[azmanam]

Question: Provide a complete, arrow-pushing mechanism for the following one-pot transformation.  Predict the product and account for stereochemistry in your response.

[Dan]

Hmmm.

Initially I think we're going to get methanolysis of the acetal.

Next, base treatment could do various Payne type rearrangements. I'm going to go for an intramolecular THP formation, leaving a terminal epoxide (although there seemed to be several plausible pathways at this point.

Finally, acid promoted intramolecular epoxide opening gives the bicyclic ether shown?

Ambitious maybe....

[azmanam]

can you be a little more specific in why you chose your ring closures out of the various epoxide-opening choices you had.

acetal removal is good, overall idea is good, based on studies running the three reactions separately, your ring opening/closings are under the wrong conditions.

[Dan]

can you be a little more specific in why you chose your ring closures out of the various epoxide-opening choices you had.

I decided to opt for the intramolecular THP formation because I thought it'd be faster than opening the epoxide with external hydroxide - but perhaps this is not the case, I certainly would expect the intramolecular reaction to dominate if we were forming a THF, but THP formation may be too slow.

Based on your clues, I propose this...

[azmanam]

no, you predicted the right product the first time.  I just wanted you to justify it - and fix your ordering of the various steps

[Dan]

Hmmm, I'm finding this very tricky. I can't really see what's going on here, the only way I have come up with to get my first predicted product is the mechanism I originally proposed.

If the THF forms before the THP then I'm having difficulty forming the THP without a tosyl migration, which I don't really like. Originally I thought the THP may form via an oxetane, but the 1,2-trans relationship between the substituents on the THF (my second structure) doesn't allow it. I'm also having trouble explaining the final acid treatment (unless it's just a neutralisation).

Anyway here it is... Not confident about this one though....

[azmanam]

I feel my attempt at hints are actually just muddying the waters.  Let me step back and try to be explicit in my question, while not giving away the answer.

The bicyclic ether is the correct product.  You note in your first guess that there are several competing pathways which are possible.  In your first guess, you went with THP formation first.  In your recent guess, you go with THF formation first.  Which do you think is more plausible and why?

Separately, what I was trying to say the first time is: the researchers did not initially attempt the one pot reaction.  First attempts were 3 sequential steps with purification between.  Thus, the product of the first acid step can be isolated, treated with base and a second intermediate can be isolated, then treated with acid again to isolate the desired product (which you've drawn).  Based on your mechanisms (first and recent), if you stopped your reaction after the acid step, you would not isolate the product the researchers do, etc.  While your transformation is correct, the elementary steps are not.

More clear, or more muddy?

[Dan]

OK. I think I'm with you now Just realised the THF formation is a Baldwin no-no...

i) HCl: Acetal hydrolysis AND THP formation. 6-Exo-trig, and NOT THF formation (5-endo-trig).

ii) NaOH: Epoxide formation

iii) HCl: THF formation. Opening of the epoxide occurs at the carbon best able to stabilise a partial positive charge (secondary vs primary).

[azmanam]

Yes.  6 exo predominates over 5 endo in the first cyclization, and 5 exo predominates over 6 endo in the second - both for the same reason: better orbital overlap (Baldwin's Rules).  Nice work.

http://dx.doi.org/10.1021/ol061339e

[wintermute]

I believe that what you call 5-endo and 6-endo are, in fact, 5-exo and 6-exo in respect to electrons of the broken bond, which do not become the part of the cycle (hence exo). Am I wrong?

[azmanam]

I believe that what you call 5-endo and 6-endo are, in fact, 5-exo and 6-exo in respect to electrons of the broken bond, which do not become the part of the cycle (hence exo). Am I wrong?

I get your point, but that does not seem to be the convention.  From Baldwin's paper (http://dx.doi.org/10.1039/C39760000734):

The rules for opening three-membered rings to form cyclic structures (13), seem to lie between those for tetrahedral and trigonal systems, generally preferring Exo modes.

see also, appx middle of page: http://euch6f.chem.emory.edu/baldwinsrules.html

[wintermute]

thanks for clearing this out for me