Preparation of Amines

Preparation — Ammonolysis & Reduction

Ammonolysis of alkyl halides: an alkyl halide reacts with ammonia (the N lone pair displaces the halide, SN2). But the 1° amine formed is itself nucleophilic and reacts further, so a mixture of 1°, 2° and 3° amines and the quaternary ammonium salt results.

R-X + NH$_3$ → R-NH$_2$ → R$_2$NH → R$_3$N → R$_4$N⁺X⁻

Using a large excess of ammonia favours the primary amine. (This route gives a mixture — for a pure 1° amine, use Gabriel synthesis instead.)

Reduction of nitro compounds: nitrobenzene is reduced (H$_2$/Ni, or Sn/Fe + HCl) to aniline. This is the main industrial route to aromatic amines.

Reduction of nitriles and amides: nitriles (R-CN) are reduced (H$_2$/Ni or LiAlH$_4$) to primary amines (R-CH$_2$-NH$_2$) — this adds one carbon. Amides (R-CONH$_2$) are reduced by LiAlH$_4$ to amines.

Key Point: ammonolysis gives a mixture (excess NH$_3$ → mostly 1°); nitrobenzene → aniline; nitriles reduce to 1° amines (adding a carbon); amides reduce with LiAlH$_4$.

Gabriel Phthalimide Synthesis

The Gabriel phthalimide synthesis gives pure primary aliphatic amines (no 2°/3° contamination). The steps:

1. Phthalimide is treated with KOH to form potassium phthalimide (the N-H is acidic).
2. This is alkylated with an alkyl halide (R-X) (SN2) to give an N-alkylphthalimide.
3. Hydrolysis (or treatment with hydrazine) releases the primary amine (R-NH$_2$).

Important limitation: Gabriel synthesis cannot be used to make aromatic primary amines (like aniline), because aryl halides do not undergo the SN2 alkylation of potassium phthalimide.

Key Point: Gabriel = phthalimide + KOH + R-X, then hydrolysis → pure 1° aliphatic amine. It does NOT work for aromatic amines (aryl halides won't react).

Hofmann Bromamide Degradation

The Hofmann bromamide degradation converts an amide into a primary amine with ONE FEWER carbon atom. An amide is treated with bromine and aqueous (or alcoholic) NaOH/KOH:

R-CO-NH$_2$ + Br$_2$ + 4 NaOH → R-NH$_2$ + Na$_2$CO$_3$ + 2 NaBr + 2 H$_2$O

The carbonyl carbon is lost as carbonate, so the amine has one carbon fewer than the amide (the alkyl group R migrates from carbon to nitrogen).

[JEE Tip] The carbon count is the giveaway: Hofmann bromamide degradation shortens the chain by one carbon (amide CnH(2n+1)CONH$_2$ → amine CnH(2n+1)NH$_2$), whereas reduction of a nitrile adds a carbon. Use this to spot the right reagent in conversion questions.

Key Point: Hofmann bromamide (R-CONH$_2$ + Br$_2$/NaOH) → R-NH$_2$ with one fewer carbon. Nitrile reduction adds a carbon — opposite directions.

Solved Examples

Example 1: Why ammonolysis gives a mixture

Why does ammonolysis of an alkyl halide not give a pure primary amine?

Solution: The 1° amine formed is itself nucleophilic and reacts with more alkyl halide to give 2° and 3° amines and the quaternary salt. A large excess of ammonia is used to maximise the primary amine.

Example 2: Gabriel synthesis product

What amine is obtained when ethyl bromide is used in the Gabriel synthesis?

Solution: Ethanamine (ethylamine, C$_2$H$_5$NH$_2$) — a pure primary amine.

Example 3: Gabriel limitation

Can aniline be prepared by the Gabriel synthesis? Explain.

Solution: No. Aniline would need an aryl halide to alkylate potassium phthalimide, but aryl halides do not undergo this SN2 substitution, so the aromatic amine cannot be made this way.

Example 4: Hofmann bromamide

What amine forms when CH$_3$CH$_2$CONH$_2$ (propanamide) undergoes Hofmann bromamide degradation?

Solution: One carbon is lost, giving ethanamine (C$_2$H$_5$NH$_2$) (a 2-carbon amine from the 3-carbon amide).

Example 5: Nitrile reduction

What amine is formed when CH$_3$CN is reduced with H$_2$/Ni?

Solution: The nitrile gains hydrogen during reduction, giving ethanamine (CH$_3$CH$_2$NH$_2$) — note that reduction of a nitrile adds one carbon relative to the alkyl group.

Example 6: Preparing aniline

How is aniline prepared industrially from benzene?

Solution: Nitrate benzene to nitrobenzene, then reduce (H$_2$/Ni or Sn/Fe + HCl) to aniline.

Example 7: Choosing add vs remove a carbon

You need to convert CH$_3$CH$_2$NH$_2$ (2 C) to CH$_3$NH$_2$ (1 C). Which method?

Solution: Use a route that removes a carbon — make the amide CH$_3$CONH$_2$ and apply Hofmann bromamide degradation → CH$_3$NH$_2$ (methanamine).

Example 8: Ascending the series

How would you convert CH$_3$Br to CH$_3$CH$_2$NH$_2$ (adding a carbon)?

Solution: CH$_3$Br + KCN → CH$_3$CN, then reduce (H$_2$/Ni or LiAlH$_4$) → CH$_3$CH$_2$NH$_2$ (ethanamine). The nitrile route adds one carbon.

Example 9: Pure primary amine

Which method gives a pure 1° amine free of 2° and 3° amines?

Solution: The Gabriel phthalimide synthesis — the phthalimide nitrogen carries only one R group, so on hydrolysis only the primary amine is released (for aliphatic amines).

Example 10: Reduction of an amide

What is the product when CH$_3$CONH$_2$ (acetamide) is reduced with LiAlH$_4$?

Solution: LiAlH$_4$ reduces the amide to the amine with the same number of carbons: ethanamine (CH$_3$CH$_2$NH$_2$). (Contrast Hofmann bromamide, which would remove a carbon to give methanamine.)