Question

Which one of the products of the following reactions does not react with Hinsberg reagent to form sulphonamide? 

• A. A
• B. B
• C. C
• D. D

Hint:

Stephen reduction is a specific method to convert nitriles into aldehydes. Remember: \(Nitrile \xrightarrow{SnCl_2/HCl} Aldehyde\). Aldehydes lack the nucleophilic nitrogen needed to react with sulphonyl chlorides.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Hinsberg reagent (Benzene sulphonyl chloride, \(C_6H_5SO_2Cl\)) is used to distinguish between primary (\(1^\circ\)), secondary (\(2^\circ\)), and tertiary (\(3^\circ\)) amines. It reacts with \(1^\circ\) and \(2^\circ\) amines to form sulphonamides. It does not react with \(3^\circ\) amines or non-amine functional groups like aldehydes.
Step 2: Key Formula or Approach:
1. \(1^\circ\) Amine + \(PhSO_2Cl \to\) Sulphonamide (soluble in alkali).
2. \(2^\circ\) Amine + \(PhSO_2Cl \to\) Sulphonamide (insoluble in alkali).
3. \(3^\circ\) Amine + \(PhSO_2Cl \to\) No reaction.
4. Non-amines (e.g., aldehydes) + \(PhSO_2Cl \to\) No reaction.
Step 3: Detailed Explanation:
Let's analyze the products of the given reactions:
- (A) \(LiAlH_4\) reduction: \(p-cyano-benzaldehyde\) reduced by \(LiAlH_4\) gives \(p-(aminomethyl)benzyl alcohol\) (contains a \(1^\circ\) amine). It will react.
- (B) Stephen reduction (\(SnCl_2/HCl\)): Nitriles are reduced to imines and then hydrolyzed to aldehydes. The product of this reaction is \(p-formylbenzaldehyde\) (or terephthalaldehyde). Since it is an aldehyde and not an amine, it does not form a sulphonamide with Hinsberg reagent.
- (C) \(H_2/Ni\) reduction: Nitriles are reduced to \(1^\circ\) amines. It will react.
- (D) Mendius reduction (\(Na/Hg\) in alcohol): Nitriles are reduced to \(1^\circ\) amines. It will react.
Step 4: Final Answer:
The product of reaction (B) is an aldehyde, which does not react with Hinsberg reagent to form a sulphonamide.