The product ‘Y’ formed in the above reaction is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → benzoic acid
When an alkylbenzene is heated with alkaline KMnO4, the whole side chain, whatever its length, is oxidised to a –\( \ce{COO^{-}} \) group attached to the ring, as long as the carbon next to the ring carries a hydrogen. Acidification then gives benzoic acid:
\( \displaystyle \ce{C6H5CH2CH2CH3} \xrightarrow[\text{heat}]{\ce{KMnO4}-\ce{KOH}} \ce{C6H5COOK}\ (\text{X}) \)
\( \displaystyle \ce{C6H5COOK} \xrightarrow{\ce{H3O+}} \ce{C6H5COOH}\ (\text{Y}) \)
Why the other options are incorrect
Option (1) 3-phenylpropanoic acid — Incorrect. Only the end carbon would have been oxidised; \( \ce{KMnO4} \) oxidises the side chain back to the ring.
Option (2) 3-phenylpropan-1-ol — Incorrect. \( \ce{KMnO4} \) is an oxidising agent; it does not stop at an alcohol.
Option (4) 4-propylbenzoic acid — Incorrect. This would need a new –COOH group on the ring, while the propyl chain is left untouched; the reverse actually happens.
NCERT (2026-27 edition):
From Alkylbenzenes
