[davidenarb]

Normally, alcohols do not undergo dehydration in the presence of a strong base, so why, in the aldol condensation, it happens?   Why does -OH group leaves?

I know that the product formed is conjugated, which is a stabilizing factor, but this does it mean that -OH IS ABLE to leave! I cannot make a connection between the fact that the product formed is stable and the OH can leave or even the fact that the intermediate is resonance-stabilized

Thanks for clarification

[davidenarb]

is is because the fact that the negative charge of OH is not being formed, but instead, it is just transferred from one location (the intermediate) to another one (the leaving group) ?

However, the point is that in the intermediate, the negative charge was resonance stabilized, after the alpha beta unsaturated product is formed, the negative charge is located in a less stabilized anion (OH-)

[quantumnumber]

Well, aldol reaction follows a E1cB mechanism. OH^- is not a leaving group in E2 reactions, but it can be a leaving
group in E1cB. As you said, the anion it is lost from is already an alkoxide, and the conjugation in the product also assists loss of HO–.

[davidenarb]

Well, aldol reaction follows a E1cB mechanism. OH^- is not a leaving group in E2 reactions, but it can be a leaving
group in E1cB. As you said, the anion it is lost from is already an alkoxide, and the conjugation in the product also assists loss of HO–.

my question is WHY a poor LG can leave? can I have the answer in a nutshell?

[quantumnumber]

The product is much more stable, so thermodinamically the reaction takes place. If a LG is good or bad depends on several factors, and one of them is the relative stability of the product that is going to be formed when the LG is expelled. In this case this is the key factor.

[spirochete]

Strongly basic leaving groups (RO- and stronger) are never formed in Sn2/Sn1/E2 and E1 reactions. Strongly basic leaving groups (barring carbonions and the hydride ion) such as RO- are formed in very exothermic, unimolecular steps such as E1CB elimination. Unimolecular reactions where one molecule splits into two molecules have less of an entropic transition state penalty than things like an Sn2 reaction.

This can also be rationalized partially with the Hammond Postulate: more exothermic reactions have more reactant like transition states, so the negative charge on the leaving group is not felt as strongly. 

Another good example of a strongly basic leaving group is the breakdown of many tetrahedral intermediates under basic conditions. The same logic used to rationalize an E1CB elimination can be used to describe this step during the basic hydrolysis of an ester, for example.

[rwiew]

I would also invoke the kinetic vs thermodynamic control (which is pretty much what you guys are saying above as well). Most subtitutions and eliminations will be under kinetic control I believe, hence having a "good" leaving group and hence a low activation energy matters. In the enol reaction, the elimination is an equilibrum effectively and under thermodynamic control - it doesn't matter much how bad the leaving group is, but the comparison of the stability of the product to that of the starting material is important.