The compound CH_3-C(CH_3)=CH-CH_3 on reaction | vedclass

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(D) The reaction of an alkene with $NaIO_4$ in the presence of $KMnO_4$ (Lemieux-von Rudloff reagent) results in the oxidative cleavage of the $C=C$ d

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$CH_3COCH_3 + CH_3COOH$

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(D) The reaction of an alkene with $NaIO_4$ in the presence of $KMnO_4$ (Lemieux-von Rudloff reagent) results in the oxidative cleavage of the $C=C$ double bond.In the compound $CH_3-C(CH_3)=CH-CH_3$,the double bond is cleaved.The substituted carbon $CH_3-C(CH_3)=$ is oxidized to a ketone,$CH_3COCH_3$ (acetone).The $=CH-CH_3$ part is oxidized to an aldehyde,which is further oxidized to a carboxylic acid,$CH_3COOH$ (acetic acid),under these conditions.Therefore,the final products are $CH_3COCH_3 + CH_3COOH$.

The compound $CH_3-C(CH_3)=CH-CH_3$ on reaction with $NaIO_4$ in the presence of $KMnO_4$ gives:

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