Luche reduction (1,2-reduction of enones): practice and mechanism
NaBH₄ with CeCl₃ selectively reduces the C=O group of enones (1,2-reduction). The C=C double bond stays and an allylic alcohol forms.
Example
Cerium(III) makes the hydride "harder" and activates the C=O group. With α,β-unsaturated ketones the carbonyl group is thus reduced selectively (1,2-addition). The C=C double bond survives and an allylic alcohol forms.
Mechanism step by step
- 1
CeCl₃ coordinates to the C=O group and makes it more electrophilic.
- 2
In methanol methoxyborohydrides form, "harder" hydride donors that prefer the hard carbonyl carbon.
- 3
The hydride attacks 1,2; after protonation the allylic alcohol forms and the C=C bond remains.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- That would be 1,4-reduction (C=C reduced, C=O kept), e.g. with copper hydrides or H₂. Cerium(III) steers the hydride to the C=O group instead.
- NaBH₄/CeCl₃ only reduces the carbonyl group. The C=C bond survives; reducing both would need e.g. H₂/Pd followed by NaBH₄.
Often confused with: 1,4-addition with cuprates (Gilman) · Catalytic hydrogenation · Michael addition
Frequently asked questions
What does NaBH₄ alone do with enones?
Often a mixture of 1,2- and 1,4-reduction, especially with cyclopentenones. CeCl₃ makes the reaction cleanly 1,2-selective.
Hard and soft, what does that mean here?
The carbonyl carbon is "hard" (strongly positive, small orbitals), the β-carbon "soft". Hard nucleophiles (Grignard, LiAlH₄, Luche) go 1,2, soft ones (cuprates, enolates, thiols) 1,4.