Reduction with NaBH4

Aldehydes and ketones can be reduced to alcohols using sodium tetrahydridoborate(III), NaBH₄, in aqueous solution. NaBH₄ acts as a source of the hydride ion, H⁻.In equations the reducing agent is written as [H]. Two [H] are added across the C=O bond:• Aldehydes are reduced to primary alcohols:RCHO + 2[H] → RCH₂OH
CH₃CH₂CHO + 2[H] → CH₃CH₂CH₂OH• Ketones are reduced to secondary alcohols:RCOR′ + 2[H] → RCH(OH)R′
CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃This is the reverse of making aldehydes and ketones by oxidising alcohols. The reaction is a nucleophilic addition: H⁻ attacks the δ+ carbonyl carbon (the mechanism is in the next lesson).NaBH₄ does not reduce C=C double bonds. The C=C bond is not polar and is rich in electrons, so it repels the H⁻ nucleophile. This means NaBH₄ can reduce the C=O in an unsaturated carbonyl compound and leave the C=C untouched.Worked example: write the equation for the reduction of butanone. Butanone is CH₃COCH₂CH₃, a ketone, so it gives a secondary alcohol: CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃, which is butan-2-ol.