Identify Product K in Multistep Reaction Sequence

To identify compound $\mathbf{K}$, we analyze the complete multi-step reaction sequence step-by-step:

1. Identification of Intermediate J and Compound I

• Perkin Reaction: The reaction of compound $\mathbf{I}$ with acetic anhydride $(\text{CH}_3\text{CO})_2\text{O}$ in the presence of sodium acetate ($\text{CH}_3\text{COONa}$) is a classic Perkin condensation.
• Properties of J ($\text{C}_9\text{H}_8\text{O}_2$):
  • Effervescence with $\text{NaHCO}_3$ confirms the presence of a carboxylic acid group ($-\text{COOH}$).
  • A positive Baeyer's test (decolorization of cold, dilute alkaline $\text{KMnO}_4$) confirms the presence of an aliphatic carbon-carbon double bond ($\text{C}=\text{C}$).
• Combining these facts, compound $\mathbf{I}$ is benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$), and compound $\mathbf{J}$ is cinnamic acid: $\text{C}_6\text{H}_5\text{CHO} + (\text{CH}_3\text{CO})_2\text{O} \xrightarrow{\text{CH}_3\text{COONa}} \underset{\mathbf{J}\ (\text{C}_9\text{H}_8\text{O}_2)}{\text{C}_6\text{H}_5\text{CH}=\text{CH}-\text{COOH}} + \text{CH}_3\text{COOH}$

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2. Conversion of Intermediate J to Compound K

• Step (i): Reduction with $\text{H}_2, \text{Pd/C}$ Catalytic hydrogenation selectively reduces the double bond of cinnamic acid without reducing the carboxylic acid group, forming 3-phenylpropanoic acid: $\text{C}_6\text{H}_5\text{CH}=\text{CH}-\text{COOH} \xrightarrow{\text{H}_2, \text{Pd/C}} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{COOH}$

• Step (ii): Chlorination with $\text{SOCl}_2$ Treatment of 3-phenylpropanoic acid with thionyl chloride converts the carboxylic acid group into an acyl chloride, giving 3-phenylpropanoyl chloride: $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{COOH} \xrightarrow{\text{SOCl}_2} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{COCl} + \text{SO}_2 \uparrow + \text{HCl} \uparrow$

• Step (iii): Intramolecular Friedel-Crafts Acylation with anhydrous $\text{AlCl}_3$ In the presence of anhydrous $\text{AlCl}_3$, 3-phenylpropanoyl chloride forms an acylium ion intermediate which undergoes an intramolecular electrophilic aromatic substitution on the adjacent position of the benzene ring. This ring closure forms a stable 5-membered cyclic ketone, 1-indanone: $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{COCl} \xrightarrow{\text{anhyd. AlCl}_3} \underset{\mathbf{K}\text{ (1-indanone)}}{\text{C}_9\text{H}_8\text{O}}$

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Conclusion

Compound $\mathbf{K}$ is 1-indanone, which corresponds to structure (C).

Correct Option: C