7
$\begingroup$

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

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:

A

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?

$\endgroup$
  • 4
    $\begingroup$ It's just an Aldol reaction. A is probably just the enol $\endgroup$ – RobChem Mar 10 '17 at 9:13
3
$\begingroup$

@Carbenoid: A year has passed since you requested a mechanism from @Mithoron. The statutes of limitation have run out. Your comment about why not do an aldol condensation (in base) is well taken. But the question was what is the structure of compound A. I have no problem with the dienol ether 5 (aka A) in spite of many people's aversion to forming 8-membered rings.

Full stereochemistry of structure 1 could have been informative but not essential. I will forgo the mechanism of the acid-catalyzed aldol condensation and defer to @ringo's contribution. The bold bonds in structure 2 form a rigid framework so that the degrees of freedom are minimized in the formation of the hemiketal 3. Effectively, the cyclization is that of a much smaller ring. This ring closure is mechanistically more sound than the SN2 displacement with an enol particularly one with an (E)-geometry when at least the (Z)-geometry would be required. The red arrows terminate with dienolether 5. Aqueous acid and microscopic reversibility via the blue arrows return dienolether 5 to structure 2. In fact, structures 2 and 3 may well be in equilibrium with each other.

enter image description here

$\endgroup$
2
$\begingroup$

The structure for A will most certainly not be that which you were shown. I would say that you were right in your original assumption, with the reaction proceeding via the mechanism:

enter image description here

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:

gross

$\endgroup$
  • 1
    $\begingroup$ Well, I can see a different mechanism in which this would be indeed an intermediate, starting with creation of tosylate. $\endgroup$ – Mithoron Mar 10 '17 at 21:34
  • $\begingroup$ @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. $\endgroup$ – SendersReagent Mar 11 '17 at 10:09
  • 1
    $\begingroup$ @Mithoron can you write out the mechanism you're thinking of? Thanks! $\endgroup$ – carbenoid Mar 11 '17 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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