When an atom with a lone pair is adjacent to a pi bond (such as an aromatic ring) there is generally a re-hybridization of the atom to put one of the lone pairs in a p orbital, as it improves orbital overlap. (Think of it as helping to improve resonance between the O and the pi bonds)
It’s less apparent in enols / phenol but for example with enamines, the nitrogen geometry is trigonal planar, not tetrahedral.

https://www.masterorganicchemistry.com/2018/01/16/a-hybridization-shortcut/

Excellent article – but I believe there may be a small error. In 4. Properties of Enols, the O in the phenol species is described as sp2 hybridised but I don’t see how that is possible.

In general the % of enol form is highest for aldehydes, then decreases for ketones and becomes essentially undetectable for esters.

My understanding is that in the absence of other factors (e.g. hydrogen bonding, resonance, aromaticity) favorability of the enol form is closely linked to the factors that govern the thermodynamic stability of alkenes. Basically there is hyperconjugative donation of adjacent substituents into the C-C pi* bond. Electron-withdrawing groups will lower the energy of the pi* orbital, whereas electron-donating groups (like OR) will raise the energy of the pi* orbital, making these stabilizing interactions less important and lowering the stability of the alkene (enol) form.

The ketone or aldehyde must have at least one C-H bond on one of its alpha carbons!

I was just wondering; when an enol tautomerizes, does it form a double bond with the most or least substituted neighboring carbon?

Many thanks in advance