If acetone () is reduced using sodium borohydride () in the presence of heavy water () as the solven — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

$NaBH_4$ provides a hydride ($H^-$) that attacks the carbonyl carbon, forming a $CH_3-CH(O^-)-CH_3$ alkoxide intermediate. The alkoxide oxygen then abstracts a deuteron ($D^+$) from the $D_2O$ solvent to form the final product $CH_3CH(OD)CH_3$.