The List-II contains products obtained from the reaction of compounds in List-I with O3/Zn-H2O followed by cyclization (via more stable enolate) in the presence of aqueous NaOH. Match each entry in List-I with appropriate entry in List-II and choose the correct option.
- P ⟶ 2; Q ⟶ 4; R ⟶ 1; S ⟶ 3
- P ⟶ 3; Q ⟶ 4; R ⟶ 5; S ⟶ 2
- P ⟶ 2; Q ⟶ 1; R ⟶ 5; S ⟶ 3
- P ⟶ 3; Q ⟶ 5; R ⟶ 4; S ⟶ 2
Answer
(3) P ⟶ 2; Q ⟶ 1; R ⟶ 5; S ⟶ 3
Solution
Ozonolysis cleaves the ring-fusion C=C, giving a ten-membered 1,6-dione with a CH₃ on a carbon α to a C=O. NaOH forms the more stable (more substituted) enolate at that α-CH(CH₃), which attacks the other C=O; the ring that closes as a 5- or 6-membered ring decides the product (C=O next to the CH₃-bearing fusion carbon, OH on the other fusion carbon).
(P) [figure] O3/Zn-H2O [figure] aq. NaOH [figure]
A five-membered ring closes, giving a 5 + 7 fused ketol with the second CH₃ in the 7-ring next to C–OH → (2)
(Q) [figure] O3/Zn-H2O [figure] aq. NaOH [figure]
A six-membered ring closes, giving a 6 + 6 ketol with the second CH₃ next to C–OH in the C=O ring → (1)
(R) [figure] O3/Zn-H2O [figure] aq. NaOH [figure]
A six-membered ring closes, giving a 6 + 6 ketol with the second CH₃ next to C–OH in the ring without C=O → (5)
(S) [figure] O3/Zn-H2O [figure] aq. NaOH [figure]
A five-membered ring closes, giving a 5 + 7 ketol with C=O in the 7-ring and the second CH₃ on the 5-ring next to C–OH → (3)
P → 2; Q → 1; R → 5; S → 3
Answer: (C)