Iodination of Aromatics with I2

by James

Description: Treatment of an aromatic such as benzene with iodine (I2) and copper (II) chloride leads to formation of the iodinated benzene.

Notes: CuCl2 is a catalyst for this reaction. The byproduct is hydroiodic acid (HI)


Notes: The byproduct of each of these reactions is HI

Mechanism: This is an electrophilic aromatic substitution reaction. The reaction begins with iodine attacking Cu (Step 1, arrow A). This takes electron density away from I2, making it a better electrophile. Attack of the aromatic π  bond on iodide (Step 2, arrows B and C) then breaks aromaticity and leads to the carbocation. Removal of a proton from the carbocation (Step 3, arrows D and E) reforms the π bond, leading to the formation of the iodobenzene product.

Notes: Note how CuCl2 is regenerated at the end of the reaction.

It’s also reasonable to show breakage of the I–I bond before the aromatic attacks, such that the aromatic is attacking I(+).


{ 5 comments… read them below or add one }


Does reaction of molecular iodine with nitric acid not produce iodobenzene in greater yield? Lesser yield? Just as good?



waleed khaled

but u forgetted that HI is reducing agent that can reverse the reaction toward the benzene ring



HI is not a reducing agent.


Simrandeep Bahal

My text said that HI is a reducing agent.(That’s why direct Iodination fails to give Iodoalkane. Prescence of oxidising ahents like HgO is required to oxidise hydrogen iodide to iodine gas)


Rajarshi Banerjee

Check the description reaction, its CuI instead of CuCl2


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