Q.Chlorobenzene is formed by reaction of chlorine with benzene in the presence of $\mathrm{AlCl_3}$. Which of the following species attacks the benzene ring in this reaction?

The key idea is electrophilic aromatic substitution — a strong Lewis acid ($\mathrm{AlCl_3}$) generates a highly reactive electrophile from chlorine.

1. $\mathrm{AlCl_3}$ accepts a lone pair from $\mathrm{Cl_2}$, polarising the bond and forming a complex: $\mathrm{Cl_2 + AlCl_3 \rightarrow Cl^\delta+ \!-\! Cl^\delta- \!-\! AlCl_3}$. …

This is a classic electrophilic aromatic substitution (EAS) reaction — the Friedel–Crafts halogenation. Benzene’s π-electron cloud is rich and nucleophilic, but it does not react directly with neutral chlorine gas at room temperature. You need a powerful electrophile to pull electrons away from the ring and form the sigma complex. That’s where the Lewis acid comes in.

The role of $\mathrm{AlCl_3}$ is to accept a lone pair from one chlorine atom of $\mathrm{Cl_2}$, creating a highly polarised complex. This weakens the Cl–Cl bond so much that it effectively breaks heterolytically, generating a $\mathrm{Cl^+}$ ion (or a strongly $\delta^+$ chlorine in the complex) that can attack the ring.

Let’s walk through the mechanism step by step.

1. Generation of the electrophile $\mathrm{AlCl_3}$ is electron-deficient (it has only six electrons in its valence shell). It coordinates to a chlorine atom of $\mathrm{Cl_2}$, forming a complex:

The Al–Cl bond in the complex pulls electron density away from the $\mathrm{Cl_2}$ molecule. This makes one chlorine strongly electrophilic — essentially a $\mathrm{Cl^+}$ equivalent.

2. Attack on the benzene ring The π-electrons of benzene attack this electrophilic chlorine, forming a delocalised carbocation intermediate (the arenium ion or sigma complex):

This step is slow and rate-determining. …

Concept: Electrophilic Aromatic Substitution (EAS) — Role of the Catalyst

Method: Identify the Active Electrophile

In the chlorination of benzene using $\mathrm{Cl_2}$ and $\mathrm{AlCl_3}$, the catalyst generates a strong electrophile that attacks the electron-rich benzene ring.

Steps:

1. Recognize the reaction type

   This is an electrophilic aromatic substitution. Benzene is electron-rich, so it needs a positive or electron-deficient species to attack it.

2. Role of $\mathrm{AlCl_3}$

   $\mathrm{AlCl_3}$ is a Lewis acid. It accepts a lone pair from chlorine in $\mathrm{Cl_2}$, polarising the $\mathrm{Cl-Cl}$ bond.

3. Formation of the attacking species

   The polarisation leads to heterolytic cleavage:

   $\mathrm{Cl_2 + AlCl_3 \rightarrow Cl^+ + [AlCl_4]^-}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Choosing $\mathrm{Cl^-}$ (option (i))

Why students pick it: They see chlorine is involved and assume the negative ion (chloride) is the attacking species, confusing nucleophilic with electrophilic substitution.

Why it’s wrong: Benzene is electron-rich (due to its delocalised π-system). It repels negative species. Chlorobenzene formation is an electrophilic aromatic substitution — the attacking species must be electron-deficient (an electrophile). $\mathrm{Cl^-}$ is a nucleophile, not an electrophile.

How to avoid: Always check the charge of the attacking species. If benzene is the substrate, the attacking species must be positive or at least neutral but electron-deficient. Memorise: Benzene loves positive things.

Mistake 2: Choosing $\mathrm{AlCl_3}$ (option (iii))

Why students pick it: They see $\mathrm{AlCl_3}$ is added as a catalyst and assume it directly attacks benzene.

Why it’s wrong: $\mathrm{AlCl_3}$ is a Lewis acid — it accepts electrons. Its role is to activate chlorine by forming a complex, not to attack benzene directly. The actual attacking species is generated from this interaction.

How to avoid: Distinguish between catalyst and attacking species. The catalyst helps create the electrophile but does not itself attack the ring. Ask: What does the catalyst do to the reagent? Here, $\mathrm{AlCl_3}$ polarises $\mathrm{Cl_2}$ to generate $\mathrm{Cl^+}$.

Mistake 3: Choosing $\mathrm{[AlCl_4]^-}$ (option (iv))

Why students pick it: They see a negative charge and think it might stabilise something, or confuse it with the attacking species.

Why it’s wrong: $\mathrm{[AlCl_4]^-}$ is a byproduct formed after $\mathrm{AlCl_3}$ accepts a chloride ion. It is negatively charged and stable — it does not attack the electron-rich benzene ring.

How to avoid: Track the reaction mechanism step-by-step:

1. $\mathrm{AlCl_3} + \mathrm{Cl_2} \rightarrow \mathrm{Cl^+} + \mathrm{[AlCl_4]^-}$
2. $\mathrm{Cl^+}$ attacks benzene.
3. $\mathrm{[AlCl_4]^-}$ is just a spectator/counterion.

Mistake 4: Not recognising $\mathrm{Cl^+}$ as the correct electrophile (option (ii)) …