MHTCET 13 Apr 2026 Shift 1 Paper

Concept:Reduction of a nitrile group $(-\mathrm{CN})$ to a primary amine $(-\mathrm{CH}_{2}\mathrm{NH}_{2})$ requires a strong reducing agent.Explanation:The $-\mathrm{CN}$ group in a nitrile can be reduced to $-\mathrm{CH}_{2}\mathrm{NH}_{2}$ by adding four hydrogen atoms.The general reaction is:$$\mathrm{R}-\mathrm{CN} + 4[\mathrm{H}] \longrightarrow \mathrm{R}-\mathrm{CH}_{2}\mathrm{NH}_{2}$$Sodium metal in ethanol $(\mathrm{Na} + \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH})$ acts as an effective reducing agent for this conversion.So the reaction becomes:$$\mathrm{R}-\mathrm{CN} \xrightarrow{\mathrm{Na} + \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \mathrm{R}-\mathrm{CH}_{2}\mathrm{NH}_{2}$$The other reagents given, such as $\mathrm{CrO}_{3}$, $\mathrm{H}_{3}\mathrm{PO}_{4}$, and $\mathrm{Al}_{2}\mathrm{O}_{3}$, are not used for reducing nitriles to amines.Therefore, the correct reagent is sodium in ethanol.Answer:$$\mathrm{Na} + \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$$Hence, Option B is correct.