Alkene X xrightarrow[Delta]{KMnO_4} (Butan-2- | vedclass

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(A) The oxidative cleavage of an alkene with hot alkaline $KMnO_4$ breaks the double bond. Each carbon atom of the double bond is oxidized to a carbon

Vedclass · Accepted Answer

$3-$methylhept$-3-$ene

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(A) The oxidative cleavage of an alkene with hot alkaline $KMnO_4$ breaks the double bond. Each carbon atom of the double bond is oxidized to a carbonyl compound. If the carbon is disubstituted $(R_2C=)$,it forms a ketone $(R_2C=O)$. If the carbon is monosubstituted $(RCH=)$,it forms a carboxylic acid $(RCOOH)$. The products are Butan$-2-$one $(CH_3CH_2COCH_3)$ and Pentanoic acid $(CH_3CH_2CH_2CH_2COOH)$. Combining these fragments at the double bond: $CH_3CH_2-C(CH_3)=CH-CH_2CH_2CH_2CH_3$. This corresponds to $3-$methylhept$-3-$ene.

Alkene $X$ $\xrightarrow[\Delta]{KMnO_4}$ (Butan$-2-$one) $+$ (Pentanoic acid). $X$ is

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