Making amines — Amines | Organic Chemistry

There is one obvious way to make an amine, and it does not work. Every method here is a way around that single failure.

The problem: alkylation cannot be stopped

Treat ammonia with an alkyl halide and the nitrogen does the S_N2 you expect, giving a primary amine. That amine is a better nucleophile than ammonia was, because the new alkyl group donates electron density to nitrogen, so it attacks the next molecule of halide faster than the ammonia still in the flask does — and the secondary amine it produces is faster again. Out comes a mixture of primary, secondary and tertiary amine plus quaternary ammonium salt, in proportions set by relative rates rather than by how much halide you weighed out. Two honest qualifications. The run is not monotonic all the way: a tertiary amine is a poorer nucleophile than the secondary amine that made it, because three alkyl groups crowd the S_N2 transition state, so it is the early steps that outrun ammonia rather than every step outrunning the last. And it is halide stoichiometry that cannot save you — a large excess of ammonia genuinely does bias the mixture toward the primary amine, and is how simple ones are made industrially. What you cannot do is control the outcome by adding less halide.

The reactions section flagged this and named exhaustive alkylation, Gabriel and reductive amination; here those are taken apart and the rest of the toolkit added.

One case welcomes the runaway: if a quaternary ammonium salt is the target, excess CH₃I run to completion gives a single product, because there is nowhere further to go.

Gabriel synthesis: alkylate something that can only react once

Phthalimide's N–H sits between two carbonyls, which makes it acidic enough (pK_a about 8.3) for KOH to deprotonate. The anion does an S_N2, and afterwards the nitrogen has no N–H left and its lone pair is delocalized into both carbonyls — an imide, not an amine, so it cannot alkylate again. Hot aqueous acid or base, or hydrazinolysis with H₂NNH₂, then releases a clean primary amine.

The catch: the alkylation is still an S_N2, so the halide should be primary or methyl and unhindered. Secondary substrates give poorer yields with elimination alongside; tertiary ones give only alkene.

Azide displacement: use a nucleophile that is not an amine

Azide, N₃⁻, is small and an excellent nucleophile, and its displacement product is an alkyl azide — not an amine and not nucleophilic, so there is nothing to over-alkylate. LiAlH₄, or H₂ over Pd, then gives the primary amine. Same S_N2 limits, fewer steps than Gabriel.

Cyanide, then reduce: the route that adds a carbon

RX + NaCN gives RCN, and LiAlH₄ reduces that to RCH₂NH₂. The nitriles section covers it; the bookkeeping is the point here. The nitrile carbon becomes the CH₂, so the amine has one more carbon than the halide did. Gabriel and azide hand back the skeleton you put in; this lengthens it.

Reductive amination: build the C–N bond by reduction instead

One failure and six escapes from it. The three that run an Sₙ² on a nitrogen surrogate — Gabriel, azide, cyanide — can only ever hand back a primary amine, and need a primary unhindered halide to do it. Reductive amination and amide reduction have neither limit, which is why they are the workhorses.Read the right-hand column first. Butanamide is the compound to keep in mind: LiAlH₄ gives butylamine and Br₂/NaOH gives propylamine, so the same starting material and the same kind of product differ by a carbon depending only on the reagent. That is why the carbon count is the first question to ask of a proposed amine synthesis and not the last — a route can be flawless step by step and still arrive one carbon short.

An aldehyde or ketone condenses with ammonia or a primary amine to an imine, or with a secondary amine to an iminium ion (see the imines and enamines section), and that C=N is reduced in the same flask. The escape is structural: the C–N bond comes from hydride adding to a C=N rather than from displacement, so the product amine has no halide left to attack, and the product is set by what you put in.

Two limits are worth naming, because they are not the S_N2 ones. The nitrogen ends up on the old carbonyl carbon, so a carbon with no hydrogen on it is unreachable this way — tert-butylamine is not a reductive amination product, and neither is an aryl amine. And making a primary amine from ammonia is the one case that does over-alkylate, because the primary amine produced competes with ammonia for the next carbonyl.

The reagent is the trick. NaBH₃CN and NaBH(OAc)₃ are attenuated hydride sources: they survive the mildly acidic pH (around 4–5) imine formation wants, and reduce the protonated imine far faster than the carbonyl, so the ketone survives to condense. NaBH₄ would simply give the alcohol.

Amide reduction: acylate once, then take the oxygen away

Acylation is self-limiting: the amide nitrogen it produces has its lone pair tied up in the carbonyl and is not nucleophilic, so one acylation gives one product. LiAlH₄ then reduces the C=O to CH₂, with no carbon lost. It is the cleanest general route to a secondary or tertiary amine: RCOCl + R′NH₂ → RCONHR′ → RCH₂NHR′.

Losing a carbon on purpose

The Hofmann rearrangement takes a primary amide with Br₂ and NaOH to a primary amine with one fewer carbon: the alkyl group migrates from the carbonyl carbon to nitrogen, giving an isocyanate that hydrolyzes and expels that carbon as CO₂. Butanamide, four carbons, gives propylamine, three. The Curtius rearrangement reaches the same isocyanate by heating an acyl azide, losing the same carbon.

Route	Gives	Carbon count
Direct alkylation	a mixture of 1°, 2°, 3°, 4°	unchanged
Gabriel	1° only	unchanged
Azide, then reduce	1° only	unchanged
Cyanide, then LiAlH₄	1° only	+1
Reductive amination	1°, 2° or 3°, your choice	gains the carbonyl's carbons
Amide, then LiAlH₄	1°, 2° or 3°	unchanged
Hofmann or Curtius	1° only	−1

Choose with two questions. Does the carbon count have to change? That cuts most of the table at once. Then: how substituted is the target nitrogen? Every route running an S_N2 on a nitrogen surrogate — Gabriel, azide, cyanide — gives only a primary amine. For anything more substituted, reductive amination or amide reduction.