How many of the following amines react with benzenesulphonyl chloride to give a sulphonamide that is soluble in alkali?
Ethanamine, N-Ethylethanamine, Triethylamine, Aniline, N-Methylaniline, Butan-1-amine
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3
Benzenesulphonyl chloride reacts with primary and secondary amines to form sulphonamides. Only the sulphonamide of a primary amine still has a hydrogen on nitrogen. The strongly electron withdrawing sulphonyl group makes this hydrogen acidic, so the sulphonamide dissolves in alkali.
\( \displaystyle \ce{C6H5SO2Cl + H2N-C2H5 -> C6H5SO2NH-C2H5 + HCl} \)
\( \displaystyle \ce{C6H5SO2Cl + HN(C2H5)2 -> C6H5SO2N(C2H5)2 + HCl} \)
Sulphonamide soluble in alkali:
Ethanamine (primary amine)
Aniline (primary amine)
Butan-1-amine (primary amine)
No alkali-soluble sulphonamide:
N-Ethylethanamine — secondary amine; its sulphonamide has no N–H and is insoluble in alkali.
Triethylamine — tertiary amine; does not react.
N-Methylaniline — secondary amine; its sulphonamide has no N–H and is insoluble in alkali.
Soluble in alkali: 3 amines.
NCERT (2026-27 edition):
9.6 Chemical Reactions – “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.”
“Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.”
“(c) Tertiary amines do not react with benzenesulphonyl chloride.”