JEE Main 2023 · 29 January · morning · Single correct · +4 / −1
Match List I with List II
Choose the correct answer from the options given below :
| List I | List II | ||
|---|---|---|---|
| Reaction | Reagents | ||
| (A) | Hoffmann Degradation | (I) | Conc.KOH, Δ |
| (B) | Clemenson reduction | (II) | CHCl₃,NaOH/H₃O ⊕ |
| (C) | Cannizaro reaction | (III) | Br₂,NaOH |
| (D) | Reimer-Tiemann Reaction | (IV) | Zn-Hg/HCl |
- (a) (A) - III, (B) - IV, (C) - I, (D) - II
- (b) (A) - II, (B) - IV, (C) - I, (D) - III
- (c) (A) - III, (B) - IV, (C) - II, (D) - I
- (d) (A) - II, (B) - I, (C) - III, (D) - IV
Answer and solution
Answer (a) (A) - III, (B) - IV, (C) - I, (D) - II
Solution
Chemistry ma’am · 3 steps
Tie each named reaction to the one reagent that defines it, and the matching falls into place.
- Recall each reagent. Every named reaction here has a signature reagent. Remember what each reaction does to the molecule, and its reagent follows.
Reaction What it does Reagent Match (A) Hoffmann degradation amide to amine with one carbon less Br₂, NaOH III (B) Clemmensen reduction C=O to CH₂ Zn-Hg/HCl IV (C) Cannizzaro reaction aldehyde with no α-hydrogen gives an alcohol and a carboxylate conc. KOH, Δ I (D) Reimer-Tiemann reaction phenol to salicylaldehyde CHCl₃, NaOH, then H₃O⁺ II - Why these reagents. In Hoffmann degradation, Br₂ in base makes an N-bromoamide that rearranges to an isocyanate, which then loses CO₂. In Reimer-Tiemann, CHCl₃ and NaOH give dichlorocarbene, which attacks the phenoxide ion. Acid work-up at the end gives the aldehyde.
- So. The matching is A-III, B-IV, C-I, D-II, which is option (a).
The idea to keep. Learn each named reaction as a reagent plus a change. Br₂/NaOH on an amide and CHCl₃/NaOH on phenol are both basic, so tell them apart by the halogen source.
Answer from the NTA final answer key. Solution by PrepWiser.