Which of the following amines reacts with Hinsberg’s reagent to form a compound which is insoluble in alkali?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → C2H5NHC2H5
Hinsberg’s reagent is benzenesulphonyl chloride, \( \ce{C6H5SO2Cl} \). Diethylamine, a secondary amine, reacts with it to give N,N-diethylbenzenesulphonamide:
\( \ce{C6H5SO2Cl + HN(C2H5)2 -> C6H5SO2N(C2H5)2 + HCl} \)
This sulphonamide has no hydrogen on nitrogen, so it is not acidic and is insoluble in alkali.
Why the other options are incorrect
Option (1) C2H5NH2 — Incorrect. A primary amine gives N-ethylbenzenesulphonamide, whose N–H hydrogen is acidic (strongly electron withdrawing sulphonyl group), so the product dissolves in alkali.
Option (3) (C2H5)3N — Incorrect. A tertiary amine has no hydrogen on nitrogen and does not react with Hinsberg’s reagent at all.
Option (4) (CH3)3N — Incorrect. Also a tertiary amine; it does not react.
NCERT (2026-27 edition):
Reaction with Arylsulphonyl Chloride – “Benzenesulphonyl chloride (C6H5SO2Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides. (a) The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide. The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.
(b) In the reaction with secondary amine, N,N-diethyl- benzenesulphonamide is formed. … Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.”