Electrophilic Aromatic Substitution – The Mechanism

by James

in Aromaticity

Last post in this series on reactions of aromatic groups we introduced activating and deactivating groups in Electrophilic Aromatic Substitution (EAS). We learned that electron-donating substituents on the aromatic ring increase the reaction rate and electron-withdrawing substituents decrease the rate. [In the fine print, we also mentioned that evidence strongly suggests that the reaction proceeds through a carbocation intermediate, and that breakage of C-H is not the slow step.]


Having established these facts, we’re now ready to go into the general mechanism of this reaction.

It’s a two-step process.

The good news is that you’ve actually seen both of the steps before (in Org 1) but as part of different reactions!

Step 1: Attack of The Electrophile (E) By a Pi-bond Of The Aromatic Ring

The first step of electrophilic aromatic substitution is attack of the electrophile (E+) by a pi bond of the aromatic ring. [Note: the identity of the electrophile E is specific to each reaction, and generation of the active electrophile is a mechanistic step in itself. We’ll cover the specific reactions soon. This post just covers the general framework for electrophilic aromatic substitution]. 

Where have we seen this type of step before? In the chapter on alkenes, we saw a whole series of reactions of pi bonds with electrophiles that generate a carbocation. A common example is the reaction of alkenes with a strong acid such as H-Cl, leading to formation of a carbocation. The reaction above is the same step, only applied to an aromatic ring.

You might recall that the second step of addition of HCl to alkenes is the attack of Cl on the carbocation, generating a new C-Cl bond. That’s not what happens in electrophilic aromatic substitution. [Note 1]

Step 2: Deprotonation Of The Tetrahedral Carbon Regenerates The Pi Bond

The second step of electrophilic aromatic substitution is deprotonation. This breaks C–H and forms C–C (π), restoring aromaticity. You may recall that this is strongly favored – the resonance energy of benzene is about 36 kcal/mol. [This is the type of phenomenon chemists like to call a “thermodynamic sink” – over time, the reaction will eventually flow to this final product, and stay there. ]
Have we seen this type of step before? Yes – it’s essentially the second step of the E1 reaction, (after loss of a leaving group) where a carbon adjacent to a carbocation is deprotonated, forming a new C-C pi bond.

Just as in the E1, a strong base is not required here. A halogen atom (such as Cl ) will usually suffice, as will any number of other weak bases, such as H2O. The exact identity of the base depends on the reagents and solvent used in the reaction.

 Putting Both Steps Together

Let’s combine both steps to show the full mechanism. Again, we won’t go into the details of generating the electrophile E, as that’s specific to each reaction.

Note that attack could have occurred at any one of the six carbons of benzene and resulted in the same product.

The Reaction Energy Diagram of Electrophilic Aromatic Substitution

What might the reaction energy diagram of electrophilic aromatic substitution look like?

First, the overall appearance is determined by the number of transition states in the process.

Recall that transition states always have partial bonds and are at the “peaks” of a reaction energy diagram,  and intermediates such as carbocations are in the “valleys” between peaks.  Intermediates can be observed and isolated (at least in theory); in contrast, transition states have a lifetime of femtoseconds, and although they may fleetingly be observed in certain cases, they can never be isolated.

Electrophilic aromatic substitution has two steps (attack of electrophile, and deprotonation) which each have their own transition state. There is also a carbocation intermediate. This means that we should have a “double-humped” reaction energy diagram.

Second, the relative heights of the “peaks” should reflect the rate-limiting step.

What’s the slow step? In other words, which of the two steps has the highest activation energy?

One clue is to measure the effect that small modifications to the starting material have on the reaction rate.

We showed in the last post that electron-donating substitutents increase the rate of reaction (“activating”) and electron-withdrawing substituents decrease the rate of reaction (“deactivating”). [Conversely, substitution of hydrogen for deuterium has very little effect on the reaction rate, which leads us to conclude that the second step is not rate-determining. ]

Since electron-donating and electron-withdrawing substitutents affect the nucleophilicity of the pi bond (through pi-donation and pi-acceptance) as well as the stability of the intermediate carbocation, the logical conclusion is that attack on the electrophile (step 1) is the rate-determining step. We therefore should depict it with the higher “hump” in our reaction energy diagram, representing its higher activation energy.

Note that this reaction energy diagram is not to scale and is more of a sketch than anything else. A truly accurate reaction energy diagram can be modelled if one had accurate energies of the transition states and intermediates, which is sometimes available through calculation.

Beyond Benzene: Electrophilic Aromatic Substitution On Substituted Aromatic Molecules

So that’s all there is to electrophilic aromatic substitution? Yes and no.

Yes, this addresses electrophilic aromatic substitution for benzene.

But, as you’ve no doubt experienced, small changes in structure can up the complexity a notch.

Imagine we start not with benzene, but with a mono-substituted derivative, such as methylbenzene (toluene).

What are the possible products of electrophilic aromatic substitution on a mono-substituted benzene derivative?

Unlike with benzene, where only one EAS product is possible due to the fact that all six hydrogens are equivalent, electrophilic aromatic substitution on a mono-substituted derivative can yield three possible products: the 1,2- isomer (also called “ortho“), the 1,3-isomer (“meta“) and the 1,4-isomer (“para“).

If we assume that the reaction obeys the laws of statistics, we might therefore expect that the product distribution should be 40% ortho. 40% meta, and 20% para.

So is that what happens?

No! Two important examples are illustrative.

In the nitration of toluene, the product distribution is far from statistical. We get much less meta (5%) than expected, and more ortho (57%) and para (37%) than expected.

In this sense we can say that the methyl group tends to act as an ortho- para- director: it “directs” the electrophile to these positions at the expense of the meta position.

Is this the case for all substituents? No. 

In the nitration of nitrobenzene, the opposite result is obtained. Much less ortho and para is produced than expected, and the meta product is major (93%).

In this case the nitro group is said to be acting as a meta- director.

Substituents on benzene tend to fall into one of two categories: orthopara directors, or meta directors.

If you’re sharp, you might have already made an intuitive leap: the ortho- para- directing methyl group is an activating group, and the meta- directing nitro group is deactivating. So, therefore,  are all activating groups ortho- para- directors and all deactivating groups meta- directors?

It’s a good guess – and almost accurate! The fly in the bourbon is the halogens (F, Cl, Br, I) which are deactivating ortho-para directors.

Why? What leads some substituents to be ortho-para directors, and others to be meta-directors?

That’s going to have to wait until the next post for a full discussion. But here’s a hint: it has to do with our old friend, “pi-donation”.

Note 1 – Why can’t the counterion attack the aromatic ring carbocation? Does that happen?

Yes, but it’s a dead end.

Let’s say we form the carbocation, and it’s attacked by a weak nucleophile (which we’ll call X).

This gives us the addition product.

However, it’s rarely a very stable product. X is typically a weak nucleophile, and therefore a good leaving group. Furthermore, loss of the leaving group will result in a highly resonance-stabilized carbocation. (Think of the first step in the SN1 or E1 reaction).

This would re-generate the carbocation, which could then undergo deprotonation to restore aromaticity. Once that aromatic ring is formed, it’s not going anywhere. : – )

To make a long story short, yes, addition could occur, but the addition product will eventually undergo E1 to form the aromatic product. 

(figure below)

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