Mechanisms for Reimer-Tiemann and Kolbe-Schmidt reaction on deuterium labelled phenol I was solving a bunch of reactions when I encountered the following:

So for the first question:

This is a Reimer-Tiemann reaction. I felt that the tautomerism would result in the deuterium getting attached to the phenolic $\ce{O^-}$. As it turns out, that was wrong. Oh well. Then I proceeded with the second question. Now this one is a Kolbe-Schmidt reaction, where the $\ce{CO2}$ liberated will react with the phenol. Going by the previous reaction, this time I decided not to add the $\ce{D}$ atom to the phenolic $\ce{O-}$. Consequently, this is how it panned out:

All of this has left me quite confused. So here's my question:

Why is it that the $\ce{D}$ atom during tautomerism adds on to the phenolic $\ce{O-}$ in the Kolbe-Schmidt reaction, but does not proceed to do so in the Reimer-Tiemann reaction?

Image credit: Created using Marvin JS
This is the mechanism of the Reimer-Tiemann reaction (source Wikipedia here)

The polyhaloform reacts with strong base to create a carbene, a highly reactive species which reacts with the phenoxide anion through the alpha carbon to give an anionic intermediate with the -ve charge on the halogenated carbon. It is this species that picks up the deuterium.
In Kolbe-Schmidt reaction, the process does not proceed through a carbene. It is nucleophilic attack (through carbon) of the phenoxide on the polyhaloform (or carbon dioxide) and the deuterium shifts to the phenol oxygen as the product re-establishes aromaticity. Mechanism from Wikipedia here