Question 8

Step 1: Kolbe's electrolysis of sodium butyrate ($CH_3CH_2CH_2COONa$) involves the dimerization of the propyl radical to form $n-hexane$.

Step 2: $n-hexane$ undergoes aromatization in the presence of $V_2O_5$ at $500 ^\circ C$ to form Benzene ($Q$).

Step 3: Friedel-Crafts acylation of Benzene with phthalic anhydride using anhydrous $AlCl_3$ produces $o-benzoylbenzoic$ acid ($R$).

Step 4: $R$ reacts with $PCl_5$ to convert the carboxylic acid to an acid chloride. Subsequent Rosenmund reduction ($H_2-Pd/BaSO_4$) reduces the acid chloride specifically to an aldehyde group, yielding $2-benzoylbenzaldehyde$ ($S$).

Step 5: $S$ (a $1,4$-dicarbonyl equivalent system) reacts with hydrazine ($NH_2NH_2$) and heat to undergo cyclization, forming $4-phenylphthalazine$ ($T$).

Analysis of Options:

(A) $S$ ($2-benzoylbenzaldehyde$) contains an aldehyde group, which reacts with Tollen's reagent (ammoniacal $AgNO_3$) to form a silver mirror. Correct.

(B) Benzene ($Q$) reacts with $Cl_2$ in UV light via addition to form benzene hexachloride ($C_6H_6Cl_6$), commonly known as gammaxane. Correct.

(C) $T$ is $4-phenylphthalazine$, which contains a pyridazine ring fused to a benzene ring. Since the ring contains nitrogen atoms, it is a heterocyclic compound. Correct.

(D) $R$ ($o-benzoylbenzoic$ acid) cyclizes to anthraquinone under acidic conditions. Clemmensen reduction ($Zn-Hg/HCl$) of anthraquinone typically yields anthracene, not $9,10-dihydroxyanthracene$. Incorrect.