Order of activating ability in electrophilic aromatic substitutions

Question

From A Guidebook to mechanism in organic chemistry by P. Sykes [1, p. 155]:

The activating effect of $\ce{Y}$ on the nucleus is found to increase, i.e. the overall rate of substitution increases, in the approximate order:

$$\ce{OCOR} < \ce{NHCOR} < \ce{OR} < \ce{OH} < \ce{NH2} < \ce{NR2}$$

$\ce{NR2}$ is more powerfully activating than $\ce{NH2}$ because of the electron-donating effect of the $\ce{R}$ groups. It should not be forgotten, however, that in acid solution, e.g. in nitration, these two groups will be con­verted into $\ce{^\oplus NHR2}$ and $\ce{^\oplus NH3},$ respectively; the nucleus will then be deactivated and substitution will be predominantly m- (cf. $\ce{^\oplus NR3},$ p. 151). The $\ce{OH}$ group is sufficiently activating to cause phenol to be brominated instantaneously to the 2,4,6-tribromo derivative (the p- and both o-positions all attacked) by bromine water at room temperature. The groups $\ce{OCOR}$ and $\ce{NHCOR}$ are less powerfully activating than $\ce{OH}$ and $\ce{NH2},$ respectively, because of the reduction in electron-availability on $\ce{O}$ and $\ce{N}$ by delocalisation over the adjacent, electron-withdrawing carbonyl group:

Why is $\ce{OR}$ a less activating group than $\ce{OH},$ but $\ce{NR2}$ is a more activating group than $\ce{NH2}?$

One would expect that that alkyl groups increase the electron density via their inductive effect and thus make the group more activating towards electrophilic aromatic substitutions (EAS) reactions, both in the case of $\ce{OR}$ as well as $\ce{NR2}.$ However, $\ce{OH}$ is a better activating group despite such an effect. Why is this so? Why would the same effect not be applicable for $\ce{NH2}?$

Reference

1. Sykes, P. A Guidebook to Mechanism in Organic Chemistry, 6th ed.; Longman; Wiley: Harlow, Essex, England; New York: New York, NY, 1986. ISBN 978-0-582-44695-3.

Answer 1

The most likely reason for this is that an $\ce{OH}$ group can act as an acid and exists in an equilibrium with the $\ce{O-}$ anion, which is significantly more activating.

$\ce{NH2},$ on the other hand, is not really that acidic. It does not benefit enough in this manner to be significant.

The loss of the anionic form is making $\ce{OR}$ less activating despite the inductive effect. $\ce{NR2},$ however, does not have as much to lose and only gains in activating ability.