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I have searched the whole internet but I could only see examples where the authors of the article took $\ce{R1}$ and $\ce{R2}$ as the alkyl groups around the carbon.
This is regarding the oxidation of carbonyl compounds (I am specifically searching about ketones) using $\ce{SeO2}$.
What I understood was $\ce{C=O}$ got introduced on the alpha-carbon and two hydrogens from that carbon were lost.
I would like to know the product obtained on reacting pentan-2-one with $\ce{SeO2}$. Will the carbon having 3 alpha-hydrogens become $\ce{C=O}$ or will the other one having 2 alpha hydrogens become $\ce{C=O}$?

References:

  1. Selenium dioxide oxidation of ketones and aldehydes. Evidence for the intermediacy of .beta.-ketoseleninic, K. Barry Sharpless and Kenneth M. Gordon Journal of the American Chemical Society, 1976, 98 (1), 300-301 DOI: 10.1021/ja00417a083
  2. https://www.adichemistry.com/organic/organicreagents/seo2/selenium-dioxide-seo2.html
Dusty_Wanderer
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The mechanism of Selenium Dioxide (Riley) oxidation proceeds via the enol tautomer of the ketone (see scheme - image from Wikipedia). If the ketone is assymetric and can enolise to either side then the reaction should give the two di-ketones in proportion to the amount of either enol formed though I cannot find any reference where this has been experimentally demonstrated. In the OP's case of 2-pentanone, a mixture of the two possible products will be formed.

enter image description here

Waylander
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  • why would it not make a double bond as it does for 1,4 diketones? –  Aug 27 '20 at 14:34
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    @https://chemistry.stackexchange.com/questions/33034/why-doesnt-the-seo2-oxidation-of-ketones-stop-at-the-hydroxyketone-stage: It's depend on the several factors including structure of starting material, solvent, and pH. You can find an answer to your question here. – Mathew Mahindaratne Aug 27 '20 at 16:40