[spirochete]

I have a doubt about an aldol cyclization. I attached a picture of it and considered two possible products. It doesn't seem obvious which one is the major product because the two possible alkenes that form are equally substituted and equally conjugated. Yet the two products do seem significantly different. One of them has the two remaining ketones far apart, making the dipole smaller I would think. I think smaller dipole would make A favored, but maybe that depends on the solvent.

Does anyone know for sure whether A or B is favored from this process? I don't have access to sci-finder or any way to search it. At least if somebody who studies these things doesn't know the answer it would make me feel better, lol.

Thank you!

[rolnor]

It will not go that way, you have a much more acidic proton in between the too carbonyls to the left. Also you can probably not control this reaction, it will be a mess with all these carbonyls and acidic protons,

[spirochete]

Thank you for your input. This example is widely used in introductory organic chemistry classes, I think because it is an easy one to make up off the top of your head, and then assign it as a practice problem. I always wondered if it would actually have a strong preference for one ring over the other, and if so, why.

I have also considered the issue with the 1,3 alpha proton being more acidic. If it were able to make either aldol cyclization product then it would happen despite that more acidic proton. I have seen a real example where the more acidic proton is present and a robinson annulation can still happen, but this was with ethyl acetoacetate, which is slightly less acidic than a 1,3-diketone. It's on page 639 of Clayden textbook and also I have seen it done in Sophomore organic chemistry labs, suggesting that the reaction actually works. In the picture it says that reaction gives 90% yield, so apparently the ring formation is so favorable that it eventually can make the ring even though there is an option for a much more stable enolate ion to form.

[rolnor]

If you look, the more acidic position in the acetoacetate is 1,4-attacking the double bond first, then the ketone-methyl attacks the other carbonyl, closing the ring. I think this is important.

[rolnor]

Its possible that the more easily formed anion in your first case is so stabilized that its a poor nucleophile, but I dont think so.

[rolnor]

You will have to form a dianion in any case.

[rolnor]

Its possible that the reaction will stop because the product will have this charge between the 1,3-carbonyls, hindering it attacking itself