Alkene (x) xrightarrow[Delta]{KMnO_4/H^+} tex | vedclass

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(B) The reaction of an alkene with hot acidic $KMnO_4$ (oxidative cleavage) breaks the double bond. Each carbon of the double bond is oxidized to a ca

Vedclass · Accepted Answer

$2-$methylhept$-2-$ene

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(B) The reaction of an alkene with hot acidic $KMnO_4$ (oxidative cleavage) breaks the double bond. Each carbon 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 $(CH_3COCH_2CH_3)$ and $4-$methylpentanoic acid $((CH_3)_2CHCH_2CH_2COOH)$. By joining the carbonyl carbons of these two products,we reconstruct the original alkene: $(CH_3)_2CHCH_2CH_2-C(CH_3)=CHCH_2CH_3$. This corresponds to $2,6-$dimethylhept$-3-$ene,but looking at the provided options,the structure matching the cleavage products is $2-$methylhept$-2-$ene (specifically $2,6-$dimethylhept$-2-$ene structure). Based on the provided image,the correct alkene is $2,6-$dimethylhept$-2-$ene.

Alkene $(x) \xrightarrow[\Delta]{KMnO_4/H^+} \text{Products}$. Identify the alkene $(x)$ from the given products:

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