JEE Main 2026 · 5 Apr · Shift 1 · Chemistry Q64

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ChemistrySingle correct+4 / -1Question 64 of 75

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

The correct structure of "P" is

S: 2-methylcyclopent-1-ene-1-carboxylic acid (cyclopentene with CH₃ and COOH on the two alkene carbons).

3,6-Dimethylcyclohexene: CH₃ groups on the two ring carbons flanking the C=C.

1-Ethylcyclohexene: an ethyl group on one alkene carbon of cyclohexene.

1-Ethyl-2-methylcyclopentene: ethyl and methyl on the two alkene carbons.

1,2-Dimethylcyclohexene: methyl groups on both alkene carbons of the ring.

Answer and worked solution

Official answer: D

Work backwards from S (2-methylcyclopent-1-ene-1-carboxylic acid).

R. The iodoform reaction turns a methyl ketone $\ce{-COCH3}$ into $\ce{-COOH}$ . So R is 1-acetyl-2-methylcyclopentene.
Q. R comes from an intramolecular aldol condensation (alkali, reflux) of octane-2,7-dione, $\ce{CH3CO(CH2)4COCH3}$ . The enolate at C3 attacks C7, closing a five-membered ring, and dehydration gives R.
P. Ozonolysis giving this single dione means P is a ring alkene with a methyl on each alkene carbon: 1,2-dimethylcyclohexene ( $\ce{C8H14}$ ).

(D)

Topics:Aldol and related condensations Alkenes Named reactions

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JEE Main 2026 · 5 Apr · Shift 1, Chemistry question 64