[darkness1102]

How does this reaction occur?

[stewie griffin]

Are you sure all the reagents needed are here??
I can get to a 7-membered ring, but my substitution pattern doesn't match your answer. Plus I have no clue how you get oxidation to the acid when there's no oxidant present 

[darkness1102]

Yes, I sure.

[orgopete]

This looks formally like an retro Claisen or Dieckmann reaction, a Michael addition, and a Claisen. However, I lack driving force for this to occur. If a Michael addition occurs first, then a retro Claisen becomes more logical, but final cyclization gives the product with the double bond conjugated to the carboxylic acid.

However, since no water is present, the retro Claisen should give a diester. Claisen condensation generates the water and hydrolysis of the isolated ester would give the acid.

[darkness1102]

But, how to get to 7-membered ring?

[movies]

I agree with orgopete.  It looks like a conjugate addition followed by an intramolecular aldol and a retro-Dieckmann reaction (and then elimination of water to install the C–C double bond).  The first two steps would give a [3.2.1]-bicycle, so the driving force for the retro-Dieckmann is probably strain release.

The one acid in the product is a little puzzling, especially is the conditions really are anhydrous.  That made me think that maybe instead of directly losing water from the retro-Dieckmann product, you instead make a lactone that then eliminates to the carboxylate.

[Heory]

I think movies is right. In some cases, just aldol condensation takes place, while in other cases it follows Michael addition to form a bicycle. I don't know why. How to predict the product without experiments? For alpha, beta-unsaturated aldehyde, the aldehyde group is very active, so how did the Michael addtion take place at first in some cases without additive such as CuX?

[jinclean]

i know a reaction named by DeMajo(maybe),the mechanism is is the [2,2]cycloaddition-retroclaisen condensation,it can form the seven member ring like that in this reaction.but i don't know how to move the aldehyde group to another position,it so strange that the with out oxidative reagent but we get a carboxy group  instead

[jinclean]

we can see this page :http://en.wikipedia.org/wiki/DeMayo_reaction

[movies]

The conjugate addition has to happen first.  If the aldol occurred first then you would lose the carbonyl activating group that you need to make the conjugate addition happen!  At any rate, the aldol could be reversible under these conditions so it might not be a problem.

As for the 1,4-selectivity, since the anion in question here is stabilized by two withdrawing groups, it behaves as a soft nucleophile (I wrote some about hard/soft theory here).  Most enolates will add to simple enones and enals in a conjugate fashion rather than the 1,2-fashion that you get with some organometallics.  It has less to do with the metal counterion than with the delocalization of the charge in an enolate.

Jinclean, I considered the De Mayo reaction as well, but it doesn't give the right substitution pattern, as you pointed out.  Additionally, you usually need light to get the [2 + 2] step to occur.  Good thinking though!

[orgopete]

I don't think it will form a bicyclic system. I really doubt dehydration as that looks like a Bredt's rule violation. I am going to attach a common retro-Claisen example from "A Handbook of Organic Chemistry Mechanisms".

The explanation I used in my classes for conjugate additions versus 1,2-addition was based upon reaction kinetics. I believe the 1,2-additions are fast, but depending on the pKa of the nucleophile, they may be reversible. If a have a nucleophile that is less basic than an alkoxide, it will reverse the 1,2-addition. Cyanide is a good example of a reversible addition to a carbonyl group. Acetoacetate I think will react similarly as it is still relatively acidic. If a slower conjugate addition occurs, protonation is more likely to occur on carbon, and the reverse reaction becomes less kinetically available. Especially as a beta keto ester will buffer the reaction away from enolate reformation.

[Heory]

orgopete, I like your explanations. But I don't quite understand your mechanism of this reaction without a scheme. And I cannot see a violation of a Bredt's rule. Maybe you misunderstood movies.
"Most enolates will add to simple enones and enals in a conjugate fashion". Did movies mean that in this reaction if the carbonyl or the ester group was removed from the reactant, it would still add to the enal in a conjuagte fashion? I cannot open the page offered by movies.

[orgopete]

Okay, here are the intermediates. You can see the Bredt's rule intermediate 5. Obviously, I don't know whether the reaction prefers to go through 4 or 9. I concede that thermodynamically, enolization of 3 would be easier than 9. Someone would have to look up whether a retro Claisen of 4 is an easy reaction or not. A carboethoxy bicyclooctan-8-one, or heptanone are probably known and whether is will undergo a retro Claisen should therefore also be known. Without making a model, I don't know if the cleavage to give the enolate would be aligned stereoelectronically. If the reaction went through intermediate 9, the stereoelectronic factors would be easy, but the thermodynamics of enolization would be more difficult, especially if the hydrolysis of the ester occurred simultaneously (by consuming base and forming the carboxylate).

This looks like a reaction I would try on paper, but not in the lab.

[Heory]

I've seen examples before. Intermediate 4 was the product in those reactions.

[orgopete]

On thinking further, I agree that the movies intermediate 4 seems likely. I also agree that there is nothing wrong with forming a bicyclic compound like this and they are reasonably common. I was more concerned with doing the retro Claisen from it. However, it seems a lot more likely to do a retro Claisen of the bicyclic compound than to do a Claisen on an ester with DBU.