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Reactions of Aromatic Molecules

By James Ashenhurst

Electrophilic Aromatic Substitutions (1) – Halogenation

Last updated: February 22nd, 2019

Now that we’ve spent ample time digging into

we’re finally ready to dig into some specific reactions in detail. First up: halogenation. How do we install Cl, Br, or I on an aromatic ring?

[quick summary here:]

[What about fluorine? If F2 wasn’t such a ravenous beast, we’d include it too, but as it stands, fluorine is best introduced to aromatic rings indirectly]

Electrophilic Aromatic Substitution: Halogenation

You may recall that alkenes react readily with  Cl2, Br2, and I2 to form dihalides.

Benzene? Not so much. And when it does react with aromatic rings, it provides substitution rather than addition products.

[Why? Recall that the resonance energy of benzene is about 36 kcal/mol, and electrophilic aromatic substitution disrupts aromaticity. ]


Lewis Acids Can Be Used To “Activate” Electrophiles

Chlorine itself will react with “activated” benzene derivatives (such as phenol and aniline) but in order for chlorination to occur with electron-neutral or electron-poor aromatics, it needs a kick in the pants.

This “kick in the pants” is provided by a Lewis acid.  FeCl3 or AlCl3 are the industry standard, but in practice many different Lewis acids can be employed.

You may recall from the chapter on alcohols that protic acids are commonly used to turn poor electrophiles (e.g. alcohols) into better electrophiles by converting an alcohol into its conjugate acid. [post: The Conjugate Acid Is Always A Better Leaving Group].  The “activated” alcohol is then able to participate in nucleophilic substitution or elimination reactions that it would not have been able to participate in otherwise:

[The leaving group may even depart entirely to give a carbocation, an even better electrophile!]

In much the same way, coordination of the Lewis acid to one of the chlorines converts it into an even better leaving group, with the net effect of weakening the Cl-Cl bond. Attack on the terminal Cl by a nucleophile results in the loss of [Cl-FeCl3] , an even better leaving group than Cl .

Sometimes we draw this as an equilibrium. I personally prefer showing the intact Cl–Cl bond, but textbooks vary.

Electrophilic Chlorination of Benzene

This “activated” electrophile can then be attacked by the nucleophile (the benzene ring) in an electrophilic aromatic substitution reaction. Here, C-Cl forms, and the C-C (pi) bond breaks!

In the next step,  a weak base removes a proton from the carbocation intermediate, breaking C-H and forming C-C (pi).

If this looks a tiny bit familiar, that’s because this second step greatly resembles the second step in the E1 reaction. As in the E1, only a very weak base is required to remove a proton adjacent to a carbocation.  The chloride ion (Cl– ) will do.

[Note: there are a lot of different ways to draw this, I’ve only shown one. See this footnote].

The final product here is chlorobenzene plus one equivalent of HCl. The FeCl3  is then free to react with another equivalent of Cl. In other words, it behaves as a catalyst in this reaction.

Electrophilic Bromination of Benzene

The same set of principles operate for electrophilic bromination. Here, the Lewis acids used are often FeBr3 or AlBr3. [Why not FeCl3 or AlCl3? Good question! See note below ]

The first step is activation of Br2, followed by attack and deprotonation, as before:

It’s essentially the same reaction as chlorination except with the halogens swapped out.

Electrophilic Iodination of Benzene

It’s also possible to iodinate benzene using I2 , but the activation step is  different.   Here, however, it turns out that a Lewis acid catalyst is not sufficient to make iodine an active enough electrophile to react with most aromatic rings.

Instead,  a stoichiometric amount of an oxidant is used to convert I2 to I. A common example cited is HNO3, which in the presence of additional acid (e.g. H2SO4) is a source of the very active oxidant [NO2]+ which converts I2 to I+ .  [see note below]

Once formed, the benzene ring then reacts with Iin the two-step electrophilic aromatic substitution mechanism to give the new carbon-iodine bond.

The exact identity of “B” will depend on the oxidant used to convert I2 into I+ . A single equivalent of water will do the trick, for example.


In the cases of chlorine, bromine, and iodine, electrophilic aromatic substitution follows three steps.

  1. Activation of the electrophile by a Lewis acid catalyst (or stoichiometric oxidant, in the case of iodine)
  2. Attack of the activated electrophile by the aromatic ring.
  3. Deprotonation to regenerate the aromatic ring.

In the next post we’ll cover two more important electrophilic aromatic substitution reactions: sulfonylation and nitration, and they will also follow this three-step pattern!

Note 1. There are a few different ways one could depict this; the drawing above shows the pair of electrons in the Fe-Cl bond acting as the base, which is essentially the same as dissociation of Cl- from FeCl4(-)   followed by it acting as a base. One could also draw a lone pair from the Cl of FeCl4 acting as a base, giving H-Cl-FeCl3, followed by breakage of the Fe-Cl bond to give FeCl3 and HCl.

Note 2. Many other oxidants have found use in this reaction. Another is copper (II) bromide (CuBr2).

Note 3. It’s not that FeCl3 or AlCl3 aren’t strong enough to do the job here; the problem is that using Br2 in the presence of FeCl3 will lead to some scrambling of the halogens, resulting in a small amount of chlorination products. Using the bromide salts eliminates this problem.

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Comment section

2 thoughts on “Electrophilic Aromatic Substitutions (1) – Halogenation

  1. Hello, love the articles. I am studying for a test including this topic, was wondering where would the Bromine attack if there was a phenyl group attached to a ketone attached to a methyl group attached to another phenyl group. I believe it attaches to the phenyl group furthest from the ketone but was not sure.

    1. One phenyl group is attached to a carbonyl (C=O). The other is attached to a CH2 attached to a carbonyl.

      Which phenyl group is more activated? : – )

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