Consider the following reaction:Alkene (A) xr | vedclass

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(A) The reaction of an alkene with hot alkaline $KMnO_4$ leads to oxidative cleavage of the double bond.The products formed are $CH_3CH_2COCH_3$ (buta

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$3-$methylhept$-3-$ene

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(A) The reaction of an alkene with hot alkaline $KMnO_4$ leads to oxidative cleavage of the double bond.The products formed are $CH_3CH_2COCH_3$ (butanone) and $CH_3CH_2CH_2CH_2COOH$ (pentanoic acid).To find the structure of the original alkene,we join the carbonyl carbon of the ketone and the carboxyl carbon of the carboxylic acid with a double bond:$CH_3CH_2-C(CH_3)=CH-CH_2CH_2CH_2CH_3$This structure corresponds to $3-methylhept-3-ene$.

Consider the following reaction:
$Alkene (A) \xrightarrow{KMnO_4, \Delta} CH_3CH_2COCH_3 + CH_3CH_2CH_2CH_2COOH$
$A$ is:

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Consider the following reaction:$Alkene (A) \xrightarrow{KMnO_4, \Delta} CH_3CH_2COCH_3 + CH_3CH_2CH_2CH_2COOH$$A$ is: