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: Identify the starting compound.
The given aromatic compound is styrene, that is, vinyl benzene \(\mathrm{C_6H_5CH{=}CH_2}\). It contains a benzene ring attached to an unsaturated side chain. Side-chain oxidation of alkyl or alkenyl benzene with strong oxidising agents usually converts the entire side chain into a carboxyl group attached to the benzene ring. Step 2: Action of \(\mathrm{KMnO_4/KOH}\).
When styrene is treated with alkaline potassium permanganate \(\mathrm{KMnO_4/KOH}\), oxidation takes place. Under alkaline conditions, the product obtained first is the potassium salt of benzoic acid, that is, potassium benzoate \(\mathrm{C_6H_5COOK}\). Therefore, \(P\) is potassium benzoate. Step 3: Conversion of \(P\) into \(Q\).
On further treatment, the potassium salt gets converted into the free acid by acidification. Thus, potassium benzoate changes into benzoic acid \(\mathrm{C_6H_5COOH}\). Hence, \(Q\) is benzoic acid. Step 4: Analysis of options.
(A) \(\mathrm{C_6H_5COOH}\) ; \(\mathrm{C_6H_5COOK}\): Incorrect. The order is reversed.
(B) \(\mathrm{C_6H_5COOK}\) ; \(\mathrm{C_6H_5COOH}\): Correct. Oxidation in alkaline medium gives potassium benzoate first, then acidification gives benzoic acid.
(C) \(\mathrm{C_6H_5COOH}\) ; \(\mathrm{C_6H_5CHO}\): Incorrect. Benzaldehyde is not formed in this oxidation sequence.
(D) \(\mathrm{C_6H_5COOK}\) ; \(\mathrm{C_6H_5CHO}\): Incorrect. The second product is benzoic acid, not benzaldehyde.
Step 5: Conclusion.
Therefore, \(P\) is potassium benzoate and \(Q\) is benzoic acid. Final Answer:\(\mathrm{C_6H_5COOK}\) and \(\mathrm{C_6H_5COOH}\).
