The products A and B formed in the following reaction scheme are respectively (i) conc. HNO3/ conc. H2SO4 323−333K (i) NaNO2,HCl,273−278K S) (ii) Sn/HCl (ii) Phenol ⟶A
(A)(see structure)
(B)(see structures)AB
(C)(see structures)AB✓ Correct
(D)(see structures)AB
Answer: (C)
Solution
Benzene is nitrated with conc. HNO3/conc. H2SO4 at 323–333 K to give nitrobenzene, which Sn/HCl reduces to aniline — this is A. Aniline is diazotised with NaNO2/HCl at 273–278 K to give the benzenediazonium salt, a stabilised electrophile. This diazonium ion couples with phenol, an electron-rich arene, at the position para to –OH (the most nucleophilic site on the phenoxide-like ring), giving the azo dye 4-hydroxyazobenzene — this is B. A is aniline and B is p-hydroxyazobenzene, C6H5–N=N–C6H4–OH.
JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option