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: Ammonolysis of benzyl chloride.
Benzyl chloride is \( \mathrm{C_6H_5CH_2Cl} \). On ammonolysis, chlorine is replaced by the amino group, forming benzylamine.
\[ \mathrm{C_6H_5CH_2Cl + NH_3 \rightarrow C_6H_5CH_2NH_2} \] So, the first amine formed is benzylamine, also called phenylmethanamine.
Step 2: Reaction of benzylamine with one mole of \( \mathrm{CH_3Cl} \).
Benzylamine is a primary amine. When it reacts with methyl chloride, one hydrogen attached to nitrogen is replaced by a methyl group, giving N-methylbenzylamine.
\[ \mathrm{C_6H_5CH_2NH_2 + CH_3Cl \rightarrow C_6H_5CH_2NHCH_3} \] Step 3: Reaction with the second mole of \( \mathrm{CH_3Cl} \).
The secondary amine formed above reacts with another mole of methyl chloride. The second hydrogen on nitrogen is also replaced by a methyl group, giving a tertiary amine: N,N-dimethylbenzylamine. Its IUPAC name is N,N-dimethylphenylmethanamine.
\[ \mathrm{C_6H_5CH_2NHCH_3 + CH_3Cl \rightarrow C_6H_5CH_2N(CH_3)_2} \] Step 4: Conclusion.
Hence, the final product formed is N,N-dimethylphenylmethanamine.
Final Answer:N, N-dimethylphenylmethanamine.