COMEDK 2025 Afternoon Solved Paper

Concept:The properties of B — quantitative $\mathrm{N}_2$ liberation with $\mathrm{NaNO}_2/\mathrm{HCl}$ and alkali‑soluble Hinsberg product — indicate B is a primary aliphatic amine.Explanation:Compound $\mathrm{A}$ ($\mathrm{C}_5\mathrm{H}_9\mathrm{N}$) is reduced by $\mathrm{Na}/\mathrm{Hg}/\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$ to give B. This reduction typically adds two hydrogen atoms to a multiple bond, converting a nitrile or imine into an amine.B reacts with $\mathrm{NaNO}_2/\mathrm{HCl}$ at $274\ \mathrm{K}$ and liberates $\mathrm{N}_2$ quantitatively. Only primary aliphatic amines ($\mathrm{RNH}_2$) form unstable alkyldiazonium salts that spontaneously release $\mathrm{N}_2$. Secondary or tertiary amines do not give this reaction.B also reacts with Hinsberg’s reagent (benzenesulfonyl chloride) to form a sulfonamide that is soluble in aqueous alkali. Primary amines produce $\mathrm{RNHSO}_2\mathrm{C}_6\mathrm{H}_5$, which has an $\mathrm{N}$—$\mathrm{H}$ bond that can be deprotonated by $\mathrm{KOH}$, making it water‑soluble. Secondary amines give insoluble sulfonamides, and tertiary amines do not react.Among the options, only $\mathrm{CH}_3\!-\!(\mathrm{CH}_2)_4\!-\!\mathrm{NH}_2$ is a primary aliphatic amine. Option A is a quaternary ammonium salt; option B is a secondary amine; option D is a tertiary amine.Answer:$\mathrm{CH}_3\!-\!(\mathrm{CH}_2)_4\!-\!\mathrm{NH}_2$ (Option C).