Any benzene side chain containing at least one benzylic hydrogen usually gets oxidised by strong oxidising agents like \(\mathrm{KMnO_4}\) into the carboxylic acid group.
Step 1: Understanding the Concept:
Strong oxidizing agents like alkaline $\mathrm{KMnO_4}$ oxidize any alkyl side chain on a benzene ring completely to a carboxylic acid group. Step 2: Formula Application:
Step 1: $\mathrm{C_6H_5CH_3} \xrightarrow{\text{alk. } \mathrm{KMnO_4}} \mathrm{C_6H_5COOK}$ ($P$).
Step 2: $\mathrm{C_6H_5COOK} \xrightarrow{\mathrm{H_3O^+}} \mathrm{C_6H_5COOH}$ ($Q$). Step 3: Explanation:
In alkaline medium, the acid formed immediately reacts with the base to form the salt (Potassium benzoate). Acidification is required in the second step to release the free Benzoic acid. Step 4: Final Answer:
P is potassium benzoate and Q is benzoic acid.
