Abstract
AbstractThe oxidation reactions of 2‐methyl‐2,4‐pentanediol upon treatment with alkaline hexacyanoferrate(III) using Ru(III) or Ru(VI) as catalysts are governed by two quasi‐identical experimental rate equations, which show that both catalysts are equally effective for the oxidation of alcohols by ${\rm Fe(CN)}_6^{3-}$. The reaction mechanism proposed involves the oxidation of 2‐methyl‐2,4‐pentanediol by the catalyst, a process that occurs through the formation of a substrate–catalyst complex. The decomposition of this complex yields Ru(IV) and a protonated ketone (owing to a hydride transfer from the α‐C—H bond of the alcohol to the oxoligand of ruthenium) in the case of Ru(VI), but a ketyl radical and Ru(II) (hydrogen transfer) for Ru(III). The role of the co‐oxidant, ${\rm Fe(CN)}_6^{3-}$, is to regenerate the catalyst. For both oxidation reactions, the rate constants of complex decomposition and catalyst regeneration have been determined. Copyright © 2004 John Wiley & Sons, Ltd.
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