Q.Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?

The key idea is that iodination is reversible because $I_2$ is a very weak electrophile, and the $HI$ byproduct reduces the product back to the arene.

Reasoning:

1. In electrophilic aromatic substitution, $I_2$ is far less reactive than $Cl_2$ or $Br_2$ — it cannot polarise sufficiently to act as an electrophile without help.
2. Even if a small amount of iodination occurs, the $HI$ produced is a strong reducing agent that rapidly reduces the aryl iodide back to the hydrocarbon. …

Why the usual method works for Cl₂ and Br₂

Chlorination and bromination of arenes proceed cleanly with just a Lewis acid catalyst (FeCl₃, AlCl₃), which polarises the halogen molecule into a strong electrophile:

The HCl or HBr released as byproduct does not attack the aryl halide product, so these reactions are effectively irreversible — no extra reagent is needed.

The problem with iodine: the reaction is reversible

Iodination is different in one decisive way. The reaction sits in an equilibrium:

The HI byproduct is a good reducing agent: it reduces the aryl iodide back to the parent arene (regenerating I₂), pulling the equilibrium backwards. It also doesn't help that I₂ is the weakest electrophile of the halogens, which makes the forward reaction sluggish to begin with — but the equilibrium is the core problem: even the product that does form is destroyed by the HI accumulating in the mixture.

The solution: oxidise away the HI

An oxidising agent — HIO₄ (periodic acid) is the one NCERT names; HNO₃ also works — is added to oxidise the HI as it forms, removing it from the equilibrium:

Concept: Reversibility of Iodination and Removal of HI

The key idea is that iodination of an arene is a reversible reaction, and the HI byproduct drives it backwards unless it is removed by oxidation.

Method: Equilibrium Analysis of Arene Halogenation

Why the problem arises:

• Chlorination and bromination (with a Lewis acid such as $\text{FeCl}_3$/$\text{AlCl}_3$) are effectively irreversible — the HCl/HBr byproduct does not attack the product, so no extra reagent is needed.
• Iodination is reversible:

The HI formed is a good reducing agent — it reduces the aryl iodide back to the arene (regenerating $\text{I}_2$). $\text{I}_2$ is also the weakest electrophile of the halogens, so the forward reaction is slow to begin with.

The solution:

Add an oxidising agent — HIO₄ (the one NCERT names) or $\text{HNO}_3$ — to oxidise the HI byproduct as it forms, removing it from the equilibrium.

Step-by-Step Reasoning

1. Electrophilic attack:

   The iodine (polarised/activated in the reaction mixture) attacks the benzene ring, forming a sigma complex.

2. Deprotonation:

   Loss of $\text{H}^+$ from the sigma complex gives the aryl iodide and $\text{HI}$.

3. The reverse reaction (the problem):

   The accumulated $\text{HI}$ reduces $\text{ArI}$ back to $\text{ArH}$, so the equilibrium yields little product.

4. Oxidation of $\text{HI}$ (the fix):

   The oxidising agent converts $\text{HI}$ back to $\text{I}_2$ — for example: …

Here’s a breakdown of the common mistakes students make on this concept, along with how to avoid each.

The Core Concept (Why the Question Exists)

The question tests your understanding of reactivity trends in electrophilic aromatic substitution (EAS) and the redox chemistry of halogens.

• For Cl₂ and Br₂: The halogen molecule is already a strong enough electrophile (when activated by a Lewis acid like FeCl₃ or AlCl₃) to attack the benzene ring.
• For I₂: Iodine is a much weaker electrophile than chlorine or bromine, and the reaction produces HI as a byproduct. HI is a good reducing agent that reduces the aryl iodide back to the arene (regenerating I₂), reversing the reaction.

The fix: An oxidizing agent (like HIO₄ or HNO₃) oxidizes the HI (the byproduct) back into I₂, preventing the reverse reaction and pushing the equilibrium forward.

Common Mistake #1: Confusing "Oxidizing Agent" with "Catalyst"

The Mistake:

Students say: "The oxidizing agent is needed because iodine is a weaker electrophile, so we need a stronger catalyst." They treat the oxidizing agent as if it’s just another Lewis acid catalyst.

Why it’s wrong:

A Lewis acid catalyst (like FeCl₃) activates the halogen by polarizing it (making it more electrophilic). An oxidizing agent does not activate iodine directly. Instead, it removes the byproduct (HI) that would otherwise destroy the aryl iodide product.

How to Avoid:

• Remember the byproduct: For every I₂ that reacts, one HI is produced. HI is a strong reducing agent.
• Trace the electron flow: HI + [O] → I₂ + H₂O. The oxidizing agent regenerates the starting material (I₂), not just activates it.
• Exam tip: If a question asks "Why is an oxidizing agent needed?" your answer must mention preventing the reduction of the aryl iodide back to the arene by HI. Do not just say "to make iodine more reactive."

Common Mistake #2: Thinking the Oxidizing Agent Makes Iodine "More Electrophilic"

The Mistake:

Students write: "The oxidizing agent increases the electrophilicity of iodine."

Why it’s wrong:

Oxidizing agents do not directly increase the positive charge or polarity of I₂. They work indirectly by removing HI. The actual electrophile in the reaction is still I₂ (or I⁺ generated in situ, but that’s a separate mechanism). The oxidizing agent doesn’t touch the I₂ molecule itself.

How to Avoid:

• Use precise language: Say "The oxidizing agent oxidizes the HI byproduct back to I₂, preventing the reverse reaction."
• Draw the equilibrium: Show the reversible arrow: $\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}$ The oxidizing agent shifts the equilibrium to the right by removing HI.

Common Mistake #3: Forgetting the Role of HI as a Reducing Agent

The Mistake:

Students say: "HI is a strong acid, so it protonates the benzene ring and stops the reaction."

Why it’s wrong:

HI is indeed an acid, but the real problem is that it reduces the aryl iodide back to the arene. The reaction is reversible, and HI is the reducing agent that drives it backward. Protonation of the ring is not the main issue here (though it can happen, it’s secondary).

How to Avoid:

• Remember the equilibrium: ArH + I₂ ⇌ ArI + HI. HI provides the electrons that reduce the ArI product back to ArH. …