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What is the mechanism of the reaction between benzene and bromine to form bromobenzene?

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The reaction between benzene and bromine to form bromobenzene is an example of electrophilic aromatic substitution. The mechanism involves the following steps:1. Formation of the electrophile: Bromine  Br2  reacts with a Lewis acid catalyst, such as iron III  bromide  FeBr3  or aluminum bromide  AlBr3 , to form a highly electrophilic bromine ion  Br+ . The catalyst polarizes the bromine molecule, making one of the bromine atoms electron-deficient.   Br2 + FeBr3  Br+ + FeBr4-2. Attack of the electrophile: The electron-rich benzene ring donates a pair of electrons from one of its pi bonds to the electrophilic bromine ion  Br+ , forming a sigma bond between the bromine and the carbon atom in the ring. This results in the formation of a positively charged cyclohexadienyl cation  also known as an arenium ion or Wheland intermediate  and disrupts the aromaticity of the benzene ring.3. Deprotonation: A base, usually the halogen anion generated in the first step  Br- , abstracts a proton from the cyclohexadienyl cation, regenerating the aromaticity of the benzene ring and forming bromobenzene as the final product.   C6H5Br+HBr-  C6H5Br + HBr4. Regeneration of the catalyst: The hydrogen bromide  HBr  formed in the previous step reacts with the iron III  bromide  FeBr4-  to regenerate the catalyst  FeBr3  and release a bromide ion  Br- .   HBr + FeBr4-  FeBr3 + Br-Overall, the reaction can be summarized as:   C6H6 + Br2  C6H5Br + HBrThe reaction mechanism involves the formation of an electrophile, attack of the electrophile on the benzene ring, deprotonation to restore aromaticity, and regeneration of the catalyst.
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