What would be the major product of ammonolysis of an alkyl halide if large excess of ammonia was taken?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Primary Amine
Ammonolysis is the nucleophilic substitution of the halogen of an alkyl halide by ammonia. The primary amine formed is itself a nucleophile (even stronger than \( \ce{NH3} \)), so it can react with more alkyl halide, step by step:
\( \displaystyle \ce{NH3 + R-X -> R-NH2 + HX} \)
\( \displaystyle \ce{R-NH2 + R-X -> R2NH + HX} \)
\( \displaystyle \ce{R2NH + R-X -> R3N + HX} \)
\( \displaystyle \ce{R3N + R-X -> R4N^+X^-} \)
So a mixture of 1°, 2°, 3° amines and the quaternary salt is formed, but with a large excess of ammonia the primary amine is the major product. An R–X molecule is then far more likely to meet an \( \ce{NH3} \) molecule than an amine molecule, so the first step dominates.
Why the other options are incorrect
Option (2) Secondary amine — Incorrect. It forms only when the primary amine reacts with a second R–X, which is suppressed by excess \( \ce{NH3} \).
Option (3) Tertiary amine — Incorrect. It needs a third alkylation step, which is even less likely with excess \( \ce{NH3} \).
Option (4) Quaternary ammonium Salt — Incorrect. It is the major product when the alkyl halide is taken in excess, the opposite condition.
NCERT (2026-27 edition):
Ammonolysis of Alkyl Halides – …
