# Mechanism of the oxidation of alcohols with KMnO4

Many oxidising agents, like chromate, dichromate, iodine in $\ce{NaOH}$ etc. seem to work via ester formation and elimination. For example, chromic acid will react with the $\ce{-OH}$ of alcohol to form a chromate ester, and then this will undergo an E2-like elimination, with a proton abstracted from carbon and loss of leaving group from oxygen.

However it seems that manganate works via a different mechanism. I think this because here: http://www.masterorganicchemistry.com/2015/05/21/demystifying-alcohol-oxidations/

It says in a footnote at the bottom:

"..main exception you’ll encounter is $\ce{KMnO4}$, which likely proceeds through a C-H abstraction/internal return type mechanism followed by collapse of the hydrate to give the new carbonyl."

I don't quite understand this brief summary of the mechanism, could someone expand on it? I don't quite understand what "internal return type mechanism" is refering to - I kind of understand it refers to something a little like the "intimate ion pair" in the $\mathrm{S_Ni}$ mechanism, but haven't been able to find online a more expanded mechanism - many places seem to show it like the case of chromic acid above, and so I am slightly confused as to what mechanism it really follows.

• March 7th ed. (p 1436) seems to imply that it is a ester formation followed by elimination, the same as it is with CrO3/H2SO4. Unfortunately, I am not quite sure what the phrasing above is referring to. – orthocresol Apr 15 '16 at 19:55
• @orthocresol I have seen ester formation followed by elimination for manganate, but the footnote seems to imply that manganate is somehow different in it's mechanism, but I couldn't understand what the phrasing of it meant either. – Swedish Architect Apr 15 '16 at 20:27

$\ce{KMnO4/OH-}$ Oxidation
$\ce{H2CrO4}$ Oxidation
1. $\ce{H2CrO4}$ Oxidation occurs in acid, $\ce{KMnO4}$ oxidation occurs in base
2. The source of the nucleophile in the removal of the metal ester. In $\ce{KMnO4}$ oxidation, the $\ce{Mn}$ ester itself extracts the $\ce{H+}$ from the alcohol carbon, while in $\ce{H2CrO4}$ oxidation the nucleophile is the solvent.