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Inorganic — Pd(II)-catalyzed oxidation

Wacker oxidation: ethylene → acetaldehyde

How the Wacker oxidation works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.

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In short. A 3-step mechanism. Typical reagents: PdCl₂ (cat.), CuCl₂, O₂, H₂O. Conditions: H₂O/DMF or HCl(aq), 50–100 °C. Bonds formed: C–O, C=O; Pd–H (transient); bonds broken: C=C π; C–H (β). Stereochemistry: Anti-nucleopalladation; Markovnikov for terminal alkenes (methyl ketone). Rate and selectivity: Pd(0) reoxidized by 2 CuCl₂; Cu(I) reoxidized by O₂.

At a glance

ReagentsPdCl₂ (cat.), CuCl₂, O₂, H₂O
Solvent / conditionsH₂O/DMF or HCl(aq), 50–100 °C
Bonds formedC–O, C=O; Pd–H (transient)
Bonds brokenC=C π; C–H (β)
StereochemistryAnti-nucleopalladation; Markovnikov for terminal alkenes (methyl ketone)
Rate / selectivityPd(0) reoxidized by 2 CuCl₂; Cu(I) reoxidized by O₂

Step 1: Water attacks the Pd-bound alkene (anti)

+PdCl₂OH₂1 · WATER ATTACKS THE PD-BOUND ALKENE (ANTI)η²-ethylene Pd(II) complexC₂H₄PdELECTROPHILEwaterH₂ONUCLEOPHILEanti to PdPd(II) makes the alkene electrophilic; water adds from the face opposite Pd.

Pd(II) coordinates the alkene, activating it toward nucleophilic attack. Water's oxygen attacks the coordinated alkene carbon from the face opposite palladium — anti nucleopalladation.

Step 2: β-Hydride elimination

OHHPdCl2 · Β-HYDRIDE ELIMINATIONβ-hydroxyethyl–PdC₂H₅OPdINTERMEDIATEβ-HPd–C σ bond becomes the C=C π bond; the β C–H adds to Pd as Pd–H, giving an enol.

This forms a Pd–alkyl species bearing a pendant hydroxyl. β-hydride elimination transfers a hydrogen from carbon to palladium, forming a new Pd–H bond and a coordinated enol.

Step 3: Enol → acetaldehyde; Pd(0) reoxidized

tautomerize+2 CuCl₂(O₂, HCl)OHOPd(0)PdCl₂3 · ENOL → ACETALDEHYDE; PD(0) REOXIDIZEDvinyl alcohol (enol)C₂H₄OINTERMEDIATEacetaldehydeC₂H₄OPRODUCTpalladium(0)PdCATALYSTPd(II) regeneratedPdCATALYSTCu(II) reoxidizes Pd(0); O₂ reoxidizes Cu(I), so only O₂ is consumed overall.

The enol tautomerizes to acetaldehyde, and Pd(0) is reoxidized to Pd(II) by Cu(II)/O₂ so the catalytic cycle continues — the basis of industrial acetaldehyde production.

Abbreviations and terms

DMF
N,N-Dimethylformamide. Polar aprotic solvent – dissolves ionic nucleophiles but does not H-bond to them, speeding SN2 and SNAr.
H2
Hydrogen gas. Reductant for catalytic hydrogenation.
HCl
Hydrochloric acid / hydrogen chloride. Strong acid; chloride is the nucleophile in additions/substitutions.
Markovnikov
Markovnikov’s rule. In HX addition, H goes to the carbon with more H’s so the more stable cation forms.

How it compares with related reactions

Wacker oxidationAssociative ligand substitution at a square-planar Pt(II) complexAsymmetric phase-transfer catalysis
ReagentsPdCl₂ (cat.), CuCl₂, O₂, H₂OK₂[PtCl₄] + NH₃ (aq)Chiral cinchona-derived Q*⁺Br⁻ (1–10 mol%), 50% aq. KOH, glycine imine tBu ester, BnBr
ConditionsH₂O/DMF or HCl(aq), 50–100 °CH₂O, 25 °CToluene/CH₂Cl₂, 0 °C, vigorous stirring
StereochemistryAnti-nucleopalladation; Markovnikov for terminal alkenes (methyl ketone)Trigonal-bipyramidal intermediate; retention of geometryHigh ee: chiral ion pair shields one enolate face
Rate / selectivityPd(0) reoxidized by 2 CuCl₂; Cu(I) reoxidized by O₂Rate = k₁[Pt] + k₂[Pt][NH₃] (associative; entering-ligand dependent)Alkylation of the ion pair is rate-limiting

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Wacker oxidation?

PdCl₂ (cat.), CuCl₂, O₂, H₂O. Conditions: H₂O/DMF or HCl(aq), 50–100 °C.

Which bonds are formed and which are broken?

Formed: C–O, C=O; Pd–H (transient). Broken: C=C π; C–H (β).

What is the stereochemical outcome?

Anti-nucleopalladation; Markovnikov for terminal alkenes (methyl ketone)

What controls the rate or the selectivity?

Pd(0) reoxidized by 2 CuCl₂; Cu(I) reoxidized by O₂

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