JEE Advanced · Chemistry · 24. Haloalkanes & Haloarenes
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The reaction of compound \(\mathbf{P}\) with \(\mathrm{CH}_{3} \mathrm{MgBr}\) (excess) in \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) followed by addition of \(\mathrm{H}_{2} \mathrm{O}\) gives \(\mathbf{Q}\). The compound \(\mathbf{Q}\) on treatment with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) at \(0^{\circ} \mathrm{C}\) gives \(\mathbf{R}\). The reaction of \(\mathbf{R}\) with \(\mathrm{CH}_{3} \mathrm{COCl}\) in the presence of anhydrous \(\mathrm{AlCl}_{3}\) in \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) followed by treatment with \(\mathrm{H}_{2} \mathrm{O}\) produces compound \(\mathbf{S}\). [Et in compound \(\mathbf{P}\) is ethyl group]

Question :
The reactions, \(\mathbf{Q}\) to \(\mathbf{R}\) and \(\mathbf{R}\) to \(\mathbf{S}\), are
- A Dehydration and Friedel-Crafts acylation
- B Friedel-Crafts alkylation, dehydration, and Friedel-Crafts acylation
- C Aromatic sulfonation and Friedel-Crafts acylation
- D Friedel-Crafts alkylation and Friedel-Crafts acylation
Answer & Solution
Correct Answer
(D) Friedel-Crafts alkylation and Friedel-Crafts acylation
Step-by-step Solution
Detailed explanation
The reaction of an ester (other than format) with Grignard’s reagent followed by hydrolysis in acidic medium produces a tertiary alcohol.

The dehydration of -alcohol at in the given compound produces an alkyl group in the benzene ring. This compound on reaction with in and shows Friedel craft's alkylation reaction

The dehydration of -alcohol at in the given compound produces an alkyl group in the benzene ring. This compound on reaction with in and shows Friedel craft's alkylation reaction
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