Ketones undergo a reduction when treated with sodium borohydride, ${\mathbf{NaBH}}_{\mathbf{4}}$. What is the structure of the compound produced by reaction of 2-butanone with ${\mathbf{NaBH}}_{\mathbf{4}}$ if it has an IR absorption at $\mathbf{3400}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$and ${\mathbf{M}}^{\mathbf{+}}\mathbf{=}\mathbf{74}$in the mass spectrum?

Sodium borohydrides are selective reducing agents.Aldehydes and ketones get converted to alcohols when sodium borohydride functions as the reagent. Aldehydes yield primary alcohols and ketones yield secondary alcohols.

The m/z value of the product is 74. The IR spectrum shows that the compound has an IR absorption of ${\text{3400 cm}}^{\text{- 1}}$.

The ketone undergoes reduction with ${\text{NaBH}}_{\text{4}}$forming an alcohol. The given compound 2-butanone undergoes reduction with ${\text{NaBH}}_{\text{4}}$to form 2-butanol.

Reaction of 2-butanone with${\text{NaBH}}_{\text{4}}$

The IR absorption value at${\text{3400 cm}}^{\text{- 1}}$indicates the hydroxyl group in 2-btanol.

The molecular weight of the compound matches with the expected compound. Hence, the structure of the compound can be given as: