JEE Advanced 2026 · Paper 1 · Chemistry Q8

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ChemistryMultiple correct+4 / -1Question 40 of 48

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)

Sodium butanoate, CH₃CH₂CH₂CO₂Na, with (1) Kolbe's electrolysis, (2) V₂O₅, 500 °C, 10-20 atm gives Q; Q with phthalic anhydride and anhydrous AlCl₃ gives R; R with (1) PCl₅, (2) H₂-Pd/BaSO₄ gives S; S with NH₂NH₂ and heat gives T.

S on warming with ammoniacal

\ce{AgNO3}

results in the formation of silver mirror.

Q on treatment with

\ce{Cl2}

(excess)/UV gives gammaxane.

T is a heterocyclic compound.

R on acid catalyzed intramolecular cyclization followed by treatment with Zn-Hg/HCl gives 9,10-dihydroxyanthracene.

Answer and worked solution

Official answer: A, B, C

Sequence.

Kolbe electrolysis of $\ce{CH3CH2CH2CO2Na}$ gives $n$ -hexane. $\ce{V2O5}$ at 500 °C and high pressure aromatizes it: Q = benzene.
Friedel-Crafts acylation with phthalic anhydride: R = 2-benzoylbenzoic acid.
$\ce{PCl5}$ gives the acid chloride, and Rosenmund reduction ( $\ce{H2}$ , Pd/ $\ce{BaSO4}$ ) gives the aldehyde: S = 2-benzoylbenzaldehyde.
Hydrazine condenses with both carbonyls: T = 1-phenylphthalazine.

(A) S has a CHO group and gives a silver mirror with Tollens' reagent. True.

(B) Benzene adds $\ce{Cl2}$ under UV to give $\ce{C6H6Cl6}$ ; its γ-isomer is gammaxane. True.

(C) Phthalazine has two ring nitrogens. True.

(D) Acid cyclization of R gives anthraquinone; Clemmensen reduction removes the C=O groups (towards anthracene / 9,10-dihydroanthracene), not 9,10-dihydroxyanthracene. False.

Topics:Named reactions Aromatic hydrocarbons and electrophilic substitution Alkanes

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JEE Advanced 2026 · Paper 1, Chemistry question 8