Identify Products V and W in Organic Reaction Sequence

To determine the identity of products $V$ and $W$ in the given reaction sequence, let us analyze each chemical transformation step-by-step:

Step 1: Formation of Product $V$

Compound $Q$ (maleic anhydride) contains a carbon-carbon double bond between two carbonyl carbons. $\text{Maleic Anhydride } (Q) \xrightarrow[\Delta]{\text{H}_2 / \text{Ni}} \text{Succinic Anhydride } (V)$ Catalytic hydrogenation ($\text{H}_2/\text{Ni}, \Delta$) selectively reduces the alkene double bond ($-\text{CH}=\text{CH}-$) to an alkane single bond ($-\text{CH}_2-\text{CH}_2-$) without affecting the cyclic anhydride group. Therefore, product $V$ is succinic anhydride.

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Step 2: Intermolecular Friedel-Crafts Acylation

When benzene reacts with succinic anhydride ($V$) in the presence of anhydrous $\text{AlCl}_3$, an intermolecular Friedel-Crafts acylation reaction occurs: $\text{Benzene} + \text{Succinic Anhydride } (V) \xrightarrow{\text{AlCl}_3 \text{ (anhydrous)}} \text{4-oxo-4-phenylbutanoic acid}$ $\text{C}_6\text{H}_6 + \text{C}_4\text{H}_4\text{O}_3 \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{COOH}$

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Step 3: Clemmensen Reduction

The intermediate 4-oxo-4-phenylbutanoic acid undergoes a Clemmensen reduction using $\text{Zn-Hg/HCl}$: $\text{C}_6\text{H}_5-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{COOH} \xrightarrow{\text{Zn-Hg/HCl}} \text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{COOH}$ The aromatic ketone ($\text{C=O}$) group is reduced completely to a methylene ($\text{-CH}_2\text{-}$) group, forming 4-phenylbutanoic acid.

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Step 4: Intramolecular Friedel-Crafts Acylation to Form $W$

Heating 4-phenylbutanoic acid with phosphoric acid ($\text{H}_3\text{PO}_4$) acts as a dehydrating acid catalyst, driving an intramolecular cyclization (Friedel-Crafts acylation): $\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{COOH} \xrightarrow{\text{H}_3\text{PO}_4} \text{1-tetralone } (W) + \text{H}_2\text{O}$ This forms a fused six-membered ring containing a carbonyl group, known as 1-tetralone ($\alpha$-tetralone).

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Conclusion:

• Product $V$: Succinic anhydride
• Product $W$: 1-tetralone ($\alpha$-tetralone)

This sequence corresponds to Option A.