Hydrocarbons
A-Level Chemistry Topic 14 9:54 English narration · English + 中文 subtitles burned in
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Two clear liquids, both nothing but carbon and hydrogen.
两种无色液体，它们都只由碳和氢组成。
And two test tubes of orange bromine water.
再准备两支装着橙色溴水的试管。
Shake the first liquid into one tube, the second into the other.
把第一种液体摇进其中一支，把第二种摇进另一支。
One tube goes clear in seconds; the other stays orange all day.
一支几秒钟就变得无色， 另一支放一整天还是橙色。
The difference is one double bond.
差别只在一个双键。
Welcome to hydrocarbons: molecules of carbon and hydrogen only.
欢迎来到碳氢化合物：只由碳和氢构成的分子。
Today, the two families, the reactions that make them, the two mechanisms you must draw, and the rule that picks the product.
今天我们要讲两大家族、制取它们的反应、 你必须会画的两个机理，以及决定哪种产物为主的那条规则。
Start with the two families.
先看这两个家族。
Alkanes are saturated: every bond single, every carbon holding all the hydrogens it can, and no functional group to attack. Alkenes are different.
烷烃是饱和的：所有的键都是单键，每个碳都尽可能多地连着氢， 而且没有官能团可以让试剂进攻。
Their carbon to carbon double bond is the functional group: it sticks out and is rich in electrons, so alkenes react readily.
烯烃不同：它的碳碳双键就是官能团—— 它突出在分子外面，电子密度又高，所以烯烃很容易反应。
Their general formulas say it in shorthand.
它们的通式说的正是同一件事。
Where do they come from?
这两个家族分别从哪里来？
Making alkanes: hydrogenation — add hydrogen to an alkene over a platinum or nickel catalyst with heat — or crack a longer alkane over aluminium oxide.
要制烷烃：在铂或镍催化剂上加热，给烯烃加氢； 或者把较长的烷烃在氧化铝上加热裂化。
Making alkenes: elimination of a hydrogen halide from a halogenoalkane using sodium hydroxide in ethanol with heat, dehydration of an alcohol over hot aluminium oxide or concentrated sulfuric acid, or crack again.
要制烯烃：用溶于乙醇的氢氧化钠并加热， 从卤代烷上消去一分子卤化氢；用热的氧化铝或浓硫酸把醇脱水；或者还是裂化长链烷烃。
In this topic the conditions carry the marks.
一定要记条件——在这个专题里，条件才是得分点。
Cracking deserves a closer look.
裂化值得再看一遍。
A heavy fraction of crude oil is heated over aluminium oxide, and one long chain snaps in two: a shorter alkane plus an alkene — both more useful than the heavy oil.
原油中较重的馏分在氧化铝上加热，一条长链断成两段： 一段较短的烷烃，加上一个烯烃。 两者都比我们最初的重油有用得多。
Now burn an alkane.
现在把烷烃点燃。
Plenty of oxygen and you get complete combustion: only carbon dioxide and water.
氧气充足时是完全燃烧，产物只有二氧化碳和水。
Too little, and the combustion is incomplete — still water, but also soot and toxic carbon monoxide.
氧气不足时燃烧不完全——仍然生成水，但还会有炭黑，以及有毒的一氧化碳。
Balance in a fixed order: carbon first, then hydrogen, oxygen last.
配平这类方程式要按固定顺序：先配碳，再配氢，最后配氧。
Take hexane: six carbons fix six carbon dioxide, fourteen hydrogens fix seven water — nineteen oxygens on the right.
以己烷为例： 六个碳定下六个二氧化碳，十四个氢定下七个水，右边一共十九个氧原子。
So the left needs nine and a half oxygen molecules, and you double it all for whole numbers.
所以左边需要九个半氧气分子——如果题目要整数系数，把整个方程式乘以二。
Alkanes burn.
烷烃能烧。
What else?
除此之外呢？
Almost nothing — and the exam wants the reason.
几乎什么都不做——而考试要的正是原因。
Alkanes are unreactive because their carbon to carbon and carbon to hydrogen bonds are strong, and those bonds have almost no polarity, so a polar reagent finds no charge to attack.
烷烃不活泼，是因为它的碳碳键和碳氢键都很强，而且这些键几乎没有极性， 所以极性试剂在上面找不到可以进攻的电荷。
Unreactive is not harmless, though: burn them in a car engine and the exhaust carries carbon monoxide, oxides of nitrogen and unburnt hydrocarbons — the environmental effects you must be able to name.
但不活泼并不等于无害： 在汽车发动机里燃烧它们，尾气中会有一氧化碳、氮的氧化物和未燃烧的碳氢化合物。
A catalytic converter cleans all three up.
催化转化器能把这三者都变成无害的气体。
Alkanes do have one reaction to know: free-radical substitution with chlorine, in ultraviolet light.
烷烃确实有一个你必须掌握的反应：在紫外线下与氯发生自由基取代。
Step one is initiation: light splits a chlorine molecule into two radicals, each with an unpaired electron.
第一步是引发：紫外线把一个氯分子劈成两个氯自由基，每个都带一个未成对电子。
Name that break: it is homolytic fission — the bond splits evenly, one electron to each atom.
这种断裂有名字——叫均裂，每个原子各得一个电子。
The opposite, heterolytic fission, sends both electrons to one atom and makes ions; that is what the polar mechanisms do. Step two is propagation, always two equations: a chlorine radical takes a hydrogen from ethane, then that ethyl radical attacks a chlorine molecule and makes a new radical.
第二步是增长，永远要写两个方程式： 一个氯自由基从乙烷上夺走一个氢，剩下的乙基自由基再去进攻另一个氯分子， 又生成一个新的氯自由基。
Step three is termination: two radicals join, the chain ends.
第三步是终止：两个自由基结合，链就断了。
Now the alkenes.
现在轮到烯烃。
Nearly all their reactions are electrophilic addition, right across the double bond. Learn the table by reagent.
它们几乎所有的反应都是亲电加成，加在双键的两个碳上。
Hydrogen, over a platinum or nickel catalyst with heat, gives an alkane. Steam, with a phosphoric acid catalyst, gives an alcohol. A hydrogen halide gas, at room temperature, gives a halogenoalkane.
按试剂来记这张表：氢气，用铂或镍催化并加热，得到烷烃；水蒸气，用磷酸催化，得到醇； 卤化氢气体，在室温下，得到卤代烷；卤素单质则在两个碳上各加一个卤原子。
And a halogen on its own adds one atom to each carbon. That last one is also the test for a C=C bond: an alkene decolourises orange bromine water, and an alkane does not.
最后这一个也是检验方法：橙色的溴水会褪成无色。
Here is the mechanism you must draw first: bromine adding to ethene.
这是你必须会画的第一个机理：溴与乙烯的加成。
An electrophile is a species that accepts a pair of electrons, and the electron-rich double bond is exactly what draws one in.
亲电试剂就是能接受一对电子的粒子，而电子密度高的双键正好把它吸引过来。
Bromine has no dipole of its own, but as it comes close the double bond pushes the near bromine atom's electrons away, so that atom becomes slightly positive.
溴分子本身没有偶极， 但当它靠近时，双键把近端溴原子的电子推开，于是那个原子带上部分正电荷。
Now the first curly arrow: it starts at the double bond and points at that bromine.
现在画第一个弯箭头：它从双键出发，指向带部分正电荷的溴原子。
At the same moment a second arrow pushes the bromine pair onto the far atom, which leaves as a bromide ion.
与此同时，第二个箭头把溴溴之间的一对电子推到远端的溴原子上，它以溴离子的形式离去。
What is left is a carbocation.
剩下的是一个碳正离子。
The bromide ion attacks it, giving dibromoethane, one bromine on each carbon.
溴离子再进攻它，生成两个碳上各带一个溴的二溴乙烷。
Same mechanism, harder question.
同样的机理，问题却更难。
When hydrogen bromide adds to propene, the hydrogen could join either carbon, and the two routes give different carbocations.
当溴化氢加到丙烯上时，氢可以连到任意一个碳上， 两条路线会生成不同的碳正离子。
Watch which one the reaction chooses.
看看反应会选择哪一条。
Why does it choose one?
它为什么会做出选择？
Because of the alkyl groups.
取决于烷基。
An alkyl group is electron-releasing: it pushes electron density towards the positive carbon and spreads the charge out, so the ion is more stable. That is the inductive effect.
烷基有推电子性：它把电子密度推向带正电的碳， 把电荷分散开，使离子更稳定，这就是诱导效应。
Count the alkyl groups — one for primary, two for secondary, three for tertiary — and stability rises along that line.
数一数烷基——伯有一个，仲有两个， 叔有三个——稳定性也沿着这条线依次升高。
Written as a rule: the hydrogen adds to the carbon that already has more hydrogens.
写成规则就是：氢加到本来氢就多的那个碳上。
Let's do one properly.
我们来认真做一道。
Hydrogen bromide is added to propene: predict the major product.
把溴化氢加到丙烯上，预测主要产物。
Pause here and try it yourself.
先暂停，自己试一试。
Ready?
好了吗？
The hydrogen adds first, and it can go two ways.
氢先加上去，而它有两种加法。
Put the hydrogen on the end carbon, and the charge lands on the middle carbon: a secondary carbocation, stabilised from two sides.
把氢加在末端的碳上，正电荷就落在中间的碳上： 这是仲碳正离子，两侧都有烷基撑着。
Put the hydrogen on the middle carbon, and you get a primary carbocation, stabilised from one side only.
把氢加在中间的碳上，得到的是伯碳正离子， 只有一侧撑着。
So the major product is 2-bromopropane — and quote that reason, never the rule alone.
所以主要产物就是2-溴丙烷——而且要写出这个理由，不能只写规则。
Alkenes are also oxidised by acidified manganate seven, and conditions decide everything.
烯烃还会被酸化的高锰酸钾氧化，而条件决定一切。
Cold and dilute, it adds two hydroxyl groups across the double bond and gives a diol.
冷的稀溶液， 在双键两端各加一个羟基，生成二醇。
Hot and concentrated, it breaks the double bond apart, and what each half becomes shows where that bond was.
热的浓溶液，则把双键彻底断开， 而两半各自变成什么，正好告诉你双键原来在哪里。
It comes down to one question: how many hydrogens was that carbon carrying?
归根到底只有一个问题： 那个碳原来带着几个氢？
None, and it stops at a ketone. One, and it is pushed on to a carboxylic acid. Two, and it is oxidised to carbon dioxide.
一个也没有，就停在酮；带一个，就继续被氧化成酸； 带两个，就一直被氧化成二氧化碳。
Now an exam question: give the products when this alkene meets hot, concentrated, acidified manganate seven.
把它变成一道考题：这个烯烃与热的、浓的、酸化的高锰酸钾一起加热，写出产物。
First, cut the molecule at the double bond and take each carbon on its own.
先在双键处把分子切开，再分别看每个碳。
The left carbon carries two methyl groups and no hydrogen, so it stops at a ketone: propanone.
左边的碳带着两个甲基、没有氢， 所以停在酮——丙酮。
The right carbon carries one methyl and one hydrogen, so it goes on to a carboxylic acid: ethanoic acid.
右边的碳带着一个甲基和一个氢，所以继续被氧化成酸——乙酸。
Write the balanced equation for it.
再把配平的方程式写出来。
Now run it backwards: a ketone means two alkyl groups, a carboxylic acid means one alkyl and one hydrogen, and carbon dioxide means the chain ended there — the strongest clue of all, because it can only come from a terminal double bond.
现在反过来做：酮说明那个碳带两个烷基，酸说明带一个烷基和一个氢， 而二氧化碳说明双键就在链的末端——这是最有力的线索， 因为它只可能来自位于末端的双键。
That is how you locate a double bond in a large molecule.
在大分子里就是这样定位双键的。
One last alkene reaction.
最后一个烯烃反应。
Many alkene molecules open their double bonds and join hand to hand into one very long chain, and nothing else is produced.
许多烯烃分子把双键打开，手拉手连成一条很长的链， 而且不生成任何其他产物。
That is addition polymerisation.
这就是加成聚合。
Name the two parts: each small starting molecule is a monomer, and the long chain it builds is the polymer.
两个名称都要会写： 每个小的起始分子叫单体，连成的长链就是聚合物。
So many ethene molecules give poly(ethene).
所以许多乙烯分子聚合， 得到的就是聚乙烯。
Four marks students throw away.
四个学生常丢的分。
First, free-radical substitution needs ultraviolet light, and all three steps: initiation, propagation, termination.
第一，自由基取代需要紫外线，而且三步都要写全——引发、增长、终止。
Second, the curly arrow starts at the double bond and points where the electrons go; an arrow from a letter earns nothing.
第二，弯箭头要从双键出发，指向电子去的地方；从字母上出发的箭头一分也拿不到。
Third, for Markovnikov, quote the reason: alkyl groups are electron-releasing, so tertiary beats secondary beats primary.
第三，马氏规则不要只写规则，还要写理由：烷基是推电子的， 所以叔比仲稳定，仲又比伯稳定。
Fourth, state the observation: bromine water decolourises, orange to colourless.
第四，要写出现象：橙色变成无色。
Get those right and this topic is yours.
这几点做对，这个专题就是你的了。
Five things to take with you.
带走五件事。
First, alkanes are unreactive because the carbon to hydrogen and carbon to carbon bonds are strong and barely polar.
第一，烷烃不活泼，因为碳氢键和碳碳键又强又几乎没有极性。
A catalytic converter still has to mop up the nitrogen oxides and carbon monoxide they make in an engine.
催化转化器仍然要清掉它们在引擎里生成的氮氧化物和一氧化碳。
Second, free-radical substitution needs ultraviolet light, and the first break is named homolytic fission.
第二，自由基取代需要紫外线，第一步断裂就叫均裂。
Third, alkenes add electrophiles across the double bond; the curly arrow starts at the carbon to carbon double bond.
第三，烯烃在双键上加亲电试剂；弯箭头从碳碳双键出发。
Fourth, Markovnikov's major product is the more stable carbocation, because alkyl groups are electron-releasing.
第四，马氏规则的主产物来自更稳定的碳正离子，因为烷基是推电子的。
Fifth, bromine water going orange to colourless is the observation that proves a carbon to carbon double bond.
第五，溴水从橙色变成无色，就是证明碳碳双键的现象。
Cold dilute manganate seven gives a diol; hot concentrated cuts the chain.
冷而稀的高锰酸钾得到二醇；热而浓的会把链切开。