Acetone on reduction with \(\mathrm{NaBH}_{4}\) gives 2 -Propanol.

In this reaction, acetone (a ketone) is being reduced by sodium borohydride (\(\mathrm{NaBH}_{4}\)) to give 2-propanol (an alcohol). The overall reaction can be written as follows: $$\text{Acetone} + \mathrm{NaBH}_{4} \rightarrow \text{2-Propanol}$$

Acetone has the structure: $$\text{CH}_{3} \text{-} \text{C} \text{=}\text{O} \text{-} \text{CH}_{3}$$ Sodium borohydride (\(\mathrm{NaBH}_{4}\)) can be represented as: $$\text{Na}^{+}[\text{BH}_{4}]^{-}$$ The expected product, 2-propanol, has the structure: $$\text{CH}_{3}\text{-}\text{CH}\text{-}\text{OH}\text{-}\text{CH}_{3}$$

Sodium borohydride (\(\mathrm{NaBH}_{4}\)) is a reducing agent. It donates a hydride ion (H\(^-\)) to the carbonyl carbon of the acetone, which reduces the carbonyl group to an alcohol group.

Here is the step-by-step mechanism of the reduction reaction: 1. The hydride ion (H\(^-\)), provided by \(\mathrm{NaBH}_{4}\), attacks the electrophilic carbonyl carbon of acetone. This results in a bond formation between the hydride ion and the carbonyl carbon (C-O bond is broken). $$\text{CH}_{3} \text{-} \text{C} \text{=}\text{O} \text{-} \text{CH}_{3} \; \text{+ H}^{-} \rightarrow \text{CH}_{3} \text{-} \text{C} \text{-} \text{OH } |- \text{CH}_{3}$$ 2. The oxygen atom of acetone undergoes protonation by accepting a proton (H\(^+\)) from the solvent or from a released hydroxide ion (OH\(^-\)) during the reaction. This leads to the formation of 2-propanol. $$\text{CH}_{3} \text{-} \text{C} \text{-} \text{OH } |- \text{CH}_{3} \; \text{+ H}^{+} \rightarrow \text{CH}_{3}\text{-}\text{CH}\text{-}\text{OH}\text{-}\text{CH}_{3}$$ So, by following these steps, acetone is reduced by \(\mathrm{NaBH}_{4}\) to give 2-propanol as the final product.