The amine that reacts with Hinsberg's reagent to give an alkali-insoluble product is:

Solution & Explanation

\textbf{Answer:} (A) A secondary amine gives an N,N-disubstituted sulfonamide that has no acidic $\mathrm{N}-\mathrm{H}$, so it is insoluble in alkali. \textbf{Solution:} Hinsberg's reagent is benzenesulfonyl chloride, $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{SO}{}_2\mathrm{Cl}$. Its behaviour distinguishes $1^{\circ }$, $2^{\circ }$, and $3^{\circ }$ amines: $\bullet$ \textbf{Primary amine}: forms an N-monosubstituted sulfonamide $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{SO}{}_2\mathrm{NHR}$. The $\mathrm{N}-\mathrm{H}$ is acidic (activated by the $-\mathrm{SO}{}_2-$ group), so it dissolves in $\mathrm{NaOH}$ (alkali) forming a soluble salt $\to$ \textbf{alkali-soluble}. $\bullet$ \textbf{Secondary amine}: forms an N,N-disubstituted sulfonamide $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{SO}{}_2\mathrm{NR}{}_2$. It has \textbf{no $\mathrm{N}-\mathrm{H}$}, hence cannot form a salt with alkali $\to$ \textbf{alkali-insoluble} (precipitate). $\bullet$ \textbf{Tertiary amine}: has no $\mathrm{N}-\mathrm{H}$ to react, so it does not form a sulfonamide (no reaction under normal conditions). Reaction for the secondary amine in option (A): $(\mathrm{CH}{}_3){}_2\mathrm{CH}-\mathrm{NH}-\mathrm{CH}(\mathrm{CH}{}_3){}_2+\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{SO}{}_2\mathrm{Cl}\to \mathrm{C}{}_6\mathrm{H}{}_5\mathrm{SO}{}_2-\mathrm{N}[\mathrm{CH}(\mathrm{CH}{}_3){}_2]{}_2+\mathrm{HCl}$ This sulfonamide lacks an acidic $\mathrm{N}-\mathrm{H}$ and is therefore insoluble in alkali. The primary amines (C, D) give alkali-soluble products and the tertiary amine (B) does not react. Hence option (A) is correct.