}^{2+}$ beats $\text{Na}^+$).
    • the ionic radius is smaller.

    Both make the attraction between the ions stronger.

    Explore · ⁨探索⁩

    The Born–Haber cycle · ⁨ボーン=ハーバーサイクル⁩

    Lattice energy can't be measured directly, so it is found from a cycle of measurable steps. · ⁨格子エネルギーは直接測定できないため、測定可能な一連の工程から求める。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    enthalpy change/enˈθælpi tʃeɪndʒ/ エンタルピー変化
    enthalpy change of atomisation/enˈθælpi tʃeɪndʒ ɒv əˌtɒmaɪˈzeɪʃn/ 原子化エンタルピー変化
    lattice energy/ˈlætɪs ˈenədʒi/ 格子エネルギー
    electron affinity/ɪˈlektrɒn əˈfɪnɪti/ 電子親和力
    Born–Haber cycle/bɔːn ˈheɪbə ˈsaɪkl/ ボルン=ハーバーサイクル
    Watch lesson · ⁨レッスンを視聴⁩
    23.2

    Enthalpies of solution and hydration

    Syllabus · ⁨シラバス⁩
    English
    1. define and use the term enthalpy change with reference to hydration, $\Delta H_{\text{hyd}}$, and solution, $\Delta H_{\text{sol}}$
    2. construct and use an energy cycle involving enthalpy change of solution, lattice energy and enthalpy change of hydration
    3. carry out calculations involving the energy cycles in 23.2.2
    4. explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of an enthalpy change of hydration
    日本語
    1. 水和、$\Delta H_{\text{hyd}}$、および溶解、$\Delta H_{\text{sol}}$を参照にしてエンタルピー変化という用語を定義し、使用する
    2. 溶解エンタルピー変化、ラチスエネルギーおよび水和エンタルピー変化を含むエネルギーサイクルを構築し、使用する
    3. 23.2.2のエネルギーサイクルを含む計算を行う
    4. イオン電荷およびイオン半径が、水化エンタルピー変化の数値に与える影響を定性的に説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    • the enthalpy change of hydration 水合焓变, $\Delta H_{\text{hyd}}$ — the energy change when one mole of gaseous ions is surrounded by water to form aqueous ions. It is exothermic.
    • the enthalpy change of solution 溶解焓变, $\Delta H_{\text{sol}}$ — the energy change when one mole of solute dissolves fully in water.

    An energy cycle links the three:

    $$\Delta H_{\text{sol}} = -\Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$$

    (To dissolve, you first pull the lattice apart, then hydrate the ions.) Like lattice energy, $\Delta H_{\text{hyd}}$ is more exothermic for ions with a higher charge and a smaller radius.

    日本語
    • the enthalpy change of hydration 水合焓变, $\Delta H_{\text{hyd}}$ — the energy change when one mole of gaseous ions is surrounded by water to form aqueous ions. It is exothermic.
    • the enthalpy change of solution 溶解焓变, $\Delta H_{\text{sol}}$ — the energy change when one mole of solute dissolves fully in water.
    An instant cold pack
    An instant cold pack feels cold because its salt dissolves endothermically (a positive $\Delta H_\text{sol}$), driven by the rise in entropy

    An energy cycle links the three:

    $$\Delta H_{\text{sol}} = -\Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$$

    (To dissolve, you first pull the lattice apart, then hydrate the ions.) Like lattice energy, $\Delta H_{\text{hyd}}$ is more exothermic for ions with a higher charge and a smaller radius.

    A triangular energy cycle linking the ionic solid, the gaseous ions and the aqueous ions by the lattice, hydration and solution enthalpies
    Dissolving energy cycle: pull the lattice apart (reverse lattice energy), then hydrate the gaseous ions, so $\Delta H_\text{sol} = -\Delta H_\text{latt} + \Delta H_\text{hyd}$
    Explore · ⁨探索⁩

    Enthalpy of solution lab · ⁨溶解エンタルピーの実験⁩

    deltaHsol = lattice + hydration terms · ⁨deltaHsol = 格子エネルギー + 水和項⁩

    Change hydration strength and see solution enthalpy shift. · ⁨水和の強さを変化させて溶解エンタルピーの変化を見る。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    enthalpy change of hydration/enˈθælpi tʃeɪndʒ ɒv haɪˈdreɪʃn/ 水和エンタルピー変化
    enthalpy change of solution/enˈθælpi tʃeɪndʒ ɒv səˈluːʃn/ 溶解エンタルピー変化
    23.3

    Entropy change, ΔS

    Syllabus · ⁨シラバス⁩
    1. define the term entropy, $S$, as the number of possible arrangements of the particles and their energy in a given system
    2. predict and explain the sign of the entropy changes that occur: (a) during a change in state, e.g. melting, boiling and dissolving (and their reverse) (b) during a temperature change (c) during a reaction in which there is a change in the number of gaseous molecules
    3. calculate the entropy change for a reaction, $\Delta S$, given the standard entropies, $S^\ominus$, of the reactants and products, $\Delta S^\ominus = \Sigma S^\ominus \text{(products)} - \Sigma S^\ominus \text{(reactants)}$ (use of $\Delta S^\ominus = \Delta S^\ominus_{\text{surr}} + \Delta S^\ominus_{\text{sys}}$ is not required)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Entropy 熵 ($S$) measures the number of ways the particles and their energy can be arranged in a system. More ways means more "disorder".

    The entropy change 熵变 ($\Delta S$) is positive when disorder increases, and negative when it falls:

    Change Sign of $\Delta S$
    solid → liquid → gas (melting, boiling), or dissolving positive
    a rise in temperature positive
    a reaction that makes more gas molecules positive
    a reaction that makes fewer gas molecules negative

    You can calculate it from standard entropies:

    $$\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})$$
    日本語

    Entropy 熵 ($S$) measures the number of ways the particles and their energy can be arranged in a system. More ways means more "disorder".

    Three boxes of particles: an ordered solid grid, a clustered liquid, and a spread-out gas, with arrows showing entropy increasing
    Entropy rises from solid to liquid to gas as the particles spread into more arrangements, so $\Delta S$ is positive

    The entropy change 熵变 ($\Delta S$) is positive when disorder increases, and negative when it falls:

    Change Sign of $\Delta S$
    solid → liquid → gas (melting, boiling), or dissolving positive
    a rise in temperature positive
    a reaction that makes more gas molecules positive
    a reaction that makes fewer gas molecules negative

    You can calculate it from standard entropies:

    $$\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})$$
    Explore · ⁨探索⁩

    Entropy change lab · ⁨エントロピー変化の実験⁩

    deltaS = products disorder - reactants disorder · ⁨deltaS = 生成物の無秩序度 - 反応物の無秩序度⁩

    Increase disorder and see entropy change become more positive. · ⁨無秩序度を増加させてエントロピー変化がより正の値になる様子を見る。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    entropy/ˈentrəpi/ エントロピー
    entropy change/ˈentrəpi tʃeɪndʒ/ エントロピー変化
    23.4

    Gibbs free energy change, ΔG

    Syllabus · ⁨シラバス⁩
    English
    1. state and use the Gibbs equation $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$
    2. perform calculations using the equation $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$
    3. state whether a reaction or process will be feasible by using the sign of $\Delta G$
    4. predict the effect of temperature change on the feasibility of a reaction, given standard enthalpy and entropy changes
    日本語
    1. ギブズ方程式 $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ を述べて適用する
    2. 方程式 $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ を用いた計算を行う
    3. 方程式 $\Delta G$ の符号を用いて、ある反応または過程が 自発的(可能) かどうかを判定する
    4. 標準エンタルピー変化および標準エントロピー変化が与えられた場合、温度変化が反応の自発性に与える影響を予測する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The Gibbs free energy 吉布斯自由能 change decides whether a reaction can happen on its own:

    $$\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$$

    Here $T$ is the temperature in kelvin. A reaction is feasible (it can happen) when $\Delta G$ is negative or zero. The feasibility 可行性 therefore depends on temperature:

    $\Delta H$ $\Delta S$ When feasible
    negative positive at all temperatures
    positive positive only at high temperature
    negative negative only at low temperature
    positive negative never

    To find the changeover temperature, set $\Delta G = 0$, which gives $T = \Delta H / \Delta S$.

    Worked example. A reaction has $\Delta H = +120\ \text{kJ mol}^{-1}$ and $\Delta S = +200\ \text{J K}^{-1}\,\text{mol}^{-1}$. Find $\Delta G$ at $298\ \text{K}$, and the temperature above which the reaction becomes feasible.

    First match the units: $\Delta S = +0.200\ \text{kJ K}^{-1}\,\text{mol}^{-1}$. Then

    $$\Delta G = \Delta H - T\Delta S = 120 - 298 \times 0.200 = +60\ \text{kJ mol}^{-1}\quad(\text{positive, so not yet feasible}).$$

    Setting $\Delta G = 0$ gives $T = \Delta H/\Delta S = 120/0.200 = 600\ \text{K}$, so the reaction is feasible above $600\ \text{K}$.

    日本語

    The Gibbs free energy 吉布斯自由能 change decides whether a reaction can happen on its own:

    $$\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$$

    Here $T$ is the temperature in kelvin. A reaction is feasible (it can happen) when $\Delta G$ is negative or zero. The feasibility 可行性 therefore depends on temperature:

    $\Delta H$ $\Delta S$ When feasible
    negative positive at all temperatures
    positive positive only at high temperature
    negative negative only at low temperature
    positive negative never
    A two-by-two grid of the signs of enthalpy change and entropy change, each cell saying when the reaction is feasible
    Whether a reaction is feasible ($\Delta G \leq 0$) depends on the signs of $\Delta H$ and $\Delta S$, and sometimes on temperature

    To find the changeover temperature, set $\Delta G = 0$, which gives $T = \Delta H / \Delta S$.

    Worked example. A reaction has $\Delta H = +120\ \text{kJ mol}^{-1}$ and $\Delta S = +200\ \text{J K}^{-1}\,\text{mol}^{-1}$. Find $\Delta G$ at $298\ \text{K}$, and the temperature above which the reaction becomes feasible.

    First match the units: $\Delta S = +0.200\ \text{kJ K}^{-1}\,\text{mol}^{-1}$. Then

    $$\Delta G = \Delta H - T\Delta S = 120 - 298 \times 0.200 = +60\ \text{kJ mol}^{-1}\quad(\text{positive, so not yet feasible}).$$

    Setting $\Delta G = 0$ gives $T = \Delta H/\Delta S = 120/0.200 = 600\ \text{K}$, so the reaction is feasible above $600\ \text{K}$.

    Explore · ⁨探索⁩

    Gibbs free energy lab · ⁨ギブズ自由エネルギー実験⁩

    deltaG = deltaH - T deltaS · ⁨⟦deltaG = deltaH - T deltaS⟦⁩

    Move temperature and see when deltaG becomes negative. · ⁨温度を変えて、deltaGが負になるタイミングを確認せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Gibbs free energy/ɡɪbz friː ˈenədʒi/ ギブズ自由エネルギー
    feasibility/ˌfiːzɪˈbɪlɪti/ 可能性
    23.4

    Exam tips

    • In a Born-Haber cycle get each step's direction and sign right (atomisation, ionisation $+$; electron affinity, lattice formation $-$) and apply Hess's law around it.
    • Lattice energy is more exothermic for smaller, more highly charged ions (higher charge density).
    • Predict the sign of $\Delta S$ from the state changes (more gas moles = more disorder).
    • Use $\Delta G = \Delta H - T\Delta S$; feasible when $\Delta G \le 0$ — convert $\Delta S$ from $\text{J K}^{-1}\,\text{mol}^{-1}$ ($\div 1000$).
  • 24

    Electrochemistry · ⁨電気化学⁩

    Watch lesson · ⁨レッスンを視聴⁩
    24.1

    Electrolysis · ⁨電気分解⁩

    Syllabus · ⁨シラバス⁩
    1. predict the identities of substances liberated during electrolysis from the state of electrolyte (molten or aqueous), position in the redox series (electrode potential) and concentration
    2. state and apply the relationship $F = Le$ between the Faraday constant, $F$, the Avogadro constant, $L$, and the charge on the electron, $e$
    3. calculate: (a) the quantity of charge passed during electrolysis, using $Q = It$ (b) the mass and/or volume of substance liberated during electrolysis
    4. describe the determination of a value of the Avogadro constant by an electrolytic method

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Electrolysis: ions discharge at the electrodes

    Electrolysis 电解 uses electricity to break down a molten or aqueous electrolyte 电解质. Positive ions move to the negative electrode and negative ions move to the positive electrode.

    At each electrode 电极 a half-reaction happens:

    • at the cathode 阴极 (negative): positive ions gain electrons (reduction).
    • at the anode 阳极 (positive): negative ions lose electrons (oxidation).

    Predicting the products

    • a molten electrolyte gives the metal at the cathode and the non-metal at the anode.
    • an aqueous electrolyte also contains water. At the cathode, a less reactive metal is released, but for a reactive metal you get hydrogen instead. At the anode you usually get oxygen, but a concentrated halide solution gives the halogen.

    Calculations

    The charge on one mole of electrons is the Faraday constant 法拉第常量, linked to the Avogadro constant 阿伏伽德罗常量 ($L$) and the charge on one electron ($e$) by $F = Le$.

    The charge passed is $Q = It$ (current $\times$ time). Then:

    1. moles of electrons $= Q / F$.
    2. use the half-equation to find moles of product.
    3. find the mass ($\times M_r$) or the gas volume.

    Measuring the mass deposited for a known charge lets you work back to a value of the Avogadro constant.

    Worked example. A current of $2.0\ \text{A}$ flows for $30$ minutes through copper(II) sulfate solution, depositing copper at the cathode: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$. Find the mass of copper deposited. ($F = 96\,500\ \text{C mol}^{-1}$, $A_r$ Cu $= 64$.)

    Charge passed: $Q = It = 2.0 \times (30 \times 60) = 3600\ \text{C}$. Moles of electrons $= Q/F = 3600/96\,500 = 0.0373\ \text{mol}$. From the half-equation, $2\ \text{mol}$ of electrons give $1\ \text{mol}$ of Cu, so $n(\text{Cu}) = 0.0187\ \text{mol}$ and

    $$m = nM = 0.0187 \times 64 \approx 1.2\ \text{g}.$$
    日本語
    電気分解:イオンが電極で放電する

    電気分解は、電気を用いて溶融または水溶液状の電解質を分解する。正イオンは負極へ、負イオンは正極へ移動する。

    各電極では半反応が起こる:

    • カソード(負極):正イオンが電子を受け取る(還元)。
    • アノード(正極):負イオンが電子を失う(酸化)。
    電気分解セル:直流電源が電解質内の2つの電極に接続され、正イオンがカソードへ、負イオンがアノードへ移動している様子
    電気分解:直流電源が正イオンをカソードへ(還元)、負イオンをアノードへ(酸化)駆動する
    スタンドに取り付けられたガラス製ホフマン・ボルタメーター、電源に配線され、両側にガス収集管がついている
    水の電気分解用のホフマン・ボルタメーター:電源が2つの電極を通って電流を流し、それぞれで生成されたガスが側面の管に集まる

    生成物の予測

    • 溶融電解質からは、カソードで金属、アノードで非金属が得られる。
    • 水溶液電解質には水も含まれる。カソードでは、より不活性な金属が析出するが、活性な金属の場合、代わりに水素が得られる。アノードでは通常酸素が得られるが、濃塩化物溶液ではハロゲンが得られる。
    溶融電解質では金属と非金属が得られ、水溶液では水も含まれる
    溶融電解質は金属と非金属を与え、水溶液は水を含む

    計算

    電子1モルの電荷量はファラデー定数であり、アボガドロ定数($L$)および電子1個あたりの電荷量($e$)とは $F = Le$ で関連している。

    通過した電荷量は $Q = It$ (電流 $\times$ × 時間)。したがって:

    1. 電子の物質量 $= Q / F$ 。
    2. 半方程式を用いて生成物の物質量を求める。
    3. 質量($\times M_r$ )または気体の体積を求める。

    既知の電荷量に対して析出した質量を測定することで、アボガドロ定数の値を逆算できる。

    ** worked example.** $2.0\ \text{A}$ の電流が $30$ 分間銅(II)硫酸水溶液を流れ、カソードで銅が析出する:$\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ 。析出された銅の質量を求めよ。($F = 96\,500\ \text{C mol}^{-1}$ 、$A_r$ Cu $= 64$ 。)

    通過した電荷量:$Q = It = 2.0 \times (30 \times 60) = 3600\ \text{C}$ 。電子の物質量 $= Q/F = 3600/96\,500 = 0.0373\ \text{mol}$ 。半方程式より、$2\ \text{mol}$ の電子が $1\ \text{mol}$ のCuを与えるため、$n(\text{Cu}) = 0.0187\ \text{mol}$ および

    $$m = nM = 0.0187 \times 64 \approx 1.2\ \text{g}.$$
    バレルメッキ装置で小型部品をメッキしている様子
    電気分解は、クロムなどの光沢のある金属の薄い層で物体をメッキするために用いられる
    Explore · ⁨探索⁩

    Faraday's first law · ⁨ファラデーの第一法則⁩

    m = k·Q

    Mass deposited is proportional to the charge passed (Faraday's law). · ⁨析出質量は比例して流過した電荷量(ファラデーの法則)に依存する。⁩

    Explore · ⁨探索⁩

    Inside an electrolysis cell · ⁨電気分解セル内部⁩

    Choose an electrolyte and watch the ions move: cations to the cathode, anions to the anode, where they are discharged. · ⁨電解液を選択するとイオンの動きが見えます:陽イオンは陰極へ、陰イオンは陽極へ移動し、そこで放電されます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    electrolysis/ɪlekˈtrɒləsɪs/ 電気分解
    electrolyte/ɪˈlektrəlaɪt/ 電気分解液
    Watch lesson · ⁨レッスンを視聴⁩
    24.2

    Standard electrode potentials and cell potentials · ⁨標準電極電位および電池電位⁩

    Syllabus · ⁨シラバス⁩
    English
    1. define the terms: (a) standard electrode (reduction) potential (b) standard cell potential
    2. describe the standard hydrogen electrode
    3. describe methods used to measure the standard electrode potentials of: (a) metals or non-metals in contact with their ions in aqueous solution (b) ions of the same element in different oxidation states
    4. calculate a standard cell potential by combining two standard electrode potentials
    5. use standard cell potentials to: (a) deduce the polarity of each electrode and hence explain/deduce the direction of electron flow in the external circuit of a simple cell (b) predict the feasibility of a reaction
    6. deduce from $E^{\ominus}$ values the relative reactivity of elements, compounds and ions as oxidising agents or as reducing agents
    7. construct redox equations using the relevant half-equations
    8. predict qualitatively how the value of an electrode potential, $E$, varies with the concentrations of the aqueous ions
    9. use the Nernst equation, e.g. $E = E^{\ominus} + (0.059/z) \log \frac{[\text{oxidised species}]}{[\text{reduced species}]}$, to predict quantitatively how the value of an electrode potential varies with the concentrations of the aqueous ions; examples include $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s})$, $\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})$
    10. understand and use the equation $\Delta G^{\ominus} = -n E^{\ominus}_{\text{cell}} F$
    日本語
    1. 用語を定義する: (a) 標準電極(還元)電位 (b) 標準電池電位
    2. 標準水素電極について説明する
    3. 標準電極電位の測定方法について説明する: (a) 水溶液中のイオンと接触している金属または非金属 (b) 異なる酸化状態にある同一元素のイオン
    4. 2つの標準電極電位を組み合わせて標準電池電位を計算する
    5. 標準電池電位を用いて: (a) 各電極の極性を導き出し、単純な電池の外回路における電子流の方向を説明・導く (b) 反応の自発性を予測する
    6. $E^{\ominus}$ の値から、元素、化合物、イオンの酸化剤または還元剤としての相対的な反応性を導く
    7. 関連する半反応式を用いて酸化還元反応式を作成する
    8. 水溶液イオンの濃度の変化に伴い、電極電位 $E$ の値がどのように変化するかを定性的に予測する
    9. ネルンスト方程式、例として$E = E^{\ominus} + (0.059/z) \log \frac{[\text{oxidised species}]}{[\text{reduced species}]}$などを用い、水溶液イオンの濃度の変化に伴う電極電位の変数を定量的に予測する。具体例には$\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s})$、$\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})$などが含まれる
    10. 方程式 $\Delta G^{\ominus} = -n E^{\ominus}_{\text{cell}} F$ を理解して適用する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The electrode potential 电极电势 ($E$) of a half-cell shows how easily it is reduced. We measure it against a reference, under standard conditions, to get the standard electrode potential 标准电极电势 ($E^{\ominus}$), always written as a reduction.

    The reference is the standard hydrogen electrode 标准氢电极: hydrogen gas at $1\ \text{atm}$ over platinum in $1\ \text{mol dm}^{-3}$ $\text{H}^+$, defined as exactly $0.00\ \text{V}$.

    To measure an $E^{\ominus}$, connect the half-cell to the standard hydrogen electrode and read the voltage. A metal sits in a solution of its ions; for two ions of the same element (such as $\text{Fe}^{3+}/\text{Fe}^{2+}$), a platinum electrode dips into a solution containing both.

    Combining half-cells

    The standard cell potential 标准电池电势 is the difference between the two standard electrode potentials:

    $$E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{more positive}) - E^{\ominus}(\text{less positive})$$

    Worked example. Find the standard cell potential of a cell built from the $\text{Zn}^{2+}/\text{Zn}$ half-cell ($E^{\ominus} = -0.76\ \text{V}$) and the $\text{Cu}^{2+}/\text{Cu}$ half-cell ($E^{\ominus} = +0.34\ \text{V}$).

    $$E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{more positive}) - E^{\ominus}(\text{less positive}) = (+0.34) - (-0.76) = 1.10\ \text{V}.$$

    From $E^{\ominus}$ values you can:

    • find the polarity: the more negative electrode is the negative terminal, and electrons flow from it through the external circuit to the positive electrode.
    • judge reactivity: a more positive $E^{\ominus}$ means a better oxidising agent 氧化剂 (easily reduced); a more negative $E^{\ominus}$ means a better reducing agent 还原剂.

    Feasibility and redox equations

    A reaction is feasible when $E^{\ominus}_{\text{cell}}$ is positive. To build the full equation, take the two half-equations 半反应方程式, reverse the one that is oxidised, and add them so the electrons cancel. This $E^{\ominus}$ test tells you the feasibility 可行性 of the reaction.

    You can also link it to free energy: $\Delta G^{\ominus} = -n E^{\ominus}_{\text{cell}} F$, where $n$ is the moles of electrons.

    The Nernst equation

    If the concentrations are not standard, the electrode potential changes. Raising the concentration of the oxidised species makes $E$ more positive. The Nernst equation 能斯特方程 gives the exact value:

    $$E = E^{\ominus} + \frac{0.059}{z} \log \frac{[\text{oxidised species}]}{[\text{reduced species}]}$$

    where $z$ is the number of electrons in the half-equation.

    日本語

    電極電位($E$ )は、半電池がどれだけ容易に還元されるかを示す。これを基準として標準条件下で測定することで、標準電極電位($E^{\ominus}$ )が得られ、常に還元として表記される。

    基準は標準水素電極である:プラチナ上に $1\ \text{atm}$ の水素ガスが通じ、$1\ \text{mol dm}^{-3}$ $\text{H}^+$ の酸溶液中において、厳密に $0.00\ \text{V}$ と定義されている。

    酸性中プラチナ電極の上を泡立ちながら通じる水素ガス、ガラス管内の全セルに0.00ボルトと記号
    標準水素電極:H$_2$ 1 atm、酸中1 mol dm$^{-3}$ H$^+$ のプラチナ上、すべての電極電位の$0.00$ V基準

    $E^{\ominus}$ を測定するには、半電池を標準水素電極に接続して電圧を読む。金属はそのイオン溶液中に置かれ、同じ元素の2種類のイオン(例えば$\text{Fe}^{3+}/\text{Fe}^{2+}$ )がある場合、白金電極を両方のイオンを含む溶液に浸す。

    半電池の組み合わせ

    標準電池電位は、2つの標準電極電位の差である:

    $$E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{more positive}) - E^{\ominus}(\text{less positive})$$
    塩橋と電圧計で結ばれた亜鉛半電池と銅半電池、導線を介して亜鉛から銅へ電子が流れている様子
    単純な電池:塩橋で結ばれた2つの半電池と電圧計。電子は負極(Zn)から正極(Cu)へ外部回路を通じて流れる

    ** worked example.** $\text{Zn}^{2+}/\text{Zn}$ 半電池($E^{\ominus} = -0.76\ \text{V}$ )と $\text{Cu}^{2+}/\text{Cu}$ 半電池($E^{\ominus} = +0.34\ \text{V}$ )から構成される電池の標準電池電位を求めよ。

    $$E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{more positive}) - E^{\ominus}(\text{less positive}) = (+0.34) - (-0.76) = 1.10\ \text{V}.$$

    $E^{\ominus}$の値から以下がわかる:

    • 極性を求める:より負の電極が負極であり、電子はそこから外部回路を通じて正極へ流れる。
    • 反応性を判断する:より正の$E^{\ominus}$ はより良い酸化剤(容易に還元)を意味し、より負の$E^{\ominus}$ はより良い還元剤を意味する。
    上部にフッ素、下部にナトリウムまでの垂直スケール、上向き矢印が強い酸化剤、下向き矢印が強い還元剤を示す
    電気化学系列:より正の$E^{\ominus}$ (上部)は強い酸化剤、より負の$E^{\ominus}$ (下部)は強い還元剤

    可能性と酸化還元方程式

    $E^{\ominus}_{\text{cell}}$が正のとき、反応は自発的である。完全な式を立てるには、2つの半反応式を取り上げ、酸化を受ける側を逆にして足し合わせ、電子を打ち消すようにする。この$E^{\ominus}$テストにより、反応の自発性が判明する。

    また、自由エネルギーとも結びつく:$\Delta G^{\ominus} = -n E^{\ominus}_{\text{cell}} F$ 、ここで $n$ は電子の物質量である。

    ネルンストの式

    濃度が標準状態ではない場合、電極電位は変化する。酸化された種の濃度を上げると、$E$がより正の値になる。ネルンスト方程式は正確な値を与える:

    $$E = E^{\ominus} + \frac{0.059}{z} \log \frac{[\text{oxidised species}]}{[\text{reduced species}]}$$

    ここで $z$ は半反応式における電子の数である。

    Explore · ⁨探索⁩

    Electrode potential lab · ⁨電極電位実験⁩

    Ecell = Eright - Eleft · ⁨⟦Ecell = Eright - Eleft⟦⁩

    Change cell potential and see oxidising power increase. · ⁨セル電位を変化させて、酸化力が増大することを確認せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    electrode/ɪˈlektrəʊd/ 電極
    cathode/ˈkæθəʊd/ カソード
    anode/ˈænəʊd/ アノード
    Faraday constant/ˈfærədeɪ ˈkɒnstənt/ ファラデー定数
    Avogadro constant/ˌævəˈɡædrəʊ ˈkɒnstənt/ アボガドロ定数
    electrode potential/ɪˈlektrəʊd pəˈtenʃl/ 電極電位
    standard electrode potential/ˈstændəd ɪˈlektrəʊd pəˈtenʃl/ 標準電極電位
    standard hydrogen electrode/ˈstændəd ˈhaɪdrədʒn ɪˈlektrəʊd/ 標準水素電極
    standard cell potential/ˈstændəd sel pəˈtenʃl/ 標準電池電位
    oxidising agent/ˈɒksɪdaɪzɪŋ ˈeɪdʒənt/ 酸化剤
    reducing agent/rɪˈdjuːsɪŋ ˈeɪdʒənt/ 還元剤
    half-equation/hɑːf ɪˈkweɪʒn/ 半反応式
    feasibility/ˌfiːzɪˈbɪlɪti/ 可能性
    Nernst equation/nɜːnst ɪˈkweɪʒn/ ネルンストの式
    Watch lesson · ⁨レッスンを視聴⁩
    24.2

    Exam tips · ⁨試験対策⁩

    English
    • $E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{more positive}) - E^{\ominus}(\text{less positive})$; a positive $E^{\ominus}_{\text{cell}}$ means the reaction is feasible.
    • Standard conditions for $E^{\ominus}$: 298 K, $1\ \text{mol dm}^{-3}$, 100 kPa, platinum electrode — state them if asked.
    • A more negative electrode potential means a stronger reducing agent; use the series to predict the direction.
    • For electrolysis quantities use $Q = It$ and moles of electrons $= Q/F$.
    日本語
    • $E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{more positive}) - E^{\ominus}(\text{less positive})$;$E^{\ominus}_{\text{cell}}$が正であれば、反応は可能である。
    • $E^{\ominus}$の標準条件:298 K、$1\ \text{mol dm}^{-3}$、100 kPa、プラチナ電極 — 問われたら答えること。
    • より負の電極電位は強い還元剤を意味する;系列を使って方向性を予測する。
    • 電気分解の量については、$Q = It$と電子の物質量 $= Q/F$を使う。
  • 25

    Equilibria · ⁨平衡⁩

    Watch lesson · ⁨レッスンを視聴⁩
    25.1

    Conjugate acids and bases

    Syllabus · ⁨シラバス⁩
    English
    1. understand and use the terms conjugate acid and conjugate base
    2. define conjugate acid–base pairs, identifying such pairs in reactions
    3. define mathematically the terms pH, $K_a$, $\text{p}K_a$ and $K_w$ and use them in calculations ($K_b$ and the equation $K_w = K_a \times K_b$ will not be tested)
    4. calculate $[\text{H}^+(\text{aq})]$ and pH values for: (a) strong acids (b) strong alkalis (c) weak acids
    5. (a) define a buffer solution (b) explain how a buffer solution can be made (c) explain how buffer solutions control pH; use chemical equations in these explanations (d) describe and explain the uses of buffer solutions, including the role of $\text{HCO}_3^-$ in controlling pH in blood
    6. calculate the pH of buffer solutions, given appropriate data
    7. understand and use the term solubility product, $K_{\text{sp}}$
    8. write an expression for $K_{\text{sp}}$
    9. calculate $K_{\text{sp}}$ from concentrations and vice versa
    10. (a) understand and use the common ion effect to explain the different solubility of a compound in a solution containing a common ion (b) perform calculations using $K_{\text{sp}}$ values and concentration of a common ion
    日本語
    1. 共役酸および共役塩基という用語を理解して適用する
    2. 共役酸-塩基対を定義し、反応式中でそのような対を特定する
    3. pH $K_a$、$\text{p}K_a$ および $K_w$ という用語を数学的に定義し、計算に使用する($K_b$ および方程式 $K_w = K_a \times K_b$ の出題はない)
    4. 以下のpHおよび$[\text{H}^+(\text{aq})]$の値を計算する: (a) 強酸 (b) 強塩基 (c) 弱酸
    5. (a) 緩衝溶液を定義する (b) 緩衝溶液の調製方法を説明する (c) 緩衝溶液がpHを制御する仕組みを化学反応式を用いて説明する (d) 緩衝溶液の用途を説明し、血液内のpH制御における $\text{HCO}_3^-$ の役割についても解説する
    6. 必要なデータを与えて、緩衝溶液のpHを計算する
    7. 溶度積 $K_{\text{sp}}$ という用語を理解して適用する
    8. $K_{\text{sp}}$ に対する表現式を書く
    9. 濃度から$K_{\text{sp}}$を計算したり、逆に濃度を計算したりする
    10. (a) 共通イオン効果を理解して、共通イオンを含む溶液中での化合物の異なる溶解性を説明する (b) $K_{\text{sp}}$ の値および共通イオンの濃度を用いた計算を行う

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Titration curve and equivalence point

    When a Brønsted acid loses an $\text{H}^+$, what is left is its conjugate base 共轭碱. When a base gains an $\text{H}^+$, it becomes its conjugate acid 共轭酸. The two species that differ by just one $\text{H}^+$ form a conjugate acid–base pair 共轭酸碱对.

    For example, in $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$, the pair is $\text{CH}_3\text{COOH}$ (acid) and $\text{CH}_3\text{COO}^-$ (its conjugate base).

    日本語
    Titration curve and equivalence point

    When a Brønsted acid loses an $\text{H}^+$, what is left is its conjugate base 共轭碱. When a base gains an $\text{H}^+$, it becomes its conjugate acid 共轭酸. The two species that differ by just one $\text{H}^+$ form a conjugate acid–base pair 共轭酸碱对.

    For example, in $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$, the pair is $\text{CH}_3\text{COOH}$ (acid) and $\text{CH}_3\text{COO}^-$ (its conjugate base).

    Ethanoic acid losing a proton to its conjugate base, and ammonia gaining a proton to its conjugate acid, with the conjugate pair bracketed
    Conjugate pairs differ by one proton: an acid loses H$^+$ to give its conjugate base; a base gains H$^+$ to give its conjugate acid
    Explore · ⁨探索⁩

    pH and H⁺ concentration

    pH = −log[H⁺]: slide it and watch [H⁺] change tenfold per unit. A conjugate acid–base pair differ by a single proton.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    conjugate base/ˈkɒndʒuːɡeɪt beɪs/ 共役塩基
    conjugate acid/ˈkɒndʒuːɡeɪt ˈæsɪd/ 共役酸
    conjugate acid–base pair/ˈkɒndʒuːɡeɪt ˈæsɪd beɪs peə/ 共役酸塩基対
    25.1

    pH and the equilibrium constants

    English
    • pH measures acidity: $\text{pH} = -\log[\text{H}^+]$, so $[\text{H}^+] = 10^{-\text{pH}}$.
    • the acid dissociation constant 酸解离常数 of a weak acid $\text{HA}$ is $K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$, and $\text{p}K_a = -\log K_a$. A larger $K_a$ (smaller $\text{p}K_a$) means a stronger acid.
    • the ionic product of water 水的离子积 is $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at $298\ \text{K}$.

    Calculating pH

    • strong acid 强酸: fully ionised, so $[\text{H}^+]$ equals the acid concentration; then take $-\log$.
    • strong alkali 强碱: find $[\text{OH}^-]$ from the concentration, then use $[\text{H}^+] = K_w / [\text{OH}^-]$.
    • weak acid 弱酸: only partly ionised, so use $[\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$.

    Worked example. Find the pH of $0.050\ \text{mol dm}^{-3}$ hydrochloric acid (a strong acid).

    It is fully ionised, so $[\text{H}^+] = 0.050\ \text{mol dm}^{-3}$, giving $\text{pH} = -\log(0.050) = 1.30$.

    Worked example. Find the pH of $0.10\ \text{mol dm}^{-3}$ sodium hydroxide (a strong base). ($K_w = 1.0 \times 10^{-14}$.)

    $[\text{OH}^-] = 0.10$, so $[\text{H}^+] = K_w/[\text{OH}^-] = (1.0 \times 10^{-14})/0.10 = 1.0 \times 10^{-13}$, giving $\text{pH} = 13.0$.

    Worked example. Find the pH of $0.10\ \text{mol dm}^{-3}$ ethanoic acid (a weak acid). ($K_a = 1.8 \times 10^{-5}$.)

    $$[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{(1.8 \times 10^{-5})(0.10)} = 1.3 \times 10^{-3}, \qquad \text{pH} = -\log(1.3 \times 10^{-3}) = 2.87.$$
    日本語
    • pH measures acidity: $\text{pH} = -\log[\text{H}^+]$, so $[\text{H}^+] = 10^{-\text{pH}}$.
    • the acid dissociation constant 酸解离常数 of a weak acid $\text{HA}$ is $K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$, and $\text{p}K_a = -\log K_a$. A larger $K_a$ (smaller $\text{p}K_a$) means a stronger acid.
    • the ionic product of water 水的离子积 is $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at $298\ \text{K}$.
    The pH scale runs from acidic through neutral to alkaline, with pH = -log of the hydrogen ion concentration
    The pH scale: pH = -log of the hydrogen-ion concentration
    pH paper strips beside a colour chart
    pH paper estimates the pH of a solution from the colour it turns

    Calculating pH

    • strong acid 强酸: fully ionised, so $[\text{H}^+]$ equals the acid concentration; then take $-\log$.
    • strong alkali 强碱: find $[\text{OH}^-]$ from the concentration, then use $[\text{H}^+] = K_w / [\text{OH}^-]$.
    • weak acid 弱酸: only partly ionised, so use $[\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$.

    Worked example. Find the pH of $0.050\ \text{mol dm}^{-3}$ hydrochloric acid (a strong acid).

    It is fully ionised, so $[\text{H}^+] = 0.050\ \text{mol dm}^{-3}$, giving $\text{pH} = -\log(0.050) = 1.30$.

    Worked example. Find the pH of $0.10\ \text{mol dm}^{-3}$ sodium hydroxide (a strong base). ($K_w = 1.0 \times 10^{-14}$.)

    $[\text{OH}^-] = 0.10$, so $[\text{H}^+] = K_w/[\text{OH}^-] = (1.0 \times 10^{-14})/0.10 = 1.0 \times 10^{-13}$, giving $\text{pH} = 13.0$.

    Worked example. Find the pH of $0.10\ \text{mol dm}^{-3}$ ethanoic acid (a weak acid). ($K_a = 1.8 \times 10^{-5}$.)

    $$[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{(1.8 \times 10^{-5})(0.10)} = 1.3 \times 10^{-3}, \qquad \text{pH} = -\log(1.3 \times 10^{-3}) = 2.87.$$
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    acid dissociation constant/ˈæsɪd dɪˌsəʊsɪˈeɪʃn ˈkɒnstənt/ 酸解離定数
    ionic product of water/aɪˈɒnɪk ˈprɒdʌkt ɒv ˈwɔːtə/ 水のイオン積
    strong acid/strɒŋ ˈæsɪd/ 強酸
    strong alkali/strɒŋ ˈælkəlaɪ/ 強塩基
    weak acid/wiːk ˈæsɪd/ 弱酸
    25.1

    Buffer solutions

    English

    A buffer solution 缓冲溶液 resists a change in pH when a small amount of acid or alkali is added. You make one from a weak acid and its conjugate base (for example ethanoic acid and sodium ethanoate).

    It works because the mixture holds a store of both partners:

    • added $\text{H}^+$ is removed by the conjugate base: $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$.
    • added $\text{OH}^-$ is removed by the weak acid: $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$.

    To find the pH, put the concentrations of the acid and its salt into the $K_a$ expression. Buffers are important in living things — for example, $\text{HCO}_3^-$ keeps the pH of blood close to $7.4$.

    日本語

    A buffer solution 缓冲溶液 resists a change in pH when a small amount of acid or alkali is added. You make one from a weak acid and its conjugate base (for example ethanoic acid and sodium ethanoate).

    It works because the mixture holds a store of both partners:

    • added $\text{H}^+$ is removed by the conjugate base: $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$.
    • added $\text{OH}^-$ is removed by the weak acid: $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$.
    A buffer store of HA and A-, with added H+ removed by A- and added OH- removed by HA
    A buffer holds a store of a weak acid and its conjugate base: added H$^+$ is mopped up by A$^-$ and added OH$^-$ by HA, so the pH barely changes

    To find the pH, put the concentrations of the acid and its salt into the $K_a$ expression. Buffers are important in living things — for example, $\text{HCO}_3^-$ keeps the pH of blood close to $7.4$.

    Explore · ⁨探索⁩

    Buffers and the titration curve

    Add alkali to acid: the flat part is where a buffer resists pH change, and the steep jump is the equivalence point.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    buffer solution/ˈbʌfə səˈluːʃn/ 緩衝溶液
    25.1

    Solubility product

    English

    For a salt that barely dissolves, the solubility product 溶度积 ($K_{\text{sp}}$) is the product of the ion concentrations in a saturated solution, each raised to the power of its number in the formula:

    $$K_{\text{sp}} = [\text{Ag}^+][\text{Cl}^-] \qquad K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^-]^2$$

    You can find $K_{\text{sp}}$ from the solubility, or the solubility from $K_{\text{sp}}$.

    The common ion effect

    The common ion effect 同离子效应 is the way a salt becomes less soluble in a solution that already contains one of its ions. The extra ion pushes the dissolving equilibrium back (Le Chatelier), so less salt dissolves. You can calculate the new solubility using $K_{\text{sp}}$ and the concentration of the common ion.

    日本語

    For a salt that barely dissolves, the solubility product 溶度积 ($K_{\text{sp}}$) is the product of the ion concentrations in a saturated solution, each raised to the power of its number in the formula:

    $$K_{\text{sp}} = [\text{Ag}^+][\text{Cl}^-] \qquad K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^-]^2$$

    You can find $K_{\text{sp}}$ from the solubility, or the solubility from $K_{\text{sp}}$.

    Stalactites hanging from a cave roof
    Stalactites grow as dissolved calcium carbonate slowly comes back out of solution — a real solubility equilibrium

    The common ion effect

    The common ion effect 同离子效应 is the way a salt becomes less soluble in a solution that already contains one of its ions. The extra ion pushes the dissolving equilibrium back (Le Chatelier), so less salt dissolves. You can calculate the new solubility using $K_{\text{sp}}$ and the concentration of the common ion.

    Explore · ⁨探索⁩

    Solubility product lab

    ionic product compared with Ksp

    Increase ion concentration and see when precipitation becomes likely.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    solubility product/ˌsɒljuːˈbɪlɪti ˈprɒdʌkt/ 難溶性塩のイオン積
    common ion effect/ˈkɒmən ˈaɪɒn ɪˈfekt/ 共通イオン効果
    25.2

    Partition coefficients

    Syllabus · ⁨シラバス⁩
    English
    1. state what is meant by the term partition coefficient, $K_{\text{pc}}$
    2. calculate and use a partition coefficient for a system in which the solute is in the same physical state in the two solvents
    3. understand the factors affecting the numerical value of a partition coefficient in terms of the polarities of the solute and the solvents used
    日本語
    1. 分配係数 $K_{\text{pc}}$ という用語の意味を述べる
    2. 溶質が2つの溶媒において同じ物理状態にある系に対して、分配係数を計算して適用する
    3. 溶質および使用された溶媒の極性という観点から、分配係数の数値に影響を与える要因を理解する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    When a solute is shaken with two solvents that do not mix, it spreads between them. The partition coefficient 分配系数 ($K_{\text{pc}}$) is the ratio of its concentrations in the two layers (at constant temperature):

    $$K_{\text{pc}} = \frac{[\text{solute in solvent 1}]}{[\text{solute in solvent 2}]}$$

    This works when the solute 溶质 is in the same physical state in both solvents. The value depends on the polarity 极性 of the solute and of each solvent 溶剂: a non-polar solute dissolves more in the non-polar solvent, while a polar solute prefers the polar solvent.

    日本語

    When a solute is shaken with two solvents that do not mix, it spreads between them. The partition coefficient 分配系数 ($K_{\text{pc}}$) is the ratio of its concentrations in the two layers (at constant temperature):

    $$K_{\text{pc}} = \frac{[\text{solute in solvent 1}]}{[\text{solute in solvent 2}]}$$

    This works when the solute 溶质 is in the same physical state in both solvents. The value depends on the polarity 极性 of the solute and of each solvent 溶剂: a non-polar solute dissolves more in the non-polar solvent, while a polar solute prefers the polar solvent.

    Two immiscible solvent layers with a solute spread between them, more in the non-polar top layer than the water below
    A solute shaken with two immiscible solvents spreads between them; the partition coefficient is the ratio of its concentrations in the two layers
    Explore · ⁨探索⁩

    Partition coefficient lab

    Kpc = concentration organic / concentration aqueous

    Change organic-layer concentration and see the partition ratio.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    partition coefficient/pɑːˈtɪʃn ˌkəʊɪˈfɪʃənt/ 分配係数
    solute/ˈsɒljuːt/ 溶質
    polarity/pəʊˈlærɪti/ 極性
    solvent/ˈsɒlvənt/ 溶媒
    25.2

    Exam tips

    • $\text{pH} = -\log[\text{H}^+]$; for a strong acid $[\text{H}^+]$ = concentration, for a weak acid use $[\text{H}^+] = \sqrt{K_a[\text{HA}]}$.
    • For a base, get $[\text{H}^+]$ from $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$.
    • For a buffer use $[\text{H}^+] = K_a \times [\text{acid}]/[\text{salt}]$ and explain how it removes added $\text{H}^+$/$\text{OH}^-$.
    • Write the $K_{sp}$ expression with the correct powers; the number of decimal places in a pH equals the significant figures of $[\text{H}^+]$.
  • 26

    Reaction kinetics · ⁨反応速度論⁩

    Watch lesson · ⁨レッスンを視聴⁩
    26.1

    Rate equations and orders

    Syllabus · ⁨シラバス⁩
    English
    1. explain and use the terms rate equation, order of reaction, overall order of reaction, rate constant, half-life, rate-determining step and intermediate
    2. (a) understand and use rate equations of the form $\text{rate} = k [\text{A}]^m[\text{B}]^n$ (for which $m$ and $n$ are 0, 1 or 2) (b) deduce the order of a reaction from concentration–time graphs or from experimental data relating to the initial rates method and half-life method (c) interpret experimental data in graphical form, including concentration–time and rate–concentration graphs (d) calculate an initial rate using concentration data (e) construct a rate equation
    3. (a) show understanding that the half-life of a first-order reaction is independent of concentration (b) use the half-life of a first-order reaction in calculations
    4. calculate the numerical value of a rate constant, for example by: (a) using the initial rates and the rate equation (b) using the half-life, $t_{\frac{1}{2}}$, and the equation $k = 0.693/t_{\frac{1}{2}}$
    5. for a multi-step reaction: (a) suggest a reaction mechanism that is consistent with the rate equation and the equation for the overall reaction (b) predict the order that would result from a given reaction mechanism and rate-determining step (c) deduce a rate equation using a given reaction mechanism and rate-determining step for a given reaction (d) identify an intermediate or catalyst from a given reaction mechanism (e) identify the rate determining step from a rate equation and a given reaction mechanism
    6. describe qualitatively the effect of temperature change on the rate constant and hence the rate of a reaction
    日本語
    1. 速度式、反応次数、総合反応次数、速度定数、半減期、律速段階、および中間体という用語を説明して適用する
    2. (a) $\text{rate} = k [\text{A}]^m[\text{B}]^n$ の形の速度式を理解して適用する(ここで $m$ と $n$ は 0, 1, または 2 である) (b) 濃度-時間グラフ、初期速度法または半減期法に関する実験データから、反応の次数を導く (c) 濃度-時間グラフおよび速度-濃度グラフを含む実験データをグラフとして解釈する (d) 濃度データを用いて初期速度を計算する (e) 速度式を作成する
    3. (a) 一次反応の半減期が濃度に依存しないことを理解する。(b) 一次反応の半減期を用いた計算を行う
    4. 速度定数kの数値を計算する、例えば:(a) 初期速度と速度式を用いる (b) 半減期 $t_{\frac{1}{2}}$ と式 $k = 0.693/t_{\frac{1}{2}}$ を用いる
    5. 多段反応について:(a) 速度式および全体反応の式に一致する反応メカニズムを提案する (b) 与えられた反応メカニズムおよび律速段階から生じる反応次数を予測する (c) 与えられた反応メカニズムおよび律速段階を用いて、特定の反応の速度式を導く (d) 与えられた反応メカニズムから中間体または触媒を同定する (e) 速度式および与えられた反応メカニズムから律速段階を同定する
    6. 温度変化が速度定数およびしたがって反応速度に与える影響を定性的に説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A rate equation 速率方程 shows how the rate depends on the concentrations of the reactants:

    $$\text{rate} = k\,[\text{A}]^m[\text{B}]^n$$
    • $m$ is the order of reaction 反应级数 with respect to A, and $n$ the order with respect to B. Each is $0$, $1$ or $2$.
    • the overall order of reaction 总反应级数 is $m + n$.
    • $k$ is the rate constant 速率常数. The rate equation can only be found by experiment, not from the balanced equation.

    Finding the order

    • initial rates method: change one concentration at a time and see how the starting rate changes. If doubling $[\text{A}]$ doubles the rate, the order in A is 1; if it quadruples the rate, the order is 2; if the rate is unchanged, the order is 0.

    Worked example. In experiments on $\text{A} + \text{B} \rightarrow$ products, doubling $[\text{A}]$ (with $[\text{B}]$ fixed) doubles the rate, and doubling $[\text{B}]$ (with $[\text{A}]$ fixed) quadruples the rate. Write the rate equation and give the overall order.

    Doubling $[\text{A}]$ doubles the rate, so first order in A. Doubling $[\text{B}]$ quadruples ($2^2$) the rate, so second order in B. Hence

    $$\text{rate} = k[\text{A}][\text{B}]^2, \qquad \text{overall order} = 1 + 2 = 3.$$
    • graphs: a concentration–time graph for a first-order reaction has a constant half-life 半衰期 (the time for the concentration to halve). A rate–concentration graph is a straight line through the origin for first order, and a curve for second order.

    Half-life and the rate constant

    For a first-order reaction the half-life is constant — it does not depend on the concentration. You can find the rate constant from it:

    $$k = \frac{0.693}{t_{\frac{1}{2}}}$$

    Worked example. A first-order reaction has a half-life of $120\ \text{s}$. Find its rate constant.

    $$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{120} = 5.8 \times 10^{-3}\ \text{s}^{-1}.$$

    You can also find $k$ by putting initial-rate data into the rate equation.

    日本語

    A rate equation 速率方程 shows how the rate depends on the concentrations of the reactants:

    $$\text{rate} = k\,[\text{A}]^m[\text{B}]^n$$
    • $m$ is the order of reaction 反应级数 with respect to A, and $n$ the order with respect to B. Each is $0$, $1$ or $2$.
    • the overall order of reaction 总反应级数 is $m + n$.
    • $k$ is the rate constant 速率常数. The rate equation can only be found by experiment, not from the balanced equation.
    A glass gas syringe
    A gas syringe measures the volume of gas made over time, which gives the rate of reaction

    Finding the order

    • initial rates method: change one concentration at a time and see how the starting rate changes. If doubling $[\text{A}]$ doubles the rate, the order in A is 1; if it quadruples the rate, the order is 2; if the rate is unchanged, the order is 0.

    Worked example. In experiments on $\text{A} + \text{B} \rightarrow$ products, doubling $[\text{A}]$ (with $[\text{B}]$ fixed) doubles the rate, and doubling $[\text{B}]$ (with $[\text{A}]$ fixed) quadruples the rate. Write the rate equation and give the overall order.

    Doubling $[\text{A}]$ doubles the rate, so first order in A. Doubling $[\text{B}]$ quadruples ($2^2$) the rate, so second order in B. Hence

    $$\text{rate} = k[\text{A}][\text{B}]^2, \qquad \text{overall order} = 1 + 2 = 3.$$
    • graphs: a concentration–time graph for a first-order reaction has a constant half-life 半衰期 (the time for the concentration to halve). A rate–concentration graph is a straight line through the origin for first order, and a curve for second order.
    Three rate-against-concentration lines: a flat line for zero order, a straight line through the origin for first order, and an upward curve for second order
    Rate against concentration: zero order is a flat line, first order a straight line through the origin, second order an upward curve

    Half-life and the rate constant

    For a first-order reaction the half-life is constant — it does not depend on the concentration. You can find the rate constant from it:

    $$k = \frac{0.693}{t_{\frac{1}{2}}}$$
    A concentration-time decay curve with three equal half-life intervals marked, the concentration halving each time
    A first-order reaction has a constant half-life: the concentration halves in the same time $t_{1/2}$ again and again, whatever the starting value

    Worked example. A first-order reaction has a half-life of $120\ \text{s}$. Find its rate constant.

    $$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{120} = 5.8 \times 10^{-3}\ \text{s}^{-1}.$$

    You can also find $k$ by putting initial-rate data into the rate equation.

    Explore · ⁨探索⁩

    Rate equations & orders · ⁨速度式 & 反応次数⁩

    [A] = [A]₀·bᵗ

    A first-order reaction decays exponentially — equal half-lives. · ⁨一次反応は指数関数的に減少し、半減期は一定である。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    rate equation/reɪt ɪˈkweɪʒn/ 速度式
    order of reaction/ˈɔːdə ɒv rɪˈækʃn/ 反応次数
    overall order of reaction/ˌəʊvəˈrɔːl ˈɔːdə ɒv rɪˈækʃn/ 全反応次数
    rate constant/reɪt ˈkɒnstənt/ 速度定数
    half-life/hɑːf laɪf/ 半減期
    26.1

    Reaction mechanisms

    English

    Most reactions happen in several steps. The slowest step is the rate-determining step 决速步骤, and it controls the overall rate. Only the species involved up to and including this step appear in the rate equation.

    • an intermediate 中间体 is a species made in one step and then used up in a later step. It is not in the overall equation.
    • you can suggest a mechanism that fits both the rate equation and the overall equation, predict the order from a given mechanism, or pick out the rate-determining step.

    If you compare the initial rate 初始速率 of different mixtures, you can deduce the rate equation, and from that work out the mechanism.

    Effect of temperature

    Raising the temperature increases the rate constant $k$ (more molecules pass the activation energy), so the rate goes up.

    日本語

    Most reactions happen in several steps. The slowest step is the rate-determining step 决速步骤, and it controls the overall rate. Only the species involved up to and including this step appear in the rate equation.

    A reaction energy profile with two humps: a tall first barrier marked as the slow rate-determining step, a dip for the intermediate, then a smaller second barrier
    A two-step profile: the slower step has the bigger barrier and is rate-determining; the dip between the two barriers is an intermediate
    • an intermediate 中间体 is a species made in one step and then used up in a later step. It is not in the overall equation.
    • you can suggest a mechanism that fits both the rate equation and the overall equation, predict the order from a given mechanism, or pick out the rate-determining step.

    If you compare the initial rate 初始速率 of different mixtures, you can deduce the rate equation, and from that work out the mechanism.

    Effect of temperature

    Raising the temperature increases the rate constant $k$ (more molecules pass the activation energy), so the rate goes up.

    Explore · ⁨探索⁩

    Organic mechanism route · ⁨有機反応メカニズムの流れ⁩

    Trace electron-pair movement from reagent to product. · ⁨試薬から生成物への電子対の移動を追跡する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    rate-determining step/reɪt dɪˈtɜːmɪnɪŋ step/ 律速段階
    intermediate/ˌɪntəˈmiːdɪət/ 中間体
    initial rate/ɪˈnɪʃl reɪt/ 今は遅い
    26.2

    Catalysts

    Syllabus · ⁨シラバス⁩
    English
    1. explain that catalysts can be homogeneous or heterogeneous
    2. describe the mode of action of a heterogeneous catalyst to include adsorption of reactants, bond weakening and desorption of products, for example: (a) iron in the Haber process (b) palladium, platinum and rhodium in the catalytic removal of oxides of nitrogen from the exhaust gases of car engines
    3. describe the mode of action of a homogeneous catalyst by being used in one step and reformed in a later step, for example: (a) atmospheric oxides of nitrogen in the oxidation of atmospheric sulfur dioxide (b) $\text{Fe}^{2+}$ or $\text{Fe}^{3+}$ in the $\text{I}^- / \text{S}_2\text{O}_8^{2-}$ reaction
    日本語
    1. 触媒は 均一系 または 不均一系 であることができることを説明する
    2. 不均一系触媒 の作用機構を記述し、反応物の 吸着、結合の弱体化、生成物の 脱離 を含む。例:(a) ハーバー法における鉄 (b) 自動車排気ガス中の窒素酸化物を触媒除去するためのパラジウム、プラチナおよびロジウム
    3. 均一系触媒の作用機構を、ある段階で使用され、後続の段階で再生される点について記述する。例:(a) 大気中の硫黄二酸化塩素の酸化における大気中の窒素酸化物 (b) $\text{Fe}^{2+}$ または $\text{Fe}^{3+}$ が関与する $\text{I}^- / \text{S}_2\text{O}_8^{2-}$ 反応

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Catalysts 催化剂 can be homogeneous or heterogeneous.

    Heterogeneous catalysts

    A heterogeneous catalyst 多相催化剂 is in a different physical state from the reactants (usually a solid with gases). It works in three stages:

    1. adsorption 吸附: reactant molecules stick to the catalyst surface.
    2. the bonds in the reactants are weakened, so they react more easily.
    3. desorption 脱附: the product molecules leave the surface.

    Examples are iron in the Haber process, and platinum, palladium and rhodium in a catalytic converter.

    Homogeneous catalysts

    A homogeneous catalyst 均相催化剂 is in the same physical state as the reactants. It is used up in one step and then reformed in a later step, so it comes back unchanged. Examples are oxides of nitrogen helping to oxidise atmospheric sulfur dioxide, and $\text{Fe}^{2+}$ or $\text{Fe}^{3+}$ speeding up the reaction between $\text{I}^-$ and $\text{S}_2\text{O}_8^{2-}$.

    日本語

    Catalysts 催化剂 can be homogeneous or heterogeneous.

    Heterogeneous catalysts

    A heterogeneous catalyst 多相催化剂 is in a different physical state from the reactants (usually a solid with gases). It works in three stages:

    1. adsorption 吸附: reactant molecules stick to the catalyst surface.
    2. the bonds in the reactants are weakened, so they react more easily.
    3. desorption 脱附: the product molecules leave the surface.
    Three stages on a catalyst surface: two reactant atoms adsorbing onto it, reacting with weakened bonds, then the product desorbing
    A heterogeneous catalyst works in three stages: the reactants adsorb onto the surface, their weakened bonds let them react, then the product desorbs
    A tray holding many small ceramic catalyst pieces in different shapes: solid cylinders, rings, ribbed rods and discs with holes
    Real heterogeneous catalysts are made as small shaped pellets, rings and perforated discs, which give a large surface area for the reactants to stick to

    Examples are iron in the Haber process, and platinum, palladium and rhodium in a catalytic converter.

    Homogeneous catalysts

    A homogeneous catalyst 均相催化剂 is in the same physical state as the reactants. It is used up in one step and then reformed in a later step, so it comes back unchanged. Examples are oxides of nitrogen helping to oxidise atmospheric sulfur dioxide, and $\text{Fe}^{2+}$ or $\text{Fe}^{3+}$ speeding up the reaction between $\text{I}^-$ and $\text{S}_2\text{O}_8^{2-}$.

    Explore · ⁨探索⁩

    How a catalyst speeds a reaction · ⁨触媒が反応を速くする仕組み⁩

    Turn the catalyst on and watch the rate jump — it gives more successful collisions per second by offering a lower-energy path. · ⁨触媒を作用させると、より低いエネルギーの経路を提供することで秒あたりの有効衝突数が増え、速度が急上昇します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    catalyst/ˈkætəlɪst/ 触媒
    heterogeneous catalyst/ˌhetrəˈdʒiːnɪəs ˈkætəlɪst/ 不均一触媒
    adsorption/ədˈsɔːpʃn/ 吸着
    desorption/dɪˈsɔːpʃn/ 脱離
    homogeneous catalyst/həˈməʊdʒnɪəs ˈkætəlɪst/ 均一触媒
    26.2

    Exam tips

    • Find orders from initial-rate data: doubling a concentration that doubles the rate is first order, quadruples it is second order, no change is zero order.
    • Write $\text{rate} = k[\text{A}]^m[\text{B}]^n$ and work out the units of $k$ from it.
    • The rate-determining step contains the species (and orders) in the rate equation — use this to test a mechanism.
    • A constant half-life means first order.
  • 27

    Group 2 · ⁨第2族⁩

    Watch lesson · ⁨レッスンを視聴⁩
    27.1

    Thermal stability of the nitrates and carbonates · ⁨ニトリート類と炭酸塩類の熱的安定性⁩

    Syllabus · ⁨シラバス⁩
    English
    1. describe and explain qualitatively the trend in the thermal stability of the nitrates and carbonates including the effect of ionic radius on the polarisation of the large anion
    2. describe and explain qualitatively the variation in solubility and of enthalpy change of solution, $\Delta H^{\ominus}_{\text{sol}}$, of the hydroxides and sulfates in terms of relative magnitudes of the enthalpy change of hydration and the lattice energy
    日本語
    1. 熱安定性 の傾向をマグネシウム炭酸塩および硝酸塩について定性的に記述・説明し、大きな陰イオンの分極に対するイオン半径の影響を含む
    2. 水化エンタルピー変化 および 格子エネルギー の相対的な大きさに基づき、水酸化物および硫酸塩の溶解度の違いおよび溶解エンタルピー変化 $\Delta H^{\ominus}_{\text{sol}}$ を定性的に記述・説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The thermal stability 热稳定性 of the Group 2 nitrates 硝酸盐 and carbonates 碳酸盐 increases down the group. Here is the reason, in terms of how the ions affect each other.

    A small cation 阳离子 with a small ionic radius 离子半径 has a high charge density. It pulls on the electrons of the nearby anion and distorts its shape — this is polarisation 极化. Distorting the large anion 阴离子 (the carbonate or nitrate ion) weakens a bond inside it, so the compound breaks down more easily.

    Going down the group, the cation gets larger. Its charge density drops, so it polarises the anion less. The anion is less distorted, so the compound is harder to break down — it is more thermally stable and needs a higher temperature to decompose.

    Worked example. Down Group 2 the hydroxides become more soluble while the sulfates become less soluble. Explain why the two trends run in opposite directions. Solubility is a contest between the lattice energy holding the solid together and the hydration energy rewarding the ions for dissolving. Both get less exothermic as the cation grows, so whichever falls faster decides the trend. With the small hydroxide ion, the lattice energy depends strongly on the cation's size, so it falls faster than the hydration energy: the lattice gets relatively easier to break and solubility rises. With the large sulfate ion, the lattice energy is already dominated by that big anion and barely changes down the group, while the cation's hydration energy still falls - so solubility falls. Name both energies and say which one falls faster; simply restating the trends earns no explanation marks.

    日本語

    第2群のニトリート類と炭酸塩類の熱的安定性は、下に行くにつれて増大する。これはイオン相互の影響によるものである。

    石灰工場内の回転炉
    石灰窯内で石灰岩(炭酸カルシウム)を加熱すると、酸化カルシウムへ分解される——これは熱分解である

    イオン半径の小さい小さなカチオンは高い電荷密度を持ちます。近傍の陰イオンの電子を引き寄せ、その形状を歪ませます——これが分極です。大きな陰イオン(炭酸イオンまたは硝酸イオン)を歪ませると、内部の結合が弱まり、化合物がより分解されやすくなります。

    族を下がるにつれて、カチオンは大きくなります。電荷密度が低下するため、陰イオンをより少ない程度で分極します。陰イオンの歪みが小さくなるため、化合物は分解されにくくなり、熱的に安定性が高くなり、分解にはより高い温度が必要になります。

    二つの状況:小さなマグネシウムカチオンが炭酸陰イオンの電子雲を涙滴状に強く歪ませる一方、大きなバリウムカチオンはほとんど歪ませない
    小さなカチオンは高い電荷密度を持つため、大きな陰イオンをより多く歪ませ(分極させ)、それを弱め、結果として化合物は不安定になる

    ** worked example. ** 第2族を下るにつれ、水酸化物は溶解度が増加し、硫酸塩は溶解度が減少します。なぜ2つの傾向が逆向きになるかを説明してください。溶解度は、固体を結びつける格子エネルギーと、イオンを溶解させることで得られる水和エネルギーとの競争です。両方ともカチオンが大きくなると発熱性が低下しますが、どちらが速く低下するかによって傾向が決まります。小さい水酸化物イオンの場合、格子エネルギーはカチオンのサイズに強く依存するため、水和エネルギーよりも速く低下します:格子は相対的に壊しやすくなり、溶解度は上昇します。大きい硫酸イオンの場合、格子エネルギーはすでにその大きな陰イオンによって支配されており、族を下ってもほとんど変化せず、カチオンの水和エネルギーだけが依然として低下するため、溶解度は低下します。両方のエネルギーの名前を挙げ、どちらが速く低下するかを述べなさい。単に傾向を繰り返すだけでは解説点数は与えられません。

    Explore · ⁨探索⁩

    Group 2 thermal stability ladder · ⁨第2族の熱安定性の梯子⁩

    Move down Group 2 and see why carbonates become harder to decompose. · ⁨第2族を下に進んで、炭酸塩が分解しにくくなる理由を見てみましょう。⁩

    Explore · ⁨探索⁩

    Group 2 solubility lab · ⁨第2族の溶解度実験⁩

    Classify Group 2 compounds by solubility trend. · ⁨溶解度の傾向に基づいて第2族化合物を分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    thermal stability/ˈθɜːml stəˈbɪlɪti/ 熱安定性
    nitrate/ˈnaɪtreɪt/ 窒素酸
    carbonate/ˈkɑːbəneɪt/ 炭酸塩
    cation/ˈkætaɪən/ カチオン
    ionic radius/aɪˈɒnɪk ˈreɪdɪəs/ イオン半径
    polarisation/ˌpəʊləraɪˈzeɪʃn/ 分極
    anion/ˈænaɪən/ アニオン
    enthalpy change of solution/enˈθælpi tʃeɪndʒ ɒv səˈluːʃn/ 溶解エンタルピー変化
    lattice energy/ˈlætɪs ˈenədʒi/ 格子エネルギー
    enthalpy change of hydration/enˈθælpi tʃeɪndʒ ɒv haɪˈdreɪʃn/ 水和エンタルピー変化
    27.1

    Solubility of the hydroxides and sulfates · ⁨水酸化物および硫酸塩の溶解度⁩

    English

    When an ionic solid dissolves, two energy changes compete. The enthalpy change of solution 溶解焓变 is the sum of them:

    $$\Delta H^{\ominus}_{\text{sol}} = -\Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$$
    • you must first pull the lattice apart (this needs the lattice energy 晶格能).
    • then water surrounds the ions (this releases the enthalpy change of hydration 水合焓变).

    If the energy released on hydration roughly matches (or beats) the energy needed to break the lattice, the solid dissolves easily.

    Going down the group, both the lattice energy and the hydration enthalpy get smaller (less negative), because the cation is larger. But they shrink at different rates, and this explains the opposite trends:

    Compound Trend in solubility 溶解度 down the group Why
    hydroxides 氢氧化物 increases the $\text{OH}^-$ ion is small, so the lattice energy falls a lot down the group; this change outweighs the fall in hydration energy
    sulfates 硫酸盐 decreases the $\text{SO}_4^{2-}$ ion is large, so the lattice energy stays almost the same; the fall in hydration energy then dominates, so dissolving becomes less favourable
    日本語

    イオン性固体が溶解する際、2つのエネルギー変化が競合します。溶解エンタルピー変化はそれらの総和です:

    第2族を下るにつれ、水酸化物は溶解度が増加するが、硫酸塩は減少する
    第2族を下るにつれ、水酸化物はより溶解しやすくなるが、硫酸塩は溶解しにくくなる
    $$\Delta H^{\ominus}_{\text{sol}} = -\Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$$
    • まず格子をばらばらにする必要がある(これには格子エネルギーがかかる)。
    • その後水がイオンを取り囲む(これにより水和エンタルピー変化が放出される)。

    水和時に放出されるエネルギーが、格子を破るために必要なエネルギーとおおよそ一致(あるいは上回る)場合、固体は簡単に溶解する。

    淡い青色のゲル状水酸化銅(II)沈殿
    格子エネルギーが水和によって償われすぎない場合、不溶性金属水酸化物がゲル状沈殿物として生成される

    族を下るにつれて、カチオンが大きくなるため、格子エネルギーと水和エンタルピーのどちらも小さく(負の値が小さく)なります。しかし、それらは異なる速度で縮小し、このことが逆向きの傾向を説明します:

    水酸化物と硫酸塩を比較する2つのボックス:水酸化物では格子エネルギーが大きく低下するのに対し、硫酸塩では一定であるため、溶解度の傾向は逆になる
    傾向が逆向きになる理由:小さいOH$^-$の格子エネルギーは族を下るにつれて大きく低下する(水酸化物はより溶けやすくなる)が、大きいSO$_4^{2-}$の格子エネルギーはほとんど変化しない(硫酸塩は溶けにくくなる)
    化合物 族を下るにつれた溶解度の傾向 理由
    水酸化物 増加 $\text{OH}^-$イオンは小さいため、格子エネルギーは族を下るにつれて大きく低下する;この変化は水和エネルギーの低下を上回る
    硫酸塩 減少 $\text{SO}_4^{2-}$イオンは大きいので、格子エネルギーはほぼ同じまま維持される;そのため水和エネルギーの低下が支配的となり、溶解は不利になる
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    solubility/ˌsɒljuːˈbɪlɪti/ 溶解度
    hydroxide/haɪˈdrɒksaɪd/ 水酸化物
    sulfate/ˈsʌlfeɪt/ 硫酸塩
    27.1

    Exam tips · ⁨試験対策⁩

    English
    • Thermal stability of carbonates and nitrates increases down the group — larger cations polarise the anion less.
    • Give the decomposition products: carbonates → oxide + $\text{CO}_2$; nitrates → oxide + $\text{NO}_2$ + $\text{O}_2$ (brown gas).
    • Hydroxides get more soluble and sulfates less soluble down the group.
    • Explain trends with cation charge density (polarising power), not just size.
    日本語
    • 炭酸塩および硝酸塩の熱的安定性は族を下るにつれて増加する——大きなカチオンは陰イオンをより少ない程度で分極する。
    • 分解生成物を答える:炭酸塩 → 酸化物 + $\text{CO}_2$;硝酸塩 → 酸化物 + $\text{NO}_2$ + $\text{O}_2$(茶色の気体)。
    • 族を下るにつれ、水酸化物は溶解度が増加し、硫酸塩は減少する。
    • 単なるサイズだけでなく、**カチオンの電荷密度(分極力)**を用いて傾向を説明する。
  • 28

    Chemistry of transition elements · ⁨遷移元素の化学⁩

    Watch lesson · ⁨レッスンを視聴⁩
    28.1

    Transition elements

    Syllabus · ⁨シラバス⁩
    English
    1. define a transition element as a d-block element which forms one or more stable ions with incomplete d orbitals
    2. sketch the shape of a $3\text{d}_{xy}$ orbital and $3\text{d}_{z^2}$ orbital
    3. understand that transition elements have the following properties: (a) they have variable oxidation states (b) they behave as catalysts (c) they form complex ions (d) they form coloured compounds
    4. explain why transition elements have variable oxidation states in terms of the similarity in energy of the 3d and the 4s sub-shells
    5. explain why transition elements behave as catalysts in terms of having more than one stable oxidation state, and vacant d orbitals that are energetically accessible and can form dative bonds with ligands
    6. explain why transition elements form complex ions in terms of vacant d orbitals that are energetically accessible
    日本語
    1. 遷移元素 を、不完全なd軌道を持つ1つ以上の安定したイオンを形成する dブロック元素 として定義する
    2. $3\text{d}_{xy}$ 軌道および $3\text{d}_{z^2}$ 軌道の形状をスケッチする
    3. 遷移元素が以下の性質を持つことを理解する:(a) 可変酸化状態 を持つ (b) 触媒 として振る舞う (c) 錯イオン を形成する (d) 有色化合物 を形成する
    4. 3dおよび4sサブシェル間のエネルギーの類似性を根拠として、遷移元素が可変酸化状態を持つ理由を説明する
    5. 遷移元素が触媒として振る舞う理由を、複数の安定した酸化状態を持ち、エネルギー的にアクセス可能な空のd軌道があり、配位子 と配位結合を形成できる点に基づいて説明する
    6. 遷移元素が錯イオンを形成する理由を、エネルギー的にアクセス可能な空のd軌道がある点に基づいて説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A transition element 过渡元素 is a d-block element that forms one or more stable ions with incomplete d orbitals. (Scandium and zinc are in the d-block but are not transition elements, because their stable ions have empty or full d orbitals.)

    The 3d orbitals 轨道 have set shapes: the $3\text{d}_{xy}$ orbital has four lobes pointing between the axes, and the $3\text{d}_{z^2}$ orbital has two lobes along the $z$-axis with a ring around the middle.

    Four key properties (and why)

    Property Reason
    variable oxidation state 氧化态 the 3d and 4s sub-shells are close in energy, so similar small amounts of energy remove different numbers of electrons
    act as a catalyst 催化剂 they have more than one stable oxidation state, and vacant d orbitals that can form dative bonds
    form complex ions vacant d orbitals can accept lone pairs
    form coloured compounds electrons move between split d orbitals (see below)
    日本語

    A transition element 过渡元素 is a d-block element that forms one or more stable ions with incomplete d orbitals. (Scandium and zinc are in the d-block but are not transition elements, because their stable ions have empty or full d orbitals.)

    The 3d orbitals 轨道 have set shapes: the $3\text{d}_{xy}$ orbital has four lobes pointing between the axes, and the $3\text{d}_{z^2}$ orbital has two lobes along the $z$-axis with a ring around the middle.

    Four key properties (and why)

    Property Reason
    variable oxidation state 氧化态 the 3d and 4s sub-shells are close in energy, so similar small amounts of energy remove different numbers of electrons
    act as a catalyst 催化剂 they have more than one stable oxidation state, and vacant d orbitals that can form dative bonds
    form complex ions vacant d orbitals can accept lone pairs
    form coloured compounds electrons move between split d orbitals (see below)
    Four properties: variable oxidation states, acting as catalysts, forming complex ions, and coloured compounds
    The four key properties of the transition elements
    A red ruby crystal
    A ruby is red because of transition-metal (chromium) ions held in its crystal lattice
    Explore · ⁨探索⁩

    Transition element property lab · ⁨遷移元素の性質実験⁩

    Sort transition-metal evidence by the property it shows. · ⁨遷移金属の証拠を、その性質によって分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    transition element/trænˈsɪʃn ˈelɪmənt/ 遷移元素
    orbital/ˈɔːbɪtl/ 軌道
    oxidation state/ˌɒksɪˈdeɪʃn steɪt/ 酸化数
    catalyst/ˈkætəlɪst/ 触媒
    28.2

    Ligands and complexes

    Syllabus · ⁨シラバス⁩
    English
    1. describe and explain the reactions of transition elements with ligands to form complexes, including the complexes of copper(II) and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions
    2. define the term ligand as a species that contains a lone pair of electrons that forms a dative covalent bond to a central metal atom/ion
    3. understand and use the terms: (a) monodentate ligand including as examples $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{Cl}^-$ and $\text{CN}^-$ (b) bidentate ligand including as examples 1,2-diaminoethane, en, $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ and the ethanedioate ion, $\text{C}_2\text{O}_4^{2-}$ (c) polydentate ligand including as an example $\text{EDTA}^{4-}$
    4. define the term complex as a molecule or ion formed by a central metal atom/ion surrounded by one or more ligands
    5. describe the geometry (shape and bond angles) of transition element complexes which are linear, square planar, tetrahedral or octahedral
    6. (a) state what is meant by coordination number (b) predict the formula and charge of a complex ion, given the metal ion, its charge or oxidation state, the ligand and its coordination number or geometry
    7. explain qualitatively that ligand exchange can occur, including the complexes of copper(II) ions and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions
    8. predict, using $E^\ominus$ values, the feasibility of redox reactions involving transition elements and their ions
    9. describe the reactions of, and perform calculations involving: (a) $\text{MnO}_4^- / \text{C}_2\text{O}_4^{2-}$ in acid solution given suitable data (b) $\text{MnO}_4^- / \text{Fe}^{2+}$ in acid solution given suitable data (c) $\text{Cu}^{2+} / \text{I}^-$ given suitable data
    10. perform calculations involving other redox systems given suitable data
    日本語
    1. 遷移元素が 配位子 と反応して 錯体 を形成する反応を記述・説明し、铜(II)およびコバルト(II)イオンの水分子およびアンモニア分子、水酸化物イオンおよび塩化物イオンとの錯体を含む
    2. 非共有電子対を含み、中心金属原子/イオンに配位結合を形成する種である 配位子 という用語を定義する
    3. 以下の用語を理解し使用する:(a) 単座配位子。例として$\text{H}_2\text{O}$、$\text{NH}_3$、$\text{Cl}^-$、$\text{CN}^-$を含む。(b) 二座配位子。例として1,2-ジアミノエタン、en、$\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$、およびエタンジオ酸イオン、$\text{C}_2\text{O}_4^{2-}$を含む。(c) 多座配位子。例として$\text{EDTA}^{4-}$を含む
    4. 錯体 という用語を、1つ以上の配位子に囲まれた中心金属原子/イオンによって形成される分子またはイオンとして定義する
    5. 遷移元素錯体の幾何学的配置(形状および結合角)を、直線形、平面四角形、四面体形、または 八面体形 で記述する
    6. (a) 配位数 が何を意味するかを述べる (b) 金属イオン、その電荷または酸化状態、配位子、およびその配位数または幾何学的配置が与えられた場合、錯イオンの化学式および電荷を予測する
    7. 配位子置換反応 が起こりうることを定性的に説明し、铜(II)およびコバルト(II)イオンの水分子およびアンモニア分子、水酸化物イオンおよび塩化物イオンとの錯体を含む
    8. $E^\ominus$ 値を用いて、遷移元素およびそのイオンに関わる酸化還元反応の進行可能性を予測する
    9. 適切なデータが与えられた場合の反応および計算を行う:(a) $\text{MnO}_4^- / \text{C}_2\text{O}_4^{2-}$ (酸性溶液中) (b) $\text{MnO}_4^- / \text{Fe}^{2+}$ (酸性溶液中) (c) $\text{Cu}^{2+} / \text{I}^-$
    10. 適切なデータが与えられた場合、他の酸化還元系に関する計算を行う

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A transition metal ion can be surrounded by complex ions 配离子. The species attached are ligands.

    A ligand 配体 is a species with a lone pair of electrons that forms a dative covalent bond 配位键 to the central metal ion. (The lone pair 孤对电子 is what it donates.) Ligands are grouped by how many such bonds they can form:

    • monodentate 单齿: one bond ($\text{H}_2\text{O}$, $\text{NH}_3$, $\text{Cl}^-$, $\text{CN}^-$).
    • bidentate 双齿: two bonds (1,2-diaminoethane "en", and the ethanedioate ion $\text{C}_2\text{O}_4^{2-}$).
    • polydentate 多齿: many bonds ($\text{EDTA}^{4-}$, which uses six).

    A complex 配合物 is a central metal atom or ion surrounded by one or more ligands. Its shape can be linear 直线形, square planar 平面正方形, tetrahedral 四面体形 or octahedral 八面体形.

    The coordination number 配位数 is the number of dative bonds from the ligands to the central ion (6 → octahedral, 4 → tetrahedral or square planar, 2 → linear). To predict the charge of a complex, add the metal's charge and all the ligand charges.

    Ligand exchange

    In ligand exchange 配体交换 one ligand replaces another, often with a colour change. For copper(II):

    $$[\text{Cu}(\text{H}_2\text{O})_6]^{2+} \;(\text{pale blue}) \;\xrightarrow{\text{NH}_3}\; [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} \;(\text{deep blue})$$

    With concentrated $\text{HCl}$ it becomes yellow $[\text{CuCl}_4]^{2-}$; cobalt(II) behaves in a similar way.

    Redox reactions of transition ions

    Use $E^{\ominus}$ values to predict whether a redox reaction is feasible. Common titrations you should be able to calculate include $\text{MnO}_4^-/\text{C}_2\text{O}_4^{2-}$ and $\text{MnO}_4^-/\text{Fe}^{2+}$ in acid (purple to colourless), and $\text{Cu}^{2+}/\text{I}^-$ (which makes iodine, then titrated with thiosulfate).

    Worked example. In acid, $\text{MnO}_4^-$ reacts with $\text{Fe}^{2+}$ in the ratio $1:5$ (see the balanced equation above). A $25.0\ \text{cm}^3$ sample of $\text{Fe}^{2+}$ solution needs $22.0\ \text{cm}^3$ of $0.0200\ \text{mol dm}^{-3}$ $\text{KMnO}_4$ to reach the end point. Find the concentration of the $\text{Fe}^{2+}$.

    Moles of $\text{MnO}_4^-$ used $= 0.0200 \times \dfrac{22.0}{1000} = 4.40 \times 10^{-4}\ \text{mol}$. Each mole of $\text{MnO}_4^-$ reacts with $5$ moles of $\text{Fe}^{2+}$, so moles of $\text{Fe}^{2+} = 5 \times 4.40 \times 10^{-4} = 2.20 \times 10^{-3}\ \text{mol}$. This was in $25.0\ \text{cm}^3$, so

    $$[\text{Fe}^{2+}] = \frac{2.20 \times 10^{-3}}{25.0/1000} = 0.0880\ \text{mol dm}^{-3}.$$
    日本語

    A transition metal ion can be surrounded by complex ions 配离子. The species attached are ligands.

    A ligand 配体 is a species with a lone pair of electrons that forms a dative covalent bond 配位键 to the central metal ion. (The lone pair 孤对电子 is what it donates.) Ligands are grouped by how many such bonds they can form:

    • monodentate 单齿: one bond ($\text{H}_2\text{O}$, $\text{NH}_3$, $\text{Cl}^-$, $\text{CN}^-$).
    • bidentate 双齿: two bonds (1,2-diaminoethane "en", and the ethanedioate ion $\text{C}_2\text{O}_4^{2-}$).
    • polydentate 多齿: many bonds ($\text{EDTA}^{4-}$, which uses six).
    A metal ion with a monodentate ligand making one bond, a bidentate ligand making two bonds, and a polydentate ligand making six
    Ligands are grouped by how many dative bonds they form: monodentate (one), bidentate (two) or polydentate (many, like EDTA's six)

    A complex 配合物 is a central metal atom or ion surrounded by one or more ligands. Its shape can be linear 直线形, square planar 平面正方形, tetrahedral 四面体形 or octahedral 八面体形.

    The coordination number 配位数 is the number of dative bonds from the ligands to the central ion (6 → octahedral, 4 → tetrahedral or square planar, 2 → linear). To predict the charge of a complex, add the metal's charge and all the ligand charges.

    Four complexes: a linear two-ligand, a tetrahedral and a square planar four-ligand, and an octahedral six-ligand, each labelled with its coordination number
    Complex shapes follow the coordination number: 2 is linear, 4 is tetrahedral or square planar, 6 is octahedral

    Ligand exchange

    In ligand exchange 配体交换 one ligand replaces another, often with a colour change. For copper(II):

    $$[\text{Cu}(\text{H}_2\text{O})_6]^{2+} \;(\text{pale blue}) \;\xrightarrow{\text{NH}_3}\; [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} \;(\text{deep blue})$$

    With concentrated $\text{HCl}$ it becomes yellow $[\text{CuCl}_4]^{2-}$; cobalt(II) behaves in a similar way.

    Three coloured boxes for copper(II): pale blue with water, deep blue after adding ammonia, yellow after adding concentrated HCl
    Ligand exchange changes the colour of copper(II): pale blue with water, deep blue with ammonia, yellow with concentrated HCl

    Redox reactions of transition ions

    Use $E^{\ominus}$ values to predict whether a redox reaction is feasible. Common titrations you should be able to calculate include $\text{MnO}_4^-/\text{C}_2\text{O}_4^{2-}$ and $\text{MnO}_4^-/\text{Fe}^{2+}$ in acid (purple to colourless), and $\text{Cu}^{2+}/\text{I}^-$ (which makes iodine, then titrated with thiosulfate).

    Worked example. In acid, $\text{MnO}_4^-$ reacts with $\text{Fe}^{2+}$ in the ratio $1:5$ (see the balanced equation above). A $25.0\ \text{cm}^3$ sample of $\text{Fe}^{2+}$ solution needs $22.0\ \text{cm}^3$ of $0.0200\ \text{mol dm}^{-3}$ $\text{KMnO}_4$ to reach the end point. Find the concentration of the $\text{Fe}^{2+}$.

    Moles of $\text{MnO}_4^-$ used $= 0.0200 \times \dfrac{22.0}{1000} = 4.40 \times 10^{-4}\ \text{mol}$. Each mole of $\text{MnO}_4^-$ reacts with $5$ moles of $\text{Fe}^{2+}$, so moles of $\text{Fe}^{2+} = 5 \times 4.40 \times 10^{-4} = 2.20 \times 10^{-3}\ \text{mol}$. This was in $25.0\ \text{cm}^3$, so

    $$[\text{Fe}^{2+}] = \frac{2.20 \times 10^{-3}}{25.0/1000} = 0.0880\ \text{mol dm}^{-3}.$$
    Explore · ⁨探索⁩

    Ligand and complex lab · ⁨配位子と錯体の実験⁩

    Identify the part of a complex ion that controls its structure. · ⁨錯イオンの構造を制御する部分を特定せよ。⁩

    Explore · ⁨探索⁩

    Ligand exchange route · ⁨配位子交換の経路⁩

    Follow one ligand replacing another around a metal ion. · ⁨金属イオン周りで1つの配位子が別の配位子に置き換わる流れを追え。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    complex ion/ˈkɒmpleks ˈaɪɒn/ 錯イオン
    ligand/ˈlɪɡænd/ 配位子
    dative covalent bond/ˈdætɪv ˈkəʊvələnt bɒnd/ 配位結合
    lone pair/ləʊn peə/ 非共有電子対
    monodentate/ˈmɒnəʊdenteɪt/ 単座
    bidentate/baɪˈdenteɪt/ 二座
    polydentate/ˌpɒlɪˈdenteɪt/ 多座配位子
    complex/ˈkɒmpleks/ このレベルにおいて contrast がよく表すものを述べよ。
    linear/ˈlɪnɪə/ 直線的である
    square planar/skweə ˈpleɪnə/ 正方形平面型
    tetrahedral/ˌtetrəˈhiːdrəl/ 四面体形
    octahedral/ˌɒktəˈhiːdrəl/ 八面体
    coordination number/kəʊˈɔːdɪneɪʃn ˈnʌmbə/ 配位数
    ligand exchange/ˈlɪɡænd eksˈtʃeɪndʒ/ 配位子交換
    28.3

    Why complexes are coloured

    Syllabus · ⁨シラバス⁩
    English
    1. define and use the terms degenerate and non-degenerate d orbitals
    2. describe the splitting of degenerate d orbitals into two non-degenerate sets of d orbitals of higher energy, and use of $\Delta E$ in: (a) octahedral complexes, two higher and three lower d orbitals (b) tetrahedral complexes, three higher and two lower d orbitals
    3. explain why transition elements form coloured compounds in terms of the frequency of light absorbed as an electron is promoted between two non-degenerate d orbitals
    4. describe, in qualitative terms, the effects of different ligands on $\Delta E$, frequency of light absorbed, and hence the complementary colour that is observed
    5. use the complexes of copper(II) ions and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions as examples of ligand exchange affecting the colour observed
    日本語
    1. 縮重 および 非縮重 d軌道という用語を定義し使用する
    2. 縮重d軌道がより高いエネルギーを持つ2つの非縮重d軌道セットに分裂することを記述し、$\Delta E$ の使用について:(a) 八面体錯体、2つ高いd軌道と3つ低いd軌道 (b) 四面体錯体、3つ高いd軌道と2つ低いd軌道
    3. 電子が2つの非縮重d軌道間で励起される際に吸収される光の周波数を根拠として、遷移元素が 有色化合物 を形成する理由を説明する
    4. 異なる配位子が $\Delta E$、吸収される光の周波数、およびしたがって観察される 補色 に与える影響を定性的に記述する
    5. 水分子およびアンモニア分子、ヒドロキシイオンおよび塩化物イオンとの銅(II)イオンおよびコバルト(II)イオンの錯体を用い、配位子交換が観測される色に影響を与える例として示す

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    In a free ion the five d orbitals are degenerate 简并 — they have the same energy. When ligands come close, they split the d orbitals into two non-degenerate 非简并 sets, separated by an energy gap $\Delta E$:

    • octahedral: three lower and two higher orbitals.
    • tetrahedral: two lower and three higher orbitals.

    A complex absorbs light whose frequency matches $\Delta E$, promoting an electron from a lower to a higher d orbital. The colour you see is the complementary colour 互补色 of the light absorbed. Different ligands give a different $\Delta E$, so they change the frequency absorbed and hence the colour — which is why ligand exchange changes the colour.

    日本語

    In a free ion the five d orbitals are degenerate 简并 — they have the same energy. When ligands come close, they split the d orbitals into two non-degenerate 非简并 sets, separated by an energy gap $\Delta E$:

    • octahedral: three lower and two higher orbitals.
    • tetrahedral: two lower and three higher orbitals.

    A complex absorbs light whose frequency matches $\Delta E$, promoting an electron from a lower to a higher d orbital. The colour you see is the complementary colour 互补色 of the light absorbed. Different ligands give a different $\Delta E$, so they change the frequency absorbed and hence the colour — which is why ligand exchange changes the colour.

    The five equal d orbitals of a free ion splitting into a lower and a higher set, separated by an energy gap, for octahedral and tetrahedral complexes
    Ligands split the five d orbitals into two sets separated by a gap $\Delta E$; the complex absorbs light of that energy, so we see the complementary colour
    Six beakers of transition-metal solutions in a row, each a strong different colour: red, orange, yellow, green, blue and violet
    Different metals and oxidation states give different colours: cobalt(II) (red), dichromate (orange), chromate (yellow), nickel(II) (green), copper(II) (blue) and permanganate (violet)
    Explore · ⁨探索⁩

    Complex colour route · ⁨錯体の色の経路⁩

    Follow light absorption from d-orbital splitting to observed colour. · ⁨d軌道の分裂から観察される色までの光吸収を追跡する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    degenerate/dɪˈdʒenəreɪt/ 縮退
    non-degenerate/nɒn dɪˈdʒenəreɪt/ 縮重していない
    complementary colour/ˌkɒmplɪˈmentəri ˈkʌlə/ 補色
    28.4

    Stereoisomerism in complexes

    Syllabus · ⁨シラバス⁩
    English
    1. describe the types of stereoisomerism shown by complexes, including those associated with bidentate ligands: (a) geometrical (cis/trans) isomerism, e.g. square planar such as $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$ and octahedral such as $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ and $[\text{Ni}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2(\text{H}_2\text{O})_2]^{2+}$ (b) optical isomerism, e.g. $[\text{Ni}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_3]^{2+}$ and $[\text{Ni}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2(\text{H}_2\text{O})_2]^{2+}$
    2. deduce the overall polarity of complexes such as those described in 28.4.1(a) and 28.4.1(b)
    日本語
    1. 錯体が示す立体異性体のタイプを記述する。二座配位子に関連するものも含む: (a) 幾何異性体(シス/トランス異性体)。例:平面正方形構造である $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$ や、八面体構造である $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ および $[\text{Ni}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2(\text{H}_2\text{O})_2]^{2+}$ (b) 光異性体。例: $[\text{Ni}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_3]^{2+}$ および $[\text{Ni}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2(\text{H}_2\text{O})_2]^{2+}$
    2. 28.4.1(a) および 28.4.1(b) で記述されたような錯体の全体的な極性を推論する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    • geometrical isomerism 几何异构 (cis 顺式 / trans 反式) appears in square planar complexes such as $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$, and in octahedral complexes such as $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$.
    • optical isomerism 旋光异构 appears in octahedral complexes with bidentate ligands, such as $[\text{Ni}(\text{en})_3]^{2+}$, which has two non-superimposable mirror images.

    You can also deduce the polarity of a complex: a cis form may be polar, while the matching trans form is often non-polar because its dipoles cancel.

    日本語
    • geometrical isomerism 几何异构 (cis 顺式 / trans 反式) appears in square planar complexes such as $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$, and in octahedral complexes such as $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$.
    A square planar platinum complex drawn twice: cis with the two ammonia ligands adjacent, trans with them opposite
    Geometrical isomerism in a square planar complex: the two identical ligands are adjacent (cis) or opposite (trans)
    • optical isomerism 旋光异构 appears in octahedral complexes with bidentate ligands, such as $[\text{Ni}(\text{en})_3]^{2+}$, which has two non-superimposable mirror images.

    You can also deduce the polarity of a complex: a cis form may be polar, while the matching trans form is often non-polar because its dipoles cancel.

    Explore · ⁨探索⁩

    Complex stereoisomer lab · ⁨錯体の立体異性実験⁩

    Classify complex isomers by ligand arrangement. · ⁨リガンドの配置により錯体異性を分類する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    geometrical isomerism/ˌdʒiːəʊˈmetrɪkl ˈaɪsəmərɪzəm/ 立体的異性体(幾何異性体)
    cis/sɪs/ シス
    trans/trænz/ トランス
    optical isomerism/ˈɒptɪkl ˈaɪsəmərɪzəm/ 光学異性
    28.5

    Stability constants

    Syllabus · ⁨シラバス⁩
    English
    1. define the stability constant, $K_{\text{stab}}$, of a complex as the equilibrium constant for the formation of the complex ion in a solvent (from its constituent ions or molecules)
    2. write an expression for a $K_{\text{stab}}$ of a complex ($[\text{H}_2\text{O}]$ should not be included)
    3. use $K_{\text{stab}}$ expressions to perform calculations
    4. describe and explain ligand exchanges in terms of $K_{\text{stab}}$ values and understand that a large $K_{\text{stab}}$ is due to the formation of a stable complex ion
    日本語
    1. 錯体イオンの生成に対する平衡定数としての錯体の安定定数 $K_{\text{stab}}$ を定義する(溶媒中において、構成イオンまたは分子から生成される)
    2. 錯体の $K_{\text{stab}}$ に対する式を記述する($[\text{H}_2\text{O}]$ は含めないこと)
    3. $K_{\text{stab}}$ の式を用いて計算を行う
    4. $K_{\text{stab}}$ の値に基づき配位子交換を記述・説明し、大きな $K_{\text{stab}}$ が安定した錯体イオンの形成によるものであることを理解する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The stability constant 稳定常数 ($K_{\text{stab}}$) is the equilibrium constant for forming a complex ion from the metal ion and its ligands in solution (water is left out of the expression).

    A large $K_{\text{stab}}$ means a very stable complex. In a ligand exchange, the position moves towards the complex with the larger $K_{\text{stab}}$ — that is why a ligand that forms a more stable complex can push out a weaker one.

    日本語

    The stability constant 稳定常数 ($K_{\text{stab}}$) is the equilibrium constant for forming a complex ion from the metal ion and its ligands in solution (water is left out of the expression).

    A large $K_{\text{stab}}$ means a very stable complex. In a ligand exchange, the position moves towards the complex with the larger $K_{\text{stab}}$ — that is why a ligand that forms a more stable complex can push out a weaker one.

    Explore · ⁨探索⁩

    Stability constant lab · ⁨安定定数実験⁩

    larger Kstab favours complex · ⁨大きいKstabは錯体を有利にする⁩

    Increase ligand binding strength and see complex formation become more complete. · ⁨リガンド結合強さを増やし、錯体形成がより完全になる様子を見る。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    stability constant/stəˈbɪlɪti ˈkɒnstənt/ 安定性定数
    28.5

    Exam tips

    • Define a transition element: it forms at least one ion with a partially filled d sub-shell — so Sc and Zn are excluded.
    • Colour comes from d-d transitions (ligands split the d orbitals); the colour seen is the complement of the light absorbed.
    • State the ligand and coordination number and predict the shape (6 = octahedral, 4 = tetrahedral or square planar).
    • Ligand exchange can change colour and coordination number; a larger stability constant = a more stable complex.
  • 29

    An introduction to A Level organic chemistry · ⁨A Level有機化学への入門⁩

    Watch lesson · ⁨レッスンを視聴⁩
    29.1

    New functional groups and naming

    Syllabus · ⁨シラバス⁩
    English
    1. understand that the compounds in the table on page 47 contain a functional group which dictates their physical and chemical properties
    2. interpret and use the general, structural, displayed and skeletal formulas of the classes of compound stated in the table on page 47
    3. understand and use systematic nomenclature of simple aliphatic organic molecules (including cyclic compounds containing a single ring of up to six carbon atoms) with functional groups detailed in the table on page 47, up to six carbon atoms (six plus six for esters and amides, straight chains only for esters and nitriles)
    4. understand and use systematic nomenclature of simple aromatic molecules with one benzene ring and one or more simple substituents, for example 3-nitrobenzoic acid or 2,4,6-tribromophenol
    日本語
    1. 47ページの表にある化合物には、その物理的および化学的性質を決定する官能基が含まれていることを理解する
    2. 47ページの表に示された化合物の種別の一般式、構造式、展開式および骨格式を読み取り、使用する
    3. 47ページの表に詳細が示された官能基を持つ単純な脂肪族有機分子の体系命名法を理解して使用する(単一環を含む環状化合物を含み、最大6個の炭素原子までとする。エステルおよびアミドについては炭素原子数6+6、エステルおよびシアノ化合物については直鎖のみ)
    4. 単一のベンゼン環と1つ以上の単純な置換基を持つ単純な芳香族分子の体系命名法を理解して使用する。例:3-ニトロ安息香酸、2,4,6-トリブロモフェノールなど

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    At A Level you meet more functional group 官能团 families, including the amide 酰胺 group and aromatic 芳香 compounds (those built on a benzene ring). As before, the functional group decides the properties, and you read it from the general, structural, displayed or skeletal formula.

    Naming aromatic compounds

    A benzene 苯 molecule is a ring of six carbons. When you name an aromatic compound, use the benzene ring 苯环 as the parent and number the positions of the substituents. For example, 3-nitrobenzoic acid has a $\text{–NO}_2$ group on carbon 3, and 2,4,6-tribromophenol has three bromine atoms on a phenol ring. You can also name cyclic compounds with a single ring of up to six carbons.

    日本語

    At A Level you meet more functional group 官能团 families, including the amide 酰胺 group and aromatic 芳香 compounds (those built on a benzene ring). As before, the functional group decides the properties, and you read it from the general, structural, displayed or skeletal formula.

    Naming aromatic compounds

    A benzene 苯 molecule is a ring of six carbons. When you name an aromatic compound, use the benzene ring 苯环 as the parent and number the positions of the substituents. For example, 3-nitrobenzoic acid has a $\text{–NO}_2$ group on carbon 3, and 2,4,6-tribromophenol has three bromine atoms on a phenol ring. You can also name cyclic compounds with a single ring of up to six carbons.

    Two numbered benzene rings: 3-nitrobenzoic acid with COOH at position 1 and a nitro group at position 3, and 2,4,6-tribromophenol with OH at position 1 and bromine at positions 2, 4 and 6
    Number the ring from the principal group (carbon 1); the substituent positions then give the name
    Explore · ⁨探索⁩

    Aromatic naming lab · ⁨芳香族化合物の名付け方実験⁩

    Classify aromatic substituents by how they are named. · ⁨名付け方に基づいて芳香族置換基を分類する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    functional group/ˈfʌŋkʃənl ɡruːp/ 官能基
    amide/əˈmaɪd/ アミド
    aromatic/ərəʊˈmætɪk/ 芳香族
    benzene/ˈbenziːn/ ベンゼン
    benzene ring/ˈbenziːn rɪŋ/ ベンゼン環
    29.2

    Two new types of mechanism

    Syllabus · ⁨シラバス⁩
    English
    1. understand and use the following terminology associated with types of organic mechanisms: (a) electrophilic substitution (b) addition–elimination
    日本語
    1. 有機反応メカニズムの種類に関連する以下の用語を理解して使用する: (a) 求電子置換反応 (b) 付加-脱離反応

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    • electrophilic substitution 亲电取代: an electrophile replaces a hydrogen atom on a benzene ring (this is how benzene reacts).
    • addition–elimination 加成消去: a molecule first adds on, then a small molecule is removed (seen with 2,4-DNPH and with acyl chlorides).
    日本語
    • electrophilic substitution 亲电取代: an electrophile replaces a hydrogen atom on a benzene ring (this is how benzene reacts).
    • addition–elimination 加成消去: a molecule first adds on, then a small molecule is removed (seen with 2,4-DNPH and with acyl chlorides).
    Electrophilic substitution and addition-elimination
    Two aromatic mechanisms: electrophilic substitution and addition-elimination
    Explore · ⁨探索⁩

    Electrophilic substitution on benzene · ⁨ベンゼンへの求電子置換反応⁩

    Step through how benzene reacts. An electrophile swaps for a hydrogen — keeping the stable ring — instead of adding across it. · ⁨ベンゼンの反応過程を確認する。求電子剤が水素原子と置き換わり、安定な環を維持したまま付加反応を起こさない。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    electrophilic substitution/ɪˌlektrəʊˈfɪlɪk ˌsʌbstɪˈtjuːʃn/ 求電子置換
    addition–elimination/əˈdɪʃn ɪˌlɪmɪˈneɪʃn/ 付加-脱離
    29.3

    The shape of benzene

    Syllabus · ⁨シラバス⁩
    English
    1. describe and explain the shape of benzene and other aromatic molecules, including $\text{sp}^2$ hybridisation, in terms of $\sigma$ bonds and a delocalised $\pi$ system
    日本語
    1. ベンゼンおよび他の芳香族分子の形状を、$\text{sp}^2$混成、$\sigma$結合、および脱局在化された$\pi$系について説明・解説する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Benzene is a flat, regular hexagon. Each carbon is sp² hybridised, using its three hybridisation 杂化 orbitals to make sigma bonds σ键 to two neighbouring carbons and one hydrogen. This gives the ring of σ bonds.

    Each carbon also has one electron left in a p orbital, standing up at right angles to the ring. These p orbitals overlap sideways all the way round, making a single delocalised 离域 pi bond π键 system — a ring of electrons above and below the plane. Because the electrons are shared evenly, all six C–C bonds are the same length, and benzene is very stable.

    How do we know benzene really is delocalised? One strong piece of evidence is its enthalpy change of hydrogenation 氢化焓变. Adding hydrogen to one C=C double bond (as in cyclohexene) releases about $120\ \text{kJ}\,\text{mol}^{-1}$, so a Kekulé ring of three separate double bonds should release about $3 \times 120 = 360\ \text{kJ}\,\text{mol}^{-1}$. Real benzene releases only $208\ \text{kJ}\,\text{mol}^{-1}$ — it is about $152\ \text{kJ}\,\text{mol}^{-1}$ more stable than the model predicts.

    日本語

    Benzene is a flat, regular hexagon. Each carbon is sp² hybridised, using its three hybridisation 杂化 orbitals to make sigma bonds σ键 to two neighbouring carbons and one hydrogen. This gives the ring of σ bonds.

    Each carbon also has one electron left in a p orbital, standing up at right angles to the ring. These p orbitals overlap sideways all the way round, making a single delocalised 离域 pi bond π键 system — a ring of electrons above and below the plane. Because the electrons are shared evenly, all six C–C bonds are the same length, and benzene is very stable.

    A benzene hexagon with a p orbital standing up at each carbon, overlapping into a shaded delocalised cloud above and below the ring
    Benzene's bonding: an sp$^2$ $\sigma$ framework makes the flat hexagon, while the p orbitals overlap into one delocalised $\pi$ system above and below the ring

    How do we know benzene really is delocalised? One strong piece of evidence is its enthalpy change of hydrogenation 氢化焓变. Adding hydrogen to one C=C double bond (as in cyclohexene) releases about $120\ \text{kJ}\,\text{mol}^{-1}$, so a Kekulé ring of three separate double bonds should release about $3 \times 120 = 360\ \text{kJ}\,\text{mol}^{-1}$. Real benzene releases only $208\ \text{kJ}\,\text{mol}^{-1}$ — it is about $152\ \text{kJ}\,\text{mol}^{-1}$ more stable than the model predicts.

    An energy-level diagram: hydrogenating the Kekulé model would release 360 kJ/mol but real benzene releases only 208, so real benzene lies about 152 kJ/mol lower in energy than the model
    Evidence for delocalisation: real benzene releases far less on hydrogenation ($-208\ \text{kJ}\,\text{mol}^{-1}$) than the Kekulé model predicts ($-360 = 3\times$ cyclohexene), so it is about $152\ \text{kJ}\,\text{mol}^{-1}$ more stable than expected
    Explore · ⁨探索⁩

    Benzene bonding lab · ⁨ベンゼンの結合実験⁩

    Follow the evidence for a planar delocalised benzene ring. · ⁨平面型非局在化ベンゼン環に関する証拠を追跡する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    hybridisation/ˌhaɪbrɪdaɪˈzeɪʃn/ 雑種交配
    sigma bond/ˈsɪɡmə bɒnd/ シグマ結合
    delocalised/dɪˈlɒkəlaɪzd/ 非局在化
    pi bond/paɪ bɒnd/ π結合
    enthalpy change of hydrogenation/enˈθælpi tʃeɪndʒ ɒv haɪˈdrɒdʒəneɪʃn/ 水素化エンタルピー変化
    29.4

    Optical isomerism

    Syllabus · ⁨シラバス⁩
    English
    1. understand that enantiomers have identical physical and chemical properties apart from their ability to rotate plane polarised light and their potential biological activity
    2. understand and use the terms optically active and racemic mixture
    3. describe the effect on plane polarised light of the two optical isomers of a single substance
    4. explain the relevance of chirality to the synthetic preparation of drug molecules including: (a) the potential different biological activity of the two enantiomers (b) the need to separate a racemic mixture into two pure enantiomers (c) the use of chiral catalysts to produce a single pure optical isomer (Candidates should appreciate that compounds can contain more than one chiral centre, but knowledge of meso compounds and nomenclature such as diastereoisomers is not required.)
    日本語
    1. 鏡像異性体は、偏光平面光を回転させる能力および潜在的な生物活性を除き、物理的および化学的性質が同一であることを理解する
    2. 光学活性およびラセミ混合物という用語を理解して使用する
    3. 単一物質の2つの光異性体が偏光平面光に与える影響を記述する
    4. 医薬品分子の合成準備におけるキラル性の関連性を説明する: (a) 2つの鏡像異性体の潜在的な異なる生物活性 (b) ラセミ混合物を2つの純粋な鏡像異性体に分離する必要性 (c) 単一の純粋な光異性体を生産するためにキラル触媒を使用する(受験生は、化合物が複数のキラル中心を持つ可能性があることを理解すべきであるが、メソ化合物やジアステレオ異性体などの命名法に関する知識は不要である)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Two enantiomers 对映体 (mirror-image isomers) have identical physical and chemical properties, with two exceptions:

    • they rotate plane polarised light 平面偏振光 in opposite directions. A substance that does this is optically active 旋光活性.
    • they may have different effects in living things (biological activity).

    A racemic mixture 外消旋混合物 is a 50:50 mix of the two enantiomers. It does not rotate plane polarised light, because the two opposite rotations cancel out.

    Why chirality matters for drugs

    Chirality 手性 is important when making medicines. A molecule with a chiral centre 手性中心 has two enantiomers, and they can behave very differently in the body — one may cure while the other does harm. So drug makers either:

    • separate a racemic mixture into the two pure enantiomers, or
    • use a chiral catalyst 手性催化剂 to make just the single enantiomer they want.

    Worked example. Which of butan-1-ol, butan-2-ol and 2-methylpropan-2-ol is chiral? A molecule is chiral if it has a carbon carrying four different groups. Take the carbons one at a time. In butan-2-ol, $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$, carbon 2 carries $\text{OH}$, $\text{H}$, $\text{CH}_3$ and $\text{C}_2\text{H}_5$ - four different groups, so it is a chiral centre and butan-2-ol exists as two optical isomers. Butan-1-ol's carbon 1 carries two hydrogens, and the central carbon of 2-methylpropan-2-ol carries two identical methyl groups, so neither is chiral. It takes only one repeated group on a carbon to destroy the chirality there - and compare groups properly: $\text{CH}_3$ and $\text{C}_2\text{H}_5$ differ, but only once you look past the first atom.

    日本語

    Two enantiomers 对映体 (mirror-image isomers) have identical physical and chemical properties, with two exceptions:

    • they rotate plane polarised light 平面偏振光 in opposite directions. A substance that does this is optically active 旋光活性.
    • they may have different effects in living things (biological activity).

    A racemic mixture 外消旋混合物 is a 50:50 mix of the two enantiomers. It does not rotate plane polarised light, because the two opposite rotations cancel out.

    Plane-polarised light passing through each enantiomer: one rotates the plane clockwise, the mirror-image enantiomer rotates it anticlockwise
    The two enantiomers rotate plane-polarised light in opposite directions; a 50:50 racemic mixture gives no net rotation

    Why chirality matters for drugs

    Chirality 手性 is important when making medicines. A molecule with a chiral centre 手性中心 has two enantiomers, and they can behave very differently in the body — one may cure while the other does harm. So drug makers either:

    • separate a racemic mixture into the two pure enantiomers, or
    • use a chiral catalyst 手性催化剂 to make just the single enantiomer they want.

    Worked example. Which of butan-1-ol, butan-2-ol and 2-methylpropan-2-ol is chiral? A molecule is chiral if it has a carbon carrying four different groups. Take the carbons one at a time. In butan-2-ol, $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$, carbon 2 carries $\text{OH}$, $\text{H}$, $\text{CH}_3$ and $\text{C}_2\text{H}_5$ - four different groups, so it is a chiral centre and butan-2-ol exists as two optical isomers. Butan-1-ol's carbon 1 carries two hydrogens, and the central carbon of 2-methylpropan-2-ol carries two identical methyl groups, so neither is chiral. It takes only one repeated group on a carbon to destroy the chirality there - and compare groups properly: $\text{CH}_3$ and $\text{C}_2\text{H}_5$ differ, but only once you look past the first atom.

    Explore · ⁨探索⁩

    Optical isomerism lab · ⁨光学異性体実験⁩

    Identify when a molecule can have non-superimposable mirror images. · ⁨分子が重なり合わない鏡像を持つことができるタイミングを特定する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    enantiomers/eˈnæntɪəməz/ 鏡像異性体
    plane polarised light/pleɪn ˈpəʊləraɪzd laɪt/ 平面偏光
    optically active/ˈɒptɪkli ˈæktɪv/ 光学活性
    racemic mixture/rəˈsiːmɪk ˈmɪkstʃə/ ラセミ混合物
    chirality/kaɪˈrælɪti/ キラル性
    chiral centre/ˈkaɪrəl ˈsentə/ キラル中心
    chiral catalyst/ˈkaɪrəl ˈkætəlɪst/ キラル触媒
    29.4

    Exam tips

    • Optical isomers need a chiral carbon (four different groups); they are non-superimposable mirror images that rotate plane-polarised light oppositely.
    • A racemic mixture forms when a planar intermediate (carbocation or $\text{C}=\text{O}$) is attacked equally from both sides.
    • Benzene's delocalised ring (equal bond lengths, planar) explains its stability versus the Kekulé model.
    • Learn the two new mechanisms — electrophilic substitution of arenes, and nucleophilic addition-elimination.
  • 30

    Hydrocarbons · ⁨炭化水素⁩

    Watch lesson · ⁨レッスンを視聴⁩
    30.1

    Arenes · ⁨アレン⁩

    Syllabus · ⁨シラバス⁩
    1. describe the chemistry of arenes as exemplified by the following reactions of benzene and methylbenzene: (a) substitution reactions with $\text{Cl}_2$ and with $\text{Br}_2$ in the presence of a catalyst, $\text{AlCl}_3$ or $\text{AlBr}_3$, to form halogenoarenes (aryl halides) (b) nitration with a mixture of concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ at a temperature between $25\text{ }^{\circ}\text{C}$ and $60\text{ }^{\circ}\text{C}$ (c) Friedel–Crafts alkylation by $\text{CH}_3\text{Cl}$ and $\text{AlCl}_3$ and heat (d) Friedel–Crafts acylation by $\text{CH}_3\text{COCl}$ and $\text{AlCl}_3$ and heat (e) complete oxidation of the side-chain using hot alkaline $\text{KMnO}_4$ and then dilute acid to give a benzoic acid (f) hydrogenation of the benzene ring using $\text{H}_2$ and $\text{Pt/Ni}$ catalyst and heat to form a cyclohexane ring
    2. describe the mechanism of electrophilic substitution in arenes: (a) as exemplified by the formation of nitrobenzene and bromobenzene (b) with regards to the effect of delocalisation (aromatic stabilisation) of electrons in arenes to explain the predomination of substitution over addition
    3. predict whether halogenation will occur in the side-chain or in the aromatic ring in arenes depending on reaction conditions
    4. describe that in the electrophilic substitution of arenes, different substituents direct to different ring positions (limited to the directing effects of $-\text{NH}_2$, $-\text{OH}$, $-\text{R}$, $-\text{NO}_2$, $-\text{COOH}$ and $-\text{COR}$)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Arenes 芳烃 are aromatic hydrocarbons, built on the benzene 苯 ring. The delocalised ring of electrons is stable and electron-rich, so benzene mostly reacts by electrophilic substitution 亲电取代 — keeping the ring — rather than by addition.

    Reactions of benzene and methylbenzene

    Reaction Reagents and conditions Product
    halogenation $\text{Cl}_2$ or $\text{Br}_2$, with $\text{AlCl}_3$ or $\text{AlBr}_3$ as a catalyst 催化剂 a halogenoarene 卤代芳烃 (aryl halide)
    nitration 硝化 concentrated $\text{HNO}_3$ + concentrated $\text{H}_2\text{SO}_4$, $25$–$60\,°\text{C}$ nitrobenzene
    Friedel–Crafts alkylation 傅克烷基化 $\text{CH}_3\text{Cl}$ + $\text{AlCl}_3$, heat methylbenzene (adds an alkyl group)
    Friedel–Crafts acylation 傅克酰基化 $\text{CH}_3\text{COCl}$ + $\text{AlCl}_3$, heat a phenyl ketone (adds an acyl group)
    side-chain oxidation hot alkaline $\text{KMnO}_4$, then dilute acid benzoic acid 苯甲酸
    hydrogenation 氢化 $\text{H}_2$, $\text{Pt/Ni}$ catalyst, heat cyclohexane

    In the side-chain oxidation, the whole side-chain 侧链 (such as the $\text{–CH}_3$ on methylbenzene) is turned into a $\text{–COOH}$ group, giving benzoic acid. In the hydrogenation, three molecules of $\text{H}_2$ add to the benzene ring 苯环 to make a saturated cyclohexane ring.

    The mechanism: electrophilic substitution

    Take nitration as the example. The acid mix makes the electrophile $\text{NO}_2^+$. Then:

    1. the delocalised electrons of the ring form a bond to the electrophile, giving an unstable intermediate.
    2. an $\text{H}^+$ is lost from that carbon, which restores the stable ring.

    Substitution wins over addition because the delocalisation 离域 (aromatic stabilisation) of the ring is kept. Addition would destroy this stable system, so it is not favoured.

    Side-chain or ring?

    Where a halogen reacts depends on the conditions:

    • with a halogen-carrier catalyst (such as $\text{AlCl}_3$) and no UV light → substitution in the ring.
    • with UV light and no catalyst → free-radical substitution in the side-chain.

    Directing effects

    A group already on the ring decides where the next group goes — its directing effect 定位效应:

    Group already present Directs the new group to
    $\text{–NH}_2$, $\text{–OH}$, $\text{–R}$ positions 2 and 4
    $\text{–NO}_2$, $\text{–COOH}$, $\text{–COR}$ position 3

    Worked example. Methylbenzene and nitrobenzene are each nitrated. Predict where the new $\text{NO}_2$ group goes, and which compound reacts faster. Look at the group already on the ring. In methylbenzene the $\text{–CH}_3$ is an alkyl group: it directs the new group to positions 2 and 4, and it releases electrons into the ring, making the ring more attractive to an electrophile - so methylbenzene nitrates faster than benzene. In nitrobenzene the $\text{–NO}_2$ directs to position 3, and it withdraws electrons from the ring - so nitrobenzene nitrates more slowly than benzene. The group already present controls both the position and the rate, and the two always travel together: 2,4-directors activate the ring, 3-directors deactivate it.

    日本語

    アレンは芳香族炭化水素で、ベンゼン環を骨格としています。電子が非局在化した環は安定で電子豊富であるため、ベンゼンは主に環を維持する求電子置換によって反応し、付加反応よりもそちらが好まれます。

    白く丸い防虫剤の山
    ナフタレンは防虫剤に使われる単純なアレンで、2つのベンゼン環が融合したものです

    ベンゼンおよびメチルベンゼンの反応

    反応 試薬および条件 生成物
    ハロゲン化 $\text{Cl}_2$ または $\text{Br}_2$、触媒として $\text{AlCl}_3$ または $\text{AlBr}_3$ を用いる ハロゲンアレン(アリルハライド)
    ニトロ化 濃 $\text{HNO}_3$ + 濃 $\text{H}_2\text{SO}_4$、$25$–$60\,°\text{C}$ ニトロベンゼン
    フリーデル=クラフツアルキル化 $\text{CH}_3\text{Cl}$ + $\text{AlCl}_3$、加熱 メチルベンゼン(アルキル基が導入される)
    フリーデル=クラフツアシル化 $\text{CH}_3\text{COCl}$ + $\text{AlCl}_3$、加熱 フェニルケトン(アシル基が導入される)
    側鎖酸化 熱い塩基性 $\text{KMnO}_4$、その後希酸 安息香酸
    水素添加 $\text{H}_2$、$\text{Pt/Ni}$ 触媒、加熱 シクロヘキサン
    フリーデル=クラフツアルキル化ではアルキル基が導入され、アシル化ではアシル基が導入される
    フリーデル=クラフツ反応:アルキル化ではアルキル基が導入され、アシル化ではアシル基が導入される
    中央にベンゼンがあり、その周囲に塩化ベンゼン、ニトロベンゼン、メチルベンゼン、フェニルケトン、シクロヘキサンへ向かう矢印が描かれ、それぞれに試薬が記載されている
    ベンゼンは安定した環を維持し、主に求電子置換反応を起こす(水素との付加反応のみが例外である)

    側鎖酸化では、メチルベンゼンの $\text{–CH}_3$ に代表される整个側鎖が $\text{–COOH}$ 基に変換され、安息香酸となる。水素添加では、3分子の $\text{H}_2$ がベンゼン環に付加して飽和したシクロヘキサン環となる。

    反応機構:求電子置換

    ニトロ化を例に示す。酸混合液により求電子体 $\text{NO}_2^+$ が生成する。次に:

    1. 環の非局在電子が求電子体と結合し、不安定な中間体を形成する。
    2. その炭素から $\text{H}^+$ が脱離し、安定した環が再生される。

    環の非局在性(芳香族安定化)が保たれるため、置換反応が優先する。付加反応はこの安定系を破壊するため、起こりにくい。

    ニトロ化の反応機構:環がNO2+に対してカール矢印で攻撃し正電荷を持つ中間体となり、H+を失って環が再生されニトロベンゼンとなる
    求電子置換(ニトロ化):環が求電子体NO$_2^+$に攻撃して不安定な中間体となり、H$^+$を失って環が再生される

    側鎖か、それとも環か?

    ハロゲンがどの部位で反応するかは条件による:

    • ハロゲンキャリア触媒($\text{AlCl}_3$ など)を使用し、紫外線なし → 環内での置換反応。
    • 紫外線ありかつ触媒なし → 側鎖内でのラジカル置換反応。
    メチルベンゼンに塩素が2つの経路で分岐:AlCl3存在下・紫外線なしでは環が置換され、紫外線あり・触媒なしでは側鎖のメチル基が置換される
    条件によって決まる:ハロゲンキャリア(AlCl₃)は環を置換し、紫外線は側鎖を置換する

    配位効果

    環上に既に存在する基が次の基の導入位置を決定する—これが配位効果である:

    2つの番号付きベンゼン環:一方はアミノ基、ヒドロキシ基、アルキル基に対して2番および4番の位置が強調され、他方はニトロ基、カルボキシル基、アシル基に対して3番の位置が強調されている
    環上の既存基が次の基の導入位置を指示する:–NH$_2$/–OH/–R は2番および4番へ、–NO$_2$/–COOH/–COR は3番へと導く
    環既に存在する基 新しい基の導入位置
    $\text{–NH}_2$、$\text{–OH}$、$\text{–R}$ 2番および4番の位置
    $\text{–NO}_2$、$\text{–COOH}$、$\text{–COR}$ 3番の位置

    計算例。 メチルベンゼンとニトロベンゼンをそれぞれニトロ化する。新しい $\text{NO}_2$ 基がどこに導入されるかを予測し、どちらの化合物の方が反応速度が速いかを答えよ。環上に既に存在する基を確認せよ。メチルベンゼンでは $\text{–CH}_3$ がアルキル基であり、新しい基を2番および4番の位置へ誘導する。また、環へ電子を供与することで求電子体に対する親和性を高め、ベンゼンよりも速くニトロ化する。ニトロベンゼンでは $\text{–NO}_2$ が3番の位置へ誘導し、環から電子を引き抜く。したがって、ベンゼンよりも遅くニトロ化する。既に存在する基は導入位置と反応速度の両方を制御しており、これらは常にセットで変動する:2,4-誘導基は環を活性化するが、3-誘導基は環を不活性化する。

    Explore · ⁨探索⁩

    Arene substitution route · ⁨アレン置換経路⁩

    Follow electrophilic substitution on benzene. · ⁨ベンゼンの求電子置換反応に従う。⁩

    Explore · ⁨探索⁩

    Directing effects lab · ⁨誘導効果実験⁩

    Classify substituents by how they affect the benzene ring. · ⁨ベンゼン環への影響に基づいて置換基を分類する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    arene/ˈæren/ アレーン
    benzene/ˈbenziːn/ ベンゼン
    electrophilic substitution/ɪˌlektrəʊˈfɪlɪk ˌsʌbstɪˈtjuːʃn/ 求電子置換
    catalyst/ˈkætəlɪst/ 触媒
    halogenoarene/ˈheɪləʊdʒnɔːren/ ハロゲンアレーン
    nitration/naɪˈtreɪʃn/ ニトロ化
    Friedel–Crafts alkylation/ˈfriːdl kræfts ˌælkɪˈleɪʃn/ フリーデル=クラフツアルキル化
    Friedel–Crafts acylation/ˈfriːdl kræfts əkɪˈleɪʃn/ フリーデル=クルプアシル化
    benzoic acid/benˈzəʊɪk ˈæsɪd/ ベンゾ酸
    hydrogenation/haɪˈdrɒdʒəneɪʃn/ 水素添加
    side-chain/saɪd tʃeɪn/ 側鎖
    benzene ring/ˈbenziːn rɪŋ/ ベンゼン環
    delocalisation/dɪˌlɒkəlaɪˈzeɪʃn/ 非局在化
    directing effect/daɪˈrektɪŋ ɪˈfekt/ 配向効果
    30.1

    Exam tips · ⁨試験対策⁩

    English
    • Benzene undergoes electrophilic substitution, not addition, because addition would destroy the stable delocalised ring.
    • Learn nitration (concentrated $\text{HNO}_3$/$\text{H}_2\text{SO}_4$, $50-60\ ^\circ\text{C}$) and halogenation (halogen + $\text{AlCl}_3$) with the electrophile-generating step.
    • Compare reactivity: benzene resists addition far more than an alkene because of delocalisation.
    日本語
    • ベンゼンは安定した非局在環を破壊してしまうため、付加反応ではなく求電子置換反応を起こす。
    • ニトロ化(濃 $\text{HNO}_3$/$\text{H}_2\text{SO}_4$、$50-60\ ^\circ\text{C}$)とハロゲン化(ハロゲン+$\text{AlCl}_3$)における求電子体生成段階を含めて学習する。
    • 反応性の比較:非局在性により、ベンゼンはアルケンに比べて付加反応に対して遥かに抵抗を示す。
  • 31

    Halogen compounds · ⁨ハロゲン化合物⁩

    Watch lesson · ⁨レッスンを視聴⁩
    31.1

    Making halogenoarenes

    Syllabus · ⁨シラバス⁩
    1. recall the reactions by which halogenoarenes can be produced: substitution of an arene with $\text{Cl}_2$ or $\text{Br}_2$ in the presence of a catalyst, $\text{AlCl}_3$ or $\text{AlBr}_3$ to form a halogenoarene, exemplified by benzene to form chlorobenzene and methylbenzene to form 2-chloromethylbenzene and 4-chloromethylbenzene
    2. explain the difference in reactivity between a halogenoalkane and a halogenoarene as exemplified by chloroethane and chlorobenzene

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A halogenoarene 卤代芳烃 (also called an aryl halide) is formed when an arene 芳烃 reacts with $\text{Cl}_2$ or $\text{Br}_2$, using $\text{AlCl}_3$ or $\text{AlBr}_3$ as a catalyst 催化剂. This is the electrophilic substitution from the arenes topic — the halogen replaces a hydrogen on the ring.

    • benzene gives chlorobenzene.
    • methylbenzene gives 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene (the methyl group directs the chlorine to positions 2 and 4).
    日本語

    A halogenoarene 卤代芳烃 (also called an aryl halide) is formed when an arene 芳烃 reacts with $\text{Cl}_2$ or $\text{Br}_2$, using $\text{AlCl}_3$ or $\text{AlBr}_3$ as a catalyst 催化剂. This is the electrophilic substitution from the arenes topic — the halogen replaces a hydrogen on the ring.

    • benzene gives chlorobenzene.
    • methylbenzene gives 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene (the methyl group directs the chlorine to positions 2 and 4).
    Benzene reacting with chlorine, using aluminium chloride as a catalyst, to give chlorobenzene and hydrogen chloride — a chlorine replaces a hydrogen on the ring
    Making a halogenoarene by electrophilic substitution: with an AlCl₃ catalyst, a chlorine replaces a hydrogen on the ring (giving HCl)
    A tractor spraying a crop in a field
    Many pesticides and herbicides are halogenoarenes — chlorine atoms bonded to a benzene ring
    Explore · ⁨探索⁩

    Halogenoarene lab · ⁨ハロゲンアレン実験⁩

    Compare halogenoarenes with halogenoalkanes. · ⁨ハロゲンアレンとハロゲンアルカンを比較する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    halogenoarene/ˈheɪləʊdʒnɔːren/ ハロゲンアレーン
    arene/ˈæren/ アレーン
    catalyst/ˈkætəlɪst/ 触媒
    31.1

    Why a halogenoarene is less reactive than a halogenoalkane

    English

    Compare chloroethane (a halogenoalkane 卤代烷) with chlorobenzene (a halogenoarene).

    A halogenoalkane reacts easily by nucleophilic substitution 亲核取代: its C–Cl bond is polar, so a nucleophile can attack the slightly positive carbon and push the halogen out.

    A halogenoarene is very unreactive towards nucleophilic substitution. There are two reasons:

    • a lone pair 孤对电子 on the chlorine overlaps sideways with the delocalised 离域 ring of electrons. This gives the C–Cl bond partial double-bond character, making it shorter and stronger, so it is much harder to break.
    • the electron-rich ring repels an approaching nucleophile 亲核试剂.

    So chlorobenzene does not react with nucleophiles such as $\text{OH}^-$ under normal conditions, while chloroethane does.

    日本語

    Compare chloroethane (a halogenoalkane 卤代烷) with chlorobenzene (a halogenoarene).

    A halogenoalkane reacts easily by nucleophilic substitution 亲核取代: its C–Cl bond is polar, so a nucleophile can attack the slightly positive carbon and push the halogen out.

    A halogenoarene is very unreactive towards nucleophilic substitution. There are two reasons:

    • a lone pair 孤对电子 on the chlorine overlaps sideways with the delocalised 离域 ring of electrons. This gives the C–Cl bond partial double-bond character, making it shorter and stronger, so it is much harder to break.
    • the electron-rich ring repels an approaching nucleophile 亲核试剂.

    So chlorobenzene does not react with nucleophiles such as $\text{OH}^-$ under normal conditions, while chloroethane does.

    Chloroethane with a nucleophile attacking its slightly positive carbon, beside chlorobenzene where the chlorine lone pair overlaps the ring and the ring repels the nucleophile
    Chloroethane reacts (a nucleophile attacks the $\delta+$ carbon), but chlorobenzene does not: the Cl lone pair strengthens the C–Cl bond and the ring repels nucleophiles
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    halogenoalkane/ˈheɪləʊdʒnəʊlkeɪn/ ハロゲン化アルキル
    nucleophilic substitution/ˌnjuːklɪəˈfɪlɪk ˌsʌbstɪˈtjuːʃn/ 求核置換
    lone pair/ləʊn peə/ 非共有電子対
    delocalised/dɪˈlɒkəlaɪzd/ 非局在化
    nucleophile/ˈnjuːklɪɒfaɪl/ 求核剤
    31.1

    Exam tips

    • A halogenoarene is less reactive than a halogenoalkane: the lone pair delocalises into the ring, giving the $\text{C-X}$ bond partial double-bond character.
    • Distinguish a halogen on the ring (needs a catalyst, unreactive to substitution) from one on a side chain (reacts like a halogenoalkane).
  • 32

    Hydroxy compounds · ⁨ヒドロキシ化合物⁩

    Watch lesson · ⁨レッスンを視聴⁩
    32.1

    Alcohols with acyl chlorides · ⁨アシル塩化物とのアルコール⁩

    Syllabus · ⁨シラバス⁩
    English
    1. describe the reaction with acyl chlorides to form esters using ethyl ethanoate
    日本語
    1. アシルクロリドとの反応を用いて、エチルエタノ酸エステルを用いてエステルが生成される反応を記述せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    At A Level you meet one more reaction of an alcohol 醇: it reacts with an acyl chloride 酰氯 to make an ester 酯. This works faster and more completely than using a carboxylic acid. For example, ethanol and ethanoyl chloride give ethyl ethanoate, plus fumes of $\text{HCl}$.

    日本語

    A-Levelでは、アルコールのもう一つの反応に触れる:それはアシル塩化物と反応してエステルを作る。これはカルボン酸を使用する場合よりも速く、より完全に進行する。例えば、エタノールとエタンアシル塩化物からエチルエタノエートと、$\text{HCl}$の白煙が得られる。

    エタノールがエタンアシル塩化物と反応してエステルエチルエタノエートと塩化水素を生成する様子
    アシル塩化物はカルボン酸よりも速く、より完全にアルコールと反応してエステルとHClの白煙を与える
    Explore · ⁨探索⁩

    Acyl chloride with alcohol route · ⁨アシルクロリドからアルコールへの経路⁩

    Follow nucleophilic acyl substitution to an ester. · ⁨求核アシル置換を経てエステルを作る。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    alcohol/ˈælkəhɒl/ アルコール
    acyl chloride/əˈkɪl ˈklɔːraɪd/ アシルクロリド
    ester/ˈestə/ エステル
    32.2

    Phenol · ⁨フェノール⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions (reagents and conditions) by which phenol can be produced: (a) reaction of phenylamine with $\text{HNO}_2$ or $\text{NaNO}_2$ and dilute acid below $10\text{ }^\circ\text{C}$ to produce the diazonium salt; further warming of the diazonium salt with $\text{H}_2\text{O}$ to give phenol
    2. recall the chemistry of phenol, as exemplified by the following reactions: (a) with bases, for example $\text{NaOH(aq)}$ to produce sodium phenoxide (b) with $\text{Na(s)}$ to produce sodium phenoxide and $\text{H}_2\text{(g)}$ (c) in $\text{NaOH(aq)}$ with diazonium salts, to give azo compounds (d) nitration of the aromatic ring with dilute $\text{HNO}_3\text{(aq)}$ at room temperature to give a mixture of 2-nitrophenol and 4-nitrophenol (e) bromination of the aromatic ring with $\text{Br}_2\text{(aq)}$ to form 2,4,6-tribromophenol
    3. explain the acidity of phenol
    4. describe and explain the relative acidities of water, phenol and ethanol
    5. explain why the reagents and conditions for the nitration and bromination of phenol are different from those for benzene
    6. recall that the hydroxyl group of a phenol directs to the 2-, 4- and 6-positions
    7. apply knowledge of the reactions of phenol to those of other phenolic compounds, e.g. naphthol
    日本語
    1. フェノールが生成される反応(試薬および条件)を思い出し、(a)フェニルアミンを$\text{HNO}_2$または$\text{NaNO}_2$および希酸と$10\text{ }^\circ\text{C}$以下の温度で反応させ、ジアゾニウム塩を生成し、さらにジアゾニウム塩を$\text{H}_2\text{O}$で温めてフェノールを得る
    2. フェノールの化学を、以下の反応によって例示し、思い出せ。(a)塩基、例えば$\text{NaOH(aq)}$との反応によりフェノキシナトリウムを生成する (b) $\text{Na(s)}$との反応によりフェノキシナトリウムおよび$\text{H}_2\text{(g)}$を生成する (c) $\text{NaOH(aq)}$中でのジアゾニウム塩との反応によりアゾ化合物を生成する (d) 室温で希$\text{HNO}_3\text{(aq)}$を用いて芳香環をニトロ化し、2-ニトロフェノールおよび4-ニトロフェノールの混合物を得る (e) $\text{Br}_2\text{(aq)}$を用いて芳香環をブロモ化し、2,4,6-トリブロモフェノールを生成する
    3. フェノールの酸性を説明せよ
    4. 水、フェノールおよびエタノールの相対的な酸性を説明・解説せよ
    5. フェノールのニトロ化および臭素化に用いる試薬および条件が、ベンゼンと異なる理由を説明せよ
    6. フェノールの水酸基は2-, 4-および6-位置への置換反応を誘導することを記憶せよ
    7. 其他フェノール性化合物(例:ナフトール)に対するフェノールの反応の知識を適用せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Phenol 苯酚 has an $\text{–OH}$ group joined directly to a benzene 苯 ring. This changes the chemistry of both the $\text{–OH}$ group and the ring.

    Making phenol

    Cool phenylamine 苯胺 with $\text{NaNO}_2$ and dilute acid below $10\,°\text{C}$ to make a diazonium salt 重氮盐. Warming this salt with water then gives phenol (and nitrogen gas).

    Reactions of phenol

    • with a base such as $\text{NaOH(aq)}$: phenol reacts to give sodium phenoxide and water. (Ordinary alcohols do not react with $\text{NaOH}$ — this shows phenol is more acidic.)
    • with sodium metal: gives sodium phenoxide and hydrogen.
    • with a diazonium salt in $\text{NaOH(aq)}$: forms a coloured azo compound 偶氮化合物 (used in dyes).
    • nitration 硝化 with dilute $\text{HNO}_3$ at room temperature: gives a mixture of 2-nitrophenol and 4-nitrophenol.
    • bromination 溴化 with bromine water at room temperature (no catalyst): gives a white precipitate of 2,4,6-tribromophenol.

    Acidity of phenol

    Phenol loses its $\text{H}^+$ to form a phenoxide ion. This ion is stabilised because its negative charge is spread (delocalised) into the ring. So phenol's acidity 酸性 is higher than that of water or ethanol:

    $$\text{ethanol} < \text{water} < \text{phenol}$$

    (Phenol is still a weak acid — weaker than a carboxylic acid.)

    Why the conditions are milder than for benzene

    A lone pair on the phenol oxygen is partly delocalised 离域 into the ring. This makes the ring more electron-rich, so it attracts electrophiles more strongly. Phenol therefore reacts faster and under much milder conditions than benzene — no catalyst is needed, and dilute reagents work at room temperature.

    The $\text{–OH}$ group directs new substituents to the 2-, 4- and 6-positions. The same ideas apply to other phenolic compounds, such as naphthol.

    Worked example. Phenol, ethanol and ethanoic acid are each shaken with aqueous $\text{NaOH}$ and then with aqueous $\text{Na}_2\text{CO}_3$. Predict what happens, and order the three by acidity. Acidity depends on how well the anion left behind is stabilised. Ethanol's ethoxide keeps its charge stuck on one oxygen, so ethanol is weakest and reacts with neither. Phenol's phenoxide spreads the charge into the ring, making phenol acidic enough to react with the strong base $\text{NaOH}$ but not with the weaker $\text{Na}_2\text{CO}_3$ - so no fizzing. Ethanoate spreads the charge over two oxygens, the most effective of the three, so ethanoic acid reacts with both and fizzes with the carbonate. Order: ethanol < phenol < ethanoic acid. The carbonate is the discriminating test - only a carboxylic acid fizzes, which is exactly how you tell phenol from an acid.

    日本語

    フェノールはベンゼン環に直接$\text{–OH}$基が結合している。これにより、$\text{–OH}$基と環の両方の化学的性質が変化する。

    ピンク色に変色したフェノールのサンプル
    フェノールは低融点固体であり、空気中で酸化されるにつれて徐々にピンク色に変化する

    フェノールの合成

    アニリンを$\text{NaNO}_2$と希酸を用いて$10\,°\text{C}$以下に冷却するとジアゾニウム塩が得られる。この塩を水とともに温めるとフェノール(および窒素ガス)が得られる。

    フェノールの反応

    • $\text{NaOH(aq)}$のような塩基との反応:フェノールは反応してフェノキシドナトリウムと水を生成する。(通常のアルコールは$\text{NaOH}$とは反応しない—これがフェノールの方が酸性であることを示す。)
    • ナトリウム金属との反応:フェノキシドナトリウムと水素ガスを生成する。
    • $\text{NaOH(aq)}$中のジアゾニウム塩との反応:有色のアゾ化合物(染料として使用)を形成する。
    • 室温での希$\text{HNO}_3$によるニトロ化:2-ニトロフェノールと4-ニトロフェノールの混合物が得られる。
    • 室温での臭素水による臭素化(触媒なし):2,4,6-トリブロモフェノールの白色沈殿が得られる。

    フェノールの酸性

    フェノールは$\text{H}^+$を失ってフェノキシドイオンとなる。このイオンは負電荷が環に分散(離域)することで安定化される。したがって、フェノールの酸性は水やエタノールよりも高い:

    $$\text{ethanol} < \text{water} < \text{phenol}$$
    フェノールがプロトンを失ってフェノキシドイオンとなり、カール矢印が負電荷を環内に広げる様子を示す図
    フェノールは水やアルコールよりも酸性が高く、H$^+$を失ってフェノキシドイオンとなり、その負電荷が環に広がって安定化される

    (フェノールは依然として弱酸であり、カルボン酸ほどではない。)

    ベンゼンよりも条件が穏やかな理由

    フェノールの酸素にある非共有電子対が一部環に離域する。これにより環は電子密度が高くなり、求電子試薬を強く引きつける。したがって、フェノールはベンゼンよりも速く、はるかに穏やかな条件下で反応する—触媒は不要であり、希試薬でも室温で機能する。

    フェノールが室温で臭素水と反応して2,4,6-トリブロモフェノール(白色沈殿)を生成する様子
    フェノールの環は活性化されているため、臭素水は触媒なしで室温において位置2, 4, 6を臭素化し、白色沈殿を与える

    $\text{–OH}$基は新しい置換基を2-, 4-, 6-位へ誘導する。ナフタロールなどの他のフェノール類にも同じ原理が適用される。

    ピンク色のカーボリックソープの棒状品
    カーボリックソープにはフェノールが含まれており、これは最初の抗生物質の一つであった

    ** worked example. フェノール、エタノール、エタン酸それぞれを$\text{NaOH}$水溶液と攪拌した後、$\text{Na}_2\text{CO}_3$水溶液を加える。起こる現象を予測し、3つを酸性の順に並べよ。酸性は残った陰イオンがどれだけ安定化できるかによって決まる。エタノールのエトキシドイオンは電荷を単一の酸素上に保持するため、エタノールは最も弱く、どちらとも反応しない。フェノールのフェノキシドイオンは電荷を環内へ広げるため、フェノールは強い塩基$\text{NaOH}$とは反応するが、弱い$\text{Na}_2\text{CO}_3$とは反応せず、泡立たない。エタン酸イオンは電荷を2つの酸素に分散させることが可能であり、3つの中で最も効果的であるため、エタン酸は両方と反応し、炭酸塩と反応して泡立つ**。順序:エタノール < フェノール < エタン酸。炭酸塩は識別テストである—炭酸のみが泡立ち、これがフェノールと酸を見分ける方法である。

    Explore · ⁨探索⁩

    Phenol reaction lab · ⁨フェノールの反応実験⁩

    Classify phenol reactions by the role of the aromatic OH group. · ⁨芳香族OH基の役割に基づいてフェノールの反応を分類する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    phenol/ˈfenɒl/ フェノール
    benzene/ˈbenziːn/ ベンゼン
    phenylamine/ˌfiːˈnaɪləmiːn/ アニリン
    diazonium salt/ˌdaɪəˈzəʊnɪəm sɒlt/ ジアゾニウム塩
    azo compound/ˈɑːzəʊ ˈkɒmpaʊnd/ アゾ化合物
    nitration/naɪˈtreɪʃn/ ニトロ化
    bromination/ˌbrəʊmɪˈneɪʃn/ ブロモ化
    acidity/æˈsɪdɪti/ 酸性度
    delocalised/dɪˈlɒkəlaɪzd/ 非局在化
    32.2

    Exam tips · ⁨試験対策⁩

    English
    • Phenol is a stronger acid than alcohols (its anion is stabilised by delocalisation) but weaker than carboxylic acids — it does not react with carbonates.
    • Phenol decolourises bromine water and gives a white precipitate without a catalyst (the ring is activated by oxygen).
    • Acyl chlorides react with alcohols/phenols to give esters readily (better than the reversible acid route); misty HCl fumes are the observation.
    日本語
    • フェノールはアルコールよりも強い酸(その陰イオンは離域によって安定化される)だが、カルボン酸よりは弱い—炭酸塩とは反応しない。
    • フェノールは臭素水を脱色させ、触媒なしで白色沈殿を与える(環は酸素によって活性化されている)。
    • アシル塩化物はアルコール/フェノールと容易に反応してエステルを与え(可逆的な酸ルートよりも優れる)、白煙状のHClが観察される。
  • 33

    Carboxylic acids and derivatives · ⁨カルボン酸及其び誘導体⁩

    Watch lesson · ⁨レッスンを視聴⁩
    33.1

    Carboxylic acids · ⁨カルボン酸⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reaction by which benzoic acid can be produced: (a) reaction of an alkylbenzene with hot alkaline $\text{KMnO}_4$ and then dilute acid, exemplified by methylbenzene
    2. describe the reaction of carboxylic acids with $\text{PCl}_3$ and heat, $\text{PCl}_5$ or $\text{SOCl}_2$ to form acyl chlorides
    3. recognise that some carboxylic acids can be further oxidised: (a) the oxidation of methanoic acid, $\text{HCOOH}$, with Fehling’s reagent or Tollens’ reagent or acidified $\text{KMnO}_4$ or acidified $\text{K}_2\text{Cr}_2\text{O}_7$ to carbon dioxide and water (b) the oxidation of ethanedioic acid, $\text{HOOCCOOH}$, with warm acidified $\text{KMnO}_4$ to carbon dioxide
    4. describe and explain the relative acidities of carboxylic acids, phenols and alcohols
    5. describe and explain the relative acidities of chlorine-substituted carboxylic acids
    日本語
    1. ベンゾ酸の生成反応を記憶せよ: (a) 熱い塩基性 $\text{KMnO}_4$ とアルキルベンゼンの反応、その後希酸処理(メチルベンゼンを例とする)
    2. カルボン酸が$\text{PCl}_3$および加熱、$\text{PCl}_5$、または$\text{SOCl}_2$と反応してアシルクロリドを形成する反応を説明する
    3. 有機酸の一部がさらに酸化されることを認識する: (a) フォルム酸 $\text{HCOOH}$ のフェーリング試薬、トレンス試薬、または酸性化された $\text{KMnO}_4$ や酸性化された $\text{K}_2\text{Cr}_2\text{O}_7$ との反応による二酸化炭素と水への酸化 (b) グリコール酸 $\text{HOOCCOOH}$ の温かい酸性化 $\text{KMnO}_4$ との反応による二酸化炭素への酸化
    4. カルボン酸、フェノールおよびアルコールの 相対的な酸性 を説明・解説せよ
    5. クロロ置換カルボン酸の相対的な酸性を説明・解説せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Making and reacting

    • an alkylbenzene 烷基苯 (such as methylbenzene) is oxidised by hot alkaline $\text{KMnO}_4$, then dilute acid, to give benzoic acid 苯甲酸. The whole side-chain becomes a $\text{–COOH}$ group.
    • a carboxylic acid reacts with $\text{PCl}_3$ and heat, $\text{PCl}_5$, or $\text{SOCl}_2$ to form an acyl chloride 酰氯.

    Acids that can be oxidised further

    Two carboxylic acids 羧酸 are special because they can still be oxidised:

    • methanoic acid ($\text{HCOOH}$) is oxidised by Fehling's or Tollens' reagent, or acidified $\text{KMnO}_4$ / $\text{K}_2\text{Cr}_2\text{O}_7$, to carbon dioxide and water.
    • ethanedioic acid ($\text{HOOCCOOH}$) is oxidised by warm acidified $\text{KMnO}_4$ to carbon dioxide.

    Relative acidities

    The acidity 酸性 order is:

    $$\text{alcohol} < \text{phenol} < \text{carboxylic acid}$$

    A carboxylic acid is the strongest because, when it loses $\text{H}^+$, the negative charge is spread over two oxygen atoms, making the ion very stable. In a phenol 苯酚 the charge spreads only into the ring, and in an alcohol 醇 (giving $\text{RO}^-$) it is not spread at all.

    Chlorine atoms make a carboxylic acid more acidic. Chlorine is electron-withdrawing 吸电子: through the inductive effect 诱导效应 it pulls electron density away, helping to spread the negative charge and stabilise the ion. So more chlorine atoms (closer to the $\text{–COOH}$) give a stronger acid.

    日本語

    全体および半分に切られたレモン *柑橘類は酸っぱい味这是因为它们含有柠檬酸,一种羧酸

    合成と反応

    • アルキルベンゼン(メチルベンゼンなど)は、熱い塩基性の$\text{KMnO}_4$で酸化され、その後希酸を加えることで安息香酸を得る。側鎖全体が$\text{–COOH}$基に変化する。
    • カルボン酸が$\text{PCl}_3$と加熱、または$\text{PCl}_5$、$\text{SOCl}_2$と反応してアシル塩化物を生成する。
    KMnO4溶液で加熱すると、環上のメチル側鎖がCOOH基に酸化され、安息香酸が生成される
    熱いKMnO4はメチル側鎖をCOOH基へ酸化します

    さらに酸化可能な酸

    2つの カルボン酸 は、まだ酸化されるという点で特別である:

    • メタノール酸($\text{HCOOH}$)は、フェリング試薬やトレンス試薬、あるいは酸性化された $\text{KMnO}_4$ / $\text{K}_2\text{Cr}_2\text{O}_7$ によって酸化され、二酸化炭素と水となる。
    • エタンジオール酸($\text{HOOCCOOH}$)は、温めた酸性の $\text{KMnO}_4$ によって酸化され、二酸化炭素となる。

    酸度の比較

    酸度 の順序は以下の通りである:

    $$\text{alcohol} < \text{phenol} < \text{carboxylic acid}$$

    カルボン酸が最も強いのは、$\text{H}^+$ を失うと負電荷が 2個 の酸素原子に分散するためイオンが非常に安定するからである。フェノール では電荷が環内にのみ分散し、アルコール($\text{RO}^-$ を形成)では全く分散しない。

    下段にアルコール、中段にフェノール、上段にカルボン酸の3段階スケール。各陰イオンが負電荷をどの程度分散させるかを示す
    アルコールからフェノール、そしてカルボン酸へと酸度は上昇する:共役塩基上の負電荷が分散すればするほどイオンは安定し、酸は強くなる

    塩素原子はカルボン酸の 酸度を高める。塩素は 電子吸引性 であり、誘起効果 を通じて電子密度を引き寄せ、負電荷の分散を助け、イオンを安定させる。したがって、より多くの塩素原子($\text{–COOH}$ に近い位置にあるもの)を持つ方が酸は強くなる。

    Explore · ⁨探索⁩

    Carboxylic acid A2 map · ⁨カルボン酸A2マップ⁩

    Follow carboxylic acids through acyl derivatives and salts. · ⁨カルボン酸からアシル誘導体や塩を通じて進める。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    carboxylic acid/ˌkɑːbəkˈsɪlɪk ˈæsɪd/ カルボン酸
    acidity/æˈsɪdɪti/ 酸性度
    phenol/ˈfenɒl/ フェノール
    alcohol/ˈælkəhɒl/ アルコール
    electron-withdrawing/ɪˈlektrɒn wɪθˈdrɔːɪŋ/ 電子吸引性
    inductive effect/ɪnˈdʌktɪv ɪˈfekt/ 誘導効果
    ester/ˈestə/ エステル
    ammonia/æˈməʊnɪə/ アンモニア
    amide/əˈmaɪd/ アミド
    amine/ˈæmaɪn/ アミン
    addition–elimination/əˈdɪʃn ɪˌlɪmɪˈneɪʃn/ 付加-脱離
    hydrolysis/haɪˈdrɒləsɪs/ 加水分解反応
    33.2

    Esters · ⁨エステル⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reaction by which esters can be produced: (a) reaction of alcohols with acyl chlorides using the formation of ethyl ethanoate and phenyl benzoate as examples
    日本語
    1. エステルの生成反応を記憶せよ: (a) アセチルクロリド(エチルアセテートおよびフェニルベンゾエートの生成を例とする)を用いたアルコールとの反応

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    An alcohol (or phenol) reacts with an acyl chloride at room temperature to give an ester 酯 and $\text{HCl}$. Examples are ethyl ethanoate (from ethanol) and phenyl benzoate (from phenol).

    日本語

    アルコール(またはフェノール)はアシルクロリドと常温で反応して エステル と $\text{HCl}$ を生成する。例として、エタノール由来のエチルエタノ酸エステル、およびフェノール由来のフェニル安息香酸エステルがある。

    アスピリン錠のボトル
    アスピリンはエステルであり、サリチル酸から合成される
    Explore · ⁨探索⁩

    Ester A2 reaction map · ⁨エステル A2 反応マップ⁩

    Compare ester formation, hydrolysis and transesterification routes. · ⁨エステル形成、加水分解、トランスエステル化の経路を比較せよ。⁩

    33.3

    Acyl chlorides · ⁨アシルクロリド⁩

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions (reagents and conditions) by which acyl chlorides can be produced: (a) reaction of carboxylic acids with $\text{PCl}_3$ and heat, $\text{PCl}_5$ or $\text{SOCl}_2$
    2. describe the following reactions of acyl chlorides: (a) hydrolysis on addition of water at room temperature to give the carboxylic acid and $\text{HCl}$ (b) reaction with an alcohol at room temperature to produce an ester and $\text{HCl}$ (c) reaction with phenol at room temperature to produce an ester and $\text{HCl}$ (d) reaction with ammonia at room temperature to produce an amide and $\text{HCl}$ (e) reaction with a primary or secondary amine at room temperature to produce an amide and $\text{HCl}$
    3. describe the addition–elimination mechanism of acyl chlorides in reactions in 33.3.2(a)–(e)
    4. explain the relative ease of hydrolysis of acyl chlorides, alkyl chlorides and halogenoarenes (aryl chlorides)
    日本語
    1. アシルクロリド の生成反応(試薬および条件)を記憶せよ: (a) カルボン酸の $\text{PCl}_3$ と加熱、または $\text{PCl}_5$ や $\text{SOCl}_2$ との反応
    2. 以下のアシルクロリドの反応を記述せよ: (a) 室温での水添加による加水分解でカルボン酸と $\text{HCl}$ を生成する (b) 室温でのアルコールとの反応でエステルと $\text{HCl}$ を生成する (c) 室温でのフェノールとの反応でエステルと $\text{HCl}$ を生成する (d) 室温でのアンモニアとの反応でアミドと $\text{HCl}$ を生成する (e) 室温での一次または二次アミンとの反応でアミドと $\text{HCl}$ を生成する
    3. 33.3.2(a)–(e)の反応におけるアシルクロリドの 付加-脱離メカニズム を説明せよ
    4. アシルクロリド、アルキルクロリドおよびハロゲン芳香族化合物(アリルクロリド)の加水分解の容易さの違いを説明せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Acyl chlorides are made from carboxylic acids (with $\text{PCl}_3$, $\text{PCl}_5$ or $\text{SOCl}_2$). They are very reactive. At room temperature they react with:

    Reactant Product (plus $\text{HCl}$)
    water the carboxylic acid
    an alcohol an ester
    phenol an ester
    ammonia 氨 an amide 酰胺
    a primary or secondary amine 胺 an amide

    The addition–elimination mechanism

    All these reactions follow an addition–elimination 加成消去 mechanism: a nucleophile first adds to the slightly positive carbonyl carbon, then $\text{HCl}$ is eliminated.

    Ease of hydrolysis

    Compare how easily three chlorides react with water (hydrolysis 水解):

    $$\text{acyl chloride} \gg \text{alkyl chloride} \gg \text{aryl chloride}$$

    An acyl chloride reacts violently with cold water; an alkyl chloride reacts slowly; an aryl chloride (halogenoarene) does not react, because its C–Cl bond is strengthened by the ring.

    Worked example. Ethanoyl chloride reacts violently with cold water, while ethyl ethanoate needs prolonged reflux with acid or alkali. Explain the difference. Both are attacked at the carbonyl carbon, so compare how open that carbon is to attack and how good the leaving group is. In ethanoyl chloride the chlorine is strongly electronegative and withdraws electrons, making the carbonyl carbon much more $\delta+$ and so far more open to a nucleophile; and $\text{Cl}^{-}$, the anion of a strong acid, is a good leaving group. In the ester the $\text{–OR}$ oxygen donates a lone pair into the carbonyl, reducing that $\delta+$ charge, and $\text{RO}^{-}$ is a poor leaving group. So the acyl chloride hydrolyses far more easily. Argue from both the $\delta+$ on the carbon and the leaving group: either one alone is usually only half the marks.

    日本語

    アシルクロリドはカルボン酸($\text{PCl}_3$、$\text{PCl}_5$ または $\text{SOCl}_2$ を用いて)から作られる。非常に反応性が高い。常温で次のものとは反応する:

    反応物 生成物(+ $\text{HCl}$ )
    水 カルボン酸
    アルコール エステル
    フェノール エステル
    -アンモニア -アミド
    primary または secondary アミン アミド
    中央にアシルクロリド、その周囲に水、アルコール、フェノール、アンモニア、アミンとの反応生成物を示す矢印
    アシルクロリドは非常に反応性が高く、水、アルコール、フェノール、アンモニア、またはアミンと反応すると、指定された生成物 plus HCl が得られる

    付加-脱離機構

    これらの反応すべては 付加-脱離機構 に従う:求核剤がまずわずかに正電荷をもつカルボニル炭素に 付加 し、その後 $\text{HCl}$ が 脱離 する。

    アシルクロリドの反応機構:求核剤がカルボニル炭素に付加して四面体型中間体を形成し、そこから塩化物イオンがHClとして脱離する過程
    付加-脱離:求核剤が $\delta+$ カルボニル炭素に付加し、C=Oが再形成され、Cl$^-$ がHClとして脱離する

    加水分解の容易さ

    3つの塩化物が水と反応する容易さ(加水分解)を比較せよ:

    $$\text{acyl chloride} \gg \text{alkyl chloride} \gg \text{aryl chloride}$$

    アシルクロリドは冷水と激しく反応するが、アルキルクロリドはゆっくり反応し、アリルクロリド(ハロゲンアレン)は反応しない。これは、環によってC–Cl結合が強固になっているためである。

    計算例。 エタノイルクロリドは冷水と激しく反応するが、エチルエタノ酸エステルは酸または塩基による長時間還流が必要である。その理由を説明せよ。両方とも カルボニル炭素 が攻撃されるので、その炭素が攻撃を受けやすいかどうか、および脱離基の性質を比較せよ。エタノイルクロリドでは、塩素が強く 電気陰性 であり電子を引き寄せるため、カルボニル炭素ははるかに $\delta+$ となり、求核剤による攻撃を受けやすくなる;さらに $\text{Cl}^{-}$ は強酸の陰イオンであり、良い脱離基 である。エステルでは、$\text{–OR}$ 酸素が非共有電子対をカルボニルに供与し、その $\delta+$ 電荷を減少させ、$\text{RO}^{-}$ は 悪い脱離基 である。したがってアシルクロリドの方がはるかに容易に加水分解する。炭素上の $\delta+$ と 脱離基の 両方の観点から議論すること:どちらか一方だけでは通常満点の半分しか得られない。

    Explore · ⁨探索⁩

    Acyl chloride reaction map · ⁨アシル塩化物反応マップ⁩

    Follow why acyl chlorides are reactive acylating agents. · ⁨アシル塩化物が反応性の高いアシル化剤となる理由を追う。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    alkylbenzene/ˈælkɪlbenziːn/ アルキルベンゼン
    benzoic acid/benˈzəʊɪk ˈæsɪd/ ベンゾ酸
    acyl chloride/əˈkɪl ˈklɔːraɪd/ アシルクロリド
    33.3

    Exam tips · ⁨試験対策⁩

    English
    • Acyl chlorides are very reactive — learn their products with water (HCl), alcohols (esters), ammonia (amides) and amines.
    • Reactivity order of derivatives: acyl chloride > ester > amide.
    • Acid strength: electron-withdrawing groups (e.g. Cl) increase it — explain via the carboxylate ion.
    日本語
    • アシルクロリド は非常に反応性が高い:水(HCl)、アルコール(エステル)、アンモニア(アミド)、アミンとの生成物を暗記すること。
    • 誘導体の反応性の順序:アシルクロリド > エステル > アミド 。
    • 酸の強さ:電子吸引基(例:Cl)は 強める :カルボキシラートイオンを用いて説明せよ。
  • 34

    Nitrogen compounds · ⁨窒素化合物⁩

    Watch lesson · ⁨レッスンを視聴⁩
    34.1

    Primary and secondary amines

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions (reagents and conditions) by which primary and secondary amines are produced: (a) reaction of halogenoalkanes with $\text{NH}_3$ in ethanol heated under pressure (b) reaction of halogenoalkanes with primary amines in ethanol, heated in a sealed tube/under pressure (c) the reduction of amides with $\text{LiAlH}_4$ (d) the reduction of nitriles with $\text{LiAlH}_4$ or $\text{H}_2/\text{Ni}$
    2. describe the condensation reaction of ammonia or an amine with an acyl chloride at room temperature to give an amide
    3. describe and explain the basicity of aqueous solutions of amines
    日本語
    1. 一次アミンおよび二次アミンが生成される反応(試薬および条件)を思い出す: (a) エタノール中で加圧加熱条件下でのハロゲンアルキルと $\text{NH}_3$ の反応 (b) エタノール中で密閉管/加圧下で加熱条件下的なハロゲンアルキルと一次アミンとの反応 (c) アミドの $\text{LiAlH}_4$ による還元 (d) ニトリルの $\text{LiAlH}_4$ または $\text{H}_2/\text{Ni}$ による還元
    2. 室温でのアンモニアまたはアミンとアシルクロリドの 縮合反応 による アミド の生成を記述せよ
    3. アミンの水溶液の 塩基性 を説明・解説せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    An amine 胺 has an $\text{–NH}_2$ (primary) or $\text{–NH}$ (secondary) group.

    Making amines

    • a halogenoalkane 卤代烷 with $\text{NH}_3$ in ethanol, heated under pressure → a primary amine.
    • a halogenoalkane with a primary amine, heated under pressure → a secondary amine.
    • reduction 还原 of an amide 酰胺 with $\text{LiAlH}_4$.
    • reduction of a nitrile 腈 with $\text{LiAlH}_4$ or $\text{H}_2/\text{Ni}$.

    An amine (or ammonia 氨) reacts with an acyl chloride in a condensation 缩合 reaction to give an amide. The reagent is an acyl chloride 酰氯.

    Basicity of amines

    The basicity 碱性 of an amine comes from the lone pair on its nitrogen, which can accept an $\text{H}^+$ from water.

    日本語

    An amine 胺 has an $\text{–NH}_2$ (primary) or $\text{–NH}$ (secondary) group.

    A primary amine has -NH2; a secondary amine has -NH between two carbons
    A primary amine has -NH2; a secondary amine has -NH-

    Making amines

    • a halogenoalkane 卤代烷 with $\text{NH}_3$ in ethanol, heated under pressure → a primary amine.
    • a halogenoalkane with a primary amine, heated under pressure → a secondary amine.
    • reduction 还原 of an amide 酰胺 with $\text{LiAlH}_4$.
    • reduction of a nitrile 腈 with $\text{LiAlH}_4$ or $\text{H}_2/\text{Ni}$.
    Two routes: a halogenoalkane with ammonia, or reducing a nitrile
    Two routes to an amine: from a halogenoalkane, or by reducing a nitrile

    An amine (or ammonia 氨) reacts with an acyl chloride in a condensation 缩合 reaction to give an amide. The reagent is an acyl chloride 酰氯.

    Basicity of amines

    The basicity 碱性 of an amine comes from the lone pair on its nitrogen, which can accept an $\text{H}^+$ from water.

    Explore · ⁨探索⁩

    Amine type lab · ⁨アミン型実験室⁩

    Classify amines by substitution at nitrogen and basic behaviour. · ⁨窒素における置換度および塩基性挙動に基づいてアミンを分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    amine/ˈæmaɪn/ アミン
    halogenoalkane/ˈheɪləʊdʒnəʊlkeɪn/ ハロゲン化アルキル
    reduction/rɪˈdʌkʃn/ 還元
    amide/əˈmaɪd/ アミド
    nitrile/ˈnaɪtraɪl/ ニトリル
    ammonia/æˈməʊnɪə/ アンモニア
    condensation/kɒndenˈseɪʃn/ 縮合
    acyl chloride/əˈkɪl ˈklɔːraɪd/ アシルクロリド
    basicity/beɪˈsɪsɪti/ 塩基性
    34.2

    Phenylamine and azo compounds

    Syllabus · ⁨シラバス⁩
    1. describe the preparation of phenylamine via the nitration of benzene to form nitrobenzene followed by reduction with hot Sn/concentrated $\text{HCl}$ followed by $\text{NaOH(aq)}$
    2. describe: (a) the reaction of phenylamine with $\text{Br}_2\text{(aq)}$ at room temperature (b) the reaction of phenylamine with $\text{HNO}_2$ or $\text{NaNO}_2$ and dilute acid below $10\text{ }^\circ\text{C}$ to produce the diazonium salt; further warming of the diazonium salt with $\text{H}_2\text{O}$ to give phenol
    3. describe and explain the relative basicities of aqueous ammonia, ethylamine and phenylamine
    4. recall the following about azo compounds: (a) describe the coupling of benzenediazonium chloride with phenol in $\text{NaOH(aq)}$ to form an azo compound (b) identify the azo group (c) state that azo compounds are often used as dyes (d) that other azo dyes can be formed via a similar route

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Making phenylamine

    Make phenylamine 苯胺 from benzene in two steps: nitrate benzene to nitrobenzene, then reduce it with hot $\text{Sn}$ and concentrated $\text{HCl}$, followed by $\text{NaOH(aq)}$.

    Reactions

    • with bromine water at room temperature → 2,4,6-tribromophenylamine (a white precipitate). The ring is activated, like phenol.
    • with $\text{HNO}_2$ (from $\text{NaNO}_2$ and dilute acid) below $10\,°\text{C}$ → a diazonium salt; warming this with water gives phenol.

    Relative basicity

    $$\text{phenylamine} < \text{ammonia} < \text{ethylamine}$$

    Ethylamine is the strongest base: its alkyl group pushes electron density onto the nitrogen, making the lone pair more available. Phenylamine is the weakest, because its lone pair is delocalised 离域 into the benzene ring, so it is less available to accept an $\text{H}^+$.

    Azo compounds

    A diazonium salt 重氮盐 (benzenediazonium chloride) couples with phenol 苯酚 in $\text{NaOH(aq)}$ to form an azo compound 偶氮化合物. The azo group is $\text{–N=N–}$. Azo compounds are brightly coloured and are often used as dye 染料s; many other azo dyes are made the same way.

    Worked example. An amino acid has an isoelectric point of pH 6.0. Give its charge, and the electrode it moves towards in electrophoresis, at pH 2, at pH 6.0 and at pH 11. Compare the solution's pH with the isoelectric point each time. At pH 2, well below it, the solution is acidic, so the $\text{–NH}_2$ gains an $\text{H}^{+}$: the amino acid is positive and moves to the cathode (negative electrode). At pH 6.0, exactly its isoelectric point, it is the zwitterion with no net charge, so it does not move. At pH 11, well above it, the $\text{–COOH}$ loses its $\text{H}^{+}$: the amino acid is negative and moves to the anode. Compare the pH with the isoelectric point, never with 7 - and note the zwitterion still carries both charges, it simply has no net charge.

    日本語

    Making phenylamine

    Make phenylamine 苯胺 from benzene in two steps: nitrate benzene to nitrobenzene, then reduce it with hot $\text{Sn}$ and concentrated $\text{HCl}$, followed by $\text{NaOH(aq)}$.

    Reactions

    • with bromine water at room temperature → 2,4,6-tribromophenylamine (a white precipitate). The ring is activated, like phenol.
    • with $\text{HNO}_2$ (from $\text{NaNO}_2$ and dilute acid) below $10\,°\text{C}$ → a diazonium salt; warming this with water gives phenol.

    Relative basicity

    $$\text{phenylamine} < \text{ammonia} < \text{ethylamine}$$

    Ethylamine is the strongest base: its alkyl group pushes electron density onto the nitrogen, making the lone pair more available. Phenylamine is the weakest, because its lone pair is delocalised 离域 into the benzene ring, so it is less available to accept an $\text{H}^+$.

    Phenylamine, ammonia and ethylamine along a basicity scale, with the ring pulling the lone pair away in phenylamine and the alkyl group pushing electrons in ethylamine
    Basicity depends on the nitrogen lone pair: an alkyl group makes it more available (ethylamine strongest), a benzene ring pulls it away (phenylamine weakest)

    Azo compounds

    A diazonium salt 重氮盐 (benzenediazonium chloride) couples with phenol 苯酚 in $\text{NaOH(aq)}$ to form an azo compound 偶氮化合物. The azo group is $\text{–N=N–}$. Azo compounds are brightly coloured and are often used as dye 染料s; many other azo dyes are made the same way.

    A small pile of bright orange methyl orange dye powder on a pale background
    Methyl orange, a bright orange azo dye; the strong colour comes from the $\text{–N=N–}$ azo group, which is why azo compounds are so widely used as dyes

    Worked example. An amino acid has an isoelectric point of pH 6.0. Give its charge, and the electrode it moves towards in electrophoresis, at pH 2, at pH 6.0 and at pH 11. Compare the solution's pH with the isoelectric point each time. At pH 2, well below it, the solution is acidic, so the $\text{–NH}_2$ gains an $\text{H}^{+}$: the amino acid is positive and moves to the cathode (negative electrode). At pH 6.0, exactly its isoelectric point, it is the zwitterion with no net charge, so it does not move. At pH 11, well above it, the $\text{–COOH}$ loses its $\text{H}^{+}$: the amino acid is negative and moves to the anode. Compare the pH with the isoelectric point, never with 7 - and note the zwitterion still carries both charges, it simply has no net charge.

    Explore · ⁨探索⁩

    Phenylamine to azo dye route · ⁨フェニルアミンからアゾ染料への経路⁩

    Follow phenylamine from diazotisation to coloured azo compound. · ⁨ジアゾ化から有色のアゾ化合物までのフェニルアミンを追う。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    phenylamine/ˌfiːˈnaɪləmiːn/ アニリン
    delocalised/dɪˈlɒkəlaɪzd/ 非局在化
    diazonium salt/ˌdaɪəˈzəʊnɪəm sɒlt/ ジアゾニウム塩
    phenol/ˈfenɒl/ フェノール
    azo compound/ˈɑːzəʊ ˈkɒmpaʊnd/ アゾ化合物
    dye/daɪ/ 染料
    34.3

    Amides

    Syllabus · ⁨シラバス⁩
    English
    1. recall the reactions (reagents and conditions) by which amides are produced: (a) the reaction between ammonia and an acyl chloride at room temperature (b) the reaction between a primary amine and an acyl chloride at room temperature
    2. describe the reactions of amides: (a) hydrolysis with aqueous alkali or aqueous acid (b) the reduction of the CO group in amides with $\text{LiAlH}_4$ to form an amine
    3. state and explain why amides are much weaker bases than amines
    日本語
    1. アミド の生成反応(試薬および条件)を記憶せよ: (a) 室温でのアンモニアとアシルクロリドの反応 (b) 室温での一次アミンとアシルクロリドの反応
    2. アミドの反応を記述せよ: (a) 水酸化アルカリ水溶液または希酸による加水分解 (b) $\text{LiAlH}_4$ によるアミド中のCO基の還元でアミンを生成する
    3. アミドがアミンに比べて why より弱い塩基 であるかを述べ、説明せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    An amide is made from ammonia or a primary amine with an acyl chloride at room temperature.

    A bottle of paracetamol tablets
    Paracetamol is a common painkiller that contains the amide group (–CONH–)

    Its reactions:

    • hydrolysis with aqueous acid or alkali, giving the carboxylic acid (or its salt) and the amine (or ammonium).
    • reduction of the C=O group with $\text{LiAlH}_4$ to give an amine.

    An amide is a much weaker base than an amine, because the nitrogen lone pair is delocalised onto the neighbouring C=O group, so it is not available to accept an $\text{H}^+$.

    Explore · ⁨探索⁩

    Amide formation route · ⁨アミド形成経路⁩

    Follow acyl chloride or acid derivative to an amide. · ⁨アシル塩化物または酸誘導体からアミドへ進む。⁩

    34.4

    Amino acids

    Syllabus · ⁨シラバス⁩
    English
    1. describe the acid/base properties of amino acids and the formation of zwitterions, to include the isoelectric point
    2. describe the formation of amide (peptide) bonds between amino acids to give di- and tripeptides
    3. interpret and predict the results of electrophoresis on mixtures of amino acids and dipeptides at varying pHs (the assembling of the apparatus will not be tested)
    日本語
    1. アミノ酸 の酸塩基性と ジオン性 の生成を記述せよ。その際に 等電点 を含む
    2. アミノ酸間の アミド結合(ペプチド結合) の形成によってジペプチドおよびトリペプチドが生成する様子を記述せよ
    3. 様々なpHにおけるアミノ酸およびジペプチド混合体の 電気泳動 の結果を解釈・予測せよ(装置の組み立ては試験対象外)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    An amino acid 氨基酸 has both a basic $\text{–NH}_2$ group and an acidic $\text{–COOH}$ group.

    Zwitterions and the isoelectric point

    The $\text{–COOH}$ can give its $\text{H}^+$ to the $\text{–NH}_2$ in the same molecule, forming a zwitterion 两性离子 ($\text{H}_3\text{N}^+\text{–CHR–COO}^-$) — an ion with both a positive and a negative end but no overall charge.

    • in acid (low pH), the amino acid gains $\text{H}^+$ and becomes positive.
    • in alkali (high pH), it loses $\text{H}^+$ and becomes negative.
    • at one special pH, the isoelectric point 等电点, it is mostly the zwitterion with no net charge.

    Peptide bonds

    Two amino acids join in a condensation reaction: the $\text{–COOH}$ of one and the $\text{–NH}_2$ of the other react, losing water and forming a peptide bond 肽键 (an amide link). Two amino acids give a dipeptide 二肽, three give a tripeptide.

    Electrophoresis

    In electrophoresis 电泳, a mixture is placed in an electric field at a chosen pH:

    • above its isoelectric point, an amino acid is negative and moves to the positive electrode.
    • below its isoelectric point, it is positive and moves to the negative electrode.
    • at its isoelectric point, it does not move. So different amino acids separate.
    日本語

    An amino acid 氨基酸 has both a basic $\text{–NH}_2$ group and an acidic $\text{–COOH}$ group.

    Zwitterions and the isoelectric point

    The $\text{–COOH}$ can give its $\text{H}^+$ to the $\text{–NH}_2$ in the same molecule, forming a zwitterion 两性离子 ($\text{H}_3\text{N}^+\text{–CHR–COO}^-$) — an ion with both a positive and a negative end but no overall charge.

    • in acid (low pH), the amino acid gains $\text{H}^+$ and becomes positive.
    • in alkali (high pH), it loses $\text{H}^+$ and becomes negative.
    • at one special pH, the isoelectric point 等电点, it is mostly the zwitterion with no net charge.
    An amino acid drawn three ways: positively charged in acid, a neutral zwitterion at the isoelectric point, and negatively charged in alkali
    An amino acid's charge depends on pH: positive in acid, the neutral zwitterion at the isoelectric point, negative in alkali

    Peptide bonds

    Two amino acids join in a condensation reaction: the $\text{–COOH}$ of one and the $\text{–NH}_2$ of the other react, losing water and forming a peptide bond 肽键 (an amide link). Two amino acids give a dipeptide 二肽, three give a tripeptide.

    Two amino acids condensing: the OH from one carboxyl and the H from the other amino group leave as water, forming a peptide bond
    Two amino acids condense: the –OH from one –COOH and the –H from the other –NH$_2$ leave as water, forming the peptide (amide) bond

    Electrophoresis

    In electrophoresis 电泳, a mixture is placed in an electric field at a chosen pH:

    • above its isoelectric point, an amino acid is negative and moves to the positive electrode.
    • below its isoelectric point, it is positive and moves to the negative electrode.
    • at its isoelectric point, it does not move. So different amino acids separate.
    A gel strip with a negative and a positive electrode: a positive amino acid moves towards the negative electrode, a negative one towards the positive, and a neutral one stays put
    Electrophoresis at a chosen pH: a positive amino acid moves to the negative electrode, a negative one to the positive, and a neutral one stays — so they separate
    Explore · ⁨探索⁩

    Amino acid lab · ⁨アミノ酸実験⁩

    Classify amino acid behaviour by the group that reacts. · ⁨反応する基によってアミノ酸の挙動を分類する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    amino acid/əˈmiːnəʊ ˈæsɪd/ アミノ酸
    zwitterion/zwɪˈtɪərɪən/ ジオニオン
    isoelectric point/ˌaɪsəʊˈlektrɪk pɔɪnt/ 等電点
    peptide bond/ˈpeptaɪd bɒnd/ ペプチド結合
    dipeptide/ˈdaɪpptaɪd/ ジペプチド
    electrophoresis/ɪˌlektrəʊfɔːˈriːsɪs/ 電気泳動
    34.4

    Exam tips

    • Phenylamine is a weaker base than aliphatic amines because the N lone pair delocalises into the ring.
    • Diazotisation ($\text{NaNO}_2$/HCl, below $10\ ^\circ\text{C}$) then coupling gives azo dyes — learn the conditions and the coloured product.
    • Amino acids are zwitterions (both $-\text{NH}_2$ and $-\text{COOH}$), so they are amphoteric; describe behaviour either side of the isoelectric point.
  • 35

    Polymerisation · ⁨重合反応⁩

    Watch lesson · ⁨レッスンを視聴⁩
    35.1

    Condensation polymerisation

    Syllabus · ⁨シラバス⁩
    English
    1. describe the formation of polyesters: (a) the reaction between a diol and a dicarboxylic acid or dioyl chloride (b) the reaction of a hydroxycarboxylic acid
    2. describe the formation of polyamides: (a) the reaction between a diamine and a dicarboxylic acid or dioyl chloride (b) the reaction of an aminocarboxylic acid (c) the reaction between amino acids
    3. deduce the repeat unit of a condensation polymer obtained from a given monomer or pair of monomers
    4. identify the monomer(s) present in a given section of a condensation polymer molecule
    日本語
    1. ポリエステル の生成を記述せよ: (a) ジオールとジカルボン酸またはジアシルクロリドの反応 (b) ハイドロキシカルボン酸の反応
    2. ポリアミド の生成を記述せよ: (a) ジアミンとジカルボン酸またはジアシルクロリドの反応 (b) アミノカルボン酸の反応 (c) アミノ酸同士の反応
    3. 与えられた モノマー もしくは モノマーの組み合わせ から得られる 縮合重合体 の 反復単位 を導き出せ
    4. 与えられた縮合重合体分子の一部に含まれる モノマー を同定せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    In condensation polymerisation 缩合聚合, monomers join into a long chain and a small molecule (such as water or $\text{HCl}$) is lost each time a new bond forms. Each monomer needs two reactive groups, so the chain can grow at both ends.

    Polyesters

    A polyester 聚酯 has many ester links along its chain. You can make one from:

    • a diol 二醇 (two $\text{–OH}$ groups) and a dicarboxylic acid 二羧酸 (two $\text{–COOH}$ groups), or a dioyl chloride.
    • a single hydroxycarboxylic acid, which has both an $\text{–OH}$ and a $\text{–COOH}$.

    Polyamides

    A polyamide 聚酰胺 has many amide links. You can make one from:

    • a diamine 二胺 (two $\text{–NH}_2$ groups) and a dicarboxylic acid or a dioyl chloride.
    • a single aminocarboxylic acid, or from amino acids 氨基酸 joining together (proteins are natural polyamides).

    Repeat units and monomers

    The repeat unit 重复单元 of a condensation polymer contains parts of both monomers, minus the atoms lost as the small molecule. To find the monomers 单体 from a section of polymer, break the chain at each ester or amide link, then add back $\text{–OH}$ and $\text{–H}$ (or $\text{–Cl}$).

    Worked example. A polymer chain contains repeating $\text{–CONH–}$ links. Name the type of polymer, identify the monomers, and give the small molecule lost. A $\text{–CONH–}$ link is an amide, so this is a polyamide made by condensation polymerisation. To find the monomers, break the chain at each amide link and give the cut ends their atoms back: the carbon side takes $\text{–OH}$, making a dicarboxylic acid, and the nitrogen side takes $\text{–H}$, making a diamine. The small molecule lost at each link is water (or $\text{HCl}$, if an acyl dichloride was used in place of the acid). Each monomer must have two functional groups, or the chain could never keep growing - if the monomer you propose has only one, you have broken the chain in the wrong place.

    日本語

    In condensation polymerisation 缩合聚合, monomers join into a long chain and a small molecule (such as water or $\text{HCl}$) is lost each time a new bond forms. Each monomer needs two reactive groups, so the chain can grow at both ends.

    A carboxyl group joining a hydroxyl to form an ester link and a carboxyl joining an amine to form an amide link, each releasing a water molecule
    Condensation links: a polyester forms ester links and a polyamide forms amide links; each new link releases a small molecule (here water)

    Polyesters

    A polyester 聚酯 has many ester links along its chain. You can make one from:

    • a diol 二醇 (two $\text{–OH}$ groups) and a dicarboxylic acid 二羧酸 (two $\text{–COOH}$ groups), or a dioyl chloride.
    • a single hydroxycarboxylic acid, which has both an $\text{–OH}$ and a $\text{–COOH}$.
    A bale of crushed clear plastic drinks bottles
    PET drinks bottles are made of a polyester — a condensation polymer, collected here for recycling

    Polyamides

    A polyamide 聚酰胺 has many amide links. You can make one from:

    • a diamine 二胺 (two $\text{–NH}_2$ groups) and a dicarboxylic acid or a dioyl chloride.
    • a single aminocarboxylic acid, or from amino acids 氨基酸 joining together (proteins are natural polyamides).
    A pink strand of nylon being pulled up out of a beaker, forming a continuous thread at the surface of the liquid
    The "nylon rope trick": nylon (a polyamide) forms where two reactant solutions meet, so a single long thread can be pulled out (it is pink here from an added indicator)

    Repeat units and monomers

    The repeat unit 重复单元 of a condensation polymer contains parts of both monomers, minus the atoms lost as the small molecule. To find the monomers 单体 from a section of polymer, break the chain at each ester or amide link, then add back $\text{–OH}$ and $\text{–H}$ (or $\text{–Cl}$).

    A polyester chain broken at its ester links and the cut ends given back -OH and -H, to recover the diol and the dicarboxylic acid monomers
    To find the monomers, break the chain at each link and add back –OH and –H to the cut ends — here giving the diol and the dicarboxylic acid

    Worked example. A polymer chain contains repeating $\text{–CONH–}$ links. Name the type of polymer, identify the monomers, and give the small molecule lost. A $\text{–CONH–}$ link is an amide, so this is a polyamide made by condensation polymerisation. To find the monomers, break the chain at each amide link and give the cut ends their atoms back: the carbon side takes $\text{–OH}$, making a dicarboxylic acid, and the nitrogen side takes $\text{–H}$, making a diamine. The small molecule lost at each link is water (or $\text{HCl}$, if an acyl dichloride was used in place of the acid). Each monomer must have two functional groups, or the chain could never keep growing - if the monomer you propose has only one, you have broken the chain in the wrong place.

    Explore · ⁨探索⁩

    Condensation polymer route · ⁨縮合重合の経路⁩

    Watch monomers join while a small molecule leaves each time. · ⁨モノマー同士が結合するたびに小分子が脱離する様子を見る。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    condensation polymerisation/kɒndenˈseɪʃn ˌpɒlɪməraɪˈzeɪʃn/ 縮加重合
    polyester/ˌpɒlɪˈestə/ ポリエステル
    diol/dɪˈɒl/ ジオール
    dicarboxylic acid/ˌdɪkɑːbəkˈsɪlɪk ˈæsɪd/ 二カルボン酸
    polyamide/ˌpɒlɪˈeɪmaɪd/ ポリアミド
    diamine/ˈdaɪəmaɪn/ ジアミン
    amino acid/əˈmiːnəʊ ˈæsɪd/ アミノ酸
    repeat unit/rɪˈpiːt ˈjuːnɪt/ 繰り返し単位
    monomer/ˈmɒnəʊmə/ モノマー
    35.2

    Predicting the type of polymerisation

    Syllabus · ⁨シラバス⁩
    English
    1. predict the type of polymerisation reaction for a given monomer or pair of monomers
    2. deduce the type of polymerisation reaction which produces a given section of a polymer molecule
    日本語
    1. 与えられた モノマー もしくは モノマーの組み合わせ に対して 重合反応 の種類を予測せよ
    2. 与えられた重合体分子の一部を生成する 重合反応 の種類を導き出せ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Clue Type
    the monomer has a C=C double bond, and nothing else is lost addition polymerisation 加成聚合
    each monomer has two functional groups, and a small molecule is lost; the chain has ester or amide links condensation polymerisation
    日本語
    Clue Type
    the monomer has a C=C double bond, and nothing else is lost addition polymerisation 加成聚合
    each monomer has two functional groups, and a small molecule is lost; the chain has ester or amide links condensation polymerisation
    Two panels comparing addition polymerisation, where a C=C opens and nothing is lost, with condensation polymerisation, where two-group monomers join and a small molecule is lost
    The two kinds of polymerisation: addition opens a C=C and loses nothing; condensation joins two-group monomers and loses a small molecule
    Explore · ⁨探索⁩

    Polymerisation type lab · ⁨重合種別の実験⁩

    Classify monomers by whether they form addition or condensation polymers. · ⁨付加重合性か付加重合性かによって単量体を分類する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    addition polymerisation/əˈdɪʃn ˌpɒlɪməraɪˈzeɪʃn/ 付加重合
    35.3

    Degradable polymers

    Syllabus · ⁨シラバス⁩
    English
    1. recognise that poly(alkenes) are chemically inert and can therefore be difficult to biodegrade
    2. recognise that some polymers can be degraded by the action of light
    3. recognise that polyesters and polyamides are biodegradable by acidic and alkaline hydrolysis
    日本語
    1. ポリアルケンが化学的に不活性であるため、生分解性に乏しく、分解されにくいことを理解する
    2. 光の作用によって分解されるポリマーがあることを認識する
    3. ポリエステルおよびポリアミドは、酸性および塩基性の加水分解により生分解性があることを認識する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    • poly(alkene)s 聚烯烃 are chemically inert — they have only strong, non-polar C–C and C–H bonds, so they are hard to biodegrade 可生物降解 and last a long time.
    • some polymers are made so that light can break them down (they are photodegradable).
    • polyesters and polyamides are biodegradable, because their ester and amide links can be broken by acidic or alkaline hydrolysis 水解.
    日本語
    • poly(alkene)s 聚烯烃 are chemically inert — they have only strong, non-polar C–C and C–H bonds, so they are hard to biodegrade 可生物降解 and last a long time.
    • some polymers are made so that light can break them down (they are photodegradable).
    • polyesters and polyamides are biodegradable, because their ester and amide links can be broken by acidic or alkaline hydrolysis 水解.
    Explore · ⁨探索⁩

    Degradable polymer route · ⁨分解性高分子の経路⁩

    Follow how polymer structure controls breakdown. · ⁨高分子構造が分解をどのように制御するかを追跡する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    poly(alkene)/ˈpɒlɪ/ ポリアルケン
    biodegradable/ˌbaɪəʊdɪˈɡreɪdəbl/ 生分解性
    hydrolysis/haɪˈdrɒləsɪs/ 加水分解反応
    35.3

    Exam tips

    • Condensation polymers (polyesters, polyamides) form with loss of a small molecule ($\text{H}_2\text{O}$ or HCl) — draw the repeat unit and the lost molecule.
    • Identify the monomers from the polymer by breaking the ester or amide link — a common question.
    • Predict the type from the monomers: a $\text{C}=\text{C}$ gives addition; two functional groups give condensation.
    • Polyesters and polyamides are hydrolysable (more degradable); addition polymers are not.
  • 36

    Organic synthesis · ⁨有機合成⁩

    Watch lesson · ⁨レッスンを視聴⁩
    36.1

    Organic synthesis · ⁨有機合成⁩

    Syllabus · ⁨シラバス⁩
    English
    1. for an organic molecule containing several functional groups: (a) identify organic functional groups using the reactions in the syllabus (b) predict properties and reactions
    2. devise multi-step synthetic routes for preparing organic molecules using the reactions in the syllabus
    3. analyse a given synthetic route in terms of type of reaction and reagents used for each step of it, and possible by-products
    日本語
    1. 複数の官能基を含む有機分子について: (a) 履修内容の反応を用いて有機官能基を特定 (b) 性質および反応を予測する
    2. 履修内容の反応を用いて有機分子を調製するための多段階合成ルートを考案する
    3. 与えられた合成ルートを反応の種類、各工程で使用される試薬、および副生成物について分析する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Like the AS synthesis topic, this asks you to join up known reactions to build a target molecule. Now you also have the A Level reactions of arenes, phenol, amines, amides and acyl chlorides.

    Identifying functional groups

    A molecule may carry several functional group 官能团 types. Use the test reactions to spot each one, then predict its behaviour. For example:

    • decolourises bromine water with no catalyst, giving a white precipitate → a phenol or a phenylamine (the ring is activated).
    • reacts violently with cold water, giving fumes of $\text{HCl}$ → an acyl chloride 酰氯.
    • gives a purple colour with neutral $\text{FeCl}_3$ → a phenol 苯酚.

    A map of the A Level reactions

    Start Reagent and conditions Product
    benzene (an arene 芳烃) $\text{HNO}_3$ / $\text{H}_2\text{SO}_4$ (electrophilic substitution 亲电取代) nitrobenzene
    nitrobenzene $\text{Sn}$ / conc $\text{HCl}$, then $\text{NaOH}$ (reduction 还原) phenylamine
    phenylamine $\text{HNO}_2$, below $10\,°\text{C}$ a diazonium salt
    diazonium salt warm water phenol; or couple with phenol → azo dye
    methylbenzene hot $\text{KMnO}_4$ (oxidation 氧化) benzoic acid
    carboxylic acid 羧酸 $\text{SOCl}_2$ or $\text{PCl}_5$ acyl chloride
    acyl chloride alcohol / phenol an ester 酯
    acyl chloride ammonia / amine 胺 an amide 酰胺
    amide or nitrile 腈 $\text{LiAlH}_4$ an amine
    halogenoalkane $\text{KCN}$ a nitrile (adds one carbon)

    Planning and analysing a route

    To devise a synthetic route 合成路线, work backwards from the target: which single reaction makes it, and from what? Repeat until you reach the starting material, then write each step with its reagent 试剂 and conditions.

    When you analyse a route, state the type of reaction for each step (for example electrophilic substitution, addition–elimination 加成消去, oxidation or reduction) and watch for likely by-products 副产物 — for example, making an amine from a halogenoalkane also gives over-substituted amines, lowering the yield.

    Worked example. Devise a route from benzene to phenylamine, $\text{C}_6\text{H}_5\text{NH}_2$. Work backwards: phenylamine comes from reducing nitrobenzene, and nitrobenzene comes from nitrating benzene - so the route is two steps. Step 1: benzene with concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ at $55\ °\text{C}$; electrophilic substitution gives nitrobenzene (keep below $55\ °\text{C}$, or further substitution follows). Step 2: reduce the nitrobenzene with tin and concentrated $\text{HCl}$, then add $\text{NaOH}$ to free the amine from its salt. There is no direct route - you cannot put an $\text{–NH}_2$ straight onto a ring, which is exactly why this nitrate-then-reduce pair is worth knowing by heart.

    日本語

    AS合成トピックと同様に、既知の反応をつなげてターゲット分子を構築することを問う。今度はA-Levelの芳香族化合物、フェノール、アミン、アミド、アシルクロリドの反応も追加された。

    官能基の特定

    1つの分子が複数の官能基種を持つことがある。検出反応を使ってそれぞれを見分け、振る舞いを予測せよ。例えば:

    • 触媒なしで臭素水を脱色し、白色沈殿を作る → フェノールまたはフェニルアミン(環が活性化されている)。
    • 冷水と激しく反応し、$\text{HCl}$の煙を出す → アシルクロリド。
    • 中性の$\text{FeCl}_3$と反応して紫色を示す → フェノール。
    A-Level試験の3列表:触媒なしの臭素水、冷水、中性塩化鉄(III)。各々における観察結果と官能基を示す
    A-Levelで追加された同定試験:各試薬は特定の官能基を示唆する特徴的な結果を与える

    A-Level反応のマップ

    出発 試薬と条件 生成物
    ベンゼン(芳香族化合物) $\text{HNO}_3$ / $\text{H}_2\text{SO}_4$(求電子置換反応) ニトロベンゼン
    ニトロベンゼン $\text{Sn}$ / 濃$\text{HCl}$、次に$\text{NaOH}$(還元) フェニルアミン
    フェニルアミン $\text{HNO}_2$、$10\,°\text{C}$未満 ジアゾニウム塩
    ジアゾニウム塩 温かい水 フェノール、またはフェノールとカップリング → アゾ染料
    メチルベンゼン 熱い$\text{KMnO}_4$(酸化) ベンゾ酸
    カルボン酸 $\text{SOCl}_2$ または $\text{PCl}_5$ アシルクロリド
    アシル塩化物 アルコール /フェノール エステル
    アシル塩化物 アンモニア / アミン アミド
    アミドまたはニトリル $\text{LiAlH}_4$ アミン
    ハロゲン化アルカン $\text{KCN}$ ニトリル(炭素が1つ増える)
    A Level反応のネットワーク:ベンゼンからニトロベンゼン、フェニルアミン、ジアゾニウム塩、フェノールまたはアゾ染料への経路と、カルボン酸からアシル塩化物、エステルおよびアミドへの経路
    A-Level反応のマップ:芳香族鎖(上)とアシルクロリド鎖(下)、およびそれぞれの投入原料。目標化合物から逆算する

    ルートの設計と分析

    合成ルートを考案するには、標的物質から逆向きに進む:それを生成する単一の反応は何であり、その原料は何か? 出発物質に達するまで繰り返した後、各ステップに試薬と条件を記す。

    ベンゼンからアゾ染料への合成チェーン、各進行方向のステップに試薬を示し、上方に大きな矢印で計画が標的物質から出発物質へ逆向きに行うことを示す
    ターゲットから逆算して、1つずつ反応を遡り、出発物質に到達するまでルートを計画する

    ルートに分析を行う際は、各ステップの反応種別(例:求電子置換、付加-除去、酸化、還元など)を明記し、予想される副産物に注意する。例えば、ハロゲン化アルカンからアミンを合成する場合、多置換アミンも生成され、収率が低下することがある。

    ** worked example.** ベンゼンからフェニルアミンへのルートを考案する、$\text{C}_6\text{H}_5\text{NH}_2$。逆向きで考える:フェニルアミンは還元によってニトロベンゼンから得られ、ニトロベンゼンはベンゼンのニトロ化によって得られる。よってルートは2ステップとなる。 ステップ1:ベンゼンに濃硫酸 $\text{HNO}_3$ と濃硝酸 $\text{H}_2\text{SO}_4$ を$55\ °\text{C}$ で反応させる。求電子置換によりニトロベンゼンが生成される($55\ °\text{C}$以下に保つ。それ以上高温だとさらに置換が進む)。 ステップ2:ニトロベンゼンをスズと濃硫酸 $\text{HCl}$ で還元し、次に $\text{NaOH}$ を加えてアミンを塩から遊離させる。直接合成のルートはない。環に直接$\text{–NH}_2$ を導入することはできないため、この「ニトロ化→還元」の組み合わせを暗記しておく必要がある。

    Explore · ⁨探索⁩

    A-Level synthesis planner · ⁨A-Level 合成プランナー⁩

    Build a route by matching functional-group changes to reagents. · ⁨官能基の変化に試薬を対応させることで経路を構築する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    functional group/ˈfʌŋkʃənl ɡruːp/ 官能基
    acyl chloride/əˈkɪl ˈklɔːraɪd/ アシルクロリド
    phenol/ˈfenɒl/ フェノール
    arene/ˈæren/ アレーン
    electrophilic substitution/ɪˌlektrəʊˈfɪlɪk ˌsʌbstɪˈtjuːʃn/ 求電子置換
    reduction/rɪˈdʌkʃn/ 還元
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    carboxylic acid/ˌkɑːbəkˈsɪlɪk ˈæsɪd/ カルボン酸
    ester/ˈestə/ エステル
    amine/ˈæmaɪn/ アミン
    amide/əˈmaɪd/ アミド
    nitrile/ˈnaɪtraɪl/ ニトリル
    synthetic route/sɪnˈθetɪk ruːt/ 合成経路
    reagent/rɪˈeɪdʒənt/ 試薬
    addition–elimination/əˈdɪʃn ɪˌlɪmɪˈneɪʃn/ 付加-脱離
    by-product/baɪ ˈprɒdʌkt/ 副生成物
    36.1

    Exam tips · ⁨試験対策⁩

    English
    • Combine AS and A-level steps and watch for benzene-ring reactions and carbon-count changes.
    • State every reagent and condition, and note when a step produces a racemic mixture.
    • Choose the shortest route that reaches the target functional group.
    日本語
    • ASおよびA-Levelのステップを統合し、ベンゼン環反応や炭素数の変化に注意する。
    • 全ての試薬と条件を明記し、ステップがラセミ混合物を生じる場合に注記する。
    • 標的官能基に至る最も短いルートを選ぶ。
  • 37

    Analytical techniques · ⁨分析技術⁩

    Watch lesson · ⁨レッスンを視聴⁩
    37.1

    Thin-layer chromatography

    Syllabus · ⁨シラバス⁩
    English
    1. describe and understand the terms (a) stationary phase, for example aluminium oxide (on a solid support) (b) mobile phase; a polar or non-polar solvent (c) $R_{\text{f}}$ value (d) solvent front and baseline
    2. interpret $R_{\text{f}}$ values
    3. explain the differences in $R_{\text{f}}$ values in terms of interaction with the stationary phase and of relative solubility in the mobile phase
    日本語
    1. (a) 固定相: 例として酸化アルミウム(固体担持体上)、(b) 移動相: 極性または非極性溶媒、(c) $R_{\text{f}}$値、(d) 溶媒前線とベースラインという用語を説明し理解する
    2. $R_{\text{f}}$値を解釈する
    3. 固定相との相互作用および移動相に対する相対的な溶解度の観点から、$R_{\text{f}}$値の違いを説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Chromatography 色谱 separates a mixture using two "phases" — one that stays still and one that moves. In thin-layer chromatography 薄层色谱 (TLC):

    • the stationary phase 固定相 stays still (for example aluminium oxide on a plate).
    • the mobile phase 流动相 moves (a polar or non-polar solvent that travels up the plate).
    • the baseline 基线 is the starting line where the spots are placed; the solvent front 溶剂前沿 is the highest level the solvent reaches.

    The $R_{\text{f}}$ value compares how far a spot moves with how far the solvent moves:

    $$R_{\text{f}} = \frac{\text{distance moved by the spot}}{\text{distance moved by the solvent front}}$$

    It is always between 0 and 1. A substance that sticks more strongly to the stationary phase, or is less soluble in the mobile phase, moves less and has a smaller $R_{\text{f}}$.

    日本語

    Chromatography 色谱 separates a mixture using two "phases" — one that stays still and one that moves. In thin-layer chromatography 薄层色谱 (TLC):

    • the stationary phase 固定相 stays still (for example aluminium oxide on a plate).
    • the mobile phase 流动相 moves (a polar or non-polar solvent that travels up the plate).
    • the baseline 基线 is the starting line where the spots are placed; the solvent front 溶剂前沿 is the highest level the solvent reaches.

    The $R_{\text{f}}$ value compares how far a spot moves with how far the solvent moves:

    $$R_{\text{f}} = \frac{\text{distance moved by the spot}}{\text{distance moved by the solvent front}}$$

    It is always between 0 and 1. A substance that sticks more strongly to the stationary phase, or is less soluble in the mobile phase, moves less and has a smaller $R_{\text{f}}$.

    A TLC plate with a baseline, a solvent front, and a spot that has moved partway up, with the spot and solvent distances marked
    Thin-layer chromatography: the $R_\text{f}$ value is how far the spot moved divided by how far the solvent front moved
    A developed TLC plate glowing under ultraviolet light
    A real TLC plate under UV light: each glowing spot is a separated component of the mixture
    Explore · ⁨探索⁩

    TLC route · ⁨TLC 経路⁩

    Follow a spot up a plate and use Rf to identify substances. · ⁨スポットをプレート上で移動させ、Rf 値を用いて物質を同定する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    chromatography/krəʊməˈtɒɡrəfi/ クロマトグラフィー
    thin-layer chromatography/θɪn ˈleɪə krəʊməˈtɒɡrəfi/ 薄層クロマトグラフィー
    stationary phase/ˈsteɪʃənəri feɪz/ 固定相
    mobile phase/ˈməʊbaɪl feɪz/ 移動相
    baseline/ˈbeɪslaɪn/ ベースライン
    solvent front/ˈsɒlvənt frʌnt/ 溶媒フロント
    37.2

    Gas/liquid chromatography

    Syllabus · ⁨シラバス⁩
    English
    1. describe and understand the terms (a) stationary phase; a high boiling point non-polar liquid (on a solid support) (b) mobile phase; an unreactive gas (c) retention time
    2. interpret gas/liquid chromatograms in terms of the percentage composition of a mixture
    3. explain retention times in terms of interaction with the stationary phase
    日本語
    1. (a) 固定相: 高い沸点の非極性液体(固体担持体上)、(b) 移動相: 不活性なガス、(c) 保持時間という用語を説明し理解する
    2. ガス/液体クロマトグラムを、混合物の組成百分率の観点から解釈する
    3. 固定相との相互作用の観点から、保持時間を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    In gas/liquid chromatography 气液色谱 (GLC):

    • the stationary phase is a high-boiling-point non-polar liquid on a solid support.
    • the mobile phase is an unreactive carrier gas.
    • the retention time 保留时间 is how long a component takes to pass through.

    The area of each peak gives the percentage of that component in the mixture. A component that interacts more with the stationary phase has a longer retention time.

    日本語

    In gas/liquid chromatography 气液色谱 (GLC):

    • the stationary phase is a high-boiling-point non-polar liquid on a solid support.
    • the mobile phase is an unreactive carrier gas.
    • the retention time 保留时间 is how long a component takes to pass through.

    The area of each peak gives the percentage of that component in the mixture. A component that interacts more with the stationary phase has a longer retention time.

    A chromatogram with three peaks at different retention times, the peak area giving the percentage of each component
    A gas–liquid chromatogram: each component gives a peak at its own retention time, and the peak area is its percentage in the mixture
    Explore · ⁨探索⁩

    Gas/liquid chromatography route · ⁨気体/液体クロマトグラフィーの流れ⁩

    Follow a volatile sample through column separation to a chromatogram. · ⁨揮発性サンプルをカラム分離を経てクロマトグラムまで追跡する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    gas/liquid chromatography/ɡæs ˈlɪkwɪd krəʊməˈtɒɡrəfi/ ガスクロマトグラフィー/液体クロマトグラフィー
    retention time/rɪˈtenʃn taɪm/ 保持時間
    37.3

    Carbon-13 NMR spectroscopy

    Syllabus · ⁨シラバス⁩
    English
    1. analyse and interpret a carbon-13 NMR spectrum of a simple molecule to deduce: (a) the different environments of the carbon atoms present (b) the possible structures for the molecule
    2. predict or explain the number of peaks in a carbon-13 NMR spectrum for a given molecule
    日本語
    1. 単純な分子の炭素-13 NMRスペクトルを解析・解釈して、以下のことを導き出す: (a) 存在する炭素原子の異なる環境、(b) 分子の可能な構造
    2. 与えられた分子について、炭素-13 NMRスペクトルのピーク数を予測または説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    NMR 核磁共振 (nuclear magnetic resonance) studies how certain nuclei behave in a strong magnetic field.

    In carbon-13 NMR, each different chemical environment 化学环境 of carbon gives one peak. So:

    • the number of peaks tells you how many different carbon environments there are (equivalent carbons share a peak).
    • the position of each peak (its chemical shift) suggests the type of carbon, which helps you deduce possible structures.
    日本語

    NMR 核磁共振 (nuclear magnetic resonance) studies how certain nuclei behave in a strong magnetic field.

    A tall cylindrical NMR spectrometer standing in a laboratory, marked with a strong magnetic field warning
    An NMR spectrometer: the large cylinder holds a superconducting magnet that makes the very strong magnetic field the technique needs

    In carbon-13 NMR, each different chemical environment 化学环境 of carbon gives one peak. So:

    • the number of peaks tells you how many different carbon environments there are (equivalent carbons share a peak).
    • the position of each peak (its chemical shift) suggests the type of carbon, which helps you deduce possible structures.
    Explore · ⁨探索⁩

    Carbon-13 NMR lab · ⁨炭素-13 NMR実験⁩

    Match carbon environments to C-13 NMR evidence. · ⁨炭素環境とC-13 NMRの証拠を対応させる。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    NMR/ˌen em ˈɑː/ NMR
    chemical environment/ˈkemɪkl enˈvaɪrənmənt/ 化学環境
    37.4

    Proton (¹H) NMR spectroscopy

    Syllabus · ⁨シラバス⁩
    English
    1. analyse and interpret a proton ($^1\text{H}$) NMR spectrum of a simple molecule to deduce: (a) the different environments of proton present using chemical shift values (b) the relative numbers of each type of proton present from relative peak areas (c) the number of equivalent protons on the carbon atom adjacent to the one to which the given proton is attached from the splitting pattern, using the $n + 1$ rule (limited to singlet, doublet, triplet, quartet and multiplet) (d) the possible structures for the molecule
    2. predict the chemical shifts and splitting patterns of the protons in a given molecule
    3. describe the use of tetramethylsilane, TMS, as the standard for chemical shift measurements
    4. state the need for deuterated solvents, e.g. $\text{CDCl}_3$, when obtaining a proton NMR spectrum
    5. describe the identification of O–H and N–H protons by proton exchange using $\text{D}_2\text{O}$
    日本語
    1. 単純な分子の水素($^1\text{H}$) NMRスペクトルを解析・解釈して、以下のことを導き出す: (a) 化学シフト値を用いて存在する水素の異なる環境、(b) 相対的なピーク面積から各タイプの水素の相対的な数、(c) 分裂パターンを用いて$n + 1$則(シングレット、ダブルト、トリプレット、クアドラット、マルチプレットに限る)に基づき、与水素が結合している炭素 adjacent の炭素上の等価水素の数、(d) 分子の可能な構造
    2. 与えられた分子の水素の化学シフトおよび分裂パターンを予測する
    3. テトラメチルシリラン, TMSを化学シフト測定の標準物質として使用する方法を説明する
    4. 水素NMRスペクトルを取得する際に、デューテル化溶媒(例: $\text{CDCl}_3$)が必要な理由を述べる
    5. $\text{D}_2\text{O}$を用いた水素交換によるO–HおよびN–H水素の同定方法を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Proton NMR looks at the hydrogen atoms (proton 质子 nuclei). From the spectrum you read off:

    • chemical environments: protons in different environments appear at different chemical shift 化学位移 values.
    • relative numbers: the relative peak areas give the ratio of each type of proton.
    • splitting 裂分: a peak is split by the protons on the neighbouring carbon, following the $n+1$ rule — $n$ equivalent neighbours split a peak into $n+1$ lines:
    Neighbours ($n$) Pattern
    0 singlet 单峰
    1 doublet 双峰
    2 triplet 三峰
    3 quartet 四峰
    many multiplet 多重峰

    Practical points

    • tetramethylsilane 四甲基硅烷 (TMS) is the standard, set at a chemical shift of $0$.
    • a deuterated solvent 氘代溶剂 (such as $\text{CDCl}_3$) is used so that the solvent itself gives no proton signal.
    • shaking the sample with $\text{D}_2\text{O}$ makes the O–H and N–H peaks disappear (their hydrogen is swapped for deuterium), which identifies those protons.

    Worked example. A compound $\text{C}_4\text{H}_8\text{O}_2$ gives three proton NMR peaks: a triplet at $\delta\ 1.2$ (area 3), a quartet at $\delta\ 4.1$ (area 2), and a singlet at $\delta\ 2.0$ (area 3). Deduce the structure. Read the areas first, then the splitting. Areas $3:2:3$ give three proton environments holding 3, 2 and 3 hydrogens. Apply the $n+1$ rule in reverse: a triplet (area 3) is a $\text{CH}_3$ with 2 neighbours, and a quartet (area 2) is a $\text{CH}_2$ with 3 neighbours - each splitting the other, which is the classic ethyl group, $\text{CH}_3\text{CH}_2-$. That $\text{CH}_2$ lies far downfield at $\delta\ 4.1$, so it is attached to an oxygen. The remaining singlet (area 3) is a $\text{CH}_3$ with no neighbours at $\delta\ 2.0$, so it sits next to the C=O. Together: $\text{CH}_3\text{COOCH}_2\text{CH}_3$, ethyl ethanoate. A triplet-and-quartet pair is almost always an ethyl group - spot it first and the rest follows.

    日本語

    Proton NMR looks at the hydrogen atoms (proton 质子 nuclei). From the spectrum you read off:

    • chemical environments: protons in different environments appear at different chemical shift 化学位移 values.
    • relative numbers: the relative peak areas give the ratio of each type of proton.
    • splitting 裂分: a peak is split by the protons on the neighbouring carbon, following the $n+1$ rule — $n$ equivalent neighbours split a peak into $n+1$ lines:
    Neighbours ($n$) Pattern
    0 singlet 单峰
    1 doublet 双峰
    2 triplet 三峰
    3 quartet 四峰
    many multiplet 多重峰
    Four NMR peak patterns: a single line, two lines, three lines and four lines, labelled singlet, doublet, triplet and quartet
    The $n+1$ rule: $n$ equivalent neighbouring protons split a peak into $n+1$ lines (singlet, doublet, triplet, quartet)
    A proton NMR spectrum of ethanol with three groups of peaks: a CH3 triplet, a CH2 quartet and an OH singlet, plus the TMS reference at zero
    Proton NMR of ethanol: three environments give a CH$_3$ triplet, a CH$_2$ quartet and an OH singlet, with areas in the ratio $3:2:1$

    Practical points

    • tetramethylsilane 四甲基硅烷 (TMS) is the standard, set at a chemical shift of $0$.
    • a deuterated solvent 氘代溶剂 (such as $\text{CDCl}_3$) is used so that the solvent itself gives no proton signal.
    • shaking the sample with $\text{D}_2\text{O}$ makes the O–H and N–H peaks disappear (their hydrogen is swapped for deuterium), which identifies those protons.

    Worked example. A compound $\text{C}_4\text{H}_8\text{O}_2$ gives three proton NMR peaks: a triplet at $\delta\ 1.2$ (area 3), a quartet at $\delta\ 4.1$ (area 2), and a singlet at $\delta\ 2.0$ (area 3). Deduce the structure. Read the areas first, then the splitting. Areas $3:2:3$ give three proton environments holding 3, 2 and 3 hydrogens. Apply the $n+1$ rule in reverse: a triplet (area 3) is a $\text{CH}_3$ with 2 neighbours, and a quartet (area 2) is a $\text{CH}_2$ with 3 neighbours - each splitting the other, which is the classic ethyl group, $\text{CH}_3\text{CH}_2-$. That $\text{CH}_2$ lies far downfield at $\delta\ 4.1$, so it is attached to an oxygen. The remaining singlet (area 3) is a $\text{CH}_3$ with no neighbours at $\delta\ 2.0$, so it sits next to the C=O. Together: $\text{CH}_3\text{COOCH}_2\text{CH}_3$, ethyl ethanoate. A triplet-and-quartet pair is almost always an ethyl group - spot it first and the rest follows.

    Explore · ⁨探索⁩

    Proton NMR lab · ⁨プロトンNMR実験⁩

    Match proton evidence to chemical environment and neighbours. · ⁨プロトンの証拠を化学環境や隣接原子と対応させる。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    proton/ˈprəʊtɒn/ 陽子
    chemical shift/ˈkemɪkl ʃɪft/ 化学シフト
    splitting/ˈsplɪtɪŋ/ 分裂
    singlet/ˈsɪŋɡlɪt/ シングレット
    doublet/ˈdʌblət/ ダブルト
    triplet/ˈtrɪplɪt/ 三重線
    quartet/kwɔːˈtet/ 四重線
    multiplet/ˈmʌltɪplɪt/ 多重線
    tetramethylsilane/ˈtetrəmiːθɪlsɪleɪn/ テトラメチルシラン
    deuterated solvent/ˈdjuːtəreɪtɪd ˈsɒlvənt/ 重水素溶媒
    37.4

    Exam tips

    • In $^{13}\text{C}$ NMR the number of peaks equals the number of carbon environments — use symmetry to count them.
    • In $^1\text{H}$ NMR use chemical shift (data booklet), integration (ratio of H's) and splitting (the $n+1$ rule): a triplet + quartet means an ethyl group.
    • TMS is the reference ($\delta = 0$); a $\text{D}_2\text{O}$ shake removes O-H and N-H peaks.
    • Combine IR, mass spectrum and NMR to deduce a structure, stating which evidence gives which feature.

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