Oxidation of olefins by palladium(II). VIII. Kinetics of the oxidation of ethylene by palladium(II) chloride in methanol
- 1 June 1976
- journal article
- Published by Canadian Science Publishing in Canadian Journal of Chemistry
- Vol. 54 (11) , 1726-1738
- https://doi.org/10.1139/v76-246
Abstract
The oxidation of ethylene in methanol at 25 °C by palladium(II) chloride plus LiCl, was found to yield mainly CH3CH(OCH3)2. If the reaction is run in CH3OD the acetal is undeuterated. In order to interpret the kinetics the equilibria between Pd(II) and chloride and Pd(II) and C2H4 were studied. The equilibria, [Formula: see text] and [Formula: see text] were found to be operative with KH = 0.010 M−1 and K1 = 8.1 at 25 °C.The value of K2 for the equilibrium [Formula: see text] was found to be less than 10−2 M. The ethylene oxidation was studied by gas uptake using quinone to prevent precipitation of Pd(0). The reaction is strongly inhibited by acid and chloride. The acid inhibition by weak acids such as methanesulfonic and acetic depends on the square root of acid concentration which is consistent with a first order acid inhibition. In the [H+] range from about 10−2 to 10−5 M the rate expression is identical to that previously found in water;−d[C2H4]/dt = kK1[PdCl42−][C2H4]/[H+][Cl−]2. The solvent isotope effect [Formula: see text] is much smaller than that found previously in water [Formula: see text] The effect of ionic strength is analogous to that previously found in water. The rate first increases with increasing ionic strength, reaches a maximum at about μ = 0.75 and then gradually decreases. The rate expression and other factors suggest a mechanism analogous to that previously proposed for the aqueous system. This mechanism has the slow step in the reaction involving cis insertion of a coordinated methoxide to coordinated ethylene to give a β-methoxyethylpalladium(II) alkyl. Evidence for the mode of decomposition of this alkyl is provided by the oxidation of C2H4 in CH3OD. The resulting dimethyl acetal of acetaldehyde contains no deuterium. This suggests a decomposition step involving Pd(II)-hydride elimination and readdition in the opposite mode to put the Pd(II) on the carbon containing the methoxy group. The Pd(II) is reductively eliminated from this adduct to give Pd(0) and the dimethylacetal product. At [H+] < 10−5 M the rate becomes independent of [H+] and at [H+] < 10−8 M the rate actually decreases with decreasing [H+]. The cause of this change in [H+] dependency is uncertain.Keywords
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