Electrophilic aromatic substitution — mechanism with benzene and Br₂/FeBr₃

medium JEE-MAIN NEET JEE Main 2023 3 min read

Question

Write the mechanism of electrophilic aromatic substitution (EAS) for the bromination of benzene using Br2/FeBr3\text{Br}_2/\text{FeBr}_3. Explain the role of the Lewis acid catalyst and why substitution occurs instead of addition.

(JEE Main 2023, similar pattern)


Solution — Step by Step

Br2\text{Br}_2 alone is not electrophilic enough to attack benzene’s stable aromatic ring. The Lewis acid FeBr3\text{FeBr}_3 polarises the Br-Br bond:

Br2+FeBr3Br++FeBr4\text{Br}_2 + \text{FeBr}_3 \rightarrow \text{Br}^+ + \text{FeBr}_4^-

The bromonium ion Br+\text{Br}^+ (or the strongly polarised BrBrFeBr3\text{Br}-\text{Br}-\text{FeBr}_3 complex) is the actual electrophile.

The π\pi-electrons of benzene attack the electrophile Br+\text{Br}^+, forming a C-Br bond. This creates a carbocation intermediate called the arenium ion (sigma complex):

The arenium ion has a positive charge delocalised over three carbon atoms (ortho, para positions relative to the point of attack). Aromaticity is temporarily lost — the ring now has only 4 π\pi-electrons.

A base (FeBr4\text{FeBr}_4^-) removes a proton from the carbon bearing the bromine:

Arenium ion+FeBr4C6H5Br+HBr+FeBr3\text{Arenium ion} + \text{FeBr}_4^- \rightarrow \text{C}_6\text{H}_5\text{Br} + \text{HBr} + \text{FeBr}_3

The catalyst FeBr3\text{FeBr}_3 is regenerated — it is truly catalytic. Aromaticity is restored in the product.

C6H6+Br2FeBr3C6H5Br+HBr\boxed{\text{C}_6\text{H}_6 + \text{Br}_2 \xrightarrow{\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br} + \text{HBr}}

Why This Works

Why substitution, not addition? In the arenium ion intermediate, the ring can either: (a) Lose H+^+ → substitution product (aromatic, stable) (b) Add Br^- → addition product (not aromatic, less stable)

Restoring aromaticity (option a) provides about 150 kJ/mol of stabilisation energy. This thermodynamic driving force overwhelmingly favours substitution. The aromatic ring “wants” to get its 6 π\pi-electron system back.

Why is a Lewis acid needed? Benzene’s π\pi-cloud is electron-rich, but Br2\text{Br}_2 is not electrophilic enough on its own. FeBr3\text{FeBr}_3 makes one bromine atom more electron-deficient, creating a strong enough electrophile to overcome the activation barrier.


Alternative Method

Other Lewis acid catalysts work too: AlCl3\text{AlCl}_3 for Friedel-Crafts reactions, H2SO4\text{H}_2\text{SO}_4 or HNO3/H2SO4\text{HNO}_3/\text{H}_2\text{SO}_4 for nitration (NO2+\text{NO}_2^+ electrophile), and SO3/H2SO4\text{SO}_3/\text{H}_2\text{SO}_4 for sulphonation. The mechanism is the same three-step pattern: (1) generate electrophile, (2) electrophilic attack to form arenium ion, (3) loss of proton to restore aromaticity.

For JEE, the EAS mechanism is the foundation for understanding directing effects (ortho/para vs meta directors) and activation/deactivation of the ring. If you understand why the arenium ion intermediate is stabilised by electron-donating groups at ortho/para positions, you can predict the product of any EAS reaction.


Common Mistake

Students sometimes write the mechanism as a one-step concerted process: C6H6+Br+C6H5Br+H+\text{C}_6\text{H}_6 + \text{Br}^+ \rightarrow \text{C}_6\text{H}_5\text{Br} + \text{H}^+. This skips the crucial arenium ion intermediate. EAS is a two-step mechanism: electrophilic addition (slow, rate-determining) followed by elimination of H+^+ (fast). The arenium ion is real and detectable — skipping it loses marks in both boards and JEE.

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