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Reduction of a ketone (R₂C=O) with NaBH₄ or LiAlH₄ produces:
AA tertiary alcohol
BA primary alcohol
CA ketone (no change)
DA secondary alcohol (R₂CH-OH)
Answer & Solution
Correct answer: D. A secondary alcohol (R₂CH-OH)
**Aldehyde → 1° alcohol**; **ketone → 2° alcohol**. NaBH₄ and LiAlH₄ deliver hydride (H⁻) to the carbonyl carbon. LiAlH₄ is stronger (reduces esters, carboxylic acids too); NaBH₄ is milder (only aldehydes and ketones, not acids/esters).
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