Question:
Identify the product obtained when phenyl ethene is oxidised using potassium permanganate in dil \(\text{H}_{2}\text{SO}_{4}\).
Identify the product obtained when phenyl ethene is oxidised using potassium permanganate in dil \(\text{H}_{2}\text{SO}_{4}\).
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Hot acidic KMnO4 cleaves the C=C bond, and the benzylic carbon becomes a carboxylic acid group.
Updated On: Nov 12, 2026
- Phthalic acid
- p-ethenyl benzoic acid
- Benzoic acid
- Salicylic acid
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Phenyl ethene (styrene) is \(\text{C}_{6}\text{H}_{5}\text{CH}=\text{CH}_{2}\). Acidic \(\text{KMnO}_{4}\) cleaves the double bond.
Step 2: Oxidation
The ring side carbon is oxidised to \(-\text{COOH}\), and the terminal \(\text{CH}_{2}\) is oxidised to \(\text{CO}_{2}\) and water.
Step 3: Product
The ring keeps one \(-\text{COOH}\) group, giving \(\text{C}_{6}\text{H}_{5}\text{COOH}\), benzoic acid. This is option (C).
Final Answer:
The product is benzoic acid. \[ \boxed{\text{(C) }\text{Benzoic acid}} \]
Phenyl ethene (styrene) is \(\text{C}_{6}\text{H}_{5}\text{CH}=\text{CH}_{2}\). Acidic \(\text{KMnO}_{4}\) cleaves the double bond.
Step 2: Oxidation
The ring side carbon is oxidised to \(-\text{COOH}\), and the terminal \(\text{CH}_{2}\) is oxidised to \(\text{CO}_{2}\) and water.
Step 3: Product
The ring keeps one \(-\text{COOH}\) group, giving \(\text{C}_{6}\text{H}_{5}\text{COOH}\), benzoic acid. This is option (C).
Final Answer:
The product is benzoic acid. \[ \boxed{\text{(C) }\text{Benzoic acid}} \]
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