JEE Advanced 2023 Paper 1 · Q07 · Aldehydes & Ketones
In the given reaction scheme, $P$ is a phenyl alkyl ether, $Q$ is an aromatic compound; $R$ and $S$ are the major products. $P \xrightarrow{HI} Q \xrightarrow{\text{(i) NaOH (ii) }CO_2\text{ (iii) }H_3O^+} R \xrightarrow{\text{(i) }(CH_3CO)_2O\text{ (ii) }H_3O^+} S$. The correct statement about $S$ is
Reveal answer + step-by-step solution
Correct answer:B
Solution
$P$ is a phenyl alkyl ether (e.g. anisole). $HI$ cleaves the ether to give phenol $Q$ and an alkyl iodide. Phenol $\xrightarrow{\text{NaOH, }CO_2,\text{ then }H_3O^+}$ salicylic acid (Kolbe–Schmitt) = $R$. Salicylic acid + acetic anhydride → aspirin (acetylsalicylic acid) = $S$. Aspirin is an NSAID; it inhibits cyclooxygenase (COX), suppressing prostaglandin synthesis — (B).
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