Claisen ester condensation equilibria-model calculations

Claisen Ester Condensation. Equilibria—. Model Calculations. Forty years ago Charles R. Hauser and co-workerseluci- dated the role of the promoting ...
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John F. Garst The University of Georgia Athens. GA 30602

Claisen Ester Condensation EquilibriaModel Calculations

Forty years ago Charles R. Hauser and co-workers elucidated the role of the promotina base in the Claisen ester condensation (11,tvpical t~x:tntplesnfwhicha r e ~ i v e nin eqns. ( 1 1 and 12).Kqt~ation13)represents the general rase. 2 CHsCOOEt s CH&OCHzCOOEt

+ EtOH

2 (CH3)2CHCOOEtt (CH3)~CHCoC(CH3)2CooEt + EtOH 2EeK+EtOH

(1)

K+B-*K-+BH EtOH + B- e EtO- + BH

(7) (8)

The equilibrium constants governing reactions 4-8 are estimated from pK, values by neglecting medium and ion aeereeation effects. We e m ~ l o vthe followine nK. values:

(2) (3)

The reactant ester E must have a t least one a hydrogen per molecule. K represents the product 8-keto ester. The success of the reaction depends on the balance of eauilihrinm 3 and others with which it is coupled. The prontoriny hase plays roles other thnn that of a truc catalyst, and its choice can be rritiral. Rrnrtim 1,ior example, is promotnl by EtO-, hut reaction 2 fails with this base. Reaction (1)is typical of cases where K has a t least one hydrogen a to COOEt. In such cases, EtO- is sufficiently basic to deprotonate K (eqn. (4)), the a hydrogen of which is exceptionally acidic. K + EtO- e K- + EtOH (4) The composite of equilibria (3) and (4) is not unfavorable, and EtO- is an adequate promoting hase. That the enolate K- is present a t equilibrium rather than the desired product K is no impediment because acidic workup protonates enolates rapidly and quenches the condensation. Reaction (2) is typical of cases where K has a t least one y, hut no a, hydrogen. The acidity of K is then insufficient for reaction 4 to he significant. Equilibrium 3 is unfavorable, and EtO- is unsuccessful as a promoting base. However, much stronger bases B-, e.g., PhaC-, succeed. Textbook explanations of the strong base effect in the latter case are interesting for their diversity: "The most acidic hydrogen [of K] is a normal proton cu to a ketone carbonyl; its pK. is therefore about 20. If a much stronger base [than EtO-] is used to catalyze the reaction, this proton can be removed and the reaction can now be observed (2). "This limitation [lack of reaction with EtO-1 can he overcome by the use of a very strong base that converts the starting ester eomoletelv . .to its enalate" (3). . . ". . . the base is now atrnne enough ,. to convert the alcohol formed in the rra