O-carboxylation in the Kolbe-Schmitt reaction
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:

$\ce{HCO3- + ROH &lt;=&gt; RCO3- + H2O}$

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.