Hydride reduction — simple carbonyl
NaBH₄ reduction of a ketone
How the NaBH₄ reduction of a ketone works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.
In short. A 3-step mechanism. Typical reagents: NaBH₄ (0.25–1 equiv). Conditions: EtOH or MeOH, 0–25 °C; aqueous workup. Bonds formed: C–H; O–H; bonds broken: C=O π; B–H. Stereochemistry: Hydride attacks the less hindered face of ring ketones. Rate and selectivity: Aldehydes > ketones; esters and amides untouched.
At a glance
Step 1: Polarized B–H hydride donor
Sodium borohydride's boron-hydride bond is polarized with hydrogen carrying significant negative character — the same hydride-donor role played by LiAlH₄ or a Grignard reagent, just far milder and selective enough to use in protic solvents like ethanol.
Step 2: Hydride adds to C=O
Hydride transfers directly to the ketone's electrophilic carbonyl carbon in a simple nucleophilic addition, exactly like the earlier hydride-addition entry, forming an alkoxide bound loosely to boron.
Step 3: Workup releases the alcohol
Protonation on workup releases the alcohol — NaBH₄'s mildness means it reduces aldehydes and ketones cleanly while leaving esters, amides, and most other functional groups untouched, unlike the much more aggressive LiAlH₄.
Abbreviations and terms
- MeOH
- Methanol. Polar protic solvent (also a nucleophile in solvolysis).
- EtOH
- Ethanol. Polar protic solvent.
- LiAlH4
- Lithium aluminium hydride. Powerful hydride (H⁻) donor; reduces esters, acids, amides to alcohols/amines.
- NaBH4
- Sodium borohydride. Mild hydride donor; reduces aldehydes and ketones but not esters.
How it compares with related reactions
| NaBH₄ reduction of a ketone | LiAlH₄ reduction of a nitrile to a primary amine | LiAlH₄ reduction of an ester to a primary alcohol | |
|---|---|---|---|
| Reagents | NaBH₄ (0.25–1 equiv) | LiAlH₄ (excess), then H₂O workup | LiAlH₄ (≥ 0.5 equiv., 2 H⁻ per ester), then H₃O⁺ or Fieser workup |
| Conditions | EtOH or MeOH, 0–25 °C; aqueous workup | Dry Et₂O or THF, 0 °C → reflux; then H₃O⁺ or Fieser workup | Dry Et₂O or THF, 0 °C to reflux |
| Stereochemistry | Hydride attacks the less hindered face of ring ketones | None (achiral CH₂NH₂) | No stereocentre (primary alcohol) |
| Rate / selectivity | Aldehydes > ketones; esters and amides untouched | Second hydride addition fast relative to the first | Fast; aldehyde reduces faster than the ester |
Test yourself
Try to answer each question before opening it.
Which reagents and conditions are typical for the NaBH₄ reduction of a ketone?
NaBH₄ (0.25–1 equiv). Conditions: EtOH or MeOH, 0–25 °C; aqueous workup.
Which bonds are formed and which are broken?
Formed: C–H; O–H. Broken: C=O π; B–H.
What is the stereochemical outcome?
Hydride attacks the less hindered face of ring ketones
What controls the rate or the selectivity?
Aldehydes > ketones; esters and amides untouched
Study guides that help with this reaction
- Carbonyl chemistry: addition, acyl substitution and enolates
- SN1, SN2, E1 or E2? How to decide which mechanism wins
- Nucleophiles, electrophiles and leaving groups: what makes a reaction go