Q.Propanal is treated with $\text{NaBH}_4$ in methanol. Identify the product and state whether $\text{LiAlH}_4$ would give a different product here.

$\text{NaBH}_4$ delivers a hydride ion to the electrophilic carbonyl carbon of propanal, $\text{CH}_3\text{CH}_2\text{CHO}$, forming an alkoxide that is protonated on aqueous work-up to give the primary alcohol propan-1-ol. Since an aldehyde is reduced to a primary alcohol regardless of whether the reducing agent is the milder $\text{NaBH}_4$ or the stronger $\text{LiAlH}_4$, the product here would be identical either way -- the two reagents differ mainly in strength and selectivity (LiAlH4 also reduces less reactive groups such as esters and carboxylic acids, which NaBH4 does not touch), not in what an aldehyde specifically becomes.

Identify the carbonyl class (aldehyde) to fix the alcohol class produced (primary), then compare the two hydride reagents only on strength/selectivity, not on outcome for this particular substrate.

Assuming LiAlH4 must give a chemically different product from NaBH4 whenever both are mentioned, when for a simple aldehyde substrate they give the identical alcohol.