Analysis of Organic Reaction Sequence involving Succinic Anhydride and Friedel-Crafts Acylation

To determine the correct statement(s), let us analyze the reaction sequence step-by-step:

Step 1: Identification of Compounds $\mathbf{P}$, $\mathbf{Q}$, $\mathbf{R}$, and $\mathbf{S}$

1. Formation of Compound $\mathbf{P}$: Benzene undergoes Friedel-Crafts acylation with succinic anhydride in the presence of anhydrous $\text{AlCl}_3$ to form $\beta$-benzoylpropionic acid (4-oxo-4-phenylbutanoic acid). $\mathbf{P} = \text{C}_6\text{H}_5-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{COOH}$

2. Formation of Compound $\mathbf{Q}$: Compound $\mathbf{P}$ undergoes Clemmensen reduction using zinc amalgam ($\text{Zn/Hg}$) and concentrated $\text{HCl}$. The keto carbonyl group ($-\text{C}(=\text{O})-$) is reduced to a methylene group ($-\text{CH}_2-$), yielding 4-phenylbutanoic acid. $\mathbf{Q} = \text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{COOH}$

3. Formation of Compound $\mathbf{R}$: Compound $\mathbf{Q}$ reacts with thionyl chloride ($\text{SOCl}_2$), converting the carboxylic acid group ($-\text{COOH}$) into an acyl chloride group ($-\text{COCl}$) to form 4-phenylbutanoyl chloride. $\mathbf{R} = \text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{COCl}$

4. Formation of Compound $\mathbf{S}$: Compound $\mathbf{R}$ undergoes intramolecular Friedel-Crafts acylation in the presence of anhydrous $\text{AlCl}_3$ to yield a six-membered cyclic ketone, 1-tetralone ($\alpha$-tetralone, or 3,4-dihydronaphthalen-1(2H)-one). $\mathbf{S} = \text{1-tetralone}$

5. Further Reduction (Hydrocarbon): Compound $\mathbf{S}$ (1-tetralone) is reduced via Clemmensen reduction ($\text{Zn/Hg, HCl}$) to yield tetralin (1,2,3,4-tetrahydronaphthalene), which is a hydrocarbon.

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Step 2: Analysis of the Options

• Option (A):

  • Compound $\mathbf{P}$ ($\text{C}_6\text{H}_5\text{COCH}_2\text{CH}_2\text{COOH}$) contains a carboxylic acid group ($-\text{COOH}$).
  • Compound $\mathbf{Q}$ ($\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CH}_2\text{COOH}$) also contains a carboxylic acid group ($-\text{COOH}$).
  • Statement (A) is CORRECT.

• Option (B):

  • Compound $\mathbf{S}$ is 1-tetralone, a stable ketone containing an aromatic ring. It lacks aliphatic alkenes or alkynes and is not activated sufficiently to decolorize bromine water.
  • Statement (B) is INCORRECT.

• Option (C):

  • Compound $\mathbf{P}$ has a ketone carbonyl group ($\text{C}_6\text{H}_5-\text{CO}-$).
  • Compound $\mathbf{S}$ (1-tetralone) is a cyclic ketone containing a carbonyl group ($>\text{C}=\text{O}$).
  • Ketones react readily with hydroxylamine ($\text{NH}_2\text{OH}$) to yield oximes ($>\text{C}=\text{N}-\text{OH}$).
  • Statement (C) is CORRECT.

• Option (D):

  • Compound $\mathbf{R}$ is an acyl chloride ($\text{R'COCl}$).
  • Acyl chlorides react with dialkylcadmium ($\text{R}_2''\text{Cd}$) to produce ketones ($\text{R'-CO-R''}$). Because dialkylcadmium is less nucleophilic than Grignard reagents, it does not react further with ketones to give tertiary alcohols.
  • Statement (D) is INCORRECT.

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Conclusion

The correct options are A and C.