The major product formed in the following reaction is
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NaBH$_4$ selectively reduces aldehydes and ketones to alcohols but does not affect carboxylic acids, esters, or amides.
Always consider hydride addition to the carbonyl carbon followed by protonation.
Step 1: Identify the reagent and type of reaction.
Sodium borohydride (NaBH$_4$) is a mild reducing agent used to reduce aldehydes and ketones to corresponding alcohols.
Here, NaBH$_4$ in methanol reduces the carbonyl group (C=O) to a hydroxyl group (C–OH).
Step 2: Mechanism of reduction.
NaBH$_4$ provides hydride ions (H$^-$), which attack the electrophilic carbonyl carbon, forming an alkoxide intermediate.
Subsequent protonation by acid (H$_3$O$^+$) produces the corresponding alcohol.
\[
\text{CH}_3\text{COCH(CH}_3\text{)CH}_2\text{CH}_3 \xrightarrow{\text{NaBH}_4}
\text{CH}_3\text{CH(OH)CH(CH}_3\text{)CH}_3
\]
Step 3: Stereochemistry.
Since the hydride attack can occur from either face of the planar carbonyl group, a racemic mixture of enantiomers is formed.
However, only one enantiomer is shown in the answer for simplicity. Step 4: Conclusion.
The product is 3-methyl-2-butanol, corresponding to **Option (A)**.
