So my textbook tells me that using a malonic ester to make a quaternary alpha carbon is impossible (I get that this might be hyperbole). However, it doesn't explain why and I don't understand why.

My thought was that after the first addition the tertiary compound would be too bulky to act as a nucleophile. However in the next section, an ethyl acetate is treated with LDA and acts as a nucleophlile 3 times to create a quaternary alpha carbon.

I also considered that you'd be making a tertiary carbon anion to make a second addition to a malonic ester's alpha carbon. However you also make a tertiary carbon anion using LDA on ethyl acetate and then alkyl halides.

I've added a screenshot of my textbook. Skip to the part in parenthesis if you want to read it.

Screenshot of the paragraph. Skip the part in parenthesis.

  • 1
    $\begingroup$ It might be useful to add which book you use and on what page the pictured excerpt is from, so that it is a proper reference. I guess it is Vollhardt & Schore. $\endgroup$
    – Jori
    Apr 10, 2015 at 10:34
  • 2
    $\begingroup$ Perhaps the author was trying to say that by using just the malonic ester synthesis you can't prepare a compound with 3 (potentially different) alkyl groups on the alpha carbon. You can only have 2 (potentially different) alkyl groups on the central carbon because the 2 other positions are occupied by the ester groups. Of course using two additional steps you can 1) decarboxylate the malonic ester and then 2) introduce the third alkyl group via alkylation. But using only the malonic ester synthesis you can't get 3 alkyl groups on the alpha carbon. $\endgroup$
    – ron
    Apr 10, 2015 at 14:33

2 Answers 2


Double alkylations of the $\ce{CH2}$ centre of malonic esters do work!

Have a look at this marvel by R. P. Mariella and R. Raube, published in Org. Synth., 1953, 33, 23

enter image description here


The use of sodium metal in a properly dried alkanol (in the case of ethanol, refluxing over magnesium ribbon is a good choice) was (and still is) an excellent and cheap method to generate sodium alkoxide in situ:

\[\ce{2Na + 2ROH -> 2NaOR + H2}\]

I don't know if nowadays commerically available sodium ethoxide and/or sodium hydride were that much used back in the days when the procedure was published. Fact is that both these can rot away in humid air when somebody leaves the bottles and cans open for too long, and this happens frequently in university labs.

With sodium metal, you're on the safe side: a freshly cut piece of metal is always good to go.


For the example given above, no further direct alkylation at that position is possible, and this is exactly, what ron pointed out in his comment to the question:

Perhaps the author was trying to say that by using just the malonic ester synthesis you can't prepare a compound with 3 (potentially different) alkyl groups on the alpha carbon.

He also suggested the viable solution, which I had outlined in another question on this site. Upon acidification and heating, $\beta$-ketoesters and 1,3-diesters undergo decarboxylation, which makes them great synthons to introduce $\ce{-CH2CO-CH3}$ and $\ce{-CH2COOH}$, respectively.

Converting a 1,3-diester to the corresponding carboxylic acid this way means that the new structure now again has an $\alpha$-proton that can be abstracted. The resulting enolate can be alkylated.

In the case of a carboxylic acid, this is usually done in a solvent like tetrahydrofurane or 1,4-dioxane, using two bases. For the deprotonation of the acid, $\ce{NaH}$ is sufficient, while the abstraction of the $\alpha$-proton, a strong base like $\ce{LDA}$ is used. (Exactly this is outlined in the textbook for the alkylation of an ester).

  • $\begingroup$ Some very neat regiochemistry there. Why was $\ce{Na}$ used? I think perhaps $\ce{NaOH}$ and certainly $\ce{NaH}$ would be sufficient Or is $\ce{NaH}$ produced in situ from the reaction of sodium with the solvent (ethanol)? $\endgroup$
    – Jori
    Apr 10, 2015 at 12:35
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    $\begingroup$ Also this does not really totally answer the question (sure it does from our perspective, but it would be good to know why the authors wrote what they wrote, assuming they don't write nonsense). $\endgroup$
    – Jori
    Apr 10, 2015 at 12:37

Traditional malonic ester synthesis post-processing includes hot acidic hydrolysis, which should lead to substituted malonic acid. However, such acids are prone to decarboxylation, leading to tertiary carbon.

In case you deviate from the tradition, there is no problem with getting quaternary carbon (though general bulkiness still applies and may prevent some synthesises)


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