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

Kolbe's electrolysis of sodium butyrate ($\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa}$) yields n-hexane.

Aromatization of n-hexane using $\text{V}_2\text{O}_5$ at $500^\circ\text{C}$ produces benzene ($Q$).

Benzene ($Q$) reacts with phthalic anhydride in the presence of anhydrous $\text{AlCl}_3$ to form o-benzoylbenzoic acid ($R$).

Treatment of $R$ with $\text{PCl}_5$ gives o-benzoylbenzoyl chloride, which upon Rosenmund reduction ($\text{H}_2-\text{Pd/BaSO}_4$) yields o-benzoylbenzaldehyde ($S$).

Condensation of $S$ with hydrazine ($\text{NH}_2\text{NH}_2$) 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 $\text{Cl}_2$ 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: $\textbf{S}$ on warming with ammoniacal $\text{AgNO}_3$ results in the formation of silver mirror.; $\textbf{Q}$ on treatment with $\text{Cl}_2$(excess)/UV gives gammaxane.; $\textbf{T}$ is a heterocyclic compound.