JEE Advanced 2026 — Paper 1 — CHEM-II solution

JEE Advanced2026Paper 1CHEM-II

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

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

A
$S$ on warming with ammoniacal AgNO $_{3}$ results in the formation of silver mirror.
B
$Q$ on treatment with Cl $_{2}$ (excess)/ UV gives gammaxane.
C
$T$ is a heterocyclic compound.
D
$R$ on acid catalyzed intermolecular cyclization followed by treatment with Zn-Hg/HCl gives 9,10-dihydroxyanthracene.

Solution

Identify the intermediates.

Kolbe electrolysis of sodium butyrate produces $n$ -hexane (CH $_{3}$ (CH $_{2}$ ) $_{4}$ CH $_{3}$ ). Aromatisation over V $_{2}$ O $_{5}$ at 500 °C and high pressure gives benzene ( $Q$ ).
Friedel–Crafts acylation of benzene with phthalic anhydride / AlCl $_{3}$ produces $o$ -benzoylbenzoic acid ( $R$ ).
$R$ with PCl $_{5}$ converts –COOH to –COCl; Rosenmund reduction (H $_{2}$ /Pd-BaSO $_{4}$ ) converts the acid chloride to an aldehyde ( $S$ , $o$ -benzoylbenzaldehyde).
$S$ + NH $_{2}$ NH $_{2}$ : the aldehyde and the ortho-ketone both condense with hydrazine, and after loss of water the product is the heterocycle 1-phenylphthalazine ( $T$ ).

(A) $S$ contains an aldehyde (–CHO), which gives a positive Tollens' test with ammoniacal AgNO $_{3}$ (silver mirror). Correct.

(B) Benzene + Cl $_{2}$ /UV gives 1,2,3,4,5,6-hexachlorocyclohexane (gammaxane / BHC) via radical addition. Correct.

(C) $T$ (phthalazine derivative) contains a six-membered ring with two ring nitrogens, hence is heterocyclic. Correct.

(D) Acid-catalysed cyclisation of $o$ -benzoylbenzoic acid gives anthraquinone, which on Clemmensen reduction (Zn-Hg/HCl) gives anthracene (not 9,10-dihydroxyanthracene). Incorrect.

Correct answers: (A), (B), (C).

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