Nitrogen compounds
A-Level Chemistry Topic 19 7:46 English narration · English + 中文 subtitles burned in
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Look at the thread in a jacket, a rope, a pair of stockings: nylon.
看看外套里的线、一根绳子、一双丝袜：尼龙。
It is built from two kinds of molecule, joined again and again — one with a nitrogen group at each end, one with an acid group at each end.
它由两种分子反复连接而成—— 一种两端各有一个氮基团，另一种两端各有一个酸基团。
The nitrogen half is made in a factory by adding hydrogen to a molecule that carries two carbon-to-nitrogen triple bonds.
工业上制造含氮的那一半， 是给一个带有两个碳氮三键的分子加氢。
Amines, nitriles, and the chemistry that turns one into the other: that is today.
胺、腈，以及把两者互相转化的化学， 就是今天的内容。
Welcome to nitrogen compounds.
欢迎来到含氮化合物。
We will build an amine and draw its mechanism, meet the cyanide ion and the two very different ways it adds a carbon, then turn a nitrile into an acid or into an amine.
我们会做出一个胺并画出它的机理，认识氰离子以及它加上一个碳的 两种完全不同的方式，再把腈变成酸或者变成胺。
Let's begin.
让我们开始吧。
Start with ammonia: a nitrogen carrying three hydrogens, and one pair of electrons left over.
先从氨开始：氮上带着三个氢，还剩下一对电子。
Replace one of those hydrogens with a carbon chain, and you have a primary amine — ethylamine, for example.
把其中一个氢换成碳链， 你就得到一个伯胺，比如乙胺。
What matters is what did not change. The lone pair is still there, and it does two jobs.
真正关键的是没有改变的东西——那对孤对电子还在， 而且它有两个本领。
It can take a hydrogen ion: amines are bases.
它可以接受一个氢离子：胺是碱。
And it can attack a slightly positive carbon, which makes it a nucleophile.
它也可以进攻带微正电的碳，这让它成为亲核试剂。
Every reaction today comes back to that one pair of electrons.
今天的每一个反应都要回到这对电子。
Take the base job first.
先看碱这个本领。
Add hydrochloric acid to ethylamine, and the lone pair takes a hydrogen ion from the acid.
把盐酸加到乙胺里，孤对电子从酸那里接过一个氢离子。
The nitrogen now carries four bonds and a positive charge, so the product is a salt: ethylammonium chloride.
现在氮带着四根键和一个正电荷，所以产物是一种盐：氯化乙铵。
Here is the exam point. An amine is a stronger base than ammonia, because the carbon chain pushes electron density towards the nitrogen.
考点在这里： 胺是比氨更强的碱，因为碳链把电子密度推向氮。
That leaves the lone pair more available, so it holds a hydrogen ion more firmly.
这让孤对电子更容易给出去，也就把氢离子抓得更牢。
Now the nucleophile job — and it is the one that makes an amine in the first place.
现在看亲核试剂这个本领——正是它造出了胺。
Take a halogenoalkane; bromoethane will do. Heat it with ammonia dissolved in ethanol, under pressure.
取一个卤代烷，溴乙烷就行， 把它和溶在乙醇里的氨一起加压加热。
The nitrogen takes the place of the bromine, and out comes ethylamine.
氮取代了溴的位置，生成乙胺。
Two details earn the marks.
有两个细节能拿分。
The solvent is ethanol, not water; in water you would make an alcohol instead.
溶剂是乙醇，不是水；在水里你得到的会是醇。
And you must use an excess of ammonia, because the amine you have just made is a nucleophile too, and it will react again.
而且必须用过量的氨，因为你刚做出的胺本身也是亲核试剂，它还会继续反应。
Before we draw it, recall what nucleophilic substitution looks like.
在画之前，先回忆亲核取代长什么样。
The nucleophile carries a lone pair.
亲核试剂带着一对孤对电子。
It attacks the carbon that the halogen has made slightly positive, and the halogen leaves, taking the bonding pair with it.
它进攻被卤素弄得微正的那个碳，卤素带着成键电子对离开。
Sometimes both happen in one step; sometimes the halogen goes first and leaves a flat, positive carbon behind.
有时这两件事在一步之内发生；有时卤素先走，留下一个平面的正碳。
Either way, change the nucleophile and you change the product.
不管走哪条路，换一个亲核试剂，就换一个产物。
Here it is for ammonia.
对氨来说是这样。
The nitrogen's lone pair attacks the slightly positive carbon, and at the same moment the bromine leaves, taking the bonding pair with it.
氮上的孤对电子进攻带微正电的碳，与此同时溴带着成键电子对离开。
Look at the nitrogen now: four bonds and a positive charge, so this is not the amine yet.
现在看这个氮：四根键，还带正电，所以这还不是胺。
A second ammonia molecule pulls that extra hydrogen away, and there is the free amine.
第二个氨分子把多出来的那个氢拉走， 自由的胺就出来了。
Remember that every curly arrow starts on a lone pair or on a bond.
记住：每一支弯箭头都必须从孤对电子或一根键上出发。
Now swap ammonia for the cyanide ion, and the same substitution does something far more useful.
现在把氨换成氰离子，同样的取代反应会做出更有用的东西。
Heat bromoethane with potassium cyanide dissolved in ethanol.
把溴乙烷和溶在乙醇里的氰化钾一起加热。
The cyanide attacks the carbon, the bromine leaves, and you have a nitrile.
氰离子进攻碳，溴离开，你就得到一个腈。
Now count the carbons.
现在数一数碳。
Bromoethane has two; the product has three, because the cyanide brought its own carbon, and that carbon is now part of the chain.
溴乙烷有两个；产物有三个，因为氰基自己带来了一个碳， 而这个碳现在是碳链的一部分。
Cyanide has a second trick, and this time nothing leaves.
氰还有第二个本领，而且这一次没有任何基团离开。
Warm an aldehyde or a ketone with hydrogen cyanide, using a little potassium cyanide as the catalyst.
把醛或酮与氰化氢一起温热， 用少量氰化钾作催化剂。
A hydrogen and a cyanide group add across the double bond between carbon and oxygen.
一个氢和一个氰基加到碳氧双键的两端。
The oxygen keeps the hydrogen as an alcohol group; the carbon keeps the cyanide.
氧留下那个氢，成为醇基；碳留下氰基。
Both groups sit on the same carbon, and we call the product a hydroxynitrile.
两个基团在同一个碳上， 这样的产物我们叫做羟基腈。
Put the two side by side, because examiners love the contrast.
把两者并排放在一起，因为考官特别爱考这个对比。
The same cyanide ion, two different reactions.
同一个氰离子，两种不同的反应。
With a halogenoalkane it is a substitution: something leaves, the bromide ion.
和卤代烷反应是取代：有东西离开，也就是溴离子。
With an aldehyde or ketone it is an addition: nothing leaves, the double bond simply opens.
和醛或酮反应是加成： 没有东西离开，双键只是打开了。
The conditions differ too.
条件也不一样。
But both do the same valuable job — K C N adds one carbon.
但两者做的是同一件有价值的事—— 氰化钾加上一个碳。
So what is a nitrile good for?
那么腈有什么用？
Two answers, and here is the first.
有两个答案，先看第一个。
Reflux it with dilute acid, and the whole cyanide group turns into a carboxylic acid group.
把它和稀酸一起回流， 整个氰基就变成羧基。
Read the equation: two waters are used up, and the nitrogen leaves as an ammonium salt.
看这个方程式：消耗了两个水分子，氮以铵盐的形式离开。
You may use dilute alkali instead, but then you get the salt of the acid, so you must add acid at the end.
你也可以改用稀碱，但那样得到的是酸的盐，所以最后必须再加酸。
Forgetting that step is a common lost mark.
忘了这一步是很常见的失分点。
Reduce it instead: a nitrile reduced this way gives a primary amine.
换成还原：腈被这样还原，得到的就是一个伯胺。
Hydrogen with a nickel catalyst will do it; so will lithium aluminium hydride.
氢气配镍催化剂可以做到；四氢铝锂也可以。
Watch the counting here, because a question will test it: the triple bond needs two molecules of hydrogen — two new hydrogens on the carbon, two on the nitrogen.
这里要留意数目，因为考题会考：三键需要两个氢分子——碳上加两个氢，氮上加两个氢。
This is how industry makes the nitrogen half of nylon.
工业上正是这样制造尼龙中含氮的那一半。
Here is the whole topic on one map.
这就是整个专题的一张地图。
The cyanide group is the hub.
氰基是中心枢纽。
You get into it in two ways: from a halogenoalkane by substitution, or from an aldehyde or ketone by addition.
进入它有两条路： 由卤代烷经取代进入，或者由醛、酮经加成进入。
Then you choose the exit.
然后你选择出口。
Hydrolysis takes you to a carboxylic acid. Reduction takes you to an amine.
水解带你到羧酸，还原带你到胺。
Let's use the map.
我们来用这张地图。
Starting from bromoethane, make propanoic acid.
从溴乙烷出发，制取丙酸。
Count the carbons first: bromoethane has two, propanoic acid has three, so a carbon must be added, which means cyanide has to appear somewhere.
先数碳：溴乙烷有两个碳， 丙酸有三个碳，所以必须加上一个碳，也就是说某一步一定要出现氰。
Step one: warm it with ethanolic K C N — potassium cyanide in ethanol; that gives propanenitrile, three carbons long.
第一步：把它和乙醇溶液中的氰化钾——乙醇中的 KCN——一起加热，得到丙腈，现在是三个碳。
Step two: reflux that nitrile with dilute hydrochloric acid, and there is your propanoic acid.
第二步：把这个腈和稀盐酸一起回流，丙酸就做出来了。
Now a real exam question, with a trap in it.
现在来看一道真题，里面藏着一个陷阱。
Butanenitrile reacts with hydrogen over a catalyst to give butylamine.
丁腈在催化剂存在下与氢气反应生成丁胺。
What volume of hydrogen, measured at room conditions, reacts with half a gram of butanenitrile?
在室温条件下，与零点五克丁腈完全反应需要多少体积的氢气？
Pause the video and try it.
先暂停，自己试一试。
Butanenitrile has four carbons, seven hydrogens and one nitrogen, so its molar mass is sixty-nine; half a gram is seven point two five times ten to the minus three moles.
丁腈有四个碳、七个氢和一个氮，所以摩尔质量是六十九；零点五克就是 七点二五乘以十的负三次方摩尔。
Here is the trap: multiply by two, because every triple bond takes two molecules of hydrogen.
陷阱在这里：要乘以二，因为每一个三键都要用掉 两个氢分子。
Then multiply by the molar gas volume, and you get about three hundred and forty-eight centimetres cubed.
再乘以气体摩尔体积，大约得到三百四十八立方厘米。
Three marks students throw away.
三个学生常丢的分。
First, a reagent on its own is not enough: give the conditions too — potassium cyanide in ethanol and heat, ammonia in ethanol under pressure, dilute acid and reflux.
第一，只写试剂是不够的，还要写条件——乙醇中的氰化钾并加热、 乙醇中的氨并加压加热、稀酸并回流。
Second, count the carbons before you plan a route; if the target has exactly one more carbon, the answer almost always goes through a nitrile.
第二，规划路线之前先数碳； 如果目标产物正好多一个碳，答案几乎总是要经过一个腈。
Third, use excess ammonia when you make an amine, or you finish with a mixture.
第三， 制备胺时要用过量的氨，否则最后得到的是混合物。