In the following mechanism of the Dieckmann Condensation, the product has been formed in the fourth step, but still the reaction proceeds and $ \ce{RO^{-}}$deprotonates the product. Then $\ce{H^{+}}$ rejoins to form the same product. Won't this result in extra expenditure of energy? Also, if the reaction proceeds through this mechanism, shouldn't there be an infinite loop?

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Can anyone explain why is there an extra step?

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    $\begingroup$ In the depicted mechanism there is consistently a charge for the $\ce{RO-}$ omitted. $\endgroup$ Mar 23, 2016 at 7:38

1 Answer 1


The first four steps (i.e. the main reaction) are conducted under basic conditions due to the presence of alkoxide. When that reaction is complete, the result is the deprotonated ketoester--the formation of that stable species is what drives the reaction to completion. Note that the first 3 steps are shown as equilibria, whereas the fourth step is essentially irreversible.

Later, during workup (shown above as the fifth step of the process), acid is added to neutralize that anion and form the final product.

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    $\begingroup$ "Stable" means the ketoester anion is not a strong enough nucleophile to attack a ester or ketone irreversibly. $\endgroup$
    – Lighthart
    Mar 22, 2016 at 14:46
  • $\begingroup$ @Ishu This is why a slight stoichiometric excess amount of base is required, where a typical NAS reaction requires only a catalytic amount of base. Here, yes, the base is catalytic for the equilibrium steps, but an equivalent of that base is neutralized by the product immediately upon formation. Experimental data supports this, which lends evidence to this mechanism. $\endgroup$ Mar 22, 2016 at 14:57
  • $\begingroup$ I would assume, that the enol-form would be predominant in this reaction. Just like it is observed for other compounds like for example here. Could you comment why the beta-ketoester should be the main product here? I don't think that the restriction of the ring is any indicator for a completely different behaviour. I also very much doubt that the first and the second step are in equilibrium, the C-C bond formation should be thermodynamically favoured over any other reaction involved. $\endgroup$ Mar 22, 2016 at 16:14
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    $\begingroup$ @Martin-マーチン Tautomer distribution depends on the solvent, probably not keto-enol in DCM, but probably in THF, right? And the C-C bond formation is reversible, moreso than a beta-keto ester deprotonation by an alkoxide. That's a pKa difference of 15 or so in DMSO, but think about the malonic ester synthesis and the C-C bond fission there. $\endgroup$ Mar 22, 2016 at 16:27
  • $\begingroup$ @DGS Reversible does not mean equilibrium. Protonation and deprotonation are usually equilibrium processes. As such, the keto-enol tautomerie is an equilibrium, too. I find it hard to believe, that under these conditions, the ring closing step is not the thermodynamic well, from an entropic perspective, this is very likely. I am also not convinced, how much the keto-enol tautomerie is dependent on the solvent. But I also don't want to worry about that any more. $\endgroup$ Mar 23, 2016 at 7:37

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