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(B) The reaction of an alkene with $KMnO_4$ in the presence of $NaIO_4$ is a form of oxidative cleavage,similar to ozonolysis.In this reaction,the dou

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$CH_3COCH_3 + CH_3COOH$

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(B) The reaction of an alkene with $KMnO_4$ in the presence of $NaIO_4$ is a form of oxidative cleavage,similar to ozonolysis.In this reaction,the double bond is broken,and each carbon atom of the double bond is oxidized to a carbonyl group.For the alkene $CH_3-C(CH_3)=CH-CH_3$:$1$. The bond between $C_2$ and $C_3$ breaks.$2$. The $C_2$ atom (which is bonded to two methyl groups) is oxidized to acetone $(CH_3COCH_3)$.$3$. The $C_3$ atom (which is bonded to one hydrogen and one methyl group) is oxidized to acetaldehyde $(CH_3CHO)$,which is further oxidized to acetic acid $(CH_3COOH)$ in the presence of $KMnO_4$.Therefore,the products are $CH_3COCH_3$ and $CH_3COOH$.

What is the product obtained when the compound $CH_3-C(CH_3)=CH-CH_3$ is treated with $KMnO_4$ in the presence of $NaIO_4$?

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