The final product 'C' in the following series of reactions

Solution

Chemistry ma’am · 4 steps

Nitro to amine, amine to diazonium, then the diazonium couples at C1 of β-naphthol, right next to the OH.

1. Reduction gives A. Sn/HCl reduces the nitro group to an amino group. So A is aniline, C₆H₅NH₂.
2. Diazotisation gives B. NaNO₂ and HCl generate nitrous acid, which converts aniline at 0 to 5 °C into benzenediazonium chloride. So B is C₆H₅N₂⁺Cl⁻.

   C₆H₅NO₂ →(Sn/HCl) C₆H₅NH₂ →(NaNO₂/HCl, 0-5 °C) C₆H₅N₂⁺Cl⁻

3. Coupling gives C. β-Naphthol has its OH at C2. In NaOH it becomes the naphthoxide ion, a strong ortho/para director. The weakly electrophilic diazonium ion attacks C1, the α-position ortho to -O⁻. Attack at C1 keeps the other benzene ring fully aromatic in the intermediate, so C1 wins over C3.
4. So. C is 1-phenylazo-2-naphthol, the orange-red dye with -N=N-C₆H₅ on C1 and OH on C2. That is option (c).

The idea to keep. Diazonium ions are weak electrophiles, so they only attack strongly activated rings. With β-naphthol, coupling always goes to C1, the α-carbon right beside the OH, and never to the other ring.

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