# Mechanism of a cyclization reaction - why does an enol ether form?

I was asked to predict compound A from the follwing organic scheme:

Looking at the formation of A, my prediction was that since TsOH is a strong acid, it would dehydrate the alcohols (like sulfuric acid does). However, looking at the next reaction of A, it seems to me that what happened was that the ketone formed its enol and did an aldol condensation with the aldehyde. But then, if this was the case, why do we have to add TsOH to get to A, then add $\ce{H3O+}$ to get to the final product when we could've gotten to the final product from the start with an aldol condensation?

I was told that the structure for A is actually:

I was not expecting this at all, and I'm not sure how this was formed. I can see that the enol ether in A is hydrolyzed in acid to get the final product, but the formation of A seems strange. What is the mechanism for the formation of A and why were my predictions wrong?

• It's just an Aldol reaction. A is probably just the enol Mar 10 '17 at 9:13

For the A you were shown, the aldol product would have to be dehydrated (why is this even considered an intermediate then?), tautomerized via the $\alpha–\beta$ unsaturated bond, and then attack the protonated alcohol with the much less nucleophilic oxygen site of the enol to form a 8-membered ring. While this isn't technically impossible, each of these steps are unfavorable, and reversible, this species would likely never be observed... at all... ever. See the rather strange looking mechanism:
• @Mithoron is right. That intermediate explains the need for aqueous $\ce{H+}$. Similarly to how the intermediate shown in the question wouldn't be my first guess (or, apparently, yours), I'm guessing the authors of whatever paper it's from didn't expect it, either. Luckily, the enol ether is easy enough to hydrolyze. Mar 11 '17 at 10:09