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(D) The reaction of an alkene with $NaIO_4$ in the presence of a catalytic amount of $KMnO_4$ is a method of oxidative cleavage of the double bond.For

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$CH_3COCH_3 + CH_3CHO$

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(D) The reaction of an alkene with $NaIO_4$ in the presence of a catalytic amount of $KMnO_4$ is a method of oxidative cleavage of the double bond.For the given compound $CH_3-C(CH_3)=CH-CH_3$ ($2$-methylbut$-2-$ene):$1$. The double bond breaks between the $C_2$ and $C_3$ carbons.$2$. The $C_2$ carbon is bonded to two methyl groups,forming acetone $(CH_3COCH_3)$.$3$. The $C_3$ carbon is bonded to one hydrogen and one methyl group,forming acetaldehyde $(CH_3CHO)$.Thus,the reaction is: $CH_3-C(CH_3)=CH-CH_3 \xrightarrow{NaIO_4/KMnO_4} CH_3COCH_3 + CH_3CHO$.

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

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