[saint_john]

Can anybody explain the mechanism that occurs here and how the epoxide gets broken up
thanks !

[Cesium-137]

Aldehyde right? No anhydride here... I may be mistaken but it looks to me like a hydrolysis. Water attacks the carbonyl carbon, pushing electrons up to the carbonyl oxygen, which come back down to form the stable double bond. The electrons bonding the carbonyl carbon to the epoxide carbon are pushed toward oxygen to form yet another stable double bond. The electrons bonding the epoxide to the tertiary carbon are pushed to attack a proton from the original water molecule. This probably occurs in one step to produce the aldehyde product and carbonic acid. Carbonic acid in water spontaneously dissociates to form CO₂ and H₂O. The driving force of this reaction (other than adding heat, obviously) is the stability of the final products. Water and carbon dioxide are two of the most stable compounds out there, so once they are formed there won't be a reverse reaction. Also, any water produced is reused in the reaction and carbon dioxide leaves as a gas shifting equilibrium towards the aldehyde. Hope this helps!

[Doc Oc]

I agree, no anhydride, but it also isn't a simple hydrolysis.  Look into malonic ester chemistry, it will give you a hint.

[Cesium-137]

Hmmm interesting... looks like it can potentially go without water! This is the mechanism I came up with after checking out the similar malonic ester decomposition... Does it look correct? If so, I'm stuck on how the epoxide can grab the carboxyl proton without water present. Or is this the dominating reaction even with water as your solvent?

[Doc Oc]

Yes, that new reaction looks right.  It's actually a little more clear what happens if you draw the 5-membered transition state between the epoxide oxygen and the carboxylic acid proton.  That may clear up some of your confusion about how the epoxide is able to grab that proton (although even in an aqueous system that proton will certainly be the first to go because it is far more acidic than water).