JEE Main 2025 · 7 April · evening · Single correct · +4 / −1
Match List - I with List - II .
Choose the correct answer from the options given below :
| List - I (Conversion) | List - II (Reagents, Conditions used) |
|---|---|
| (I) Warm H₂O | |
| (II) (a) NaOH, 368K; (b) H₃O⁺ | |
| (III) (a) NaOH, 443K; (b) H₃O⁺ | |
| (IV) (a) NaOH, 623K, 300 atm; (b) H₃O⁺ |
- (a) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
- (b) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
- (c) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
- (d) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Answer and solution
Answer (d) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Solution
Chemistry ma’am · 3 steps
Every extra nitro group ortho or para to Cl makes the swap to OH easier, so the conditions get gentler.
- Why chlorobenzene resists. The C–Cl bond in a haloarene has partial double-bond character from resonance, so OH⁻ cannot replace Cl easily. Plain chlorobenzene needs the harshest conditions.
- What nitro groups do. A -NO₂ group at the ortho or para position pulls electron density out of the ring and stabilises the negative intermediate formed when OH⁻ attacks. Each extra ortho or para nitro group lowers the temperature needed.
Conversion Nitro groups ortho/para to Cl Conditions (A) chlorobenzene to phenol 0 (IV) NaOH, 623 K, 300 atm; then H₃O⁺ (B) 4-nitrochlorobenzene to 4-nitrophenol 1 (III) NaOH, 443 K; then H₃O⁺ (C) 2,4-dinitrochlorobenzene to 2,4-dinitrophenol 2 (II) NaOH, 368 K; then H₃O⁺ (D) 2,4,6-trinitrochlorobenzene to picric acid 3 (I) warm H₂O - So. The match is (A)-(IV), (B)-(III), (C)-(II), (D)-(I), which is option (d).
The idea to keep. In haloarenes, nitro groups at ortho or para positions activate nucleophilic substitution. More such groups means milder conditions, down to just warm water with three.
Answer from the NTA final answer key. Solution by PrepWiser.