Methyl chloride on treatment with excess of ammonia yields

The correct answer is Option (1) → Methylamine

When an alkyl halide is heated with ethanolic ammonia in a sealed tube, ammonia acts as a nucleophile and replaces the halogen. This is ammonolysis. The ammonium salt formed gives the free amine with a base:

    \( \displaystyle \ce{CH3-Cl + NH3 -> CH3-NH3^+Cl^-} \)

    \( \displaystyle \ce{CH3-NH3^+Cl^- + NaOH -> CH3-NH2 + NaCl + H2O} \)

The methylamine formed is itself a nucleophile, so it can react with more methyl chloride and give the secondary and tertiary amines and finally the quaternary salt:

    \( \displaystyle \ce{CH3NH2} \xrightarrow{\ce{CH3Cl}} \ce{(CH3)2NH} \xrightarrow{\ce{CH3Cl}} \ce{(CH3)3N} \xrightarrow{\ce{CH3Cl}} \ce{(CH3)4N^+Cl^-} \)

When ammonia is taken in large excess, a molecule of methyl chloride is far more likely to meet an ammonia molecule than a methylamine molecule, so further alkylation is suppressed and the primary amine is obtained as the major product. So methyl chloride with excess ammonia yields methylamine.

Why the other options are incorrect

• Option (2) Dimethylamine — Incorrect. This needs a second methylation of methylamine, which an excess of ammonia suppresses.

• Option (3) Trimethylamine — Incorrect. This needs three methyl groups on nitrogen; it becomes important only when the alkyl halide is in excess.

• Option (4) Mixture of methylamine, dimethylamine and trimethylamine — Incorrect. A mixture is what ammonolysis gives when the reagents are not in large excess on the ammonia side; the excess of ammonia is used to avoid it.

NCERT (2026-27 edition):
Ammonolysis of Alkyl Halides – …