Krapcho's decarboxylation is a reaction involving esters with a electron-withdrawing group $\beta$ to the carbonyl group and halide ions. It is typically conducted with $\ce {LiCl}$ and $\ce {DMSO}$ at high temperatures. However, in one of Krapcho's original papers, $\ce {NaCN}$ was also used alongside $\ce {DMSO}$.

What interests me is the first step of the reaction which involves the nucleophilic subsitution, likely by an $\ce {S_N2}$ pathway, of the alkyl group of the ester functionality. This is seemingly weird as the $\ce {C-O}$ bond usually does not break unless in acid catalysis. This is the case for the cleavage of ethers using $\ce {HX}$. I suppose the use of $\ce {LiCl}$ was to allow the $\ce {Li^+}$ to coordinate to the oxygen atom, allowing the $\ce {C-O}$ bond to cleave more easily. Why is it that $\ce {NaCN}$ can also do the job?

  • 1
    $\begingroup$ CN- is much stronger nucleophile. $\endgroup$
    – Mithoron
    Feb 17, 2019 at 20:50

1 Answer 1


So as mentioned the Krapcho-Decarboxylation follows a SN2-mechanism and is therefore working best with a methylester.

The solvent choice for DMSO (polar aprotic) is due to low stabilisation of the chloride anion rendering it highly reactive for the nucleophilic attack. The resulting methyl chloride is gaseous at room temperature and leaves the reaction mixture (le Chatelier -> formation of product strongly favoured) The leaving group of this substitution reaction decays into CO2 and the decarboxylated product. (Again: formation of a gas)

I think that the destabilisation of the cyanide in DMSO and the formation of stable acetonitrile can be considered as the driving force in the second reaction.


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