Reaction Sequence Involving Kolbe Electrolysis and Phthalic Anhydride Derivatives

To determine the correct statements, let us analyze the given reaction sequence step-by-step:

Step 1: Identification of Compound Q

1. Kolbe's Electrolysis: The starting reactant is sodium butyrate, $\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa}$. Electrolysis of this sodium carboxylate salt results in decarboxylative dimerization: $2\,\text{CH}_3\text{CH}_2\text{CH}_2\text{COO}^- \xrightarrow{-2e^-,\, -2\text{CO}_2} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \quad (\text{n-hexane})$

2. Aromatization: Heating $n$-hexane with vanadium pentoxide ($\text{V}_2\text{O}_5$) at $500\,^\circ\text{C}$ and $10\text{--}20\text{ atm}$ causes catalytic dehydrocyclization (reforming) to yield benzene: $\text{CH}_3(\text{CH}_2)_4\text{CH}_3 \xrightarrow[\text{10--20 atm}]{\text{V}_2\text{O}_5,\, 500\,^\circ\text{C}} \text{C}_6\text{H}_6 \quad (\mathbf{Q})$

• Evaluation of Option (B): Benzene ($\mathbf{Q}$) on treatment with excess $\text{Cl}_2$ under UV light undergoes free-radical addition to give benzene hexachloride ($\text{C}_6\text{H}_6\text{Cl}_6$ or BHC). The $\gamma$-isomer of BHC is commonly known as gammaxane (lindane). $\text{C}_6\text{H}_6 + 3\text{Cl}_2 \xrightarrow{h\nu} \text{C}_6\text{H}_6\text{Cl}_6 \quad (\text{Gammaxane})$ Therefore, Statement (B) is correct.

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Step 2: Identification of Compound R

Benzene ($\mathbf{Q}$) reacts with phthalic anhydride in the presence of anhydrous $\text{AlCl}_3$ via a Friedel-Crafts acylation reaction: $\text{C}_6\text{H}_6 + \text{Phthalic Anhydride} \xrightarrow{\text{anhyd. }\text{AlCl}_3} o\text{-C}_6\text{H}_4(\text{COOH})(\text{CO-C}_6\text{H}_5) \quad (\mathbf{R})$ Thus, $\mathbf{R}$ is 2-benzoylbenzoic acid.

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Step 3: Identification of Compound S

1. Reaction of $\mathbf{R}$ (2-benzoylbenzoic acid) with $\text{PCl}_5$ converts the carboxylic acid group ($-\text{COOH}$) into an acyl chloride group ($-\text{COCl}$), forming $2\text{-benzoylbenzoyl chloride}$.
2. Subsequent treatment with $\text{H}_2/\text{Pd-BaSO}_4$ (Rosenmund reduction) selectively reduces the acyl chloride group ($-\text{COCl}$) to an aldehyde group ($-\text{CHO}$): $o\text{-C}_6\text{H}_4(\text{COOH})(\text{CO-C}_6\text{H}_5) \xrightarrow{\text{PCl}_5} o\text{-C}_6\text{H}_4(\text{COCl})(\text{CO-C}_6\text{H}_5) \xrightarrow{\text{H}_2\text{-Pd/BaSO}_4} o\text{-C}_6\text{H}_4(\text{CHO})(\text{CO-C}_6\text{H}_5) \quad (\mathbf{S})$ Thus, $\mathbf{S}$ is 2-benzoylbenzaldehyde.

• Evaluation of Option (A): Since $\mathbf{S}$ contains an aldehyde group ($-\text{CHO}$), warming it with ammoniacal $\text{AgNO}_3$ (Tollen's reagent) reduces the $\text{Ag}^+$ ions to metallic silver, forming a silver mirror. Therefore, Statement (A) is correct.

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Step 4: Identification of Compound T

Reaction of $2\text{-benzoylbenzaldehyde}$ ($\mathbf{S}$) with hydrazine ($\text{NH}_2\text{NH}_2$) upon heating results in condensation with both carbonyl groups (the aldehyde and the ketone) to form a fused six-membered ring containing two adjacent nitrogen atoms: $o\text{-C}_6\text{H}_4(\text{CHO})(\text{CO-C}_6\text{H}_5) + \text{NH}_2\text{NH}_2 \xrightarrow{\Delta} \text{1-phenylphthalazine} \quad (\mathbf{T})$

• Evaluation of Option (C): 1-phenylphthalazine ($\mathbf{T}$) contains a bicyclic aromatic system with nitrogen atoms present in the ring structure. Hence, it is a heterocyclic compound. Therefore, Statement (C) is correct.

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Step 5: Evaluation of Option (D)

• Acid-catalyzed intramolecular cyclization of 2-benzoylbenzoic acid ($\mathbf{R}$) using concentrated $\text{H}_2\text{SO}_4$ yields 9,10-anthraquinone.
• Subsequent treatment of 9,10-anthraquinone with $\text{Zn-Hg/HCl}$ (Clemmensen reduction) reduces both carbonyl groups ($>\text{C}=\text{O}$) completely to methylene groups ($-\text{CH}_2-$), giving anthracene, not 9,10-dihydroxyanthracene. Therefore, Statement (D) is incorrect.

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Conclusion:

The correct statements are A, B, and C.