# Ozonolysis of alkene

(from Wikipedia)

I am having trouble in the last step where the alkene ozonide is reduced to ketones / aldehydes specifically in the presence of water. My teacher is telling me that water will be abstracting an oxygen atom to form $$\ce{H2O2}$$.

But why? Both oxygen atoms already have a complete octet, and further are repelled by their electron clouds. Why does water abstract an oxygen atom at all?

• I was always under the impression that you needed a mild reducing agent, like dimethyl sulfide. I could see the same mechanism working with water, but am rather skeptical, because water is a pretty bad reducing agent, and on top of that if you put H2O2 in there you will get carboxylic acids (that's the "oxidative workup")... Aug 6, 2020 at 5:41
• Agree with orthocresol, a workup with just water makes no sense. It´s probably going to happen if you leave the trioxolane on the shelf with water for a week, and you get a wild mixture of ketones/aldehydes and acids. Not synthetically useful.
– Karl
Aug 6, 2020 at 6:13
• Not as far as I'm aware. You need to introduce a reducing agent, usually a sulfide, sometimes a phosphine. I think your teacher is in error. Aug 6, 2020 at 6:42

When the ozonide is reduced to Ketone/Aldehyde, We're concerned with what is called the "reductive workup" . We can summarize it like this:

If we look a little further into the mechanism (of a different alkene):

Step (e) is what we're concerned with. Water is not having an impact on this reaction in any manner. The main compound here that's assisting this reaction is $$\ce{(CH3)_2S}$$ which is a mild reducing agent.

Source of the pictures:
Reaction summary
Reaction mechanism

• I agree. If you work up an ozonolysis reaction mixture with just water, you get the ozonide. DO NOT DO THIS - ozonides are explosive. Aug 6, 2020 at 8:18
• I'll keep that in mind when I enter the lab (I won't be doing so in the near future :P ) Aug 6, 2020 at 8:25