Luche reduction selectively converts an α,β-unsaturated ketone into an allylic alcohol. The reaction is named after French chemist Jean-Louis Luche, who reported it in 1978. ^[1–4]

General Reaction

An α,β-unsaturated ketone contains a carbonyl group conjugated with a carbon–carbon double bond. The carbonyl group is reduced, while the conjugated carbon–carbon double bond generally remains unchanged. The product is called an allylic alcohol because the carbon bearing the hydroxyl group is directly adjacent to a carbon–carbon double bond.  ^[5,6]

Reduction of an α,β-unsaturated carbonyl compound can occur by either 1,2–reduction or 1,4-reduction. In the Luche reduction, 1,2-reduction of the carbonyl group is favored, while the conjugated carbon–carbon double bond generally remains unchanged.  ^[5,6]

Luche reduction commonly uses sodium borohydride (NaBH₄) together with cerium(III) chloride, often as cerium(III) chloride heptahydrate (CeCl₃·7H₂O). Methanol is commonly used as the solvent.

Examples

Mechanism of Luche Reduction

The reaction mixture contains several species formed from cerium salts, borohydride, and the alcohol solvent. Therefore, the following mechanism is simplified.

Cerium(III) ions promote the reaction of borohydride with methanol, increasing the formation of methoxyborohydride species. These species are proposed to favor hydride transfer to the carbonyl carbon and therefore promote 1,2-reduction.  ^[7]

Step 1: Formation of Methoxyborohydride Species Promoted by Ce³⁺ 

In methanol, sodium borohydride undergoes alcoholysis. Methoxy groups replace one or more hydrogens attached to boron:

NaBH₄ + nMeOH → Na⁺[BH_(4−n)(OMe)_n]^– + nH₂

For the methoxy-substituted reducing species shown here, n may be 1, 2, or 3. At least one B–H bond must remain because it supplies the hydride used for reduction.

Step 2: Hydride Transfer to the Carbonyl Group

The substrate is an α,β-unsaturated ketone. A hydride from a B–H bond is transferred directly to the carbonyl carbon.

At the same time:

• The C=O π electrons move onto the oxygen.
• The carbonyl carbon forms a new C–H bond.
• The conjugated carbon–carbon double bond remains unchanged.

This is therefore a 1,2–reduction of the carbonyl group through an alkoxide intermediate.

Step 3: Protonation of the Alkoxide Intermediate

The negatively charged oxygen of the alkoxide intermediate accepts a proton from methanol. The electrons from the methanol O–H bond remain on the methanol oxygen, producing methoxide.

The carbonyl group is consequently converted into an alcohol group.