Electrophilic Aromatic Substitution Reactions - Bromination | Chemistry Optional Notes for UPSC PDF Download

Introduction

Halogenation is an example of electrophillic aromatic substitution. In electrophilic aromatic substitutions, a benzene is attacked by an electrophile which results in substition of hydrogens. However, halogens are not electrophillic enough to break the aromaticity of benzenes, which require a catalyst to activate.

A Mechanism for Electrophilic Substitution Reactions of Benzene

• A two-step mechanism has been proposed for these electrophilic substitution reactions. In the first, slow or rate-determining, step the electrophile forms a sigma-bond to the benzene ring, generating a positively charged arenium intermediate. In the second, fast step, a proton is removed from this intermediate, yielding a substituted benzene ring. The following four-part illustration shows this mechanism for the bromination reaction. Also, an animated diagram may be viewed.

• The bromine molecule is plarized so that one end is electrophilic and the other nucleophilic. Although the electrophilic end reacts easily with simple alkene and dienes, it fails to react with more stable and weaker nucleophilic π electron system of benzene. 

• Ferric bromide and other Lewis acids enhance the electrophilic strength of bromine by forming a complex anion, in this case FeBr₄Θ At the same time, this complexation creates the strongly electrophilic bromine cation, which reacts with nucleophiles. 

Preliminary step: Formation of the strongly electrophilic bromine cation

The π-electrons of the benzene ring are polarized by the electrophile, and two electrons are diverted to form a a-bond to the bromine atom. The positive charge is thereby relocated on the six-carbon ring, where it is conjugated with the remaining double bonds. The resulting intermediate is a benzenonium cation. Since the aromatic character of benzene is lost, the activation energy (ΔE₁^t) for this reaction is large. Note that the charge alternates so it is greatest ortho and pars to the location of the bromine attack. 

Step 1: The electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate

In the second step of the bromination reaction, a base abstracts the hydrogen proton bonded to the only sp³ hybridizied carbon atom in the ring. Any base will serve for this purpose, but the most likely candidate in this case if bromine ion or it complex with ferric bromide (FeBr4^Θ. The remaining electron pair immediately bonds to one of the adjacent positively charged carbons to reform the aromatic ring, which now bears a bromine substituent. She the proton transfer gives a very stable product, its activation energy (ΔE₂t) is small and the reaction is fast. 

Step 2: A proton is removed from this intermediate, yielding a substituted benzene ring

This mechanism for electrophilic aromatic substitution should be considered in context with other mechanisms involving carbocation intermediates. These include SN1 and E1 reactions of alkyl halides, and Brønsted acid addition reactions of alkenes.

To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:

• The cation may bond to a nucleophile to give a substitution or addition product.
• The cation may transfer a proton to a base, giving a double bond product.
• The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.

S_N1 and E1 reactions are respective examples of the first two modes of reaction. The second step of alkene addition reactions proceeds by the first mode, and any of these three reactions may exhibit molecular rearrangement if an initial unstable carbocation is formed. The carbocation intermediate in electrophilic aromatic substitution (the arenium ion) is stabilized by charge delocalization (resonance) so it is not subject to rearrangement. In principle it could react by either mode 1 or 2, but the energetic advantage of reforming an aromatic ring leads to exclusive reaction by mode 2 (ie. proton loss).

Example 1: What reagents would you need to get the given product?

Ans: Cl₂ and AlCl₃ or Cl₂ and FeCl₃

Example 2: Draw the mechanism of the reaction between Cl⁺ and a benzene.
Ans: