When reading about $\ce{SeO_2}$ I read this following reaction:


In the following reaction it was stated that $\ce{CH3CH2COCHO}$ was the major product. But I fail to understand why ?
The two possible reasonings I can think of are that the terminal hydrogen is either less sterically hindered or that it is more acidic. But I am not quite sure of either.

What is the reason for the same ?


To understand the major factor we have to consider the mechanism of the reaction first. The reaction you mentioned is popularly know as Riley oxidation. The mechanism is as follows:

Riley oxidation

As you can observe the RDS of the reaction is the tautomerization of the keto form into an enol. Thus the extent of enolization determines the major product in this reaction.

As the reaction takes place in an neutral environment the acidity of the alpha-hydrogen plays the major role in deciding the major enol formed and so the corresponding oxidized product is major.

In short, the major product in Riley oxidation is the product associated with the major enol content which in the given conditions is determined by the acidity of respective alpha-hydrogens.


This is known as the Riley Oxidation.

The Wikipedia entry states that ketones with two alpha methylene positions react more quickly at the least hindered site which references Riley's original paper1.


  1. Selenium dioxide, a new oxidising agent. Part I. Its reaction with aldehydes and ketones by Riley et.al. J. Chem. Soc., 1932, 1875-1883, DOI: https://doi.org/10.1039/JR9320001875

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