Hydrocarbons (A2): arenes
A-Level Chemistry Topic 30 7:42 English narration · English + 中文 subtitles burned in
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Orange bromine water is the classic test for a double bond.
橙色的溴水是检验双键的经典方法。
Shake it with an alkene, and the orange colour vanishes at once.
把它和烯烃摇一摇，橙色立刻消失。
Now shake the very same bromine water with benzene. Nothing. The colour just stays.
现在用同样的溴水去摇苯——什么也没有发生，颜色一直留着。
And that is strange, because on paper benzene looks like it has three carbon to carbon double bonds.
这很奇怪， 因为在纸面上，苯看起来含有三个碳碳双键。 那么苯为什么不肯反应？
So why does benzene refuse to react? That one answer runs the whole topic.
把这个问题弄明白，这一章的其余内容就都通了。
Welcome to arenes.
欢迎来到芳烃。
Today: why the benzene ring is so stable, the mechanism that earns three marks every year, the reagent for each reaction, and how to predict where the next group lands.
今天我们会弄清楚苯环为什么这么稳定，画出每年都考、值三分的反应机理， 记住每个反应的试剂，并且学会预测下一个基团会接在环的哪个位置。
Let's begin.
让我们开始吧。
Benzene is six carbon atoms in one flat ring.
苯是六个碳原子构成的一个平面环。
Each carbon keeps one electron in an orbital above and below that ring.
每个碳都在环的上下各留一个电子在轨道里。
Those orbitals overlap all the way round, so the six electrons belong to all six carbons at once.
这些轨道沿着整个环重叠起来，于是这六个电子不再属于某一对碳原子， 而是同时属于全部六个碳。
We call that delocalisation.
我们把这叫做离域。
So the ring is electron rich, and anything that loves electrons is pulled towards it. But the ring is also unusually stable, and benzene works hard to keep it.
由此有两个结果： 环是富电子的，所以任何"喜欢电子"的粒子都会被吸引过来； 但环同时也异常稳定，苯会想尽办法保住它。
Now the key idea of the topic.
现在是整章最关键的想法。
An alkene reacts by addition: the double bond opens and two groups join on.
烯烃发生的是加成：双键打开，两个新基团接上去。
If benzene did the same, the delocalised ring would be broken and all that stability thrown away.
如果苯也这样做，离域的环就会被破坏，那份额外的稳定性也就白白丢掉了。
So benzene takes the other road: one hydrogen leaves, and the new group takes its place.
所以苯走了另一条路：环上离去一个氢原子，新基团顶替它的位置，环本身毫发无损。
That is electrophilic substitution, and it is why benzene behaves nothing like an alkene.
这就是亲电取代，也正是苯的表现完全不像烯烃的原因。
Every one of these reactions begins the same way: you must make the electrophile first.
这些反应的开头都一样：你必须先制造出亲电试剂。
Take nitration. The conditions are concentrated nitric acid with concentrated sulfuric acid, between twenty five and sixty degrees Celsius. Here is what happens.
以硝化为例， 条件是浓硝酸与浓硫酸混合，温度保持在二十五到六十摄氏度之间。
Sulfuric acid is the stronger acid, so it gives a proton to the nitric acid, water is pushed out, and what is left is the nitronium ion — the attacker.
过程是这样的：硫酸是更强的酸，它把一个质子给硝酸，水被挤出去， 剩下的就是硝鎓离子。 这个离子就是进攻苯环的亲电试剂。
Learn this equation. It is one mark all by itself, nearly every year.
把这个方程式记住——它本身就是一分，而且几乎年年都考。
Here is the mechanism, worth three separate marks.
这就是反应机理，它一共值三个独立的分。
Mark one: a curly arrow from inside the hexagon to the electrophile — the delocalised electrons reach out and make the new bond.
第一分：一支弯箭头，从六元环内部出发， 指向亲电试剂——离域电子伸出来，形成新的键。
Mark two: the intermediate.
第二分：中间体。
Draw a horseshoe, not a full circle, because delocalisation now covers only five carbons, and put a positive charge inside.
要画马蹄形，而不是完整的圆圈，因为离域现在只覆盖五个碳， 并且要在环内标出正电荷。
That top carbon now holds a hydrogen and the new group.
顶端那个碳同时连着一个氢和新基团。
Mark three: a second curly arrow, from that carbon hydrogen bond back into the ring.
第三分：第二支弯箭头，从那根碳氢键指回环里。
The hydrogen leaves as a positive ion, and you have nitrobenzene.
氢以正离子的形式离去， 完整的圆圈回来了，你就得到了硝基苯。
Two arrows, one intermediate, and the ring finishes exactly as it began.
两支箭头、一个中间体， 而环最后和开始时一模一样。
Halogenation uses the same mechanism with a halogen, but bromine alone is not a strong enough electrophile.
同样的机理也适用于卤素，但溴本身还不是足够强的亲电试剂。
So you add a halogen carrier: aluminium bromide, or iron bromide.
所以要加入一个卤素载体：溴化铝，或者溴化铁。
The carrier pulls a bromide ion away and leaves a positive bromine, which attacks the ring in the two steps you just saw.
载体把一个溴离子拉走， 留下一个带正电的溴。
The product is a halogenoarene — bromobenzene, an aryl halide.
它按你刚才看到的那两步进攻苯环，产物是溴苯。
Then the hydrogen ion joins the carrier, giving hydrogen bromide and handing the catalyst straight back — so you only need a little.
还有一行是考试要的：被推下来的氢离子和载体结合，生成溴化氢， 并把催化剂原样还回来。 所以催化剂只需要一点点。
Next, two reactions that build a side chain onto the ring, both named after Friedel and Crafts.
接下来是两个在环上"接出侧链"的反应，都以傅列德尔和克拉夫茨命名。
Alkylation uses a chloroalkane with aluminium chloride and heat: benzene becomes methylbenzene.
烷基化用氯代烷加三氯化铝并加热：催化剂把氯拉走，烷基进攻苯环，苯就变成甲苯。
Acylation uses an acyl chloride with the same catalyst and heat, and gives a phenyl ketone.
酰基化用酰氯加同样的催化剂并加热：这次接上去的是酰基，产物是一个苯基酮。
In the exam, always name the reagent and the catalyst together. One without the other scores nothing.
考试时，试剂和催化剂一定要一起写出来，只写其中一个是不给分的。
Two more reactions, and neither substitutes the ring.
还有两个反应，而且它们都不是环上的取代。
The first one attacks the side chain: warm methylbenzene with hot alkaline potassium manganate seven, then dilute acid, and the whole side chain becomes an acid group — so the product is benzoic acid.
第一个进攻的是侧链： 把甲苯与热的碱性高锰酸钾一起加热，再加稀酸，整条侧链就被氧化成一个酸基。 不管侧链原来是什么，产物都是苯甲酸。
The second is hydrogenation: hydrogen, with a platinum or nickel catalyst and heat, adds three hydrogen molecules across the ring, and benzene becomes cyclohexane.
第二个进攻的是苯环本身： 氢气在铂或镍催化剂下加热，三个氢分子直接加到环上，苯就变成环己烷。
That is the one addition reaction benzene really does, and it destroys the delocalisation doing it.
这是苯真正会发生的那一个加成反应，代价是离域被破坏，环变成饱和的环己烷环。
Put chlorine and methylbenzene together: where does the chlorine go?
那么把氯和甲苯放在一起，氯会去哪里？
The conditions decide, and examiners love this.
由条件决定，而考官很喜欢考这一点。
With a halogen carrier catalyst and no ultraviolet light, the chlorine goes onto the ring, by electrophilic substitution.
有卤素载体催化剂、没有紫外光时，氯以亲电取代的方式接到环上。
With ultraviolet light and no catalyst, the chlorine goes onto the side chain instead, as a free radical reaction.
有紫外光、没有催化剂时，光把氯分解成自由基，氯以自由基取代的方式接到侧链上。
Same chemicals, different products.
同样的两种物质，产物却完全不同。
So read the conditions before you write anything. They are the answer.
所以动笔之前先看条件——条件不是摆设。
The last skill: directing effects. Where does the next group land?
最后一项技能：下一个基团会接在哪里？
The group already on the ring decides — that is its directing effect.
由环上已有的基团决定。 从那个基团开始给环编号。
Number the ring starting from that group. Amino, hydroxyl and alkyl groups push electron density into the ring: they send the new group to positions two and four, and the ring reacts faster.
氨基、羟基和烷基会把电子密度推入环中： 它们把新基团送到 2 位和 4 位，并让环比苯反应得更快。
Nitro, carboxyl and acyl groups pull electron density out of the ring: they send it to position three, and the ring reacts slower.
硝基、羧基和酰基会把电子密度从环中拉出：它们把新基团送到 3 位， 并让环反应得更慢。
Notice that the position and the rate always travel together.
注意，位置和速率总是同时出现——答对一个，另一个就白送。
Let's use that.
我们来用一用。
Methylbenzene and nitrobenzene are each nitrated. Where does the new nitro group go, and which reacts faster?
把甲苯和硝基苯分别硝化，预测新的硝基接在哪里， 并说明哪一个反应更快。
Start with methylbenzene. The group already there is an alkyl group, so it directs to positions two and four, and it releases electrons, so methylbenzene nitrates faster than benzene. Now nitrobenzene.
先看甲苯：环上已有的是甲基，而烷基定位到 2 位和 4 位； 它同时向环中释放电子，所以甲苯硝化比苯快。
The nitro group directs to three, and it withdraws electrons, so nitrobenzene nitrates more slowly than benzene.
再看硝基苯：硝基定位到 3 位， 它把电子从环中拉走，所以硝基苯硝化比苯慢。
One rule, two answers — always look at the group that is already on the ring.
一条规则，两个答案——永远先看环上已经有的那个基团。
A real exam question.
现在来一道真题。
Start from methylbenzene and make four-nitrobenzoic acid.
原料是甲苯，要制得 4-硝基苯甲酸。
Two steps are needed: nitrate the ring, and oxidise the side chain.
需要两步：环上硝化， 以及氧化侧链。 顺序重要吗？
Does the order matter?
完全重要。
Completely. Nitrate first. A methyl group directs to position four, so the nitro group lands where you want it.
先硝化：甲基定位到 4 位， 硝基正好接在你想要的位置。
Then oxidise that methyl group with hot alkaline potassium manganate seven and dilute acid.
然后用热的碱性高锰酸钾和稀酸把那个甲基氧化， 产物就是 4-硝基苯甲酸。
The product is four-nitrobenzoic acid. Now the other way round. Oxidise first, and the ring carries a carboxyl group, which directs to three.
再反过来试试：先氧化，环上带的就是羧基， 而羧基定位到 3 位；这时再硝化，得到的是 3-硝基苯甲酸。
Nitrate now and you get three-nitrobenzoic acid.
同样的两步，化合物却错了，一分也没有。
Same two steps, wrong compound, no mark — the order decides the product.
环上已有的基团决定位置， 所以步骤的顺序决定产物。
Four marks students throw away.
四个学生常丢的分。
First, never say benzene adds. It substitutes, and the reason is delocalisation: the stable ring is kept.
第一，不要说苯发生加成——它发生的是取代， 原因是离域：稳定的环被保留了下来。
Second, the mechanism is three separate marks — an arrow from inside the ring, a horseshoe intermediate with a plus, and an arrow from the carbon hydrogen bond back into the ring.
第二，机理是三个独立的分： 一支从环内部出发的箭头、一个带马蹄形和正电荷的中间体， 以及一支从碳氢键指回环里的箭头，三样都要画。
Draw all three. Third, when you give a position, give the reason too.
第三，答位置时要一并给出原因： 这一分要的是位置加上电子效应。
Fourth, write the full conditions: reagent, catalyst and temperature.
第四，条件要写全——试剂、催化剂和温度。
Get those four right, and this topic is yours.
把这四点做对，这个专题就是你的了。