1-Methylcyclohexene xrightarrow[text{cold}]{a | vedclass

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(A) $1$. The reaction of $1-methylcyclohexene$ with cold alkaline $KMnO_4$ (Baeyer's reagent) is a syn-hydroxylation reaction,which converts the alken

Vedclass · Accepted Answer

$1-$Methylcyclohexane$-1,2-$diol and $2-$hydroxy$-2-$methylcyclohexanone

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(A) $1$. The reaction of $1-methylcyclohexene$ with cold alkaline $KMnO_4$ (Baeyer's reagent) is a syn-hydroxylation reaction,which converts the alkene into a vicinal diol.$2$. Thus,$A$ is $1-methylcyclohexane-1,2-diol$.$3$. The reaction of a vicinal diol with $CrO_3$ in $CH_3COOH$ is a selective oxidation. In this case,the secondary alcohol group is oxidized to a ketone,while the tertiary alcohol group remains unaffected.$4$. Therefore,$B$ is $2-hydroxy-2-methylcyclohexanone$.

$1-Methylcyclohexene$ $\xrightarrow[\text{cold}]{alkaline\,KMnO_4} A$ $\xrightarrow[CH_3COOH]{CrO_3} B$
$A$ and $B$ are

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$1-Methylcyclohexene$ $\xrightarrow[\text{cold}]{alkaline\,KMnO_4} A$ $\xrightarrow[CH_3COOH]{CrO_3} B$$A$ and $B$ are