Hex-4-en-2-ol on treatment with PCC gives 'A' | vedclass

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(C) $1$. $PCC$ (Pyridinium chlorochromate) oxidizes the secondary alcohol $CH_3-CH=CH-CH_2-CH(OH)-CH_3$ to the ketone $CH_3-CH=CH-CH_2-CO-CH_3$ (Compo

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$2-$butene

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(C) $1$. $PCC$ (Pyridinium chlorochromate) oxidizes the secondary alcohol $CH_3-CH=CH-CH_2-CH(OH)-CH_3$ to the ketone $CH_3-CH=CH-CH_2-CO-CH_3$ (Compound $'A'$).$2$. Compound $'A'$ contains a methyl ketone group,so it undergoes the iodoform reaction with sodium hypoiodite $(NaOI)$ to form the carboxylic acid $CH_3-CH=CH-CH_2-COOH$ (Compound $'B'$) and iodoform $(CHI_3)$.$3$. Decarboxylation of the carboxylic acid $CH_3-CH=CH-CH_2-COOH$ with soda lime $(NaOH + CaO)$ removes $CO_2$ to yield $CH_3-CH=CH-CH_3$ (Compound $'C'$),which is $but-2-ene$.

Hex$-4-$en$-2-$ol on treatment with $PCC$ gives $'A'.$ $'A'$ on reaction with sodium hypoiodite gives $'B',$ which on further heating with soda lime gives $'C'.$ The compound $'C'$ is $....... .$

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