[jj74]

Hello
I have a problem working on the mechanism of the oxidative cleavage of alkenes with permanganate.
The alkene in a basic solution of permanganate at room temperature gives the 1,2 diol. If, however, the basic solution is heated or if it’s acidified, the reaction will break also the sigma C-C bond breaking the alkene and the products are ketone and carboxylic acid; terminal alkenes give CO₂ as product.
Unfortunately there’s no mechanism explained in the books that I have at disposal and nothing I found on the net; I’ve tried on my own but I can’t go on after the intermediate Syn-diol formation.

Here are the reaction that I’ve found (sorry no software here at work, hope it’s clear):

(CH₃CH₂)(CH₃)C = CHCH₃ [3,methyl-2-pentene]      (CH₃CH₂)CO(CH₃) [Methylethylketone] + CH₃COO^-
CH₃CH₃CH = CH₂  CH₃CH₂COOH + CO₂


Could someone please give me a hint ?

[movies]

The mechanism is probably through a series of one-electron transfers and radical intermediates.  It tends to go that way with most of the manganese oxidants.  Since those mechanisms are very difficult to study and draw definitively, it is hard to give a straight answer. 

If you are an organic chemist and you like pretty arrow pushing mechanisms, then you could draw a C–C cleavage mechanism akin to that of NaIO₄ with a diol.  It is a possibility with KMnO₄ as well, but I think it is probably not what is actually going on.

[jj74]

hey thanks,
I tried with a mechanism similar to the periodate cleavage but it was a no go 

Anyway my real problem is: is there a regioselectivity rule for this reaction ?
As you can see from both my examples, the two sp² carbons of the double bond both end up oxidated in the product but to a different degree: in the first one carbon is C=O (oxidation nr = 2) and the other is COO (ox nr = 3), in the second example the terminal carbon (bound to two Hs in the substrate) becomes CO₂ (ox nr = 4) and the other one becomes COO (ox nr = 3). I'm getting a little confused here 

[Arctic-Nation]

I can't help you with the exact mechanism, but the extent of the final oxidation can be explained easily by looking at the number of hydrogen atoms present on the olefin carbon atoms. While a 1,1-disubstituted olefin is oxidized to a ketone (on the 1,1 side), a singly substituted olefin is oxidized first to the aldehyde (presumably), and aldehydes are pretty unstable in regard to strong oxidizers. Thus, they are turned into carboxylic acids (compare with the Jones oxidation, which shows a similar result). Formaldehyde can be considered to be a 'super-aldehyde', and as such is oxidized twice, to carbon dioxide.

[movies]

Actually, I am a little surprised that the terminal alkene would give CO₂.  I would have expected it to stop at formic acid which is reasonably stable stuff.  Anyway, Arctic-Nation summed it up in terms of the degree of oxidation.  The only exception would be if you have an activated carbon undergoing oxidation, like a benzylic carbon.  These will proceed to the carboxylic acid even if it requires the cleavage of a C–C bond.

[Arctic-Nation]

Formic acid is not entirely uncommon to be used as a reducing agent, as under the right conditions it is a good hydride source. In this case, it is first oxidized to carbonic acid, H₂CO₃, which quickly converts to carbon dioxide.

[movies]

Hehe – I am quite familiar with the reducing capability of formic acid from my own research.

The usual mechanism of hydride transfer from formic acid is quite different than what you would imagine for a typical oxidation by a metal oxide.  I guess I am not entirely surpised to hear that it would ultimately go to CO₂, just I had never considered that before.

[jj74]

thank you all for the replies, now I got it