Select the reagents that reduce nitriles to primary amines. A. (i) \ce{LiAlH4}; (ii) \ce{H2O} B. \ce{Sn + HCl} C. \ce{H2/Ni} D. \ce{Na(Hg)/C2H5OH} E. \ce{Br2/aq.\ NaOH} Choose the correct answer from the options given below.

Solution & Explanation

\textbf{Answer:} (B) \ce{LiAlH4}, \ce{H2/Ni}, and \ce{Na(Hg)/C2H5OH} all reduce \ce{-C#N} to \ce{-CH2-NH2}; \ce{Sn/HCl} and \ce{Br2/NaOH} do not. \textbf{Solution:} Reduction of a nitrile adds two molecules of hydrogen across the \ce{C#N} triple bond to give a primary amine: \ce{R-C#N ->[reduction] R-CH2-NH2} Checking each reagent: A. \ce{LiAlH4} then \ce{H2O} (work-up): a strong hydride reducing agent $\rightarrow$ reduces nitrile to $1^\circ$ amine. \textbf{Yes}. B. \ce{Sn + HCl}: a classic reagent for reducing \textbf{nitro} groups (\ce{-NO2 -> -NH2}); it is not used to reduce nitriles to amines. \textbf{No}. C. \ce{H2/Ni} (catalytic hydrogenation): reduces \ce{-C#N} to \ce{-CH2NH2}. \textbf{Yes}. D. \ce{Na(Hg)/C2H5OH} (sodium amalgam, dissolving-metal reduction): reduces nitriles to $1^\circ$ amines. \textbf{Yes}. E. \ce{Br2/aq.\ NaOH}: this is the \textbf{Hofmann bromamide} reaction, which converts an \textbf{amide} \ce{R-CONH2} to an amine \ce{R-NH2} (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).