Which of the following is true for the given reaction?
\((CH_3)_3N + C_6H_5SO_2Cl \longrightarrow\) No reaction
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Tertiary amines do not contain any replaceable 'H' on amino group
Benzenesulphonyl chloride (\( \ce{C6H5SO2Cl} \), Hinsberg’s reagent) reacts with primary and secondary amines to form sulphonamides. In this reaction a hydrogen atom on the nitrogen is replaced by the \( \ce{C6H5SO2} \)– group, and HCl is eliminated:
\( \ce{C6H5SO2Cl} \) + \( \ce{H-N(C2H5)2} \) → \( \ce{C6H5SO2N(C2H5)2} \) + HCl
Trimethylamine, \( \ce{(CH3)3N} \), is a tertiary amine: its nitrogen carries three alkyl groups and no hydrogen, so no stable sulphonamide can form and there is no reaction. This different behaviour is used to distinguish primary, secondary and tertiary amines.
Why the other options are incorrect
Option (2) It is a very bulky due to many \(-CH_3\) group — Incorrect. Methyl groups are small; the reason is the absence of an N–H hydrogen, not crowding.
Option (3) Lone pair of electrons on nitrogen does not allow this reaction — Incorrect. The lone pair on nitrogen is what attacks the sulphonyl group in primary and secondary amines too, so it cannot be the reason for no reaction.
Option (4) \(SO_2\) is eliminated and stops the reaction — Incorrect. No \( \ce{SO2} \) is released in Hinsberg’s test; the by-product with primary and secondary amines is HCl.
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.”
Reaction with Arylsulphonyl Chloride – “… (c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines.”