JEE Main 2024 · 1 February · evening · Single correct · +4 / −1
Match List - I with List - II .
Choose the correct answer from the options given below :
| List I (Reactants) | List II (Product) |
|---|---|
| (A) Phenol, Zn/Δ | (I) Salicylaldehyde |
| (B) Phenol, CHCl₃, NaOH, HCl | (II) Salicylic acid |
| (C) Phenol, CO₂, NaOH, HCl | (III) Benzene |
| (D) Phenol, Conc. HNO₃ | (IV) Picric acid |
- (a) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
- (b) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
- (c) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
- (d) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Answer and solution
Answer (b) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
Solution
Chemistry ma’am · 3 steps
Name each reaction first: zinc dust, Reimer–Tiemann, Kolbe, nitration, and the products follow.
- Zinc dust. Heating phenol with zinc dust removes the oxygen: C₆H₅OH + Zn → C₆H₆ + ZnO. So (A) gives benzene, (III).
Reactants Reaction Product (A) Phenol, Zn, heat reduction by zinc dust benzene (III) (B) Phenol, CHCl₃, NaOH, then HCl Reimer–Tiemann, -CHO goes ortho salicylaldehyde (I) (C) Phenol, CO₂, NaOH, then HCl Kolbe, -COOH goes ortho salicylic acid (II) (D) Phenol, conc. HNO₃ nitration at 2, 4 and 6 picric acid (IV) - Aldehyde or acid. Both B and C start from sodium phenoxide and both put a group ortho to -OH. With CHCl₃ the group is -CHO, an aldehyde. With CO₂ the group is -COOH, an acid.
- So. The matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV), which is option (b).
The idea to keep. Learn each phenol reagent with its product: CHCl₃ gives -CHO, CO₂ gives -COOH, conc. HNO₃ gives picric acid, Zn dust gives benzene.
Answer from the NTA final answer key. Solution by PrepWiser.