Organic synthesis (A2)
A-Level Chemistry Topic 36 7:52 English narration · English + 中文 subtitles burned in
Chapters
Transcript
Three very different things: a bright orange dye, a painkiller, and a tough plastic.
三样完全不同的东西：一种鲜橙色的染料、一片止痛药、一块结实的塑料。
All three begin as the same colourless liquid — benzene.
它们都源自同一种无色液体——苯。
But no single reaction turns benzene into any of them.
但没有任何一个反应能把苯直接变成它们中的任何一个。
Chemists get there in steps, and every step is a reaction you already know.
化学家是一步一步走过去的，而每一步都是你已经学过的反应。
At A2 the exam does the same: here is a target molecule — find the road to it.
A2 考试也一样：它给你一个目标分子，让你找出通往它的路线。
This is organic synthesis at A2.
这就是 A2 的有机合成。
You already own the AS toolkit.
AS 的工具箱你已经掌握了。
Today we add the benzene ring, phenol, amines, amides and acyl chlorides, join all of it into one map, and then plan real synthetic routes with it.
今天我们再加上苯环、苯酚、胺、酰胺和酰氯，把它们连成一张完整的图， 然后用它来规划真正的合成路线。
Let's begin.
让我们开始吧。
So what is actually new?
那么，究竟新增了什么？
At AS you built chains: an alkene to a halogenoalkane to an alcohol, then up the oxidation ladder.
在 AS 阶段你搭的是链：烯烃到卤代烷，卤代烷到醇， 再沿着氧化阶梯往上走。
A2 bolts on the ring: nitration and Friedel-Crafts on arenes, and the activated ring of phenol.
A2 在此之上接上了苯环： 芳烃的硝化和傅克反应，以及被活化的苯酚环。
It adds amines, amides, and acyl chlorides, the most reactive tool you own.
此外还有胺、酰胺， 以及你手上最活泼的工具——酰氯。
The marks sit where the new half joins the old half.
分数就落在新旧两半衔接的地方。
First, know what you are holding.
在规划之前，你必须知道手里拿的是什么。
A2 adds three quick tests.
A2 增加了三个快速检验。
One: bromine water with no catalyst.
第一：不加催化剂的溴水。
If it decolourises and a white precipitate appears, the ring is activated: a phenol or a phenylamine.
如果溴水褪色并出现白色沉淀，说明环被活化了， 那就是苯酚或苯胺。
Two: cold water.
第二：冷水。
A violent reaction with steamy fumes of hydrogen chloride means an acyl chloride.
剧烈反应并冒出氯化氢白雾，说明是酰氯。
Three: neutral iron three chloride.
第三：中性三氯化铁。
Purple means a phenol.
出现紫色，说明是苯酚。
Learn the observation, not only the reagent.
要记住现象，而不只是试剂。
Now the chain the exam loves most.
现在来看考试最爱的那条链。
Benzene, with concentrated nitric acid and concentrated sulfuric acid, gives nitrobenzene.
苯用浓硝酸和浓硫酸处理，生成硝基苯。
Reduce that with tin and concentrated hydrochloric acid, then sodium hydroxide: phenylamine.
再用锡和浓盐酸还原，然后加氢氧化钠，就得到苯胺。
Cool it with sodium nitrite and hydrochloric acid below ten degrees, and you get a diazonium salt.
把苯胺与亚硝酸钠和盐酸在十摄氏度以下冷却反应，就得到重氮盐。
That salt goes two ways: warm it with water for phenol, or couple it with phenol in alkali for an azo dye.
这个盐有两条去路：与水共热得到苯酚，或在碱性条件下与苯酚偶联，得到有颜色的偶氮染料。
You nitrate first because you cannot put an amine straight onto a ring.
之所以要先硝化，是因为你没办法把氨基直接装到环上。
The second new chain runs through the acyl chloride.
第二条新链要经过酰氯。
Treat any carboxylic acid with thionyl chloride or phosphorus pentachloride, and the hydroxyl group is swapped for chlorine.
把任何羧酸用二氯亚砜或五氯化磷处理， 羟基就被换成氯。 得到的酰氯是你手上最活泼的衍生物。
It is the most reactive derivative you own. At room temperature, give it an alcohol or a phenol and you get an ester. Give it ammonia or an amine and you get an amide.
在室温下，给它一个醇或苯酚，就得到酯；给它氨或胺，就得到酰胺。
Every one of those is addition-elimination.
这些反应全都是加成消去。
And reduce the amide with lithium aluminium hydride, and you are back at an amine.
再用氢化铝锂还原酰胺，你就又回到了胺。
Now a habit that saves whole questions: count the carbon atoms.
再养成一个能救下整道题的习惯：数碳原子。
Most reactions leave the number alone. Three do not.
大多数反应不改变碳数，但有三个会改变。
Potassium cyanide on a halogenoalkane adds one carbon as a nitrile, and that nitrile goes on to an acid or an amine.
氰化钾与卤代烷反应，以腈的形式多加一个碳；这个腈再往下可以变成酸或胺。
Friedel-Crafts adds a whole alkyl or acyl group to the ring.
傅克反应给环加上一整个烷基或酰基。
And hot alkaline potassium manganate cuts any side chain back to a single carboxyl carbon.
而热的碱性高锰酸钾把任何侧链都砍回到一个羧基碳。
Here is the method; it never changes.
方法就在这里，而且永远不变。
Do not start at the beginning; start at the target and walk backwards.
不要从起点开始，要从目标出发往回走。
First, name the functional group in the target.
第一，说出目标分子里的官能团。
Second, ask which single reaction makes that group, and from what.
第二，问自己：哪一个反应能生成这个官能团，从什么原料生成。
Third, that starting material becomes your new target, and you go round again.
第三，把那个原料当作新的目标，再来一遍。
Last, reverse the whole thing and write it forwards, with reagent and conditions on every arrow.
最后，把整条路线反过来，正着写出来，每一个箭头上都写清试剂和条件。
Let's run a real one, from a recent A Level paper.
我们来做一道真题，出自最近的一份 A Level 试卷。
The target is 4-aminobenzoic acid: a benzene ring with a carboxyl group and, opposite it, an amine group.
目标是 4-氨基苯甲酸：一个苯环上带着羧基，对位上带着氨基。
You start from plain benzene.
起点是纯苯。
Step one puts a methyl group on the ring: chloromethane with aluminium chloride, a Friedel-Crafts alkylation.
第一步在环上装上一个甲基：用氯甲烷加三氯化铝，这是傅克烷基化。
Step two nitrates it: concentrated nitric acid with concentrated sulfuric acid, twenty-five to sixty degrees.
第二步硝化：浓硝酸加浓硫酸，温度在二十五到六十摄氏度之间。
The methyl group is a two-four director, so the nitro group lands opposite it.
甲基是邻对位定位基，所以硝基落在它的对位。
Two steps left, and each changes a different group.
还剩两步，每一步改变的都是不同的基团。
Step three oxidises the methyl side chain to a carboxyl group: hot alkaline potassium manganate, then dilute acid.
第三步把甲基侧链氧化成羧基： 用热的碱性高锰酸钾，再加稀酸。
That product is 4-nitrobenzoic acid.
这一步的产物是 4-硝基苯甲酸。
Step four reduces the nitro group to an amine: tin and hot concentrated hydrochloric acid, then sodium hydroxide.
第四步把硝基还原成氨基：用锡和热的浓盐酸，再加氢氧化钠。
Notice what those reagents do not do: they leave the carboxyl group untouched.
注意这些试剂不会做什么：它们完全不碰羧基。
Four steps, four reagents, and the target is built.
四步，四组试剂，目标分子就搭好了。
Now the twist the same paper asked.
接下来是同一份试卷问的那个转折。
Make compound Y instead: the same two groups, but the amine sits at position three, not four.
改做化合物 Y：同样的两个基团，但氨基这次在 3 位，而不是 4 位。
Same reactions, same reagents; you change nothing but the order. Oxidise the methyl group before you nitrate. Why?
反应相同，试剂相同，你要改的只有顺序：先把甲基氧化，再硝化。
Because the group already on the ring decides where the next one lands.
为什么？ 因为环上已有的基团决定了下一个基团落在哪里。
A methyl group sends it to two and four; a carboxyl group sends it to three.
甲基把它送到 2 位和 4 位；羧基则把它送到 3 位。
The exam also hands you a route and asks you to analyse it.
考试也会给你一条不是你设计的路线，让你分析它。
That question always has the same shape: for each step, name the type of reaction and give the reagent.
这类题的形式永远一样：对每一步，写出反应类型并给出试剂。
A group added to a ring is electrophilic substitution.
往环上装一个基团，是亲电取代。
Nitro to amine, or amide to amine, is reduction.
硝基变胺、酰胺变胺，是还原。
A side chain cut back to a carboxyl group is oxidation.
侧链被砍回羧基，是氧化。
A nucleophile replacing a halogen is nucleophilic substitution.
亲核试剂取代卤素，是亲核取代。
And anything an acyl chloride does is addition-elimination.
而酰氯做的任何反应，都是加成消去。
Now the by-products.
再来看副产物。
Ammonia with a halogenoalkane does not stop at the primary amine: the amine you just made is a nucleophile itself, so it attacks again, and you finish with secondary and tertiary amines and a poor yield.
氨与卤代烷反应不会停在伯胺： 你刚做出来的胺本身也是亲核试剂，于是它继续进攻， 最后得到仲胺和叔胺的混合物，产率很低。
Use a large excess of ammonia to push it back.
用大量过量的氨可以把反应压回去。
On an activated ring the same greed appears: phenol with bromine water gives the tribromo product.
在被活化的环上也有同样的贪心：苯酚与溴水反应得到的是三溴产物。
One more by-product hides in the shape of the molecule.
还有一种副产物藏在分子的形状里。
Whenever a step turns a flat carbon into a carbon with four different groups, the reagent can attack from either side.
只要某一步把一个平面碳变成连着四个不同基团的碳， 试剂就可以从任意一边进攻。
You get both mirror images in equal amounts: a racemic mixture.
于是你得到等量的两种镜像异构体，也就是外消旋混合物。
Say so — the mark scheme asks for it by name.
遇到这种情况一定要写出来，因为评分标准会点名要这个词。
Second worked example, the classic two-step.
第二道例题，是经典的两步法。
The target is an azo dye, so look for the azo group: two nitrogen atoms joined by a double bond, between two rings.
目标是一种偶氮染料， 所以先找偶氮基：两个氮原子以双键相连，夹在两个环之间。
It is made in exactly one way.
它只有一种做法。
Step one: phenylamine with sodium nitrite and hydrochloric acid, below ten degrees, gives the diazonium salt.
第一步：苯胺与亚硝酸钠和盐酸反应，保持在十摄氏度以下，得到重氮盐。
Keep it cold, or it breaks down to phenol and nitrogen gas.
一定要保持低温，否则它会分解成苯酚和氮气。
Step two: add phenol in aqueous sodium hydroxide.
第二步：加入溶于氢氧化钠溶液的苯酚。
The two halves couple, and the colour appears.
两半偶联起来，颜色就出现了。
Three marks students throw away.
三个学生常丢的分。
First, write the conditions, not just the reagent: concentrated or dilute, hot or cold, below ten degrees, then sodium hydroxide.
第一，要写条件，不能只写试剂：浓还是稀、热还是冷、 十摄氏度以下、之后再加氢氧化钠。
Second, count the carbons before you write anything; if the target has one more carbon, a cyanide step has to be in the route.
第二，动笔之前先数碳：如果目标比原料多一个碳，路线里就一定要有氰化物那一步。
Third, when a route could run in two orders, choose the order the directing effects allow, and say which group is doing the directing.
第三，当一条路线可以有两种顺序时，选定位效应允许的那一种，并说明是哪个基团在定位。