How come bromine adds meta to phenol in this reaction, which is catalysed by HSbF6 [duplicate]








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[Electrophilic Aromatic Substitution: Are protonated OH-groups ortho-/para- or meta-directing?](/questions/8509/electrophilic-aromatic-substitution-are-protonated-oh-groups-ortho-para-or-me)

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[![enter image description here](https://i.stack.imgur.com/NesER.png)](https://i.stack.imgur.com/NesER.png)


This is a rxn in the synthesis of Tramadol. I found in in IChO prep problem 2001, Q19...


Edit: I have seen in clayden a rxn with HSbF6 whereby the benzene ring gets protonated so I followed that path of thinking and got stuck. Thanks for the answers and enlightenment!




Fluoroantimonic acid - $\ce{HSbF6}$ is made by combining $\ce{HF}$ and $\ce{SbF5}$. In this system, $\ce{HF}$ releases its proton ($\ce{H+}$) concomitant with the binding of $\ce{F-}$ by antimony pentafluoride, which is a Lewis acid. The resulting anion ($\ce{SbF6-}$) is both a weak nucleophile and an extraordinarily weak base.


Two things to note are:


1. $\ce{HSbF6}$ is a very strong acid.
2. Its conjugate base is a very poor nucleophile.


The hydroxyl group of phenol attacks $\ce{H+}$ and makes phenylhydronium ion which would behave like a nitro group. Therefore, the substitution would occur preferably on the 3 position.


Since $\ce{HSbF6}$ is such a strong acid, you can safely assume that virtually all of the phenol is protonated all the time.


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