JEE Main 2021 · 26 August · morning · Single correct · +4 / −1
Among the following compounds I-IV, which one forms a yellow precipitate on reacting sequentially with (i) NaOH (ii) dil. HNO₃ (iii) AgNO₃?
- (a) II
- (b) IV
- (c) I
- (d) III
Answer and solution
Answer (b) IV
Solution
Chemistry ma’am · 3 steps
Ask which halogen NaOH can actually pull off: aryl halides hold on, benzyl halides let go.
- What the test needs. AgNO₃ only reacts with free halide ions. So the halogen must first leave the molecule as X⁻ when warmed with NaOH. Dilute HNO₃ then removes the excess OH⁻, which would otherwise give a brown Ag₂O precipitate.
- Aryl versus benzyl. A halogen bonded directly to the benzene ring has partial double-bond character from resonance, so NaOH cannot replace it. A halogen on a CH₂ next to the ring is easily replaced, because the benzyl carbocation is resonance-stabilised.
Compound Halogen and where it sits Halide ion released by NaOH? With AgNO₃ I Cl on the ring (aryl) No no precipitate II Cl on the ring (aryl) No no precipitate III Br on the ring (aryl) No no precipitate IV I on CH₂ (benzyl), Cl on the ring Yes, I⁻ only yellow AgI - So. In IV, NaOH converts -CH₂I to -CH₂OH and frees I⁻, which gives yellow AgI with AgNO₃. The answer is (b), IV.
The idea to keep. Before you look at the halide's colour, check whether the halogen can come off at all. Halogens on the benzene ring stay put with NaOH; benzyl and alkyl halogens leave as ions.
Answer from the NTA final answer key. Solution by PrepWiser.