Electrophilic Aromatic Substitution: Are protonated OH-groups ortho-/para- or meta-directing?

In Electrophilic Aromatic Substitution ($\text{S}\_{\text{E}}\text{Ar}$) reactions groups that can donate an electron pair into the aromatic ring, e.g. $\ce{-OH}$ or $\ce{-NH2}$, are *ortho*-/*para*-directing.


![](https://i.stack.imgur.com/66MCG.png)


In the case of $\ce{-NH2}$-groups one has to be cautious because if the $\text{pH}$ value of the solution is too low the $\ce{-NH2}$-group will be present mostly in its protonated form, $\ce{-NH3^+}$, which lacks a free electron pair and has a *meta*-directing effect because of its strong $-I$ effect. This brings up the question: What is the situation like with protonated $\ce{-OH}$ groups, $\ce{-OH2^{+}}$? In contrast to $\ce{-NH3^+}$ this group still has a free electron pair which could, in principle, interact with the aromatic ring ($\Rightarrow$ *ortho*-/*para*-directing) but it should also have a very powerful $-I$ effect ($\Rightarrow$ *meta*-directing).


![](https://i.stack.imgur.com/bCOdy.png)


I'd say, if the electron pair can be effeciently delocalised into the aromatic ring then its *ortho*-/*para*-directing power should outweigh the meta-directing power of the inductive effect. But I wonder if there even is any appreciable degree of delocalisation present because the mesomeric structures don't look very favourable:


![](https://i.stack.imgur.com/V0Zgb.png)


Thinking about it from a frontier orbital perspective I would say: The orbital of the $\ce{-OH2^{+}}$ group's free electron pair will be much lower in energy than that of the $\ce{-OH}$ group's free electron pair. Thus, the energy difference between free electron pair's orbital and the aromatic ring's $\pi$-system might be too large to facilitate a good interaction. Therefore, I'd expect a protonated $\ce{-OH}$ group to be *meta*-directing.


But what is the actual situation when phenols or phenol ethers undergo ($\text{S}\_{\text{E}}\text{Ar}$) reactions in strongly acidic solutions? Does one get the *ortho/para* or the *meta* product?


**Edit:**


I looked up the $\mathrm{pK}\_{\mathrm{a}}$ value for protonated phenol: It is approximately $-7$. This makes me wonder if there are any reasonable (real world) situations where the solution is acidic enough to protonate phenol (or phenol ethers) completely (even $\ce{HI}$ with its $\mathrm{pK}\_{\mathrm{a}}$ of $-10$ shouldn't be enough for that, right). I'm not sure if even sulfonation or nitration reach such low $\text{pH}$ values.
But @Uncle Al's answer made me aware that I forgot about Lewis acids and I'd say that Lewis acids like $\ce{AlCl3}$ that coordinate strongly to oxygen might lead to a state that will be very similar to a $\ce{-OH2^{+}}$ group (please, correct me if I'm wrong and the coordination of Lewis acids is not strong enough to be compared to the $\ce{O-H}$-bond). If this is the case, then what about Friedel-Crafts reactions which make heavy use of Lewis acids: Would those Lewis acids interfere with the *ortho*-/*para*-directing properties of phenols? Does anyone have experience with the directional properties of a $\ce{-OH}$ group in the presence of a strong Lewis acid, maybe even in the context of Friedel-Crafts reactions?




O-protonated phenols direct in meta, according to an [article on that topic](http://link.springer.com/article/10.1007%2FBF03155678#page-1). As it occurs, phenols can be, depending on acid, protonated on oxygen, but also on carbon. They can even be diprotonated, when most powerful [mixtures of HF and SbF5](https://en.wikipedia.org/wiki/Fluoroantimonic_acid) are used, but not twice on the same atom. O-protonated forms can be brominated, hydroxylated and alkylated. C-protonated forms don't posses such reactivity and are essentially [Meisenheimer's complexes](https://en.wikipedia.org/wiki/Meisenheimer_complex).


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A four-coordinate $\ce{sp^2}$ oxygen with two positive charges is a very high energy state. Aniline is a much weaker base than an aliphatic primary amine for leaking electron density into the ring. Phenol in turn will be a much weaker base than a primary alcohol. Deactivating the phenol will require a very Brønsted acidic medium indeed. 


Lewis acids would be more effective. Anhydrous $\ce{AlCl3}$ is a powerful oxophile. Adding a mole would turn off all the phenols. One might have quite a time getting the phenol free of the aluminum during workup. Note the large number of LAH workup recipes to avoid intractable aluminum oxohydroxide gels during workup.


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