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
Both X and Y are oxygen containing compounds.
B
Y on heating with CHCl $_{3}$ /KOH forms isocyanide.
C
Z reacts with Hinsberg's reagent.
D
Z is an aromatic primary amine.

Solution

Trace the sequence:

1,2,3,4-Tetrahydronaphthalene is oxidised by hot acidic KMnO $_{4}$ at the benzylic ring carbons to give phthalic acid (benzene-1,2-dicarboxylic acid). With NH $_{3}$ and heat (loss of 2 H $_{2}$ O), this becomes phthalimide. Hence X = phthalimide (contains C=O).

Phthalimide treated with strong heat, ethanolic KOH and R-Br gives N-alkyl phthalimide (this is the alkylation step of the Gabriel synthesis). Hence Y = N-alkyl phthalimide (also contains C=O groups).

NaOH hydrolysis of N-alkyl phthalimide releases the disodium phthalate (aromatic compound) and a primary amine R-NH $_{2}$ as Z.

Now assess the statements:

(A) Both X (phthalimide) and Y (N-alkyl phthalimide) contain oxygen. Correct.

(B) The carbylamine reaction (CHCl $_{3}$ /KOH giving isocyanide) requires a primary amine. Y has no N-H, so it does not undergo the carbylamine test. Incorrect.

(C) Z is a primary amine and reacts with Hinsberg's reagent (PhSO $_{2}$ Cl) to give an alkali-soluble sulphonamide. Correct.

(D) Z = R-NH $_{2}$ is a primary alkyl amine (R came from R-Br), not an aromatic amine. Incorrect.

$∴$ Correct options are (A) and (C).

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