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

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(A) The reaction of an alkene with hot acidic $KMnO_4$ causes oxidative cleavage of the double bond.An alkene of the form $R_2C=CH-R'$ will be cleaved

Vedclass · Accepted Answer

$3-$methylhept$-2-$ene

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(A) The reaction of an alkene with hot acidic $KMnO_4$ causes oxidative cleavage of the double bond.An alkene of the form $R_2C=CH-R'$ will be cleaved to form a ketone $(R_2C=O)$ and a carboxylic acid $(R'-COOH)$.The products given are butan$-2-$one $(CH_3-CH_2-CO-CH_3)$ and pentanoic acid $(CH_3-CH_2-CH_2-CH_2-COOH)$.Combining these fragments at the carbonyl carbons,the original alkene must be $3-$methylhept$-2-$ene $(CH_3-CH_2-C(CH_3)=CH-CH_2-CH_2-CH_3)$.

Alkene $(x)$ $\xrightarrow[\Delta ]{KMnO_4/H^+}$ gives the following products:
$x$ is

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Alkene $(x)$ $\xrightarrow[\Delta ]{KMnO_4/H^+}$ gives the following products:$x$ is