Preparation of Amines
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Direct answer
Four JEE-favourite routes build amines. Ammonolysis of alkyl halides (RX + NH3) climbs the ladder primary → secondary → tertiary → quaternary salt, giving a mixture — fine for MCQ reasoning, poor for synthesis. Gabriel's phthalimide synthesis delivers clean primary aliphatic amines (phthalimide alkylated, then hydrolysed) but fails for aryl halides, which refuse SN2. Reduction covers nitriles (R-CN from RX + KCN — one carbon longer), nitro compounds (nitrobenzene → aniline with Sn/HCl or H2/Pd) and amides (LiAlH4 keeps the carbon count). The Hofmann bromamide degradation, RCONH2 + Br2 + KOH, hands back RNH2 with one carbon fewer — the counting trick at the heart of most JEE questions here.
What you must remember
- Ammonolysis ladder: excess ammonia biases the primary amine; excess alkyl halide drives to the quaternary salt — product control by stoichiometry.
- Gabriel rules: primary aliphatic amines only; aryl halides do not undergo the required substitution; hydrazine (N2H4) cleaves the N-alkyl phthalimide more cleanly than alkali.
- Cyanide duality: KCN (ionic) gives R-CN, later reduced to RCH2NH2; AgCN (covalent) gives isocyanide R-NC — the classic reagent-pair comparison.
- Amide reductions: LiAlH4 on RCONH2 gives RCH2NH2 (same chain); on R-CN gives RCH2NH2 (chain plus one, since KCN added a carbon).
- Hofmann bromamide: RCONH2 + Br2 + 3KOH → RNH2 + 2KBr + K2CO3 + H2O; the carbonyl carbon exits as carbonate — amine one carbon shorter.
- Aromatic entry: nitration of benzene then reduction (Sn/HCl or H2/Pd) is the only practical route to aniline; direct amination of benzene does not exist.
- Syllabus note: diazonium chemistry was trimmed from the JEE Main listing in the 2023 rationalisation, but aryl-amine interconversions survive in JEE Advanced habits — verify against the current year's bulletin.
Three routes to the same amine, counted
Target propylamine, C3. Route one: bromoethane + KCN → propionitrile (now three carbons) → LiAlH4 or H2/Ni → propan-1-amine — the chain grew by one at the cyanide step. Route two: propanamide + LiAlH4 → propan-1-amine — the chain is untouched. Route three: butanamide + Br2 + KOH → propan-1-amine — the chain shrank by one as the carbonyl left as carbonate. Same bottle on the shelf, three different starting lengths, and JEE tests exactly this bookkeeping: given the halide, name the amine; given the amine, choose the amide.
Add the purity angle. Ammonolysis of 1-bromopropane gives a mixture of propylamine, dipropylamine and tripropylamine in one pot; the Gabriel route on the same halide gives propylamine alone. When a question says "pure primary amine", Gabriel or nitrile-reduction is the answer; when it says "mixture", ammonolysis is the mechanism being probed.
The examiner's angle
Two traps dominate. First, the Gabriel-aryl blind spot: making aniline by Gabriel synthesis is impossible, because the phthalimide anion cannot attack an sp2 carbon — options offering it exist purely to be eliminated. Second, the Hofmann carbon count: students deliver an amine the same length as the starting amide; the carbonyl carbon is gone, migrated into K2CO3, and the amine is shorter — check it against the balanced equation before answering. JEE Main also probes reduction conditions: Sn/HCl work on nitro groups but leave aryl halides alone, while LiAlH4 is indiscriminate enough that substrates must be chosen, not assumed. On the aromatic side, remember that the nitration step precedes reduction — and the amine so formed is strongly activating, so any further substitution on the ring must be planned around it.
Frequently asked questions
Why is the Gabriel synthesis restricted to aliphatic primary amines?
The phthalimide anion attacks by SN2, which aryl halides do not undergo, so no N-aryl phthalimide forms and no aryl amine can be released.
How does the Hofmann bromamide degradation shorten the chain?
The carbonyl carbon of the amide is expelled as carbonate in the alkaline medium, so RNH2 contains one fewer carbon than RCONH2.
Why does KCN give nitriles while AgCN gives isocyanides?
Ionic KCN attacks through carbon, but covalent AgCN leaves only nitrogen accessible — the same alkyl halide therefore lengthens toward R-CN or R-NC depending on the reagent.
How is aniline prepared industrially relevant for exams?
Nitrate benzene to nitrobenzene, then reduce with Sn/HCl, Fe/HCl or catalytic hydrogenation — the two-step classical sequence.
What is the main drawback of ammonolysis as a synthesis?
Sequential alkylation past the primary amine gives a mixture of secondary, tertiary and quaternary products requiring separation.