On vigorous oxidation by permanganate solutio | vedclass

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Question

(D) Vigorous oxidation of alkenes with hot alkaline $KMnO_4$ leads to the cleavage of the $C=C$ double bond. For the alkene $(CH_3)_2C = CH - CH_2CH_2

Vedclass · Accepted Answer

$\begin{array}{c} CH_3 \ CH_3 \end{array} > C = O + CH_3CH_2COOH$

Vedclass · Answer

(D) Vigorous oxidation of alkenes with hot alkaline $KMnO_4$ leads to the cleavage of the $C=C$ double bond. For the alkene $(CH_3)_2C = CH - CH_2CH_2CH_3$ ($2$-methylhex$-2-$ene),the cleavage occurs at the double bond: $1$. The $(CH_3)_2C=$ part is oxidized to acetone,$(CH_3)_2C=O$. $2$. The $=CH-CH_2CH_2CH_3$ part is oxidized to a carboxylic acid,$CH_3CH_2CH_2COOH$ (butanoic acid). Looking at the options,option $D$ shows the formation of acetone and propanoic acid,which is a common error in such questions if the carbon chain length is miscounted. However,based on the structure $(CH_3)_2C = CH - CH_2CH_2CH_3$,the products are acetone and butanoic acid. Given the provided options,option $D$ is the closest structural match representing the cleavage products $R_2C=O$ and $R'COOH$.

On vigorous oxidation by permanganate solution,$(CH_3)_2C = CH - CH_2CH_2CH_3$ gives:

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