The azo-dye (Y) formed in the following reactions is
Sulphanilic acid +NaNO₂+CH₃COOH→ X.

Solution

Chemistry ma’am · 4 steps

Diazotise sulphanilic acid first, then let it couple para to the NH₂ of 1-naphthylamine.

1. Make X. NaNO₂ with acetic acid gives nitrous acid, which diazotises the -NH₂ of sulphanilic acid. So X is the diazonium ion HO₃S-C₆H₄-N₂⁺, with the -SO₃H still para to the diazo group.
2. Find the attack site. The diazonium ion is a weak electrophile. It attacks the ring that carries the activating -NH₂ group of 1-naphthylamine. The free position para to NH₂ on that ring is C-4, so the azo link forms there.
3. Check the options. Only (c) has the azo group at C-4 of the ring that bears NH₂, with one sulphophenyl group on the other side of -N=N⁻. Option (d) puts NH₂ on the other ring, (b) has lost the NH₂ altogether, and (a) is a cyclic structure that one coupling cannot form.
4. So. Y is 4-amino-1-naphthylazo-benzene-4′-sulphonic acid, option (c). This red dye is the basis of the Griess–Ilosvay test for nitrite.

The idea to keep. In azo coupling, the diazonium ion attacks the ring carrying -OH or -NH₂, para to that group. Diazotise the amine first, then look for the free para position on the partner.

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