Ph-CH=CH-C(=O)CH_3 to Ph-CH=CH-CO_2HThe above | vedclass

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(B) The conversion of a methyl ketone $(R-CO-CH_3)$ to a carboxylic acid $(R-COOH)$ while preserving the carbon-carbon double bond is achieved via the

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$I_2/NaOH$ followed by $H^+$

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(B) The conversion of a methyl ketone $(R-CO-CH_3)$ to a carboxylic acid $(R-COOH)$ while preserving the carbon-carbon double bond is achieved via the Haloform reaction (specifically the Iodoform test).$1$. $I_2/NaOH$ (Iodoform reagent) selectively oxidizes the methyl ketone group to a carboxylate salt $(R-COO^-)$ and $CHI_3$,without affecting the $C=C$ double bond.$2$. Subsequent acidification with $H^+$ yields the carboxylic acid $(R-COOH)$.$3$. $KMnO_4$ is a strong oxidizing agent that would cleave the $C=C$ double bond.$4$. $H_2/Pt$ would reduce the $C=C$ double bond.$5$. $LiAlH_4$ would reduce the ketone to a secondary alcohol.

$Ph-CH=CH-C(=O)CH_3 \to Ph-CH=CH-CO_2H$
The above conversion can be achieved by:

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$Ph-CH=CH-C(=O)CH_3 \to Ph-CH=CH-CO_2H$The above conversion can be achieved by: