Introduction-
If you bubble hydrogen chloride gas into a solution of styrene, the HCl adds to the double bond of the alkene forming 1-chloro-1-phenylethane as shown in Equation 1. Similarly, if you mix a solution of styrene with dibromine, as outlined in Equation 2, the red-orange color of the dibromine disappears almost instantly, indicating a rapid reaction between the two reactants. The product of the reaction is 1,2-dibromo-1-phenylethane. Equations 1 and 2 both describe electrophilic addition reactions.
Electrophilic addition reactions are characteristic of alkenes and alkynes. They are not, as Equations 1 and 2 make apparent, characteristic of aromatic rings. In other words, under conditions where electrophilic reagents add to the pi system of an alkene or alkyne, they do not react with the pi system of the aromatic ring. If you mix a solution of benzene with dibromine, reaction 3 does not occur.
It is reasonable to conclude from a comparison of Equations 2 and 3 that the pi electrons of benzene, and by extension, of aromatic rings in general, are less reactive than those of simple alkenes and alkynes. What this really means is that the activation energy for reaction 3 is much higher than it is for reaction 2.
One ploy that chemists use to promote a reaction that has a high activation energy is to add a catalyst to the reaction mixture. In the case of reaction 3, if the mixture is heated in the presence of a small amount of FeBr3, the red-orange color of the dibromine slowly disappears. While this indicates that a reaction has occurred, product analysis demonstrates that it is not reaction 3. For one thing, one of the products of the reaction is HBr. For another, elemental analysis shows that the molecular formula of the other product is C6H5Br, not C6H6Br2. Equation 4 summarizes this result.
Clearly this reaction is not an addition reaction. It is a substitution. One of the hydrogen atoms on the aromatic ring has been replaced by a bromine atom. Such a process is characteristic of the reactivity pattern of aromatic rings with electrophilic reagents. This pattern is called electrophilic aromatic substitution. In the rest of this topic we will examine various aspects of this reaction mechanism.
Electrophilic Aromatic Substitution Reactions
Benzene undergoes electrophilic aromatic substitution reactions with a variety of electrophiles. Table 1 provides an overview of the most common situations.
Table 1
Electrophilic Aromatic Substitution Reactions
| Type of Reaction | Equation | "Electrophile" |
| chlorination |  | "Cl+" |
| bromination |  | "Br+" |
| alkylation |  | "CH3+" |
| acylation |  |  |
| nitration |  | "NO2+" |
| sulfonation |  | SO3 |
In the third column, the heading "Electrophile" is within quotation marks to emphasize that each electrophilic reagent acts as though it were a source of the electrophilic species shown in quotes. As we will see, the actual electrophilic species more often than not is different from the structure shown in quotes.
Catalysts
Most of the reactions shown in Table 1 require the addition of a catalyst to the reaction mixture. When dibromine is used as the electrophilic reagent, ferric bromide, FeBr3, is often added as a catalyst. How does ferric bromide catalyse reaction 4? Like most metal halides, FeBr3 is a Lewis acid, i.e. an electron pair acceptor. Lewis acids have at least one empty orbital they can use to accept an electron pair. In the case of FeBr3 the iron atom has empty 3d orbitals. Equation 5 illustrates a Lewis acid-Lewis base reaction between FeBr3 and Br2. The electron deficient nature of the iron atom is indicated by the blue color, while the electron rich nature of the bromine is symbolized by red.
Exercise 1 Aldrich Chemical Company sells ferric bromide. According to their catalog, this high melting solid (mp = 684oC) is hygroscopic, i.e. reacts readily with water, and is an irritant. Write an equation similar to Equation 5 showing the interaction of water and FeBr3. Write another equation showing how the product of the first reaction might react further to produce the species that is responsible for the irritating properties of ferric bromide.
Note that reaction 5 produces a complex in which one of the bromine atoms bears a positive charge. Having a positive charge on an electronegative atom is an unfavorable situation. The structure has high potential energy. It is so reactive that even benzene reacts with it. Figure 1 compares the reaction profiles of the catalysed and uncatalysed reactions. Figure 1
Getting Over the Hump
The figure suggests that the catalysed reaction proceeds faster the uncatalyzed one because the potential energy of the system is higher when FeBr3 is one of the reactants than when it is not. By the same token, the difference in potential energy of the reactants and the transition state, i.e. E'a, is lower in the presence of FeBr3. Therefore the catalyzed reaction proceeds faster.
The first step of that reaction is shown in Equation 6.
As the positive charge develops on the bromine atom in reaction 5, the Coulombic attraction between the complex and a molecule of benzene increases as indicated by the arrow labeled 1. Formation of a bond between a carbon of the aromatic ring and the uncharged bromine atom of the complex is accompanied by cleavage of the Br+-Br bond as indicated by arrow 2. This sequence of bonding changes reduces the charge on the bromine from +1 to 0. The effect of complexing Br2 with FeBr3, is to create a good leaving group, FeBr4-. This is analogous to the effect of protonating an alcohol in an Sn2 reaction, i.e. it converts a relatively poor leaving group HO-, to a good one, H2O.
The intermediate produced in reaction 6 is called a cyclohexadienyl cation. It is involved in all electrophilic aromatic substitution reactions. We have seen that carbocations generally suffer one of three fates
- Addition- A nucleophile adds to the trivalent carbon.
- Elimination- A base removes a proton from the carbon atom adjacent to the positively charged carbon.
- Rearrangement- A group migrates to the positively charged carbon.
In the case of a cyclohexadienyl carbocation, elimination is the only path followed. See Equation 7.
Note the following points about this reaction:
- The formation of the product is accompanied by the regeneration of FeBr3, making this reagent catalytic.
- The formation of the product is accompanied by the evolution of HBr gas.
- The hydrogen is eliminated from the same carbon atom that the bromine bonded to originally.
- The elimination regenerates the resonance stabilized, aromatic 6-p electron system.
Exercise 2 The positive charge in the cyclohexadienyl cation shown in Equation 6 is stabilized by resonance. Draw two additional resonance structures for this ion.
Exercise 3 Aluminum chloride is frequently used as a catalyst in the chlorination of benzene. Write equations similar to 5, 6, and 7 to show the conversion of benzene to chlorobenzene.
Exercise 4 The methylation of benzene with chloromethane is an example of a reaction called the Friedel-Crafts alkylation. It is catalysed by AlCl3 in a manner similar to chlorination. Write equations analogous to 5, 6, and 7 to show the conversion of benzene to methylbenzene.
Exercise 5 The preparation of acetophenone by treatment of benzene with acetyl chloride in the presence of a catalytic quantity of AlCl3 is an example of a process called Friedel-Crafts acylation. Write equations analogous to 5, 6, and 7 to show the conversion of benzene to acetophenone.
Exercise 6 Draw a valid Lewis structure for the nitronium ion, NO2+.
