Oxidation using selenium dioxide

[maverick280857]

Hello

My question is regarding the oxidative function of selenium dioxide (when reacted with carbonyl compounds).

Selenium Dioxide converts an active methylene group to a carbonyl group -C=O, so the reaction of butan-2-one with selenium dioxide should form two products (corresponding to two active methylene groups):

2MeCOEt + SeO2 -> MeCOCOMe + HCOCOEt + Se

I know that MeCOCOMe is the Major Product whereas HCOCOEt is the minor product but I want to know why this is so. Perhaps a mechanistic explanation may help. I have been unable to locate the exact mechanistic steps of this reaction. I only know that SeO2 delivers nascent oxygen.

Help would be greatly appreciated.

Thanks and cheers
Vivek

 

[chem_tr]

Don't try to find the solution far away; the solution is right in the structure. Terminal methyl groups are more vulnarable to selenium dioxide attack than partly shielded ethyl group's alfa methylene. It is possible though, some product is formed with the action of SeO₂.

 

[movies]

But what Maverick said is that the more hindered side is oxidized more often.

I think that it has to do with the more stable enolate of butanone.

Corey has proposed a mechanism that begins by forming selenous acid by adding a water molecule to the SeO2, then the ketone oxygen attacks the selenium displacing a water molecule. The activated carbonyl can then enolize. The oxidation step is though to be a 2,3 shift, where one of the remaining oxygens on selenium attacks the enol double bond forcing the electrons of the enol back into the carbonyl and then forcing the electrons from the Se-O bond onto selenium (so selenium is reduced). The final step is deprotonation of the carbon bearing the O-Se-OH group to form either Se(OH)2 or selenium zero and two water molecules.

Sharpless has proposed a different mechanism where you form the enolate first and then the enolate attack the selenium of selenous acid to make a C-Se bond. Then he proposes that the carbonyl group is again enolized, but this time instead of protonating again the electrons from the enol form a double bond to selenium and eject a water molecule off of the selenium to give a species that looks sort of like this: O=C-C=Se=O . This species can give the product by attack of water at the carbon of the C=Se bond to form the ketone, selenium zero, and a water molecule after some electron pushing. Sharpless also mentions the posibility that the oxygen of the Se=O could attack the carbon of the C=Se bond to make a three membered ring that could fragment open to give the product and selenium zero.

Are those mechanistic explanations clear? I might be able to make up a pictoral representation of that if you want.

Here are the references, if you're interested:

Corey, J. Am. Chem. Soc. 1960, 82, 917
Sharpless, J. Am. Chem. Soc. 1976, 98, 300

 

[maverick280857]

Thanks :-)

 

[chem_tr]

Oops, you are right, Movies. Thanks for the references, and Maverick, if you need these, I can get them for you as my library has a subscription to American Chemical Society.

Movies, you are right about that more stable enolate is formed principally (and thermodynamically) but the side product may be explained with kinetic easiness of the reaction. I've reviewed it and methylene forms a secondary carbanion where terminal methyl forms a primary one.

And I realize that trying to contribute to the forum while exhausted does not produce useful results.

 

[movies]

Yep, the minor product could be explained by kinetic enolate formation.