In the Kolbe–Schmitt reaction, we know that the pi electrons of the benzene ring attack the carbon in carbon dioxide, but why doesn't the reaction occur on oxygen instead (as shown in the diagram below)?

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    $\begingroup$ Try to think the other way around: If such an attack would occur, what is the fate of the resulting intermediate, a carbonate half ester? With other words: Such a reaction might take place, but it is reversible and does not lead to a product that con be isolated. $\endgroup$ Feb 14, 2016 at 19:40

3 Answers 3


The resulting product would be a carbonate hemiester monoanion, specifically monophenyl carbonate. Similar species have been observed in carbonated alcoholic beverages. (Instead of phenol, the alcohol involved is ethanol so the carbonate hemiester is monoethyl carbonate.) There is an equilibrium between bicarbonate and the hemiester:

$$\ce{HCO3- + ROH <=> RCO3- + H2O}$$

According to the paper I linked the equilibrium constant is about 2.0, i.e.

$$K_{eq} = \frac{[\ce{H2O}][\ce{RCO3-}]}{[\ce{HCO3-}][\ce{ROH}]} \approx 2.$$

Under aqueous conditions, the concentration of water is very large and the concentration of bicarbonate is much lower, so usually the equilibrium concentration of the carbonate hemiesters is very low, on the order of a few mM.

In contrast, under the conditions of the Kolbe–Schmitt reaction, the carbon-carbon bond formation is effectively irreversible. There is no equilibrium, and the desired product can be obtained in high yield.


Carbon in the benzene ring are more prone to give their electrons than oxygen atom to CO2 because oxygen is more electronegative than carbon and thus carbon gives electron to carbon dioxide more easily. Comparing these effects with the help of electronegativity is not bad, because carbon and oxygen belongs to same period.

  • $\begingroup$ Sorry, but this is too simplistic. You're not comparing carbon anions with oxygen anions; both the carbon and the oxygen are part of a single pi system which can react at two points (or even three, if you consider ortho/para). $\endgroup$ Aug 23, 2022 at 13:26
  • $\begingroup$ @orthocresol Thanks for suggestion . Can you please elaborate more the problem so that this answer can be improved? $\endgroup$ Aug 23, 2022 at 13:30
  • $\begingroup$ @orthocresol Please elaborate which carbon and oxygen you are talking of : of the benzene or the CO2 $\endgroup$ Aug 23, 2022 at 13:31
  • $\begingroup$ First, you should consider whether you are addressing the kinetic product or the thermodynamic product: the latter is already covered by the accepted answer, so there's nothing more for me to elaborate there. And probably the question is more interested in the thermodynamic product anyway, because that is what is observed in practice. And if you are trying to address the kinetic product, I'm afraid it really requires a rather deeper analysis than what you've posted. $\endgroup$ Aug 23, 2022 at 13:36
  • $\begingroup$ @orthocresol Thanks, it looks like I am not much experienced to answer so deeply. I will edit it as soon as I will learn such higher chemistry. $\endgroup$ Aug 23, 2022 at 13:39

The phenoxide doesn't attack on oxygen because the oxygen on the carbon dioxide is electron withdrawing leaving an electron deficient region on the carbon atom.

  • $\begingroup$ You misunderstood the question. $\endgroup$
    – Jan
    Aug 18, 2016 at 23:12

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