# Can we prepare aryl halides by heating with halogen acids?

My book says, we can't prepare aryl halides ($\ce{RX}$, where $\ce{R}$ is aryl group and $\ce{X}$ is any halogen) by treating phenol with halogen acids ($\ce{HX}$, where $\ce{X}$ is halogen). It is said to be, because of partial double bond character of carbon-oxygen bond in phenols ($\ce{C6H5OH}$), which is difficult to break (I don't know how there is partial double bond character). What I thought was, if we assume there is no partial double bond character between $\ce{C}$ and $\ce{O}$, we can heat the reactants (i.e. phenol and halogen acid), so that $\ce{OH}$ bond eliminates from the phenol (which has the least bond dissociation energy than $\ce{CH}$), and $\ce{X}$ can get dissociated from the $\ce{HX}$, then it can add up to $\ce{R}$ (aryl group) to form aryl halide. So, can we prepare aryl halides by heating with halogen acids?

• This Electrophilic aromatic substitution- no substitutions are ever simple (always involve catalysts) since the resonance of the aromatic (inelegantly expressed as partial double bonds) makes the side chain far more stable than its equivalent aliphatic e.g. hydroxyl-cyclohexane in your case. A quick wikipedia search will show the complexity- its not much different between substituting a hydrogen atom or hydoxyl group- either way it first needs to separated from it parent chain- the most difficult step with resonant structures. – user2617804 Nov 5 '13 at 2:15
• Can you propose mechanism for the presented reaction? – ssavec Nov 5 '13 at 7:03
• It sounds like the questioner wants to perform a nucleophilic substitution on the aromatic ring. Conceivably, the acid can protonate the phenol, creating a relatively good leaving group (oxonium) which could be replaced by the nucleophilic halide ion. However, this is probably unlikely to happen without very energetic conditions, as the aromatic ring strongly repels nucleophiles' electron pairs. The only reaction of that class I know uses aryl diazonium salts. The diazonium group is just about the best leaving group there is, far better than oxonium. Only then can it be replaced. – Nicolau Saker Neto Nov 6 '13 at 0:55
• OH-group in some strongly electron-deficient phenols may be replaced with halogen using $\ce{PCl5}$. As an immediate example I can come with picric acid, but probably other examples exist. The reaction, however, happens as nucleophilic substitution with oxygen firstly bound to phosphorous by replacing one chlorine. The net reaction is $\ce{Ar-OH + PCl5 = ArCl + HCl + POCl3}$ – permeakra Dec 31 '15 at 22:46

The partial double bond character of the C-O bond in phenol becomes obvious from the resonance structures. Contributions from the structures with $\ce{C=O}$ double bond explain why the character of the C-O bond is in between single and double bond, and also show that oxygen lone pairs are less available for protonation to form $\ce{Ph-OH2+}$.

Protonated phenols are therefore fairly acidic, with a $pK_a$ around -6. $\ce{HBr}$ and $\ce{HI}$ have significantly lower $pK_a$ values (-9 and -10, source), so they would indeed be able to protonate phenol. However, $\ce{H2O}$ is unlikely to be eliminated from this compound due to the instability of the resulting phenyl cation. Its positive charge cannot be delocalized into the aromatic system as it resides in an empty $sp^2$ hybrid orbital which is perpendicular to the delocalized $\pi$ bonds (source). Formation of the phenyl cation by elimination therefore requires much better leaving groups like $\ce{N2}$, which can be eliminated from aryl diazonium cations.

Substitution of the protonated OH group by halides with an $S_N2$-like mechanism is hindered by Coulomb repulsion between the $\pi$ electron cloud of the benzene ring and the lone pairs of the nucleophile. This reaction only works with aromatic compounds in which the electron density in the $\pi$ system is decreased by multiple electron-withdrawing substituents, like nitro groups.

We cannot. There are two mechanisms possible for transforming alkanols into alkyl halides, commonly known as $\mathrm{S_N1}$ and $\mathrm{S_N2}$. For the former, a leaving group (in our case: something from the hydroxy group) must dissociate away to create a carbocation which can then be captured by the halide. In the second case, the halide attacks from the rear side of the carbon-hydroxyl bond and displaces it.

Both mechanisms are only possible if the hydroxy group was initially protonated, because $\ce{OH-}$ is a terrible leaving group (but $\ce{OH2}$ is okay).

In case of phenol, the $\mathrm{S_N2}$ mechanism is not practical, since the rear side of the $\ce{C-OH}$ bond is blocked by the phenyl ring. An $\mathrm{S_N1}$ reaction is also strongly disfavoured since the intermediate would be the very unstable (with respect to other carbocations) phenyl cation — the charge is sitting in an $\mathrm{sp^2}$ hybrid orbital in the ring plane and cannot be delocalised across the ring. Hence you can protonate phenol as much as you want, water will never displace itself.

Finally, there is also the $\mathrm{S_NAr}$ mechanism which is essentially a two-step process: attack of the nucleophile onto the carbon in question with loss of aromaticity in the ring followed by displacement of the leaving group regaining aromaticity. This mechanism requires very electron-poor aromatic rings — at least one nitrogen in the ring itself, more are better. In the case of phenol, the aromatic system is generally considered electron-rich (nucleophilic) not electron-poor (electrophilic), thus an electrophilic substitution is possible on the ring but not a nucleophilic one.