Class 12 NEET Chemistry Organic Compounds Containing Nitrogen

Concept: Reduction of nitriles ($R-\mathrm{CN}$) to primary amines ($R-\mathrm{CH}_2-\mathrm{NH}_2$) requires a strong reducing agent that adds hydrogen across the $\mathrm{C}\equiv\mathrm{N}$ triple bond. Chemical Equation / Formula: $R-\mathrm{CN} \xrightarrow{\mathrm{LiAlH}_4 \;/\; \mathrm{H}_2\mathrm{O}} R-\mathrm{CH}_2-\mathrm{NH}_2$ $R-\mathrm{CN} \xrightarrow{\mathrm{H}_2 / \mathrm{Ni}} R-\mathrm{CH}_2-\mathrm{NH}_2$ $R-\mathrm{CN} \xrightarrow{\mathrm{Na(Hg)} / \mathrm{C}_2\mathrm{H}_5\mathrm{OH}} R-\mathrm{CH}_2-\mathrm{NH}_2$ b $R-\mathrm{CN} \xrightarrow{\mathrm{Sn} + \mathrm{HCl}} R-\mathrm{CHO} \quad (\text{not an amine})$ $R-\mathrm{CN} \xrightarrow{\mathrm{Br}_2 / \mathrm{aq. NaOH}} \text{no reduction to amine}$ Explanation: • Reagent A: $\mathrm{LiAlH}_4$ followed by water gives primary amine. • Reagent B: $\mathrm{Sn} + \mathrm{HCl}$ reduces nitrile to aldehyde ($R-\mathrm{CHO}$), not to amine. • Reagent C: Catalytic hydrogenation ($\mathrm{H}_2 / \mathrm{Ni}$) directly yields primary amine. • Reagent D: Sodium amalgam in alcohol reduces nitrile to primary amine. • Reagent E: $\mathrm{Br}_2 / \mathrm{aq. NaOH}$ is used for Hofmann rearrangement of amides, not for nitrile reduction. Thus reagents A, C, and D are correct. Shortcut: Remember: $\mathrm{LiAlH}_4$, $\mathrm{H}_2/\mathrm{metal}$, and $\mathrm{Na(Hg)/alcohol}$ reduce nitriles to primary amines; $\mathrm{Sn/HCl}$ gives aldehydes; $\mathrm{Br}_2/\mathrm{NaOH}$ does not apply. Answer: Option B: A, C and D only