The azo-dye (Y) formed in the following reactions is Sulphanilic acid + NaNO2 + CH3COOH X. [figure]
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Answer
(3) [figure]
Solution
Diazotisation: HO3S-C6H4-NH2 NaNO2/CH3COOH HO3S-C6H4- N N (X)
Coupling: X (weak electrophile) attacks 1-naphthylamine at C-4, para to the strongly activating –NH2 group:
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Y = red azo dye (Griess–Ilosvay test for nitrite)
Options (A), (B): two azo groups, no free para coupling ⇒ wrong
Option (D): –N=N– and –NH2 on different rings (not para) ⇒ wrong
Answer: (3)