organic chemistry - 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.

Answer

$\ce{KMnO4/OH-}$ Oxidation

$\ce{H2CrO4}$ Oxidation

The two main differences are

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.