The products A and B formed in the following reaction scheme are respectively

Solution

Chemistry ma’am · 4 steps

Nitro reduces to aniline, aniline becomes a diazonium salt, and phenol grabs it at the para position.

1. Nitrate, then reduce. Concentrated HNO₃/H₂SO₄ at 323 to 333 K puts one -NO₂ group on benzene to give nitrobenzene. Sn/HCl then reduces -NO₂ to -NH₂, so A is aniline, C₆H₅NH₂.
2. Diazotise in the cold. NaNO₂ + HCl makes nitrous acid, which turns aniline into benzenediazonium chloride, C₆H₅N₂⁺Cl⁻. The temperature is kept at 273 to 278 K because this salt decomposes when warm.
3. Couple with phenol. The diazonium ion is a weak electrophile. It attacks the activated phenol ring at the para position, the least hindered site, and the -N=N⁻ bridge stays in the product.

   C₆H₅N₂⁺Cl⁻ + C₆H₅OH → C₆H₅-N=N-C₆H₄-OH (p) + HCl

4. So. B is p-hydroxyazobenzene, the orange azo dye. Aniline and p-hydroxyazobenzene is option (b).

The idea to keep. In azo coupling the diazonium nitrogens are kept, giving an -N=N⁻ link, and the attack goes to the para position of phenol or aniline. Ortho coupling happens only when para is blocked.

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