Reduction with NaBH4 or LiAlH4 of a ketone yields which product?

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Multiple Choice

Reduction with NaBH4 or LiAlH4 of a ketone yields which product?

Reducing a ketone with hydride donors like NaBH4 or LiAlH4 adds a hydride to the carbonyl carbon, breaking the C=O bond and forming a tetrahedral alkoxide intermediate. After aqueous workup, that alkoxide is protonated to give an alcohol. Because the carbonyl carbon in a ketone is attached to two carbon-containing groups, the resulting alcohol has the OH-bearing carbon connected to two carbon substituents, making it a secondary alcohol. If the starting material were an aldehyde, the product would be a primary alcohol; here you’re converting a ketone, so the product is a secondary alcohol. Ether is not formed by this reduction. If the ketone has two different substituents, the product can be chiral and appear as a pair of enantiomers.

Reducing a ketone with hydride donors like NaBH4 or LiAlH4 adds a hydride to the carbonyl carbon, breaking the C=O bond and forming a tetrahedral alkoxide intermediate. After aqueous workup, that alkoxide is protonated to give an alcohol. Because the carbonyl carbon in a ketone is attached to two carbon-containing groups, the resulting alcohol has the OH-bearing carbon connected to two carbon substituents, making it a secondary alcohol. If the starting material were an aldehyde, the product would be a primary alcohol; here you’re converting a ketone, so the product is a secondary alcohol. Ether is not formed by this reduction. If the ketone has two different substituents, the product can be chiral and appear as a pair of enantiomers.

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