"P" is a hydrocarbon of molecular formula:- C₈H₁₄. On ozonolysis, "P" forms "Q". "Q" on treatment with alkali under reflux condition produces "R", which on treatment with I₂/NaOH gives a yellow precipitate. Acidification of the solution gives "S". The structure of "S" is given below:-
The correct structure of "P" is

Solution

Chemistry ma’am · 4 steps

Work backwards: aldol builds the ring in S, so Q is a 1,6-diketone cut open from a cyclohexene.

1. Undo the iodoform. The yellow precipitate is CHI₃, so R has a CH₃CO⁻ group, and I₂/NaOH turns it into -COOH. Put the CH₃ back on the COOH of S: R is 1-acetyl-2-methylcyclopentene, C₈H₁₂O.
2. Undo the aldol. The ring C=C in R comes from aldol condensation, which loses one H₂O. Open that bond and add the water back: Q is octane-2,7-dione, CH₃CO(CH₂)₄COCH₃, C₈H₁₄O₂. Its two C=O groups are in a 1,6 relationship, and the enolate of the CH₂ next to one C=O attacks the other C=O to close a five-membered ring.

   CH₃CO(CH₂)₄COCH₃ →(OH⁻, Δ) 1-acetyl-2-methylcyclopentene →(I₂/NaOH, H₃O⁺) S + CHI₃

3. Undo the ozonolysis. Both carbonyl groups sit in one chain, so they came from a single C=C inside a ring. Join the two carbonyl carbons with a double bond. This gives a six-membered ring with a methyl group on each alkene carbon: 1,2-dimethylcyclohexene, C₈H₁₄, which matches P.
4. So. P is 1,2-dimethylcyclohexene, option (d).

The idea to keep. Ozonolysis of a ring alkene gives one open-chain dicarbonyl. A 1,6-dicarbonyl closes by aldol to a five-membered ring, and a 1,5-dicarbonyl closes to a six-membered ring.

Answer from the NTA final answer key. Solution by PrepWiser.