Unidade 4: Cinética, equilíbrios e química orgânica avançada
Pearson Edexcel · International A-Level · Chemistry Tópico 4 30:51 Narração em inglês · Legendas em inglês + 中文 gravadas
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Transcrição
Rate concerns speed; total entropy concerns feasibility; K concerns equilibrium composition. Use Unit 1 mole ratios and Unit 2 energy, collision and equilibrium models.
速率关乎速度。 总熵变关乎可行性。 平衡常数关乎平衡组成。 请运用第一单元的化学计量比，以及第二单元的能量、碰撞和平衡模型。
Doubling A doubles the rate, so it is first order in A.
将A加倍，速率加倍：对A为一级。
Doubling B quadruples the rate, so it is second order in B.
将B加倍，速率变为四倍：对B为二级。
The rate equals k times A times B squared; the overall order is three, independent of the overall coefficients.
r等于k乘以A乘以B的平方；总级数为三，与总化学计量系数无关。
k equals 0.00800 divided by the product of 0.100 and 0.200 squared, which equals 2.00.
k 等于 0.00800 除以 0.100 与 0.200 的平方之积，结果为 2.00。
For a third order reaction, the units are dm to the sixth power per mole squared per second.
对于三级反应，单位为 dm⁶ mol⁻² s⁻¹。
Temperature must be held fixed for comparisons.
比较时必须保持温度恒定。
A straight concentration–time line supports zero order.
一条直线形的浓度-时间曲线支持零级反应。
Constant successive half-lives support first order.
连续半衰期恒定支持一级反应。
A tangent gives instantaneous rate; a chord gives an interval average.
切线给出瞬时速率，而弦线给出区间平均速率。
Doubling A quadruples rate; tripling B triples rate.
A加倍，反应速率变为原来的四倍。 B变为三倍，反应速率也变为三倍。
Which law fits?
哪条速率方程符合？
Pause and justify your choice.
请先暂停，完成这道原创练习再继续。
B. The rate equals k times A squared times B, overall order three.
答案 B：速率等于 k 乘以 A 的平方再乘以 B，总反应级数为三。
A gives the wrong order for A; C gives the wrong orders; D ignores the dependence on B.
选项 A 给出的 A 的级数错误；选项 C 给出的级数均错误；选项 D 忽略了对 B 的依赖。
A proposed mechanism must pass two tests: its steps must add up to the overall equation, and it must agree with the measured rate law.
一个提出的机理必须通过两项检验：各步骤相加必须得到总反应方程式，并且必须与测得的速率方程一致。
S N 1 is a slow ionisation; S N 2 is one concerted step in which the nucleophile attacks as the leaving group departs.
SN1 是慢速电离；SN2 是一个协同步骤，亲核试剂进攻的同时离去基团离去。
Agreement supports a mechanism but does not prove it is the only one.
一致只能支持某一机理，而不能证明它是唯一的。
For the supplied equation, natural log of k equals natural log of A minus E sub a divided by R times T, the gradient against one over T is negative E sub a divided by R.
对于给定的方程，ln k 等于 ln A 减去 E_a 除以 R 乘 T，以 1/T 为横坐标作图时斜率为负 E_a 除以 R。
Use kelvin and consistent units of joules or kilojoules.
请使用开尔文温标，并确保能量单位统一使用焦耳或千焦。
Heterogeneous catalysts adsorb reactants, and poisons can block active sites.
多相催化剂吸附反应物，而毒物会占据活性位点。
Identify the order in A for graphs I, II and III.
确定图I、图II和图III中A的反应级数。
Use the change in rate when concentration doubles; restrict claims to the measured range.
利用浓度加倍时速率的变化来判断。 结论仅限于测量范围内。
I: rate quadruples when A doubles, second order.
I，当A加倍时速率变为四倍，二级反应。
II: rate stays fixed, zero order over the measured range.
II，在测量范围内速率保持不变，零级反应。
III: rate doubles, first order.
III，速率加倍，一级反应。
Read two successive half-lives from the curve.
从曲线上读出两个连续半衰期。
Draw a tangent at P and estimate the positive disappearance rate.
在点P处画一条切线，并估算正消失速率。
The concentration changes from 0.120 to 0.060 to 0.030, taking thirty seconds and then another thirty seconds, which supports first order.
浓度从 0.120 降至 0.060 再降至 0.030，分别耗时 30 秒和 30 秒，这支持一级反应。
The tangent gradient is approximately negative 0.00139 moles per cubic decimetre per second. The disappearance rate is approximately positive 0.00139.
切线斜率约为每立方分米每秒负 0.00139 摩尔；消失速率约为正 0.00139。
Find the plotted gradient using two points on the line.
利用直线上的两个点求出绘制的斜率。
The horizontal variable is 1000/T: calculate Ea with R=8.31J mol⁻¹ K⁻¹.
横坐标变量为 1000 除以 T。 使用 R 等于 8.31 焦耳每摩尔开尔文计算 E a。
The plotted gradient is negative 5.5 plus 3.1, over 3.3 minus 2.9, which is negative 6.0.
绘制的斜率等于负5.5加3.1，除以3.3减2.9，结果为负6.0。
E a equals negative 1000 R times the gradient, which is 49,860 joules per mole, or 49.9 kilojoules per mole.
活化能Ea等于负1000乘以R乘以斜率，得出49860 J/mol，即49.9 kJ/mol。
Omitting the 1000 makes the answer one thousand times too small.
省略1000会使答案小一千倍。
Use propanone, P, its oxygen-protonated form, P H plus, and the enol, E.
使用丙酮 P、其氧质子化形式 PH 正离子以及烯醇 E。
Propose rapid protonation, slow enol formation, then fast iodination.
提出快速质子化、慢速形成烯醇，然后快速碘化的机理。
Match a rate equal to k times P times H plus, and zero order in iodine; check atom and charge balances and the net acid.
使其符合速率等于 k 乘以 P 乘以氢离子浓度，且对碘为零级。 检查原子和电荷平衡以及净酸。
Fast equilibrium: P plus H plus, double arrow, PH plus.
快速平衡：P + H⁺ ⇌ PH⁺。
Slow enol formation: PH plus, single right arrow, E plus H plus.
慢速烯醇形成：PH⁺ → E + H⁺。
Fast iodination: E plus I two, single right arrow, CH three CO CH two I plus H plus plus I minus.
快速碘化反应：E + I₂ → CH₃COCH₂I + H⁺ + I⁻。
Rate equals k sub two, times the concentration of P H plus, which equals k sub two, times K, times the concentration of P, times the concentration of H plus.
速率等于k下标2乘以P H加离子的浓度，即等于k下标2乘以K乘以P的浓度和H加离子的浓度。
This is first order in P and acid, and zero order in iodine.
该反应对P和酸是一级，对碘是零级。
The overall equation is P plus I sub 2 yields C H 3 C O C H 2 I plus H plus plus I minus.
总反应方程式为P加I下标2生成CH3COCH2I加H加离子加I负离子。
In this reaction, P H plus and E cancel out.
在此反应中，PH加离子和E相互抵消。
The catalytic acid is regenerated, and iodination also generates acid.
催化酸被再生，且碘化反应也会生成酸。
Agreement between rate and overall equation is not unique proof.
速率方程与总反应方程式一致并非唯一证明。
NaOH hydrolysis of an ester replaces mobile OH⁻ with less mobile ethanoate.
氢氧化钠对酯的水解用高迁移率的氢氧根离子替换为低迁移率的乙酸根离子。
Calibrate conductivity against composition; control temperature, probe geometry, mixing and solvent background.
需根据组分校准电导率，并控制温度、探头几何形状、混合条件及溶剂背景。
A closed reaction producing no gas cannot be followed by gas volume or mass loss.
在密闭体系中不产生气体的反应，无法通过气体体积或质量损失来追踪。
The entropy change of the system is the sum of product entropies minus reactant entropies, with coefficients.
体系的熵变等于生成物熵之和减去反应物熵之和，并乘以化学计量系数。
The entropy change of the surroundings is negative delta H over T, using joules with entropy in joules per mole per kelvin.
环境的熵变等于负的焓变除以温度，焓变用焦耳表示，熵的单位为焦耳每摩尔每开尔文。
Add both terms before judging the sign.
判断符号前，需把两项相加。
At 298 kelvin, delta H is negative 40.0 kilojoules per mole and the system entropy change is negative 100 joules per mole per kelvin.
在 298 开尔文时，焓变为每摩尔负 40.0 千焦，体系熵变为每摩尔每开尔文负 100 焦耳。
The surroundings change is 40,000 over 298, which is plus 134 joules per mole per kelvin.
环境熵变为 40000 除以 298，即每摩尔每开尔文正 134 焦耳。
The total is plus 34 joules per mole per kelvin, so the forward direction is favoured, though not necessarily fast.
总熵变为每摩尔每开尔文正 34 焦耳，因此正反应方向是有利的，但不一定快。
ΔH = +36.0 kJ mol⁻¹; ΔS(system) = +120 J mol⁻¹ K⁻¹.
ΔH = +36.0 kJ mol⁻¹；ΔS(系统) = +120 J mol⁻¹ K⁻¹。
Which condition gives positive total entropy?
哪种条件能使总熵变为正值？
Pause and justify your choice.
请先暂停，论证你的选择。
B: 120 − 36000/T > 0 when T > 300 K.
当T大于300开尔文时，B等于120减去36000除以T大于零。
At 300 K total entropy is zero.
在300开尔文时，总熵为零。
This boundary treats ΔH and system ΔS as approximately constant.
该边界将ΔH和系统ΔS视为近似恒定。
Describe warming, melting and boiling from a perfect crystal at 0 K.
描述从零开尔文完美晶体开始加热、熔化和沸腾的过程。
Why can ionic dissolution have either system-entropy sign?
为什么离子溶解的系统熵变符号可能为正也可能为负？
Why does gas spread?
为什么气体会扩散？
Perfect-crystal entropy is zero at 0 K; warming and melting/boiling increase it.
完美晶体的熵在零开尔文时为零。 升温以及熔化或沸腾都会使其增加。
Ions disperse on dissolution, but hydration can order water.
离子在溶解时会分散，但水合作用可能使水分子有序化。
Gas spreading increases accessible arrangements; the spontaneous direction increases total entropy.
气体扩散会增加可及排列数。 自发过程会使总熵增加。
Lattice formation: gaseous ions to one mole of solid, negative.
晶格形成：气态离子生成一摩尔固体，该过程为放热。
Reverse it for lattice dissociation.
逆向即为晶格解离。
A Born-Haber route must produce the correct numbers and charges of gaseous ions.
Born-Haber循环必须产生正确数量与电荷的气态离子。
ΔH(solution) = −ΔH(lattice formation) + total hydration enthalpy.
ΔH(溶液) = −ΔH(晶格形成) + 总水合焓。
−850 and −820 give +30 kJ mol⁻¹.
−850 加 −820 等于 +30 kJ mol⁻¹。
Extra stabilisation relative to a purely ionic model supports covalent character; solubility also needs entropy.
相对于纯离子模型的额外稳定化作用支持共价特征；溶解度还需要熵。
Draw the formation of calcium fluoride through gaseous atoms and gaseous ions, including electron counts.
画出CaF₂经由气态原子和气态离子的生成过程，并包含电子数。
Supplied values in kilojoules per mole are formation, negative 1220; calcium atomisation, 178; fluorine atomisation, 79; calcium ionisations, 590 and 1145; fluorine electron affinity, negative 328.
提供的数值单位为千焦每摩尔：生成焓负1220；钙原子化178；氟原子化79；钙电离能590和1145；氟电子亲和能负328。
Atomising calcium and two fluorine atoms takes 178 plus 2 times 79, which is 336 kilojoules per mole.
钙和两个氟原子的原子化：178 + 2 × 79 = 336 kJ/mol。
The two calcium ionisations take 590 plus 1145, which is 1735 kilojoules per mole.
两次钙的电离能：590 + 1145 = 1735 kJ/mol。
Two fluorine electron affinities give 2 times minus 328, which is minus 656 kilojoules per mole.
两次氟的电子亲和能：2 × (−328) = −656 kJ/mol。
So the gaseous-ion route totals 336 plus 1735 minus 656, which is 1415 kilojoules per mole.
气态离子途径：336 + 1735 − 656 = 1415 kJ/mol。
Negative 1220 equals 1415 plus the lattice enthalpy, so lattice formation is negative 2635 kilojoules per mole.
负 1220 等于 1415 加晶格焓，因此晶格形成焓为每摩尔负 2635 千焦。
Both routes form one mole of solid calcium fluoride; two first electron affinities are not one second electron affinity.
两条路径都生成一摩尔固态氟化钙；两个第一电子亲和能不等于一个第二电子亲和能。
Construct the reaction of solid magnesium chloride forming aqueous magnesium ions and two aqueous chloride ions through gaseous ions.
通过气态离子构建氯化镁固体生成水合镁离子和两个水合氯离子的反应。
Use lattice formation energy of negative two thousand five hundred twenty-three, magnesium ion hydration energy of negative one thousand nine hundred twenty, and each chloride ion hydration energy of negative three hundred eighty-one kilojoules per mole.
使用晶格形成能负2523，镁离子水合能负1920，每个氯离子水合能负381千焦每摩尔。
Define each molar process.
定义每个摩尔过程。
Reverse the lattice formation, giving plus 2523 kilojoules per mole, and hydrate both chlorides.
将晶格形成反向，得到每摩尔正 2523 千焦，并使两种离子水合。
Total hydration is negative 1920 plus 2 times negative 381, which is negative 2682; the solution enthalpy is negative 159 kilojoules per mole.
总水合焓等于负 1920 加 2 乘以负 381，即负 2682；溶解焓为每摩尔负 159 千焦。
For A plus B in equilibrium with C, starting with 0.500 moles each of A and B produces 0.200 moles of C.
对于 A 加 B 可逆生成 C 的反应，初始时 A 和 B 各为 0.500 摩尔，生成 0.200 摩尔 C。
Equilibrium amounts are 0.300, 0.300 and 0.200 moles.
平衡时的物质的量分别为 0.300、0.300 和 0.200 摩尔。
In 2.00 cubic decimetres, K c equals 0.100 over 0.150 squared, which is 4.44 cubic decimetres per mole.
在 2.00 立方分米中，Kc 等于 0.100 除以 0.150 的平方，即每摩尔 4.44 立方分米。
Partial pressure equals mole fraction times total pressure.
分压等于摩尔分数乘以总压。
For dinitrogen tetroxide in equilibrium with two nitrogen dioxide, K p equals the partial pressure of nitrogen dioxide squared, over the partial pressure of dinitrogen tetroxide.
对于四氧化二氮与二氧化氮之间的平衡，Kp 等于二氧化氮分压的平方除以四氧化二氮的分压。
Only temperature changes K; catalysts change the speed of approach.
只有温度会改变 K 值；催化剂只改变达到平衡的速度。
Conjugate pairs differ by one H⁺.
共轭酸碱对相差一个H⁺。
Strong/weak describes dissociation; concentrated/dilute describes amount per volume.
强与弱描述的是解离程度；浓与稀描述的是单位体积内的物质的量。
pH = −log₁₀ ; Kw = at the specified temperature.
pH等于负以10为底 的对数。 Kw等于 乘以 ，且该值取决于指定温度。
is approximately equal to the square root of K a times c, which equals the square root of 1.80 times 10 to the minus 5, times 0.100, which equals 1.34 times 10 to the minus 3 moles per cubic decimetre.
≈ √(Kₐc) = √(1.80×10⁻⁵ × 0.100) = 1.34×10⁻³ mol dm⁻³。
pH is approximately equal to 2.87; dissociation fraction is 1.34 percent.
pH ≈ 2.87；解离度为 1.34%。
Core practical 11: calibrate pH metre, know concentration and control temperature.
核心实验 11：校准 pH 计，掌握浓度并控制温度。
Added H plus reacts with A minus; added O H minus reacts with H A.
加入的氢离子与 A 负离子反应；加入的氢氧根离子与 HA 反应。
React the moles first, then use the hydrogen ion concentration, approximately K a times H A over A minus.
先按物质的量进行反应计算，再用氢离子浓度约等于 Ka 乘以 HA 浓度除以 A 负离子浓度。
Both components must remain; buffer capacity is finite.
两种组分都必须保留；缓冲容量是有限的。
0.0100 mol HA receives 0.00400 mol OH⁻.
0.0100 mol HA 接受 0.00400 mol OH⁻。
Which HA:A⁻ ratio enters the buffer expression?
哪个 HA 与 A⁻ 的比值会代入缓冲表达式？
Pause and justify your choice.
请先暂停，完成这道原创练习再继续。
B. H A remaining is 0.00600 moles; A minus formed is 0.00400 moles.
答案 B：剩余 HA 为 0.00600 摩尔；生成的 A 负离子为 0.00400 摩尔。
At half-neutralisation, the pH equals the p K a; beyond equivalence, use the excess strong reagent.
在半中和点，pH 等于 pKa；超过等当点后，按过量的强试剂计算。
An indicator transition must lie in the steep region of the actual curve.
指示剂的变色范围必须落在实际滴定曲线的陡峭区域。
H A starts at 0.100 molar, pH 3.00; after a 1000-fold dilution, the pH is 4.57.
HA 初始浓度为 0.100 摩尔每升，pH 为 3.00；稀释 1000 倍后，pH 为 4.57。
Find the hydrogen ion concentration and the percentage dissociation.
求氢离子浓度及解离百分比。
Is c minus x approximately equal to c still valid?
c 减 x 约等于 c 的近似是否仍然成立？
Final c equals 1.00 times 10 to the minus 4 M.
最终浓度 c 等于 1.00 × 10⁻⁴ M。
The concentration of H plus equals 10 to the minus 4.57, which is 2.69 times 10 to the minus 5 M.
氢离子 等于 10⁻⁴·⁵⁷，即 2.69 × 10⁻⁵ M。
Dissociation is approximately 26.9 percent.
解离度约为 26.9%。
Equilibrium HA is appreciably below initial c.
平衡时 HA 的浓度显著低于初始浓度 c。
Repeated half-log predictions are not reliable at this dilution.
在此稀释度下，重复半对数预测并不可靠。
The target pH is 5.00 and K a is 2.00 times 10 to the minus 5 molar; the final total of H A and A minus is 0.120 molar.
目标 pH 为 5.00，Ka 为 2.00 乘以 10 的负 5 次方摩尔每升；最终 HA 与 A 负离子的总浓度为 0.120 摩尔每升。
Make 250.0 cubic centimetres from 40.0 cubic centimetres of acid stock and 60.0 cubic centimetres of salt stock.
用 40.0 立方厘米酸储备液和 60.0 立方厘米盐储备液配制 250.0 立方厘米溶液。
Find both stock concentrations.
求两种储备液的浓度。
The ratio of A minus to H A is 2 to 1; the final concentrations are 0.0800 and 0.0400 molar.
A 负离子与 HA 之比为二比一；最终浓度分别为 0.0800 和 0.0400 摩尔每升。
The acid stock is 0.0400 times 250 over 40, which is 0.250 molar.
酸储备液浓度等于 0.0400 乘以 250 除以 40，即 0.250 摩尔每升。
The salt stock is 0.0800 times 250 over 60, which is 0.333 molar; dilute the mixed aliquots to 250.0 cubic centimetres.
盐储备液浓度等于 0.0800 乘以 250 除以 60，即 0.333 摩尔每升；将混合的等分试样稀释至 250.0 立方厘米。
25.0 cubic centimetres of 0.100 molar base B receives 0.100 molar hydrochloric acid; the pH at 12.5 cubic centimetres is 9.25.
向 25.0 立方厘米 0.100 摩尔每升的碱 B 中加入 0.100 摩尔每升的盐酸；加入 12.5 立方厘米时，pH 为 9.25。
Identify the acid whose K a is measured; find the pH at 20.0 cubic centimetres.
请确定所测 Ka 对应的酸，并求加入 20.0 立方厘米时的 pH。
At half-equivalence B equals B H plus, so the p K a of B H plus is 9.25, and K a is 5.62 times 10 to the minus 10 molar.
在半等当点，B 等于 BH 正离子，因此 BH 正离子的 pKa 为 9.25，Ka 为 5.62 乘以 10 的负 10 次方摩尔每升。
At 20.0 cubic centimetres, B is 0.000500 moles and B H plus is 0.00200 moles.
在 20.0 立方厘米处，B 为 0.000500 摩尔，BH 正离子为 0.00200 摩尔。
The pH is 9.25 plus the log of 0.000500 over 0.00200, which is 8.65; B consumes the added H plus.
pH 等于 9.25 加上 0.000500 除以 0.00200 的对数，即 8.65；B 消耗了加入的氢离子。
25.0 cubic centimetres of 0.0600 molar diprotic acid H 2 A, or of dibasic B, receives 0.100 molar titrant.
向 25.0 立方厘米 0.0600 摩尔每升的二元酸 H2A 或二元碱 B 中加入 0.100 摩尔每升的滴定剂。
Calculate both equivalence volumes; sketch the rising acid curve and the falling base curve.
计算两个等当点体积，并画出上升的酸曲线和下降的碱曲线。
Use the worksheet 4.4.4 data.
使用工作表 4.4.4 的数据。
The solute amount is 0.00150 moles, so the equivalence amounts fall at 15.0 and 30.0 cubic centimetres.
溶质的物质的量为 0.00150 摩尔，因此等当点分别在 15.0 和 30.0 立方厘米处。
Plot the supplied points smoothly; the stages need not have identical jump sizes.
平滑地描出所给的点；各阶段的突跃幅度不必相同。
HCO₃⁻ consumes H⁺; H₂CO₃ consumes OH⁻.
碳酸氢根离子消耗氢离子。
Stable pH supports enzyme conditions.
碳酸消耗氢氧根离子。 稳定的pH值支持酶的作用条件。
Buffers are consumed; blood also needs respiratory and renal regulation.
缓冲剂会被消耗。 血液还需要呼吸和肾脏调节。
Food buffering limits pH changes from microbial activity; it does not sterilise food.
食品缓冲作用限制微生物活动引起的pH变化，但不能使食品灭菌。
2-bromobutane has Br, H, CH₃ and C₂H₅ at carbon 2.
2-溴丁烷在2号碳上连有Br、H、CH₃和C₂H₅。
Swapping two groups gives the opposite configuration.
交换其中两个基团会得到相反的构型。
A racemate is inactive, but an achiral substance is also inactive.
外消旋体无光学活性，但非手性物质同样无光学活性。
Both aldehydes and ketones give 2,4-DNPH precipitates.
醛和酮都能与2,4-二硝基苯肼反应生成沉淀。
Aldehydes give Tollens’ silver or warm Fehling’s brick-red precipitate; ordinary ketones do not.
醛能与托伦试剂或温热的斐林试剂反应，分别产生银镜或砖红色沉淀；普通酮则不能。
LiAlH₄ in dry ether reduces aldehydes to primary and ketones to secondary alcohols, after work-up.
在无水乙醚中，四氢铝锂可将醛还原为伯醇、将酮还原为仲醇，经后处理即可得到产物。
Cyanide attacks through carbon; the carbon–oxygen pi pair moves to oxygen.
氰根离子通过碳原子进攻。 C=O π键的电子对转移到氧原子上。
The O minus then takes H plus from H C N, regenerating cyanide.
带负电的氧随后从HCN中夺取质子，再生氰根离子。
Planar ethanal can be attacked from either face, making a racemate under achiral conditions.
平面构型的乙醛可从任一面被进攻，从而在非手性条件下生成外消旋体。
Enantiopure chiral halogenoalkane gives a structurally chiral alcohol with zero rotation.
光学纯的手性卤代烷生成具有手性结构但旋光度为零的醇。
Which simple substitution model is supported?
这支持哪种简单的取代模型？
Contrast with inversion and explain the structural control.
与构型翻转进行对比，并解释其结构控制机制。
A chiral product with zero rotation supports equal enantiomers; ideal SN1 allows two-face attack.
具有零旋光度的手性产物支持外消旋体。 理想的SN1反应允许两面进攻。
Simple SN2 gives inversion and an enantiopure product; zero rotation alone could also mean achiral structure.
简单的SN2反应给出构型翻转和光学纯产物。 仅凭零旋光度也可能意味着分子为非手性结构。
Carbonyl O accepts a hydrogen bond from water.
羰基氧接受来自水分子的氢键。
Simple aldehydes/ketones lack an O–H donor.
简单的醛和酮缺乏O–H供体。
Acids form hydrogen-bonded associations; compare similar-sized molecules for boiling points.
酸会形成通过氢键连接的缔合结构。 请比较分子大小相似的物质的沸点。
A longer non-polar chain can reduce water solubility without removing the polar group.
更长的非极性链会降低在水中的溶解度，而不会移除极性基团。
U gives a D N P H derivative, no Tollens' silver and a positive iodoform test; the derivative melts at 125 to 127 degrees Celsius.
U与DNPH反应，不与Tollens试剂产生银镜反应，且碘仿试验呈阳性。 该衍生物的熔点为125至127摄氏度。
Illustrative reference values are propanal 146 degrees, propanone 126 degrees and butan-2-one 116 degrees.
参考值如下：丙醛为146摄氏度，丙酮为126摄氏度，丁-2-酮为116摄氏度。
Select and justify.
请选择并说明理由。
Propanone fits the supplied 126 degree Celsius derivative reference and the ketone test behaviour.
丙酮符合所给的 126 摄氏度衍生物参考值以及酮的检验现象。
Butan-2-one also gives iodoform; that test alone cannot distinguish the two.
丁-2-酮也能发生碘仿反应；仅凭该检验无法区分两者。
A sharp range supports purity; coincident reference ranges still limit identification.
熔程尖锐说明纯度高；参考范围重合仍会限制鉴定。
Draw propanal + HCN with CN⁻ catalyst, including all four electron-pair arrows.
画出丙醛与HCN在CN⁻催化下的反应，包含全部四个电子对箭头。
Show O⁻, proton transfer from HCN, regenerated CN⁻ and the optical result.
标出O⁻、来自HCN的质子转移、再生的CN⁻以及光学结果。
Cyanide attacks through its carbon lone pair; the carbon–oxygen pi pair moves to oxygen.
氰根离子通过碳上的孤对电子进攻；碳氧 π 键电子对转移到氧原子上。
The intermediate keeps the ethyl group, the hydrogen and the new nitrile carbon.
中间体保留乙基、氢原子和新的腈基碳。
O minus lone pair, arrow to H of HCN. H dash C bond pair, arrow to cyanide carbon.
O⁻孤对电子指向HCN中的H；H–C键电子对指向氰基碳。
CH three CH two CH OH CN has a chiral centre. Equal attack on two planar faces gives a racemate.
CH₃CH₂CH(OH)CN具有手性中心；对两个平面面的等量进攻生成外消旋体。
Acid plus alcohol, reversible reaction to ester plus water, with acid catalyst and heat.
酸与醇反应，可逆生成酯和水，需酸催化及加热。
Acyl chloride plus two NH three yields amide plus NH four Cl.
酰氯与两分子氨反应生成酰胺和氯化铵。
Alkaline ester hydrolysis produces carboxylate salt and alcohol.
碱性酯水解生成羧酸盐和醇。
A diacid and diol form a polyester with ester links and water.
二元酸与二元醇反应生成聚酯，产物包含酯键并释放水。
Keep both skeletons and open bonds through repeat-unit brackets.
在重复单元括号内保留两个骨架及未成键的连接点。
Carbonate gas, derivative melting range and spectra provide different evidence, not interchangeable proofs.
碳酸气体、衍生物熔点范围及光谱提供的是不同证据，而非可互换的证明。
Write acid and alkaline hydrolysis equations for CH₃CH₂CN.
写出丙腈的酸性和碱性水解方程式。
Give conditions, then obtain free acid from the alkaline product.
给出反应条件，然后从碱性产物中获得游离酸。
In acid, propanenitrile plus 2 waters plus H plus yields propanoic acid plus ammonium.
酸性条件下，丙腈加两分子水和氢离子生成丙酸和铵离子。
In alkali, propanenitrile plus water plus hydroxide yields propanoate plus ammonia.
碱性条件下，丙腈加水和氢氧根离子生成丙酸根和氨。
Heat under reflux with the aqueous reagent, and acidify the carboxylate afterwards.
用水溶液试剂回流加热，之后将羧酸盐酸化。
Carbon-13 counts environments, not atoms.
碳-13 统计的是化学环境，而非原子数量。
Ethyl ethanoate: 3H singlet, 2H quartet, 3H triplet.
乙酸乙酯：3H 单峰，2H 四重峰，3H 三重峰。
Use integration, splitting, shifts, IR and formula together.
需结合积分、裂分、化学位移、红外光谱及分子式综合判断。
Rf uses distances from the same baseline and has no units.
Rf 使用来自同一基线的距离，且无单位。
Retention-time matches need matched conditions; one peak can conceal co-elution.
保留时间匹配需要匹配的实验条件，一个峰可能掩盖共洗脱现象。
GC–MS adds mass/fragment evidence; standards, blanks and provenance remain essential.
气相色谱-质谱联用技术增加了质量和碎片证据，但标准品、空白样品和来源信息仍然至关重要。
At 298 K: strong/strong equivalence near 7; weak acid/strong base above 7; strong acid/weak base below 7.
在298 K时，强酸与强碱的等当点接近7；弱酸与强碱的等当点在7以上；强酸与弱碱的等当点在7以下。
Weak/weak usually has no sharp ordinary-indicator endpoint.
弱酸与弱碱滴定通常没有普通指示剂能显示的明显终点。
Label initial pH, added-volume axis, equivalence amount and excess-titrant region.
请标出初始pH、加入体积轴、等当点以及过量滴定剂区域。
25.0 cubic centimetres of 0.100 molar H 2 A uses 25.0 cubic centimetres, then 50.0 cubic centimetres in total, of 0.100 molar sodium hydroxide.
25.0 立方厘米 0.100 摩尔每升的 H2A 先消耗 25.0 立方厘米、累计共消耗 50.0 立方厘米 0.100 摩尔每升的氢氧化钠。
Each step removes one proton; a dibasic base takes up protons in the reverse direction.
每一步移除一个质子；二元碱则沿相反方向结合质子。
Two stoichiometric stages need not produce two sharp visible jumps.
两个化学计量阶段不一定产生两个明显的突跃。
Methyl propanoate is CH₃CH₂COOCH₃: three acid-side carbons and one alcohol-side carbon.
丙酸甲酯的化学式为 CH3CH2COOCH3，包含三个酰基侧碳原子和一个醇基侧碳原子。
Displayed formulae show all atoms/bonds; skeletal formulae imply carbon and its H but show heteroatoms.
展开式显示所有原子和化学键；骨架式隐含碳原子及其氢原子，但会标出杂原子。
Count carbonyl carbon in the parent chain and keep the ester oxygen between the two groups.
在母体链中计入羰基碳，并将酯氧保留在两个基团之间。
Calculate accurate masses of C 3 H 8 O and C 2 H 4 O 2, with carbon 12.0000, hydrogen 1.0078 and oxygen 15.9949.
取碳 12.0000、氢 1.0078、氧 15.9949，计算 C3H8O 和 C2H4O2 的精确质量。
The measured mass is 60.0210; there is an acid O–H and a C=O, and two carbon-13 signals.
测得质量为 60.0210；存在羧酸的羟基和羰基，以及两个碳-13 信号。
Suggest a structure.
请推测其结构。
C₃H₈O:60.0573; C₂H₄O₂:60.0210.
C3H8O，60.0573。 C2H4O2，60.0210。
Difference 0.0363 resolves the formula.
差值0.0363可确定分子式。
CH₃COOH fits the acid IR and two distinct carbon environments.
CH3COOH符合酸类红外光谱特征及两种不同的碳环境。
Accurate mass selects a formula; the other evidence constrains its structure.
精确质量测定用于筛选分子式；其他证据则用于限定其结构。
For butan-2-one, give the proton sets, their areas and simple n plus 1 splitting.
对于丁-2-酮，给出各组质子、峰面积和简单的 n 加 1 裂分。
Use the supplied proton regions of 0.8 to 1.5 and 2.0 to 2.7 p p m, and carbon regions of 8 to 50 and 190 to 220 p p m.
使用所给的氢谱区域 0.8 到 1.5 及 2.0 到 2.7 ppm，以及碳谱区域 8 到 50 及 190 到 220 ppm。
3H acyl singlet; 2H quartet; 3H terminal triplet.
3H 酰基单峰；2H 四重峰；3H 末端三重峰。
Equivalent methyl H do not split each other.
等价的甲基氢不会相互裂分。
Acyl CH₃/CH₂ lie at 2.0–2.7ppm; terminal CH₃ at 0.8–1.5ppm in supplied data.
酰基 CH₃/CH₂ 位于 2.0到2.7 ppm；末端 CH₃ 在提供的数据中位于 0.8到1.5 ppm。
Four carbon environments: three alkyl signals and one ketone carbonyl at 190–220ppm.
四个碳环境：三个烷基信号和一个酮羰基，位于 190到220 ppm。
T L C distances of 1.8 and 4.2 centimetres, with a front at 6.0 centimetres, give R f values of 0.30 and 0.70.
薄层色谱中，斑点移动距离分别为1.8厘米和4.2厘米，溶剂前沿为6.0厘米时，对应的Rf值分别为0.30和0.70。
A lower R f means greater relative retention by the stationary phase under the same conditions.
在相同条件下，较低的Rf值意味着对固定相的相对保留更强。
One G C peak or reference match can hide co-elution; add mass spectrometry evidence and matched standards and blanks.
气相色谱中一个峰与标准品匹配可能掩盖共洗脱现象；需补充质谱证据以及匹配的标样或空白对照。
C 3 H 6 O has two carbon-13 signals, gives iodoform and no Tollens' silver.
C三H六O有两个碳-13信号，能发生碘仿反应且不与托伦试剂反应。
Which structure fits?
哪种结构符合？
Pause and justify your choice.
请先暂停，说明你的选择理由。
B: propanone has equivalent methyl carbons, a CH₃CO– group and ketone behaviour.
B：丙酮具有等价的甲基碳、CH3CO基团和酮的性质。
Its HCN product has two identical methyl groups and is achiral.
其HCN产物含有两个相同的甲基，且为非手性分子。
Reject alternatives using all observations, not one familiar test.
应依据所有观察结果排除其他选项，而非仅凭一个熟悉的测试。