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What is the mechanism of the reaction between benzene and bromine in the presence of an iron catalyst? Describe the intermediates formed during the process and explain the role of the catalyst in the reaction.

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The reaction between benzene and bromine in the presence of an iron catalyst is an example of electrophilic aromatic substitution, specifically bromination. The overall reaction can be represented as:C6H6 + Br2  C6H5Br + HBrThe mechanism of this reaction involves several steps:1. Formation of the electrophile:The iron catalyst  usually FeBr3  reacts with bromine  Br2  to form a complex, which generates the electrophile, the bromonium ion  Br+ . The iron catalyst polarizes the bromine molecule, making one of the bromine atoms more electrophilic.FeBr3 + Br2  FeBr4- + Br+2. Electrophilic attack on benzene:The electrophilic bromine atom  Br+  attacks the benzene ring, breaking one of the pi bonds and forming a sigma bond with a carbon atom. This results in the formation of a positively charged intermediate called a sigma complex or arenium ion. The aromaticity of the benzene ring is disrupted in this step.3. Deprotonation and regeneration of aromaticity:A base, usually the FeBr4- ion, abstracts a proton from the carbon atom that is bonded to the bromine atom in the sigma complex. This step restores the aromaticity of the benzene ring and generates the bromobenzene product.C6H5Br-FeBr4  C6H5Br + HFeBr44. Regeneration of the catalyst:The iron catalyst is regenerated when the HFeBr4 complex dissociates into FeBr3 and HBr, which can be used in further reactions.HFeBr4  FeBr3 + HBrIn summary, the iron catalyst plays a crucial role in the bromination of benzene by generating the electrophilic bromonium ion  Br+  and facilitating the formation of the sigma complex. The reaction proceeds through an electrophilic aromatic substitution mechanism, with the formation of a sigma complex intermediate and the regeneration of the aromaticity in the final product, bromobenzene.
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