含氮化合物

An amine 胺 has an $\text{–NH}_2$ group (the nitrogen replaces a hydrogen of ammonia).

Making a primary amine

Heat a halogenoalkane 卤代烷 with ammonia dissolved in ethanol, under pressure:

This is a nucleophilic substitution 亲核取代: the lone pair on the nitrogen of ammonia attacks the slightly positive carbon and pushes out the halogen. You use an excess of ammonia, or the amine made can react again.

Why an excess of ammonia. The amine made still has a lone pair on its nitrogen, so it is a nucleophile too. If there is little ammonia, the ethylamine attacks another bromoethane:

This gives a secondary amine 仲胺, which can react again to give a tertiary amine. A large excess of ammonia makes it much more likely that each bromoethane meets an ammonia molecule rather than an amine, so the main product is the primary amine.

Making a nitrile

Heat a halogenoalkane with potassium cyanide ($\text{KCN}$) in ethanol:

This is also a nucleophilic substitution, with the $\text{CN}^-$ ion as the nucleophile. It is useful because it adds one carbon to the chain. The product is a nitrile 腈.

Making a hydroxynitrile

Add $\text{HCN}$ (with $\text{KCN}$ as catalyst, and heat) to an aldehyde 醛 or ketone 酮. The $\text{H}$ and $\text{CN}$ add across the C=O bond to give a hydroxynitrile 羟基腈. The reagent is hydrogen cyanide 氰化氢, and the mechanism is nucleophilic addition 亲核加成.

Hydrolysis of nitriles

Warm a nitrile with dilute acid (or dilute alkali, then acidify). This hydrolysis 水解 turns the $\text{–CN}$ group into a $\text{–COOH}$ group, giving a carboxylic acid 羧酸:

A nitrile can also be reduced by hydrogen and a catalyst to form an amine, which is the reduction 还原 route to a longer-chain amine.

Worked example. Starting from bromoethane, make propanoic acid. Compare the carbons first: bromoethane has 2, propanoic acid has 3, so a carbon must be added - and the $\text{KCN}$ step is the reaction that does it. Step 1: warm bromoethane with ethanolic $\text{KCN}$; nucleophilic substitution gives propanenitrile, $\text{CH}_3\text{CH}_2\text{CN}$, which now has 3 carbons because the $\text{CN}$ carbon joins the chain. Step 2: reflux the nitrile with dilute $\text{HCl}$; hydrolysis gives propanoic acid. Count the carbons before you plan: whenever the target has exactly one more than the starting material, the nitrile route is almost always the intended answer, and remember the $\text{CN}$ carbon is part of the new chain.

Common mistakes.

• Using aqueous KCN. The solvent must be ethanol; in water the hydroxide ions present give an alcohol instead.
• Naming a nitrile without counting the CN carbon. $\text{CH}_3\text{CH}_2\text{CN}$ has three carbons, so it is propanenitrile, not ethanenitrile.
• Forgetting that the HBr made reacts with the excess ammonia (or amine) to give a salt, $\text{NH}_4\text{Br}$. The equation above is the simple form.
• Mixing up the two cyanide reactions. KCN with a halogenoalkane is substitution; HCN with an aldehyde or ketone is addition.

• Amines are bases (the N lone pair accepts $\text{H}^+$); aliphatic amines are stronger bases than ammonia.
• Making amines: reduce a nitrile, or react a halogenoalkane with excess ammonia.
• KCN adds one carbon (halogenoalkane → nitrile) — track the carbon count carefully in synthesis routes.
• State reagents and conditions precisely for each conversion.