Which of the following reaction/s will not give p-aminoazobenzene?

Solution

Chemistry ma’am · 3 steps

Coupling needs a diazonium salt, and that needs an -NH₂ group first. Check each route.

1. What builds the target. p-Aminoazobenzene, C₆H₅-N=N-C₆H₄-NH₂, forms when benzenediazonium chloride couples with aniline at the para position. So a route works only if it makes C₆H₅N₂⁺Cl⁻ before aniline is added.

   Route	Step i gives	Diazonium salt formed?	p-Aminoazobenzene?
   A	Sn/HCl reduces -NO₂ to -NH₂, giving aniline	Yes, HNO₂ diazotises the aniline	Yes
   B	NaBH₄ does not reduce an aromatic -NO₂ group, so nitrobenzene stays as it is	No, there is no -NH₂ and no HNO₂	No
   C	HNO₂ turns aniline into C₆H₅N₂⁺Cl⁻	Yes	Yes, by coupling with aniline in mild acid

2. Why B fails. NaBH₄ is a mild hydride donor that reduces aldehydes and ketones, not nitroarenes. With no amine there is no diazonium ion, and NaOH followed by aniline cannot form an -N=N⁻ link.
3. So. Routes A and C both give p-aminoazobenzene. Only B does not, which is option (d).

The idea to keep. For any azo dye, trace back to the diazonium salt: you need an aromatic -NH₂ and cold HNO₂. NaBH₄ will not reduce -NO₂, but Sn/HCl will.

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