Match the major products obtained in the reactions given in List-I with the corresponding structures in List-II and choose the correct option.
- P → 2; Q ⟶ 1; R → 5; S ⟶ 4
- P → 1; Q ⟶ 2; R → 4; S ⟶ 5
- P → 1; Q ⟶ 2; R → 3; S ⟶ 4
- P → 2; Q ⟶ 1; R → 3; S ⟶ 5
Answer
(2) P → 1; Q ⟶ 2; R → 4; S ⟶ 5
Solution
In all four compounds Br is para to NO2, so it is activated for SNAr.
(P) [figure] [-H2O] OH- [figure] [- Br-] OH- (SNAr) [figure]
The aldoxime H is anti to OH, so base eliminates H2O to give the nitrile; hot NaOH then replaces the activated Br (ortho-CN, para-NO2) by OH → (1)
(Q) [figure] (CH3CO)2O [figure] [-CH3COOH]Na2CO3 [figure]
The O-acetyl oxime loses CH3COOH (good leaving group) under mild base to give the nitrile; mild Na2CO3 does not displace Br → (2)
(R) [figure] OH- [figure] [- Br-]intramolecular SNAr [figure]
The ketoxime has no aldehydic H (no nitrile); its O⁻ is syn to the ring and attacks the C–Br carbon, closing 3-methyl-5-nitro-1,2-benzisoxazole → (4)
(S) [figure] aq. Na2CO3 [figure]
Aqueous Na2CO3 only hydrolyses the oxime ester back to the ketoxime; here O is anti to the ring, so no cyclisation → (5)
P → 1; Q → 2; R → 4; S → 5
Answer: (B)