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

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(C) The reaction of an alkene with $KMnO_4$ (a strong oxidizing agent) leads to the oxidative cleavage of the double bond.For the compound $CH_3-C(CH_

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$CH_3COCH_3 + CH_3COOH$

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(C) The reaction of an alkene with $KMnO_4$ (a strong oxidizing agent) leads to the oxidative cleavage of the double bond.For the 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,so it forms acetone $(CH_3COCH_3)$.$3$. The $C_3$ carbon is bonded to one hydrogen and one methyl group,so it forms acetic acid $(CH_3COOH)$ upon further oxidation.Thus,the 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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