JEE Main 2025 · 22 January · evening · Single correct · +4 / −1
Match the Compounds (List - I) with the appropriate Catalyst/ Reagents (List - II) for their reduction into corresponding amines.
Choose the correct answer from the options given below :
| List - I (Compounds) | List - II (Catalyst/Reagents) | ||
|---|---|---|---|
| (A) | R-C(∥(O))-NH₂ | (I) | NaOH (aqueous) |
| (B) | (II) | H₂/Ni | |
| (C) | R-C≡ N | (III) | LiAlH₄,H₂O |
| (D) | (IV) | Sn,HCl |
- (a) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
- (b) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
- (c) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
- (d) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
Answer and solution
Answer (b) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Solution
Chemistry ma’am · 4 steps
Match each group to its famous reagent, starting with the ones that can have only one partner.
- Read the structures. (B) is nitrobenzene, C₆H₅NO₂. (D) is an N-alkylphthalimide, the intermediate of the Gabriel synthesis.
- Fix the sure ones. An amide RCONH₂ needs the strong hydride LiAlH₄ to become RCH₂NH₂, because catalytic H₂/Ni does not do this. The N-alkylphthalimide releases its amine only on hydrolysis with aqueous NaOH, which is not a reduction at all. That places (A)-(III) and (D)-(I).
Compound Reagent Amine formed (A) RCONH₂ (III) LiAlH₄, H₂O RCH₂NH₂ (B) C₆H₅NO₂ (IV) Sn, HCl C₆H₅NH₂ (C) RC≡N (II) H₂/Ni RCH₂NH₂ (D) N-alkylphthalimide (I) aqueous NaOH RNH₂ and sodium phthalate - Remaining two. Sn/HCl is the classic metal-acid reagent for turning a nitro group into -NH₂, so (B)-(IV). The nitrile takes catalytic hydrogenation, RC≡N + 2H₂ →(Ni) RCH₂NH₂, so (C)-(II).
- So. The matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I), which is option (b).
The idea to keep. Amides need LiAlH₄, nitro groups go with metal and acid like Sn/HCl, nitriles take H₂/Ni, and the Gabriel phthalimide frees its amine by alkaline hydrolysis.
Answer from the NTA final answer key. Solution by PrepWiser.