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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 is posted below.

reaction mechanism

image source

When an allylic alcohol with stereochemistry (cis, trans) around the double bond is oxidized, the stereochemistry around the double is preserved.

rxn example

image source

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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  • $\begingroup$ 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, 2014 at 22:58
  • $\begingroup$ The link does not work $\endgroup$ Mar 2, 2020 at 15:24
  • $\begingroup$ @SchwarzKugelblitz I've added the mechanism directly to the answer. $\endgroup$
    – ron
    Mar 3, 2020 at 18:20

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