Product (D) in the above reaction is: | vedclass

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(A) $1$. The starting material is $2-(hydroxymethyl)cyclohexanol$. Treatment with $PCC$ (Pyridinium chlorochromate) selectively oxidizes the primary a

Vedclass · Accepted Answer

$2-$($1$-hydroxy$-1-$methylcyclohexyl)methanol

Vedclass · Answer

(A) $1$. The starting material is $2-(hydroxymethyl)cyclohexanol$. Treatment with $PCC$ (Pyridinium chlorochromate) selectively oxidizes the primary alcohol to an aldehyde,forming $2-formylcyclohexanone$ $(A)$. $2$. Reaction with ethylene glycol in the presence of $H^+$ forms a cyclic acetal at the aldehyde group,protecting it to give $2-(1,3-dioxolan-2-yl)cyclohexanone$ $(B)$. $3$. Addition of $MeMgBr$ followed by $H_3O^+$ attacks the ketone carbonyl to form a tertiary alcohol,yielding $2-(1,3-dioxolan-2-yl)-1-methylcyclohexanol$ $(C)$. $4$. Finally,the acetal is hydrolyzed (implied in the workup or subsequent step) and the resulting aldehyde is reduced by $NaBH_4$ to a primary alcohol,resulting in $2-(1-hydroxy-1-methylcyclohexyl)methanol$ $(D)$.

Product $(D)$ in the above reaction is:

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