Q.Acetone is treated with $\text{NaBH}_4$ in methanol. Identify the product and state whether $\text{LiAlH}_4$ would give a different outcome for this particular substrate.

$\text{NaBH}_4$ delivers a hydride ion to the electrophilic carbonyl carbon of acetone, giving a tetrahedral alkoxide that is protonated (by the methanol solvent) to give the secondary alcohol, propan-2-ol. $\text{LiAlH}_4$ is a considerably more powerful and reactive hydride donor overall, and must be handled under strictly anhydrous, aprotic conditions -- but for a simple substrate like acetone, which contains only the one carbonyl group and no other easily-reduced functionality (such as an isolated $\text{C=C}$ double bond, an ester, or a nitrile) for the reagents' differing reactivity/selectivity to distinguish between, both reagents simply reduce the carbonyl to the alcohol and give …

Recognise that both common hydride reagents reduce a plain, isolated carbonyl group to the corresponding alcohol identically; their practical differences (solvent compatibility, selec …

Assuming $\text{LiAlH}_4$, being the 'stronger' reagent, must give a DIFFERENT product from $\text{NaBH}_4$ -- on a substrate with only one simple, isolated carbonyl group and not …