Select the reagents that reduce nitriles to primary amines. A. (i) $\mathrm{LiAlH}{}_4$; (ii) $\mathrm{H}{}_2\mathrm{O}$ B. $\mathrm{Sn}+\mathrm{HCl}$ C. $\mathrm{H}{}_2/\mathrm{Ni}$ D. $\mathrm{Na}(\mathrm{Hg})/\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}$ E. $\mathrm{Br}{}_2/\mathrm{aq}\cdot \mathrm{NaOH}$ Choose the correct answer from the options given below.

Solution & Explanation

\textbf{Answer:} (B) $\mathrm{LiAlH}{}_4$, $\mathrm{H}{}_2/\mathrm{Ni}$, and $\mathrm{Na}(\mathrm{Hg})/\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}$ all reduce $-\mathrm{C}\equiv \mathrm{N}$ to $-\mathrm{CH}{}_2-\mathrm{NH}{}_2$; $\mathrm{Sn}/\mathrm{HCl}$ and $\mathrm{Br}{}_2/\mathrm{NaOH}$ do not. \textbf{Solution:} Reduction of a nitrile adds two molecules of hydrogen across the $\mathrm{C}\equiv \mathrm{N}$ triple bond to give a primary amine: $\mathrm{R}-\mathrm{C}\equiv \mathrm{N}\stackrel{\mathrm{reduction}}{\to }\mathrm{R}-\mathrm{CH}{}_2-\mathrm{NH}{}_2$ Checking each reagent: A. $\mathrm{LiAlH}{}_4$ then $\mathrm{H}{}_2\mathrm{O}$ (work-up): a strong hydride reducing agent $\to$ reduces nitrile to $1^{\circ }$ amine. \textbf{Yes}. B. $\mathrm{Sn}+\mathrm{HCl}$: a classic reagent for reducing \textbf{nitro} groups ($-\mathrm{NO}{}_2\to -\mathrm{NH}{}_2$); it is not used to reduce nitriles to amines. \textbf{No}. C. $\mathrm{H}{}_2/\mathrm{Ni}$ (catalytic hydrogenation): reduces $-\mathrm{C}\equiv \mathrm{N}$ to $-\mathrm{CH}{}_2\mathrm{NH}{}_2$. \textbf{Yes}. D. $\mathrm{Na}(\mathrm{Hg})/\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}$ (sodium amalgam, dissolving-metal reduction): reduces nitriles to $1^{\circ }$ amines. \textbf{Yes}. E. $\mathrm{Br}{}_2/\mathrm{aq}\cdot \mathrm{NaOH}$: this is the \textbf{Hofmann bromamide} reaction, which converts an \textbf{amide} $\mathrm{R}-\mathrm{CONH}{}_2$ to an amine $\mathrm{R}-\mathrm{NH}{}_2$ (with one carbon lost); it does not reduce a nitrile. \textbf{No}. The correct reagents are A, C and D only, which is option (B).