Identify the major product $C$ formed in the following reaction sequence: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{I}\stackrel{\mathrm{NaCN}}{\to }\mathrm{A}\underset{\mathrm{partial}\mathrm{hydrolysis}}{\overset{\mathrm{OH}{}^{-}/\mathrm{NaOH}}{\to }}\mathrm{B}\stackrel{\mathrm{Br}{}_2/\mathrm{NaOH}}{\to }\mathrm{C}$

Solution & Explanation

\textbf{Answer:} (D) Nitrile $\to$ amide $\to$ Hofmann bromamide degradation gives an amine with one fewer carbon: propylamine. \textbf{Solution:} $n$-Propyl iodide undergoes $\mathrm{S}{}_{\mathrm{N}}2$ substitution with cyanide to give butanenitrile: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{I}+\mathrm{NaCN}\to \mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{CN}$ ($A$, four carbons). Partial (controlled) alkaline hydrolysis stops at the amide stage, butanamide: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{CN}\stackrel{\mathrm{OH}{}^{-}}{\to }\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{CONH}{}_2$ ($B$). Treatment with $\mathrm{Br}{}_2/\mathrm{NaOH}$ is the Hofmann bromamide degradation, which converts a primary amide to a primary amine with loss of one carbon (as carbonate): $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{CONH}{}_2\stackrel{\mathrm{Br}{}_2/\mathrm{NaOH}}{\to }\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{NH}{}_2$. The product $C$ is propylamine (3 carbons) (D). Note butylamine (A) would require an amine-forming step with no carbon loss; the Hofmann reaction always removes the carbonyl carbon.