2
$\begingroup$

A "reaction map" in my lecture handout suggests that the product of this reaction is: enter image description here

So, basically overall it just merges two ethylacetoacetate molecules together at the central carbon. What is the mechanism for this? My thoughts are to swap the acidic proton for an Iodine atom by making the enol (in acidic conditions to avoid iodoform reaction) and react with $I_2$. Then, the enol of another molecule of EAA substitutes the Iodine to make the product. However, I don't know whether this is $S_N1$ or $S_N2$. Typically alkylation is $S_N2$ (I don't know why, please explain if you can) but in this case its a bulky secondary substituent and the solvent is protic (acidic conditions). Surely it's $S_N1$, right?

$\endgroup$
2
$\begingroup$

Are you certain that the coupling itself isn't a radical reaction?

The bond energy of an ordinary $\ce{C-I}$ bond is in the range of $210\,\mathrm{kJ\cdot\,mol^{-1}}$, which isn't a hell of a lot. Moreover, the resulting radical of an 1,3-dicarbonyl compound is nicely stabilized.

$\endgroup$
  • $\begingroup$ Iodine is definitely involved. $\endgroup$ – RobChem Apr 8 '15 at 22:37
  • 1
    $\begingroup$ Sorry if I didn't make myself clear enough :) I meant: homolytical cleavage of the iodinated $\beta$-ketoester is the first step, yielding an iodine atom and a stabilized radical. $\endgroup$ – Klaus-Dieter Warzecha Apr 8 '15 at 22:44
  • 1
    $\begingroup$ On a side note, these radicals of 1,3-dicarbonyl compounds can also be generated (this time without the iodine) using $\ce{Mn(III)}$acetate. Barry Snider publishes a lot on that. $\endgroup$ – Klaus-Dieter Warzecha Apr 8 '15 at 22:49
  • 1
    $\begingroup$ @RobChem My pleasure! $\endgroup$ – Klaus-Dieter Warzecha Apr 8 '15 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.