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Consider the molecule Ph-CO-CH2-CH2-CO2Me (methyl 4-oxo-4-phenylbutanoate), which contains both a ketone group and a methyl ester group. Which of the following reagents will selectively reduce only the ketone to give Ph-CH(OH)-CH2-CH2-CO2Me, leaving the ester intact?

1. LiAlH4
2. BH3, THF
3. H2, Raney Ni
4. NaBH4

Correct answer: NaBH4

Solution

NaBH4 selectively reduces the ketone carbonyl to give the secondary alcohol while leaving the methyl ester untouched, because NaBH4 lacks the reactivity to attack the less electrophilic ester carbonyl under standard conditions.

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