Identify A and B in the following reaction:

The correct answer is Option (3) → \(A = C_6H_5CH_2NH_2, B = C_6H_5CH_2-N(CH_3)_2\)

Benzyl chloride is an alkyl-type halide (Cl on an sp³ carbon), so it undergoes nucleophilic substitution. In ammonolysis, ammonia replaces the chlorine and gives a primary amine:

    \( \displaystyle \ce{C6H5CH2Cl + NH3 -> C6H5CH2NH2 + HCl} \quad (\text{A, benzylamine}) \)

An amine is itself a nucleophile, so it can be alkylated further. Two molecules of \( \ce{CH3Cl} \) replace both hydrogens on nitrogen, one after the other:

    \( \displaystyle \ce{C6H5CH2NH2 + 2CH3Cl -> C6H5CH2N(CH3)2 + 2HCl} \quad (\text{B, N,N-dimethylbenzylamine}) \)

So A is C₆H₅CH₂NH₂ and B is C₆H₅CH₂N(CH₃)₂.

Why the other options are incorrect

• Option (1) — Incorrect. A is right, but B carries only one methyl group; two moles of \( \ce{CH3Cl} \) introduce two.

• Option (2) — Incorrect. Ammonia does not reduce C–Cl to C–H, so toluene cannot be A; B would also need a C–C bond to form.

• Option (4) — Incorrect. The same error for A, and \( \ce{CH3Cl} \) does not form C–C bonds with toluene under these conditions.