The reaction sequence is given as:1-methylcyc | 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 adds two hydroxyl

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$1-$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 adds two hydroxyl groups across the double bond to form a vicinal diol.$2$. Thus,$A$ is $1-methylcyclohexane-1,2-diol$.$3$. The subsequent reaction of $A$ with $CrO_3$ in $AcOH$ is an oxidation reaction. $CrO_3$ in acetic acid is a selective oxidizing agent that can oxidize secondary alcohols to ketones.$4$. In $1-methylcyclohexane-1,2-diol$,the hydroxyl group at the $C-2$ position is secondary,while the hydroxyl group at the $C-1$ position is tertiary (attached to a carbon already bonded to a methyl group).$5$. Therefore,$CrO_3$ selectively oxidizes the secondary alcohol at $C-2$ to a ketone,resulting in $2-hydroxy-2-methylcyclohexanone$ as product $B$.

The reaction sequence is given as:
$1-methylcyclohexene$ $\xrightarrow[alkaline \ KMnO_4]{cold} A$ $\xrightarrow[AcOH]{CrO_3} B$
Identify $A$ and $B$.

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The reaction sequence is given as:$1-methylcyclohexene$ $\xrightarrow[alkaline \ KMnO_4]{cold} A$ $\xrightarrow[AcOH]{CrO_3} B$Identify $A$ and $B$.