Most reaction mechanisms use $\ce{EtO-}$ as a base to deprotonate the alpha hydrogen in the malonic ester sythesis. But in the equilibrium there are still esters that haven't been deprotonated. Why won't the enolate do a nucleophilic attack on the ester, just like in the Claisen reaction?

Why isnt LDA used instead, which would deprotonate all the ester molecules, preventing Claisen reactions?

Enolate formation with ethoxide as base

  • 2
    $\begingroup$ I suspect steric hindrance plays a part. The enolate is a fairly bulky nucleophile and would be attacking a bulky ester group so I imagine the kinetic barrier is high. The wikipedia page for malonic ester synthesis indicates that diakylation is a problem so clearly this route is much more favourable than the one you are suggesting. $\endgroup$
    – bon
    Commented Sep 11, 2015 at 18:11
  • $\begingroup$ LDA should also work as expected $\endgroup$
    – K_P
    Commented Sep 12, 2015 at 0:50

2 Answers 2


Understanding how this reaction is actually performed in the lab is key to answering your question.

Typically in the malonic ester synthesis, a full equivalent (or 2 equivalents if you are carrying out a dialkylation) of base is used, so little, if any, ester remains with the $\ce{\alpha}$-proton still in place.

enter image description here

(image source)

Also, keep in mind that both the alkylation and Claisen reaction have an activation energy. This means that it is usually necessary to heat the solution to bring about reaction (see here for a lab example).

The reaction is carried out by first forming the deprotonated malonate ester at room temperature and then dripping it into a refluxing solution of the alkylating agent ($\ce{RX}$ in the above diagram). This procedure achieves the following:

  • concentration of non-deprotonated ester is very low
  • deprotonated ester is formed at room temperature, so even if any non-deprotonated ester remains there is not enough energy (at room temperature) to make it over the energy barrier and react
  • the deprotonated ester is dripped into the solution of refluxing alkylating agent; therefore, now that there is enough thermal energy to bring about reaction, the relative concentration of alkylating agent (the entire solution) is much higher than the relative concentration of deprotonated ester (added drop by drop) making the desired alkylation reaction much more probable than reaction of deprotonated ester with protonated ester (which, for reasons explained above, is in very low concentration in any case).

It's unnecessary.

In the reaction you are suggesting, the enolate would have to attack the carbonyl double bond from above. The optimum angle of attack is roughly $107^\circ$ but since both the attacking enolate and the diester substrate are reasonably bulky, there I suspect there will be a considerable amount of steric hindrance here, creating a large kinetic barrier to the reaction. If you compare this to attack by an alkyl halide then you can see that the latter reaction is strongly kinetically favoured.

enter image description here

Additionally, the problem may not be as bad as you think. According to Bordwell's $pK_\mathrm{a}$ table, diethylmalonate has a DMSO $pK_\mathrm{a}$ of 16.4. If you compare this to ethanol with a DMSO $pK_\mathrm{a}$ of 29.8, you can see that the malonate ester is actually quite acidic, and so ethoxide is easily a strong enough base to achieve facile deprotonation. Even in a protic solvent such as water (this reaction is probably done in ethanol but the effect is similar) which greatly reduces the ethoxide ion's basicity, Evans' $pK_\mathrm{a}$ table gives a value of 13 for dimethylmalonate and 15.5 for methanol (no data given for the ethyl variants) so the malonate ester is still a couple of orders of magnitude more acidic.

  • 1
    $\begingroup$ In addition: using a base like a methanolate (or an ethanolate) yields after the deprotonation methanol (or ethanol) that may be distilled off easily as azeotrope, for example with toluene as a co-solvent, favouring the advancement of the reaction. $\endgroup$
    – Buttonwood
    Commented Sep 11, 2015 at 21:52

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