# Thermodynamics behind organic chemical reactions? [closed]

I took organic chemistry a few years ago at my university, and although I went through the motions to pass the exams, I didn't understand the thermodynamic reasoning behind why the reactions actually occurred. For instance, I memorized that Claisen Condensation reactions occur between two esters in the presence of a strong base, but I don't fully grasp WHY the reaction happens in the first place. Why are the products more stable than the reactants in cases like these? Does this reaction always occur whenever these components are in a mixture, or is there a sort of catalyst to trigger it? I feel similarly confused for many other organic reactions (e.g., Aldol condensations, Fischer esterification, Sn2, etc). I suppose I'm wondering if there is a general thermodynamic reasoning I'm missing in organic chemistry.

## closed as too broad by Buck Thorn, Nilay Ghosh, Todd Minehardt, Mithoron, TyberiusMay 14 at 1:16

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• Generally speaking, we would refer to the concept of the Gibbs free energy change (en.wikipedia.org/wiki/Gibbs_free_energy) of reaction when talking about the thermodynamic driving force of reactions. It essentially relates to the 2nd Law of Thermodynamics, which tells us that the total entropy of the universe always increases. This is the essential idea behind all reactions, not just those in organic chemistry. – Tan Yong Boon May 13 at 15:04
• Alkoxides form ester enolates that add to electrophilic carbonyls of neutral esters with loss of alkoxide to form beta-ketoesters. Because beta-ketoesters are much more acidic than the conjugate acid of the alkoxide, they are deprotonated by the alkoxide which drives the reaction to completion. Thus, the reaction is at least stoichiometric in base. In your example, 2 beta-keto esters will form. The other one is where the red and blue esters switch places. – user55119 May 13 at 20:14

Whether or not the product is "more stable" would really depend on the sense in which you consider its reactivity. The $$\mathrm{p}K_\mathrm{a}$$ of a $$\beta$$-keto ester is about 11, so in that sense, it is much more reactive than either of the reactants. Once the $$\beta$$-keto ester is formed it is quickly deprotonated. It remains predominantly in this form because the anion is so stable, and the reaction is incapable of reversing. Thermodynamics does not always determine the result of an organic reaction. Many times, kinetics is the deciding factor.
• I suppose that’s fair, but another equivalent of alkoxide is generated ($\ce{OR'^-}$). – ringo May 13 at 15:43
• Yes but no; if you check the mechanism very carefully (took me a while to do), overall the transformation is $$\ce{2RCH2COOR' + OR'- -> RCH2C(O)C(R)^-CO2R' + 2ROH}$$ The first equiv of base is used to deprotonate the ester starting material. After addition of enolate into ester, the tetrahedral intermediate kicks out one equiv of base, so we are at net zero base consumption/production here. However, you then need one equiv of base to deprotonate the b-ketoester. – orthocresol May 13 at 16:15