JEE Advanced 2025 — Paper 2 — CHEM-II solution

JEE Advanced2025Paper 2CHEM-II

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

A
P is optically active.
B
S gives Bayer's test.
C
Q gives effervescence with aq. NaHCO $_{3}$ .
D
R is an alkyne.

Solution

The starting $δ$ -lactone (six-membered ring) bearing a -CH $_{2}$ -Ph group at C-4 (chirality centre) is reduced by LiAlH $_{4}$ to a diol, since lactone $\to$ diol. The product P is Ph-CH $_{2}$ -CH(-)-CH $_{2}$ -CH $_{2}$ -CH $_{2}$ -OH with an additional -CH $_{2}$ -OH (a 1,5-diol with the benzyl branch on C-2 of the chain).

(A) Although the starting lactone was chiral at the carbon bearing -CH $_{2}$ Ph, in P that same carbon now has two identical chains terminating in -CH $_{2}$ OH on either side (after the lactone ring opens). The stereocentre is lost, so P is not optically active. Incorrect.

(B) Dehydration of P with H $_{2}$ SO $_{4}$ at 443 K eliminates water from both alcohol groups, producing a diene S. A diene decolourises cold dilute alkaline KMnO $_{4}$ (Baeyer's test). Correct.

(C) Oxidation of P with CrO $_{3}$ -H $_{2}$ SO $_{4}$ (Jones oxidation) converts both primary alcohols to carboxylic acids, giving a dicarboxylic acid Q. Carboxylic acids give effervescence (CO $_{2}$ ) with aq. NaHCO $_{3}$ . Correct.

(D) Decarboxylation of the dicarboxylic acid Q with NaOH/CaO (soda-lime) removes both -COOH groups as CO $_{2}$ , leaving an alkane (specifically isopropylbenzene-type / cumene-like alkylbenzene, not an alkyne). Incorrect.

$∴$ Correct options are (B) and (C).

← Back to full paper

Practice more CHEM-II

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →