The compound (CH3)2C=CH-CO-CH3 (4-methylpent-3-en-2-one) can be oxidised to (CH3)2C=CH-COOH (4-methylpent-3-enoic acid) by which of the following reagents?

Question

Accepted Answer

KMnO4. The starting material is an alpha,beta-unsaturated ketone with a CH3CO group. Conversion to the COOH group (same carbon count, just -CO-CH3 -> -COOH losing one carbon... actually: (CH3)2C=CH-CO-CH3 has C6; (CH3)2C=CH-COOH has C5 carbon chain - so a CH3 group is removed). This is a haloform-like oxidation or oxidative cleavage. NaOI (hypoiodite) would do haloform reaction on the methyl ketone giving COOH + CHI3, but would give (CH3)2C=CH-COO⁻ (carboxylate). Actually NaOI cleaves CH3CO- group: gives (CH3)2C=CH-COOH and CHI3. But this would work too. However, KMnO4 is the more standard answer for direct oxidation of the ketone to acid. The most correct answer for oxidation preserving the

The compound (CH3)2C=CH-CO-CH3 (4-methylpent-3-en-2-one) can be oxidised to (CH3)2C=CH-COOH (4-methylpent-3-enoic acid) by which of the following reagents?

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