Q17 / 150 · MHT CET 2025 (27 Apr Shift 2) · Chemistry
Which of the following is Hofmann exhaustive alkylation?
- Correct answerA$\mathrm{R} \mathrm{CONH}_2 \xrightarrow[\Delta]{\mathrm{Br}_2, \mathrm{KOH}_{(\mathrm{aq})}} \mathrm{R}-\mathrm{NH}_2$
- B$\mathrm{R}-\mathrm{NH}_2 \xrightarrow[\text { excess }]{\mathrm{R}-\mathrm{X}} \mathrm{R}_4 \stackrel{+}{\mathrm{N}} \overline{\mathrm{X}}$
- C$\mathrm{R}_4 \stackrel{+}{\mathrm{N}} \overline{\mathrm{X}} \xrightarrow[\Delta]{\text { moist } \mathrm{Ag}_2 \mathrm{O}}$ Alkene $+\mathrm{R}_3 \mathrm{~N}$
- D$\mathrm{R} \mathrm{CN} \xrightarrow{\mathrm{Na} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \mathrm{R} \mathrm{CH}_2 \mathrm{NH}_2$
Solution
Hofmann exhaustive alkylation describes the reaction of an amine with excess alkyl halide to form a quaternary ammonium salt through successive alkylation steps.
Option A shows Hofmann bromamide degradation, converting an amide to a primary amine with one less carbon. Option C depicts Hofmann elimination, decomposing a quaternary ammonium salt to form an alkene and tertiary amine. Option D represents nitrile reduction to a primary amine (Mendius reduction).
Option B corresponds exactly to Hofmann exhaustive alkylation: $\mathrm{R}-\mathrm{NH}_2 \xrightarrow[\text{excess}]{\mathrm{R}-\mathrm{X}} \mathrm{R}_4 \stackrel{+}{\mathrm{N}} \overline{\mathrm{X}}$.
The reaction proceeds through successive nucleophilic substitutions where the primary amine is progressively alkylated to secondary, tertiary, and finally quaternary ammonium salt.
Final answer: $\boxed{B}$
Hence, the correct answer is option: 'A'.