JEE Advanced 2026 — Paper 2 — CHEM-II solution

JEE Advanced2026Paper 2CHEM-II

In the following reaction sequence, $P, Q, S$ and $T$ are the major products.

The correct statement(s) about $P, Q, S$ and $T$ is (are)

A
(A) $Q$ on treatment with ethanol generates an aromatic aldehyde.
B
(B) $S$ gives positive phthalein dye test.
C
(C) $P$ is a dinitro compound.
D
(D) $T$ is a coloured compound.

Solution

Identify each product.

Aniline first reacts with acetic anhydride/pyridine to give acetanilide (the -NHCOCH $_{3}$ group is a milder ortho/para director than -NH $_{2}$ and protects the amine from oxidation). Nitration with conc. HNO $_{3}$ /H $_{2}$ SO $_{4}$ at 288 K then introduces one NO $_{2}$ predominantly at the para position. So $P$ = p-nitroacetanilide (a mono-nitro compound).

Acid hydrolysis (H $_{3}$ O $^{+}$ ) removes the acetyl group to give p-nitroaniline; treatment with NaNO $_{2}$ /HCl at 273-278 K diazotises the amine. So $Q$ = p-nitrobenzenediazonium chloride.

Cumene undergoes the cumene process (autoxidation with O $_{2}$ then acid cleavage) to give phenol. Reaction of phenol with NaOH and CO $_{2}$ (Kolbe-Schmitt) followed by acid workup gives salicylic acid (2-hydroxybenzoic acid). So $S$ = salicylic acid.

$Q$ then couples with $S$ in aqueous NaOH (the phenolate activates ortho/para to OH) to give an azo dye: p-nitrophenyl-azo-salicylic acid, which is coloured.

(A) Reaction of an aryl diazonium salt with ethanol typically gives the arene (replacement of N $_{2}^{+}$ by H) or, under different conditions, the aryl ethyl ether, not an aromatic aldehyde. Incorrect.

(B) Salicylic acid is a phenol derivative and gives a positive phthalein dye test (condensation with phthalic anhydride to form a phthalein dye). Correct.

(C) $P$ is mono-nitro (p-nitroacetanilide), not dinitro. Incorrect.

(D) $T$ is an azo dye containing the -N=N- chromophore and is intensely coloured. Correct.

$∴$ Correct options are (B) and (D).

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