Benzyl chloride is heated with an ethanolic solution of ammonia, and the amine so formed is then treated with two moles of CH3Cl. Which of the following is the final product?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \ce{C6H5CH2N(CH3)2} \)
Step 1 – Ammonolysis: the C–Cl bond of a benzyl halide is easily broken by a nucleophile. Ammonia replaces the chlorine atom and gives benzylamine (phenylmethanamine), a primary amine:
\( \displaystyle \ce{C6H5CH2Cl + NH3} \xrightarrow{\text{ethanol}} \ce{C6H5CH2NH2 + HCl} \)
Step 2 – Alkylation with two moles of CH3Cl: benzylamine is itself a nucleophile. Each mole of CH3Cl replaces one hydrogen atom on nitrogen by a methyl group:
\( \displaystyle \ce{C6H5CH2NH2 + CH3Cl -> C6H5CH2NHCH3 + HCl} \) (first mole: secondary amine)
\( \displaystyle \ce{C6H5CH2NHCH3 + CH3Cl -> C6H5CH2N(CH3)2 + HCl} \) (second mole: tertiary amine)
Final product: after two moles of CH3Cl, both hydrogen atoms of the –NH2 group have been replaced. The product is the tertiary amine N,N-dimethylphenylmethanamine, \( \ce{C6H5CH2N(CH3)2} \) (Option 2). Its formula fits the reaction in three ways:
the –CH2– group stays between the ring and nitrogen, because the chlorine of benzyl chloride was on that carbon;
nitrogen carries two methyl groups, one from each mole of CH3Cl, and no hydrogen;
it is a neutral amine, not a salt, because a third mole would be needed to form the quaternary ammonium salt.
Why the other options are incorrect
Option (1) \( \ce{C6H5N(CH3)2} \) — Incorrect.
This is N,N-dimethylaniline, in which nitrogen is bonded directly to the ring.
In benzyl chloride the chlorine is on the –CH2– carbon, so the –CH2– group stays between the ring and nitrogen.
Option (3) \( \ce{C6H5CH2NHCH3} \) — Incorrect. This secondary amine is formed after one mole of CH3Cl; the second mole converts it into the tertiary amine.
Option (4) \( \ce{[C6H5CH2N(CH3)3]^+Cl^-} \) — Incorrect. This quaternary ammonium salt needs a third mole of CH3Cl, which reacts with the tertiary amine.
NCERT (2026-27 edition):
9.4 Preparation of Amines – “…an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH2) group. … The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt.”
Example 9.1 – “(ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of CH3Cl.”
