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I am currently reading chapter 26 of Clayden's Organic Chemistry, and a diagram for the elementary steps of the Claisen condensation with ethyl acetate in ethoxide is provided. It is obvious why the net reaction is favorable. The 1,3-dicarbonyl compound has a pKa of ~10 while the pKa of ethanol is 16 (so the more acidic 1,3-dicarbonyl compound will be the product in its deprotonated form). Even though the first few elementary steps may not favor the product, the last step is essentially irreversible.

I am having trouble understanding why the second equilibrium arrow claims to favor the formation of the reactants and not the products.

The first equilibrium arrow is obvious, the pKa of ethanol is 16, and the pKa of the ester enolate is ~25 (when protonated), so the less acidic ester will be favored in equilibrium. However, in the second equilibrium arrow, the pKa of the starting enolate is ~25 (when protonated) while the pKa (when protonated) of the product alkoxide is likely somewhere around 16. Wouldn't this mean the product for the second equilibrium arrow would be favored thermodynamically? I have considered that this step does result in a decrease in entropy, but I wouldn't expect this to have a drastic impact on the direction of the equillibrium.

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    $\begingroup$ Eli: I think you are reading too much into the magnitude of the equilibrium arrows. If the reaction is conducted in the presence of ethanol, then the arrows are likely pointing in the right directions. Notice that the three sets of arrows are all the same and are probably pre-set as a symbol. Realize that the reaction is not catalytic in base. $\endgroup$
    – user55119
    Dec 6 '20 at 22:18
  • $\begingroup$ @user55119 Is the direction of the equilibrium arrow due to the fact that this is a bimolecular process occurring in dilute solvent (i.e. more molecules are dissociating than coming together due to the dilution)? I know it is not that important to the net reaction, but I am just curious! $\endgroup$
    – Eli Jones
    Dec 6 '20 at 23:00
  • $\begingroup$ I think that there is little thought to the magnitude of the arrows. Based on pKa's they are probably correct. $\endgroup$
    – user55119
    Dec 6 '20 at 23:36
  • $\begingroup$ The second step is the nucleophilic attack of enolate on the ester, it is not a deprotonation, so pKa values would not affect the equilibrium of that step. $\endgroup$
    – S R Maiti
    Apr 11 at 12:06

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