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My attempt: $\ce{MeO-}$ preferably gives SN2 or E2, being a strong base and good nucleophile. But which reaction proceeds in this case?


1 Answer 1


I've given a strong hint by changing the title of your question!

  • Methoxide as a base can abstract a proton from an acidic $\alpha$-position of an ester, that is from $\ce{-C\color\red{H}_2COOMe}$.

  • The resulting carbanion/enolate can add to the other ester group of your starting material.

  • The cyclic intermediate eventually loses methoxide to form ...

This Dieckmann condensation is the intramolecular variant of a Claisen condensation.

I suggest to do it on paper!

Dieckmann condensation of methyl adipate

  • $\begingroup$ The only problem here is OMe- is not a good leaving group. So why does it leave? I think it owes to great stability of C=O bond. $\endgroup$
    – miyagi_do
    Commented May 9, 2015 at 1:15
  • $\begingroup$ OMe is able to leave in this instance because the reaction is under basic conditions, therefore OMe is already in solution (so it is stabilized by these conditions). $\endgroup$
    – Angelo A
    Commented May 4, 2017 at 6:30

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