An amine on reaction with benzensulphonyl chloride produces a compound insoluble in alkaline solution. This amine can be prepared by ammonolysis of ethyl chloride. The correct structure of amine is :

Solution

Chemistry ma’am · 3 steps

Two checks: Hinsberg says secondary amine, and ethyl chloride says an ethyl group must sit on N.

1. Read the Hinsberg result. With C₆H₅SO₂Cl, a primary amine gives C₆H₅SO₂NHR. Its N–H is acidic, so it dissolves in alkali. A secondary amine gives C₆H₅SO₂NR₂, which has no H on N and stays insoluble in alkali. So the amine is secondary.
2. Read the ammonolysis clue. Ammonolysis of C₂H₅Cl means an N–H group attacks the alkyl halide. That attaches a C₂H₅ group to nitrogen. For example, CH₃CH₂CH₂NH₂ + C₂H₅Cl → CH₃CH₂CH₂NHC₂H₅ + HCl. So the amine must carry an ethyl group on N.

   Option	Type of amine	Ethyl group on N?
   (a) CH₃CH₂NH₂	primary	yes
   (b) CH₃CH₂CH₂NHCH₃	secondary	no, it is methyl
   (c) CH₃CH₂CH₂NHCH₂CH₃	secondary	yes
   (d) C₆H₅NHCH₂CH₂CH₃	secondary	no, phenyl and propyl

3. So. Only (c), N-ethylpropan-1-amine, passes both checks. It is secondary, and its ethyl group can come from ethyl chloride.

The idea to keep. A Hinsberg product that is insoluble in alkali means a secondary amine, because no acidic N–H is left. Then match the alkyl groups on N to the alkyl halide you are told was used.

Answer from the NTA final answer key. Solution by PrepWiser.