In the Claisen condensation, an alkoxide $\ce{RO-}$ acts as a base. However, in trans-esterification, an alkoxide $\ce{R'O-}$ attacks the carbonyl carbon of an ester as a nucleophile. Is this because a different 'R' group is used in trans-esterification?
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asked Jul 7, 2016 at 10:36
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$\begingroup$ RO- doesn't act as a nucleophile in the claisen condensation. It's like an Aldol but the alkoxy group leaves giving a beta keto ester. $\endgroup$– RobChemCommented Jul 7, 2016 at 12:23
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$\begingroup$ my bad, I meant it acted as a base $\endgroup$– MrObjectOrientedCommented Jul 7, 2016 at 14:55
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1 Answer
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In the Claisen condensation, notice that the alkoxide base usually matches the ester given. For example, if an ethyl ester is used, the base will be ethoxide. The reason for this is that trans-esterification is occuring, except that the product is the same as the reactant.
In general, the Claisen condensation is slow compared to trans-esterification. The rate limiting step of the Claisen is the C-C bond formation. Keep in mind that there will be very little of the active ester enolate nucleophile present, because the equilibrium of alkoxide deprotonating an ester's alpha-position will lie dramatically to the ester (by ~ 10 billion fold).
If one tried to run a Claisen condensation where the base does not match the ester (e.g., an ethyl ester with methoxide as the base), I would expect a mixture of methyl and ethyl esters as the products.
answered Jul 7, 2016 at 16:21
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$\begingroup$ Figures to be added when I get a chance later. $\endgroup$ Commented Jul 7, 2016 at 16:22