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

Solution

Chemistry ma’am · 3 steps

Three reagents take -CN to -CH₂NH₂, but Stephen reduction stops at the aldehyde.

1. What Hinsberg needs. Hinsberg reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl. It forms a sulphonamide only with an amine that has an N–H bond, that is, a primary or secondary amine. A product with no amine nitrogen cannot react.
2. Reduce each nitrile. Na/Hg in ethanol (Mendius reduction), LiAlH₄ and H₂/Ni all reduce -C≡N fully to -CH₂NH₂. SnCl₂ + HCl (Stephen reduction) stops at the imine, -CH=NH, which is hydrolysed to -CHO.

   Option	Reagent	Product	Hinsberg reaction?
   (a)	Na/Hg, C₂H₅OH	-CN becomes -CH₂NH₂ (1° amine)	Yes
   (b)	SnCl₂ + HCl	C₆H₅CHO, benzaldehyde	No, no amine
   (c)	LiAlH₄, then H₃O⁺	HOCH₂-C₆H₄-CH₂NH₂ (1° amine)	Yes
   (d)	H₂/Ni	CH₃-C₆H₄-CH₂NH₂ (1° amine)	Yes

3. So. Only the Stephen reduction in (b) gives a product without an N–H amine. Benzaldehyde cannot form a sulphonamide, so the answer is (b).

The idea to keep. Nitriles with Na/Hg, LiAlH₄ or H₂/Ni go all the way to -CH₂NH₂. With SnCl₂/HCl (Stephen) they stop at the aldehyde. Hinsberg needs an N–H amine.

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