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I am told that $\ce{MnO_2}$ oxidizes allylic and benzylic alcohols. However, what is the mechanism of this oxidation? All the professor had to offer me was that the metal ion, manganese 2+, chelates the substrate; there is some sort of interaction between a pi bond, the metal ion, and the electrophilic carbon atom bearing the hydroxyl group. No further explanation was given, and a Google search hasn't unearthed anything of use. What, therefore, is the mechanism of this oxidation? I would like to know more about the mechanism to better understand why $\ce{MnO_2}$ only touches allylic and benzylic alcohols rather than any vanilla alcohol.

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This is one of those heterogenous reactions occurring on a surface which are mechanistically messy. As expected for a reaction that occurs on the surface of a reagent, factors such as surface area and reagent preparation (freshness, surface contamination) are significant variables. A free radical mechanism is generally accepted and a reasonable mechanism may be found here, see page 363. When an allylic alcohol with stereochemistry (cis, trans) around the double bond is oxidized, the stereochemistry around the double is preserved.

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This suggests that we are not really dealing with a free, long-lived radical intermediate, but rather a free radical held in close association with the manganese ion.

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  • $\begingroup$ FYI, Your link doesn't seem to work. $\endgroup$ – Geoff Hutchison Sep 18 '14 at 21:49
  • $\begingroup$ Hmm, it works for me. Let me put the url here again in case anyone else has a problem. citab.utad.pt/refbase/files/1890_JI-DONGLOU2011.pdf $\endgroup$ – ron Sep 18 '14 at 22:58
  • $\begingroup$ looks fine now - I was getting a link back to Chemistry.SE not the PDF. Thanks! $\endgroup$ – Geoff Hutchison Sep 19 '14 at 12:41

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