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In the synthesis of N-ethoxycarbonyl L-proline methyl ester from L-proline and ethyl chloroformate, I use $\ce{K2CO3}$ to neutralize $\ce{HCl}$. Can I use $\ce{CH3CO2K}$ instead of $\ce{K2CO3}$? I think I cannot use it because it reacts with $\ce{HCl}$ forming acetic acid which can hydrolyse the ester. Is it correct?

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    $\begingroup$ The important thing the pot carb is doing is stopping the proline nitrogen from getting protonated as the reaction progresses by the HCl that is generated by the reaction. Pot. acetate will not do this. $\endgroup$
    – Waylander
    Aug 4, 2020 at 15:42
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    $\begingroup$ Think of this way: $\mathrm{p}K_\mathrm{b}$ of proline $\ce{NH}$ is 3.60, thus it is already protonated (zwitterionic). $\mathrm{p}K_\mathrm{b}$ of $\ce{AcO-}$ is 9.25 (weakly basic) and that of $\ce{CO3^2-}$ is 3.67 (much better base than $\ce{AcO-}$ and comparable with proline). Thus, you must use $\ce{K2CO3}$ as the base here. $\endgroup$ Aug 4, 2020 at 16:25

1 Answer 1

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Stay with the carbonate.

In the course of the reaction $\ce{HCl}$ is formed. The aim is to remove this from the reaction mixture both as soon as well as complete as reasonably possible. Compare

  • The addition of sodium acetate will increase the complexity of your reaction mixture. $\ce{HCl}$ is soluble in methanol (reference), and upon dissociation ($\ce{HCl <=> H+ + Cl-}$), the $\ce{H+}$ and the acetate $\ce{OAc-}$ join each other to generate the buffer system of $\ce{HOAc <<=> H+ + OAc-}$. Think about the $\ce{HOAc}$ as a reservoir, it may release $\ce{H+}$ at any moment you probably do not intend; a prominent example may be an aqueous workup.

    The quantity of $\ce{HOAc}$ generated may be small -- in comparison of the methanol used as solvent of reaction. Still it may complicate an aqueous workup and following extraction, because this organic substance solutes so well in water, but fairly enough in organic solvents such as hexanes, or chloroform, too. So your extraction of the product may be affected. A distillation (bp 118 C (pure acetic acid) instead of 65 C (pure methanol) at ambient pressure) ahead of the workup may remove this from your reaction mixture.

with the alternative

  • addition of $\ce{K2CO3}$. Darn efficient removal of $\ce{H+}$ for once and forever, because the quench along $\ce{K2CO3 + 2H+ -> 2K+ + H2O + CO2}\uparrow$ is practically irreversible. In addition, it allows you to monitor the advancement of the reaction since $\ce{K2CO3}$ is not well soluble in methanol.
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