Question

What will be the name of product on ammonolysis of benzyl chloride and reaction of amine so formed with two moles of \( \mathrm{CH_3Cl} \)?

• A. N, N-dimethylphenylmethanamine
• B. N, N-diphenylmethanamine
• C. N, N-diphenyl ethanamine
• D. N-methyl, N-phenyl methanamine

Hint:

In alkylation of amines, each mole of alkyl halide can replace one hydrogen attached to nitrogen. A primary amine can therefore form secondary and then tertiary amines on further alkylation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Ammonolysis involves the reaction of an alkyl/benzyl halide with ammonia to form a primary amine. Subsequent reaction with alkyl halides leads to alkylation of the amine.
Step 2: Formula Application:
1. Ammonolysis: $C_6H_5CH_2Cl + NH_3 \to C_6H_5CH_2NH_2$ (Benzylamine). 2. Alkylation (1st mole): $C_6H_5CH_2NH_2 + CH_3Cl \to C_6H_5CH_2NH(CH_3)$. 3. Alkylation (2nd mole): $C_6H_5CH_2NH(CH_3) + CH_3Cl \to C_6H_5CH_2N(CH_3)_2$.
Step 3: Explanation:
The starting amine is phenylmethanamine (benzylamine). After reacting with two moles of methyl chloride, two methyl groups replace the two hydrogens on the nitrogen. The IUPAC name is N, N-dimethylphenylmethanamine.
Step 4: Final Answer:
The product is N, N-dimethylphenylmethanamine.