The following reaction came in a recent test:

reaction in test

According to this reaction, the acidic hydrogen would protonate the ylide, and the standard wittig reaction would not take place. This source agrees with the same results.

However, according to El-Batta et al.1, the Wittig reaction in water can take place even in the presence of acidic groups, such as carboxylic acids and phenols:

It is also interesting that the Wittig reaction in water works well with substrates having acidic unprotected functional groups present; hence the presence of a carboxylic acid group (vide infra) or phenolic groups, such as p-hydroxybenzaldehyde, gives the corresponding cinnamate product 16 in 92% after heating the reaction for 1 h at 90 °C (entry 16). In sharp contrast, the same reaction in refluxing CH2Cl2(4 h) has been reported to provide only 8% of 16

reaction 2

reaction 3

Which one of these references is correct? Does the Wittig reaction really proceed in the presence of acidic groups? I'm looking for a good reference, along with an explanation on what occurs when a carboxylic acid group is present on the substrate.


  1. El-Batta, Amer, et al. “Wittig Reactions in Water Media Employing Stabilized Ylides with Aldehydes. Synthesis of α,β-Unsaturated Esters from Mixing Aldehydes, α-Bromoesters, and $\ce{Ph3P}$ in Aqueous $\ce{NaHCO3}$.” The Journal of Organic Chemistry, vol. 72, no. 14, July 2007, pp. 5244–59. doi:10.1021/jo070665k.
  • $\begingroup$ Here exactly 1 equivalent of the ylide is given so I guess that matters. $\endgroup$ Commented Sep 21, 2020 at 9:14
  • $\begingroup$ By the principle of least motion, one would expect the deprotonation to occur faster than the Wittig reaction, as the former requires less nuclear rearrangement, especially considering the fact that you have a single equivalent of the ylide, so it would be anticipated theoretically that the deprotonation occurs preferentially in this case $\endgroup$ Commented Sep 21, 2020 at 9:24
  • 4
    $\begingroup$ You are comparing an unstabilized ylid, Ph3P=CH2, with a stabilized ylid, Ph3P=CHCO2Et. The latter can be stored as a solid; the former can't. The former is protonated by RCO2H. The latter, if protonated, is easily deprotonated by a weak base. $\endgroup$
    – user55119
    Commented Sep 21, 2020 at 13:50

1 Answer 1


This works because the acidic starting material is going to be deprotonated as there is excess bicarbonate present. Because of the adjacent carboxylate group, the pKa of the phosphonium species is going to be around 6-7 pKa Table here so within the range where bicarbonate can deprotonate it. The reaction is between the anion of the starting material and the anion of the phosponium species in aqueous phase. This will only work for readily deprotonated phosphonium species and aqueous soluble aldehydes in the presence of excess base.

  • $\begingroup$ This and @user55119 's comment clarified it for me. Thanks! $\endgroup$ Commented Sep 22, 2020 at 7:31

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