Hydride transfer — reverse of MPV
Oppenauer oxidation
How the Oppenauer oxidation 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: Al(OiPr)₃ (or Al(OtBu)₃), acetone (large excess). Conditions: Toluene/acetone, reflux. Bonds formed: C=O (ketone); C–H (iPrOH); bonds broken: C–H (carbinol); C=O (acetone π). Stereochemistry: Stereocentre at carbinol carbon destroyed; other centres untouched. Rate and selectivity: Equilibrium: driven by excess acetone; hydride transfer rate-limiting.
At a glance
Step 1: Al³⁺ binds alcohol O and acetone O
Aluminum isopropoxide coordinates the oxygen of a secondary alcohol substrate, while also binding a large excess of acetone as a hydride acceptor — setting up the same six-membered cyclic transition state as the Meerwein-Ponndorf-Verley reduction, just run with the roles reversed.
Step 2: Six-membered hydride transfer (MPV in reverse)
A hydride transfers directly from the substrate's carbinol carbon to acetone's carbonyl carbon, intramolecularly through the aluminum — the identical mechanism as MPV reduction, just with the thermodynamics pushed the opposite direction by using a large excess of the hydride acceptor.
Step 3: Ketone + isopropanol
The substrate alcohol is oxidized to a ketone while acetone is reduced to isopropanol — because MPV and Oppenauer are mechanistically the same equilibrium, which direction dominates is controlled entirely by which carbonyl partner is present in excess.
Abbreviations and terms
- R
- R group. Any carbon substituent (alkyl, aryl…) whose identity does not matter for the mechanism.
- TS
- Transition state. The highest-energy point on the pathway between two minima; it cannot be isolated.
How it compares with related reactions
| Oppenauer oxidation | Meerwein-Ponndorf-Verley reduction | Borch reductive amination | |
|---|---|---|---|
| Reagents | Al(OiPr)₃ (or Al(OtBu)₃), acetone (large excess) | Al(OiPr)₃ (cat. to stoich.), iPrOH (excess) | Amine, ketone/aldehyde, NaBH₃CN |
| Conditions | Toluene/acetone, reflux | Reflux; distil off acetone | MeOH, pH 6–7 (AcOH), 25 °C |
| Stereochemistry | Stereocentre at carbinol carbon destroyed; other centres untouched | Chair-like TS; reversible (Oppenauer is the reverse) | New stereocentre usually racemic (unless chiral amine/ketone) |
| Rate / selectivity | Equilibrium: driven by excess acetone; hydride transfer rate-limiting | Driven by distilling acetone / excess iPrOH | Iminium formation is rate-limiting; iminium reduced much faster than ketone at this pH |
Test yourself
Try to answer each question before opening it.
Which reagents and conditions are typical for the Oppenauer oxidation?
Al(OiPr)₃ (or Al(OtBu)₃), acetone (large excess). Conditions: Toluene/acetone, reflux.
Which bonds are formed and which are broken?
Formed: C=O (ketone); C–H (iPrOH). Broken: C–H (carbinol); C=O (acetone π).
What is the stereochemical outcome?
Stereocentre at carbinol carbon destroyed; other centres untouched
What controls the rate or the selectivity?
Equilibrium: driven by excess acetone; hydride transfer rate-limiting
Study guides that help with this reaction
- SN1, SN2, E1 or E2? How to decide which mechanism wins
- Nucleophiles, electrophiles and leaving groups: what makes a reaction go