Reduction of ketone to secondary alcohol Steps:

• Propanone (CH3COCH3) is a ketone with a carbonyl group.
• NaBH4 is a mild reducing agent that adds hydride to the carbonyl carbon.
• This reduces the C=O bond, forming a C-OH bond and yielding a secondary alcohol.
• The product is CH3CH(OH)CH3, which is propan-2-ol.

Why C is correct:

• NaBH4 reduces ketones to secondary alcohols by delivering H- to the carbonyl, protonating the oxygen to form -OH at the central carbon.

Why the others are wrong:

• A: Propan-1-ol (CH3CH2CH2OH) is a primary alcohol from aldehyde reduction, not ketone.
• B: Propanal (CH3CH2CHO) is an aldehyde, the precursor to propan-1-ol, not produced here.
• D: Propane (CH3CH2CH3) is a saturated hydrocarbon from complete reduction, not this selective reaction.

Final answer: C