0
$\begingroup$

In the mechanism of SeO2 I have seen that a more stable keto form first converts to a less stable enol form and then yields a diketone. I have seen many other reaction where the reactant converts to an unstable form first and then reacts. My doubt is why does the reactant converts itself rather than just simply not react ?

$\endgroup$
1
  • 2
    $\begingroup$ Well... If you take a look at the mechanism: en.wikipedia.org/wiki/Riley_oxidation. The reaction cannot occur by starting with a ketone. It can only begin with the enol form. Also, it is not the case where the enol form doesn't already exists in the first place. Although the carbonyls exist predominantly in the keto form, some of them are in the enol form. These enols can take part in the reaction. Even if these enols were consumed, the equilibrium would shift towards forming more enols, gradually the keto form would also deplete. $\endgroup$ Commented Mar 19, 2019 at 10:10

1 Answer 1

1
$\begingroup$

We can think of it an an equilibrium that is disturbed by another, irreversible reaction.

Think of a more familiar reaction. When you try to dissolve magnesium hydroxide in water you get only a little solubility; the dissolved magnesium and hydroxide ions are only a minor species at equilibrium. But when hydrochloric acid is added the acid reacts selectively with the hydroxide ions, neutralizing them; in response more magnesium hydroxide keeps dissolving into the water (to form magnesium and hydroxide ions, the latter then reacting with more acid) until you have a salt solution containing the accumulated magnesium ions and the chloride from the acid. Reaction continues until you've dissolved all of the magnesium hydroxide or neutralized all of the acid.

Similarly, the enol form is only a minor species in the equilibrium, but when you add the selenium dioxide it reacts selectively with the enol form. This draws more of the carbonyl compound into the enol form in the same way that the acid in the example above draws more magnesium hydroxide into its dissolved form. Thereby the reaction keeps going until you run out of one reactant or another.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.