JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

Kolbe's electrolysis of sodium butyrate ( CH _3 CH _2 CH _2 COONa ) yields n-hexane. Aromatization of n-hexane using V _2 O _5 at 500^ C produces benzene (Q). Benzene (Q) reacts with phthalic anhydride in the presence of anhydrous AlCl _3 to form o-benzoylbenzoic acid (R). Treatment of R with PCl _5 gives o-benzoylbenzoyl chloride, which upon Rosenmund reduction ( H _2- Pd/BaSO _4) yields o-benzoylbenzaldehyde (S). Condensation of S with hydrazine ( NH _2 NH _2) forms 1-phenylphthalazine (T). Evaluating the options: (A) S contains an aldehyde group and gives a positive Tollens' test (silver mirror). (B) Q (benzene) reacts with excess Cl _2 under UV light to form benzene hexachloride (gammaxane). (C) T (1-phenylphthalazine) is a heterocyclic compound. (D) Acid-catalyzed cyclization of R yields anthraquinone, which on Clemmensen reduction (Zn-Hg/HCl) gives anthracene, not 9,10-dihydroxyanthracene. Answer: S on warming with ammoniacal AgNO _3 results in the formation of silver mirror.; Q on treatment with Cl _2(excess)/UV gives gammaxane.; T is a heterocyclic compound.