The product formed in the following reaction is: $4\text{-}\mathrm{methyl}\text{-}2\text{-}(2\text{-}\mathrm{methoxy}\text{-}2\text{-}\mathrm{oxoethyl})\mathrm{cyclohexan}\text{-}1\text{-}\mathrm{one}\underset{\text{C2H5OH}}{\overset{\text{NaBH4}}{\to }}?$ (A cyclohexanone ring bearing a $-\mathrm{CH}{}_2\mathrm{COOCH}{}_3$ methyl-ester side chain and a $\mathrm{CH}{}_3$ substituent.)

Solution & Explanation

\textbf{Answer:} (B) ring $\mathrm{C}=\mathrm{O}$ reduced to ring $-\mathrm{OH}$, methyl-ester side chain unchanged \textbf{Solution:} Sodium borohydride $\mathrm{NaBH}{}_4$ is a mild reducing agent: it reduces aldehydes and ketones to alcohols but does not reduce esters or carboxylic acids. Hence only the ring ketone is reduced to a secondary alcohol (a cyclohexanol ring), while the $-\mathrm{CH}{}_2\mathrm{COOCH}{}_3$ methyl-ester side chain remains intact.