Identify Major Organic Products Q R and S in Multi Step Reaction Sequence

To determine the correct structures of the major organic products $\mathbf{Q}$, $\mathbf{R}$, and $\mathbf{S}$, we analyze the multi-step reaction sequence step by step:

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Step 1: Formation of Product $\mathbf{Q}$

• Reaction: Compound $\mathbf{P}$ is subjected to Friedel-Crafts acylation using succinic anhydride in the presence of Lewis acid $\text{AlCl}_3$.
• Regioselectivity: Compound $\mathbf{P}$ contains an aromatic ring substituted with a bulky tert-butyl group $-\text{C}(\text{CH}_3)_3$ and a fused five-membered ring having a gem-dimethyl group $-\text{C}(\text{CH}_3)_2-$ at one benzylic position and a methylene group $-\text{CH}_2-$ at the other. Electrophilic aromatic substitution occurs at the position that is least sterically hindered (ortho to the less bulky $-\text{CH}_2-$ group of the indan ring).
• Product: The electrophilic attack of succinic anhydride yields a keto-acid: $\mathbf{Q} = \text{Ar}-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{COOH}$ This matches the structure shown in Option (B).

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Step 2: Formation of Product $\mathbf{R}$

• Step (i) - Clemmensen Reduction ($\text{Zn-Hg / HCl}$): The keto-acid $\mathbf{Q}$ undergoes reduction of the aryl ketone group ($-\text{C}(=\text{O})-$) to a methylene group ($-\text{CH}_2-$), giving the 4-arylbutanoic acid intermediate: $\text{Ar}-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{COOH} \xrightarrow{\text{Zn-Hg / HCl}} \text{Ar}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{COOH}$

• Step (ii) - Intramolecular Friedel-Crafts Acylation ($\text{H}_3\text{PO}_4$): Heating with phosphoric acid ($\text{H}_3\text{PO}_4$) promotes intramolecular cyclization of the carboxylic acid onto the adjacent open position of the benzene ring, forming a new six-membered ring fused to the aromatic core. The carbonyl group formed is located at the benzylic position: $\text{Ar}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{COOH} \xrightarrow{\text{H}_3\text{PO}_4} \mathbf{R} \quad (\text{a tetralone derivative})$ This matches the structure shown in Option (C).

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Step 3: Formation of Product $\mathbf{S}$

• Step (i) & (ii) - Grignard Addition and Acidic Workup ($\text{CH}_3\text{MgBr}$, then $\text{H}_3\text{O}^+$): Methylmagnesium bromide ($\text{CH}_3\text{MgBr}$) undergoes nucleophilic addition to the carbonyl group of tetralone $\mathbf{R}$. Subsequent acidic workup yields a tertiary benzylic alcohol bearing a methyl group at the reaction site: $\text{Ketone } (\mathbf{R}) \xrightarrow{\text{(i) }\text{CH}_3\text{MgBr}, \text{(ii) }\text{H}_3\text{O}^+} \text{Tertiary Benzylic Alcohol}$

• Step (iii) - Acid-Catalyzed Dehydration ($\text{H}_2\text{SO}_4 / \Delta$): Protonation of the tertiary alcohol followed by loss of a water molecule generates a stable benzylic carbocation. Elimination of a proton gives the thermodynamically favored, highly stable endocyclic alkene conjugated with the aromatic ring, which retains the methyl group attached to the double bond: $\text{Tertiary Alcohol} \xrightarrow{\text{H}_2\text{SO}_4 / \Delta} \mathbf{S} \quad (\text{conjugated endocyclic alkene with a methyl group})$

This matches the structure shown in Option (A), while Option (D) is incorrect because it omits the methyl group introduced by the Grignard reagent.

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Conclusion:

The correct options are A, B, and C.