"P" is hydrocarbon of molecular formula : $\mathrm{C}_{8} \mathrm{H}_{14}$. On ozonolysis, "P" forms "Q". "Q" on treatment with alkali under reflux condition produces. "R", which on treatment with $\mathrm{I}_{2} / \mathrm{NaOH}$ gives a yellow precipitate. Acidification of the solution gives "S". The structure of " S " given below : (see structure) The correct structure of " P " is

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Hydrocarbons

"P" is hydrocarbon of molecular formula :

C_{8} H_{14}

. On ozonolysis, "P" forms "Q". "Q" on treatment with alkali under reflux condition produces. "R", which on treatment with

I_{2} / NaOH

gives a yellow precipitate. Acidification of the solution gives "S". The structure of " S " given below :

(see structure)

The correct structure of " P " is

(A)(see structure)
(B)(see structure)
(C)(see structure)
(D)(see structure)
✓ Correct

Answer: (D)

Solution

The molecular formula

C_{8} H_{14}

carries two degrees of unsaturation, met by one ring and one double bond, so P is a cyclic alkene. Taking P as 1,2-dimethylcyclohex-1-ene, ozonolysis cleaves the ring double bond to open the ring and furnish the diketone Q, octane-2,7-dione,

C H_{3} COC H_{2} C H_{2} C H_{2} C H_{2} COC H_{3}

. Warming with alkali drives an intramolecular aldol condensation that closes a five-membered ring, the carbon

α

to one carbonyl joining the other carbonyl carbon and losing water to give R, 1-acetyl-2-methylcyclopent-1-ene; its methyl ketone group answers the

I_{2} /NaOH

test with a yellow iodoform precipitate. Acidification after that test replaces the

C H_{3} CO -

group by

- COOH

, giving S, 2-methylcyclopent-1-ene-1-carboxylic acid, exactly the structure shown. The structure of P is 1,2-dimethylcyclohex-1-ene.

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Hydrocarbons

Solution

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