If acetone () is reduced using sodium borohydride () in the presence of heavy water () as the solven β Alcohols Phenols and Ethers Chemistry Question
Question
If acetone ($CH_3COCH_3$) is reduced using sodium borohydride ($NaBH_4$) in the presence of heavy water ($D_2O$) as the solvent, what is the exact structure of the final product?
Answer: A
π‘ 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 $CH_3CH(OD)CH_3$.
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