Q.Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark.

The key idea is that direct halogenation of benzene requires a Lewis acid to polarise the halogen molecule, making it a strong enough electrophile to attack the aromatic ring — this is electrophilic aromatic substitution (not a free-radical process, which would need light).

Reasoning:

1. In the dark, no free radicals form, so the reaction cannot proceed via a radical chain mechanism (which would give side-chain substitution).
2. A Lewis acid (e.g., $\ce{FeBr3}$ or $\ce{AlCl3}$) accepts a lone pair from the halogen, generating a highly polarised complex: $\ce{Br2 + FeBr3 -> Br^{\delta+}...FeBr4-}$. …

The direct halogenation of benzene with Br₂ or Cl₂ does not occur in the dark — or even in diffuse light — because benzene’s delocalised π-electron cloud is too stable to act as a nucleophile toward a non-polarised halogen molecule. The key is to generate a stronger electrophile. This is where a Lewis acid comes in.

A Lewis acid is an electron-pair acceptor. When you add FeBr₃ (or AlCl₃) to bromine, the iron atom accepts a lone pair from bromine, polarising the Br–Br bond heavily. The result is a complex that behaves as if it were “Br⁺” — a powerful electrophile that can attack benzene’s π-system.

Let’s walk through the mechanism step by step.

1. Generation of the electrophile The Lewis acid (say FeBr₃) coordinates with one bromine atom of Br₂. This pulls electron density away from the Br–Br bond, making the far bromine strongly electron-deficient.

In effect, we get a complex that can be thought of as $\text{Br}^+$ (the electrophile) paired with a $\text{FeBr}_4^-$ counterion.

2. Electrophilic attack on benzene The π-electron cloud of benzene donates two electrons to the electrophilic bromine. This forms a sigma bond between bromine and one carbon of the ring, breaking the aromaticity and creating a resonance-stabilised carbocation intermediate (the arenium ion or σ-complex).

3. Restoration of aromaticity The arenium ion is unstable. It loses a proton (H⁺) from the carbon that attacked bromine. The pair of electrons from the C–H bond moves back into the ring, restoring the aromatic sextet. The H⁺ then combines with the $\text{FeBr}_4^-$ to reform FeBr₃ and HBr. $\text{C}_6\text{H}_6\text{Br}^+ + \text{FeBr}_4^- \longrightarrow \text{C}_6\text{H}_5\text{Br} + \text{FeBr}_3 + \text{HBr}$ …

Concept: Electrophilic Aromatic Substitution (EAS) — Role of Lewis Acids in Halogenation

Method: Lewis Acid Catalysis in Aromatic Halogenation

The direct halogenation of benzene with $Br_2$ or $Cl_2$ is very slow in the dark because benzene's $\pi$-electrons are not nucleophilic enough to attack the halogen molecule directly. A Lewis acid is needed to polarise the halogen, making it a stronger electrophile.

Steps of the Method

1. Activation of the Halogen The Lewis acid (e.g., $FeBr_3$, $AlCl_3$, $FeCl_3$) accepts a lone pair from the halogen molecule. For bromination:

This creates a highly polarised $Br$–$Br$ bond, or even a free $Br^+$ (bromonium ion) in extreme cases.

2. Electrophilic Attack The activated halogen (now a strong electrophile) is attacked by the $\pi$-electrons of benzene, forming a sigma complex (arenium ion):

3. Deprotonation The sigma complex loses a proton to the $FeBr_4^-$ (or $AlCl_4^-$) regenerating the Lewis acid and giving the aryl halide:

Key Exam Points …

Common Mistakes: Role of Lewis Acids in Aryl Halide Preparation (Dark Reaction)

This question is about the electrophilic aromatic substitution mechanism for preparing aryl bromides and chlorides — specifically, why a Lewis acid (like $\ce{FeBr3}$ or $\ce{AlCl3}$) is needed when the reaction is carried out in the dark.

Mistake 1: Thinking the Lewis acid is a catalyst that "activates" the halogen molecule by simple coordination

Why it's wrong:

Many students say: "The Lewis acid polarises the $\ce{Br2}$ molecule, making the bromine more electrophilic." While this is partially true, it misses the key point — in the dark, there is no radical initiation. The Lewis acid actually generates a stronger electrophile via a different mechanism.

Correct understanding:

In the dark, the reaction follows an electrophilic substitution path (not radical). The Lewis acid (e.g., $\ce{FeBr3}$) accepts a lone pair from bromine, forming a complex like $\ce{Br2-FeBr3}$. This complex polarises the $\ce{Br-Br}$ bond so strongly that it can ionise to give $\ce{Br^+}$ (or a $\ce{Br}$-$\delta+$ species). This $\ce{Br^+}$ is the actual electrophile that attacks the benzene ring.

How to avoid:

• Always distinguish: in light → radical mechanism (no Lewis acid needed). In dark → electrophilic substitution (Lewis acid essential).
• Write the step:

Mistake 2: Claiming the Lewis acid is "not required" for chlorination

Why it's wrong:

Chlorine is more reactive than bromine, but in the dark, even chlorination of benzene requires a Lewis acid (usually $\ce{AlCl3}$ or $\ce{FeCl3}$). Without it, the reaction is extremely slow.

Correct understanding:

Chlorine is more electrophilic than bromine, but benzene's $\pi$ electrons are still too stable to react directly with $\ce{Cl2}$ in the dark. The Lewis acid is needed to generate $\ce{Cl^+}$ (or a highly polarised $\ce{Cl}$-$\delta+$ species).

How to avoid:

• Memorise: Both bromination and chlorination in the dark require a Lewis acid.
• The only difference is that chlorination can sometimes proceed without Lewis acid under light (radical) conditions — but the question specifically says in the dark.

Mistake 3: Confusing the role of Lewis acid with a "dehydrating agent" or "catalyst that lowers activation energy vaguely"

Why it's wrong:

Some students write generic statements like "Lewis acid speeds up the reaction" without explaining the specific chemical role.

Correct understanding:

The Lewis acid forms a complex with the halogen, making it a much stronger electrophile. It does not just "catalyse" — it changes the reactive species.

How to avoid:

• Use the exact phrase: "generates the electrophile $\ce{X^+}$" (or a highly polarised $\ce{X}$-$\delta+$).
• Show the complex formation:

Mistake 4: Forgetting that the Lewis acid is consumed and regenerated — but not a true catalyst in the usual sense

Why it's wrong:

Students often say "Lewis acid is a catalyst" without noting that it forms a stable salt ($\ce{FeBr4^-}$) that must be decomposed.

Correct understanding:

After the electrophilic attack, the $\ce{FeBr4^-}$ donates a $\ce{Br^-}$ back to the arenium ion, regenerating $\ce{FeBr3}$ and $\ce{HBr}$. So it is catalytic in net, but the mechanism involves stoichiometric complex formation.

How to avoid:

• Write the regeneration step: $\ce{ArHBr^+ + FeBr4^- -> ArBr + HBr + FeBr3}$ …