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They ask us to give the structure of the intermediate "A" and the mechanisms of both steps

My attempt: I found the name of the started and end molecules on reaxys.com

NaOEt could deprotonate an alpha carbon of one of the two carbonyl groups. This alpha carbon would then attack the carbonyl carbon of the other (!) carbonyl group. This would create a six membered ring. So the intermediate "A" would be a ring where the ester groups are still attached to the carbons.

Would this make sense in this case ?

EDIT: after googling a bit more, maybe this reaction above could be a "Dieckmann Condensation", is this possible ? The alpha carbon is deprotonated, and attacks the carbonyl carbon of the other ester group. This is exactly what could be happening above, or am I missing something ? The second step would be hydrolysis of ester, followed by decarboxylation. Is that correct ?

  • 1
    $\begingroup$ Well, yeah, that's the thing. $\endgroup$
    – Mithoron
    Commented May 3 at 22:33
  • $\begingroup$ @Mithoron Are you sure ? After googling, maybe this makes sense but I prefer to ask. Feel free to post an answer if you want $\endgroup$
    – wengen
    Commented May 3 at 23:32
  • 2
    $\begingroup$ You are correct. This is a classic Dieckmann $\endgroup$
    – Waylander
    Commented May 4 at 6:21

1 Answer 1


This is a good question but there are few similar kind in CSE web base (see here and here for examples). However, it did not catch our reviewers' attention because of the unique nature on its starting material. Yet, I should give Mithoron and Waylander the credit who deserved to answer the question since they both agreed on OP's opinion identifying the name reaction belong to the scheme. Therefore I waited few days before to post my answer.

The original reaction (first step in the question) is the intramolecular version of Claisen condensation, which is a carbon–carbon bond forming (intermolecular) reaction that occurs between two (enolizable) esters or one (enolizable) ester and another carbonyl compound in the presence of a strong base to give a $\beta$-keto ester as the product. The major requirement to undergo this reaction is at least one of the reagents must be enolizable, meaning it has an $\alpha$-proton and be able to undergo deprotonation to form the enolate anion (other requirement is to use a non-nucleophilic base such as ethoxide in a little excess over stochiometric amount). The mechanism of the reaction is suggested as follows (see Wikipedia or Organic Chemistry Portal):

Mechanism of Claisen condensation

Basically, the last step is neutralization, adding enough acid to convert all enol-form to keto-form. No heating involved.

As pointed above, the intra-molecular version of Claisen condensation is called Dieckmann Condensation. Discovery and development of this reaction are generally credited to the German chemist Walter Dieckmann, who found that heating an adipic or a pimelic ester with sodium and a trace alcohol led to cyclization with formation of a cyclopentanone or a cyclohexanone. The mechanism of the reaction is suggested as follows (see Wikipedia or Chemist.Swiswiswift):

Mechanism of Dieckmann condensation

Keep in mind that at least first four steps are in equilibrium (read comments here for valuable conversation).

The following scheme schematically demonstrates the use of Dieckmann condensation for the synthesis of 4-piperidones, which is the question above:

Synthesis of 4-piperidones

According to the scheme, your intermediate product $\bf{A}$ (before hydrolysis) is ethyl 1-methyl-4-oxopiperidine-3-carboxylate, which can be isolated if desired. The hydrolysis of $\bf{A}$ under hot acidic condition leads to decarboxylation at $\beta$-position to give the final product, 1-methyl-4-oxopiperidine (for mechanism for this reaction, check decarboxylation of $\beta$-keto carboxylic acids).


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