JEE Advanced 2025 · Paper 2 · Chemistry Q8

ChemistryMultiple correct+4 / -2Question 40 of 48

For the reaction sequence given below, the correct statement(s) is(are)

P is optically active.

S gives Bayer's test.

Q gives effervescence with aq.

\ce{NaHCO3}

R is an alkyne.

Answer and worked solution

Official answer: B, C

The compounds. The lactone is 4-benzyltetrahydropyran-2-one.

P: $\ce{LiAlH4}$ opens it to the diol $\ce{HOCH2CH2CH(CH2Ph)CH2CH2OH}$ (3-benzylpentane-1,5-diol).
Q: Jones oxidation gives 3-benzylpentanedioic acid.
R: Soda-lime decarboxylation removes both $\ce{CO2}$ groups, giving $\ce{PhCH2CH(CH3)2}$ .
S: Dehydrating P with acid gives unsaturated products.

Options.

(A) False. C3 of P carries two identical $\ce{CH2CH2OH}$ arms, so P is achiral and optically inactive.
(B) True. S contains C=C, so it decolourises Baeyer's reagent.
(C) True. Q is a dicarboxylic acid, so it releases $\ce{CO2}$ with $\ce{NaHCO3}$ .
(D) False. R is an alkane (isobutylbenzene), not an alkyne.

JEE Advanced 2025 · Paper 2, Chemistry question 8