Q.Write the structures and names of the compounds formed when compound ‘A’ with molecular formula, $\mathrm{C_7H_8}$ is treated with $\mathrm{Cl_2}$ in the presence of $\mathrm{FeCl_3}$.

Concept: Electrophilic Aromatic Substitution (Toluene Chlorination) — The reaction is electrophilic aromatic substitution (chlorination) on the methylbenzene ring.

Reasoning:

1. $\mathrm{C_7H_8}$ is toluene (methylbenzene). The methyl group is an ortho/para director.
2. $\mathrm{Cl_2}$ in the presence of $\mathrm{FeCl_3}$ generates a strong electrophile ($\mathrm{Cl}^+$), which attacks the aromatic ring. …

Concept and Intuition: Why This Reaction Works

The molecular formula $\mathrm{C_7H_8}$ is a classic giveaway — it fits toluene (methylbenzene). The reagent $\mathrm{Cl_2}$ in the presence of $\mathrm{FeCl_3}$ is the standard condition for electrophilic aromatic chlorination. Here, $\mathrm{FeCl_3}$ acts as a Lewis acid catalyst, generating the active electrophile $\mathrm{Cl^+}$ (or a polarized $\mathrm{Cl_2–FeCl_3}$ complex).

Why does this matter? Because the methyl group ($\mathrm{-CH_3}$) on toluene is an activating and ortho/para-directing group. It donates electron density into the benzene ring via hyperconjugation and inductive effect, making the ortho and para positions more nucleophilic. So when the $\mathrm{Cl^+}$ attacks, it preferentially goes to those positions.

A common pitfall: students often confuse this with free-radical chlorination (using $\mathrm{Cl_2}$/light or heat), which would attack the side-chain methyl group. But here, the presence of $\mathrm{FeCl_3}$ ensures electrophilic substitution on the ring, not radical substitution.

Step-by-Step Solution

1. Identify the starting compound ‘A’.

The formula $\mathrm{C_7H_8}$ has a degree of unsaturation:

Four degrees of unsaturation strongly suggest a benzene ring (which accounts for 4 DU) plus one extra carbon. The only common aromatic compound with $\mathrm{C_7H_8}$ is toluene ($\mathrm{C_6H_5CH_3}$). So compound A is toluene.

2. Recognize the reaction type.

$\mathrm{Cl_2/FeCl_3}$ is the classic electrophilic aromatic chlorination system. The $\mathrm{FeCl_3}$ polarizes the chlorine molecule:

The $\mathrm{Cl^+}$ (or the polarized complex) acts as the electrophile.

3. Predict the directing effect of the methyl group.

The $\mathrm{-CH_3}$ group is an ortho/para director because it stabilizes the intermediate carbocation (arenium ion) formed during attack at ortho and para positions. Attack at meta position gives a less stable intermediate.

4. Write the products.

The major products are: …

Concept: Electrophilic Aromatic Substitution (EAS) – specifically, chlorination of toluene

Method: Electrophilic aromatic halogenation (chlorination) of an activated ring

(Note: although a Lewis-acid catalyst is used, ring halogenation is not a Friedel–Crafts reaction — that name is reserved for alkylation and acylation.)

Why this method?

Compound ‘A’ with formula $\mathrm{C_7H_8}$ is toluene (methylbenzene). The methyl group ($-\mathrm{CH_3}$) is an activating and ortho/para-directing group. In the presence of $\mathrm{FeCl_3}$ (a Lewis acid catalyst), $\mathrm{Cl_2}$ generates an electrophile ($\mathrm{Cl}^+$), which substitutes a hydrogen on the benzene ring.

Steps:

1. Identify the substrate

   • Molecular formula $\mathrm{C_7H_8}$ → toluene: $\mathrm{C_6H_5CH_3}$

2. Recognize directing effect

   • The $-\mathrm{CH_3}$ group activates the ring and directs incoming $\mathrm{Cl}$ to ortho and para positions.

3. Write the reaction

   • Toluene + $\mathrm{Cl_2}$ + $\mathrm{FeCl_3}$ → mixture of ortho-chlorotoluene and para-chlorotoluene (major products).
   • Minor meta product is negligible due to activation.

4. Draw the structures

   

   …

Here are the common mistakes students make when tackling this specific electrophilic aromatic substitution problem, along with how to avoid each.

1. Mistake: Forgetting to Identify Compound ‘A’ First

• The Error: Students jump straight to writing the chlorination products without first identifying what $\mathrm{C_7H_8}$ is. They might treat it as an alkane or an unknown cyclic compound.
• Why It Happens: The molecular formula $\mathrm{C_7H_8}$ is not immediately obvious. Students often fail to calculate the Degree of Unsaturation (DU) .
• How to Avoid:
  • Always calculate DU first. For $\mathrm{C_7H_8}$:

- A DU of 4 strongly suggests an aromatic ring (benzene ring has DU = 4). The only common aromatic compound with $\mathrm{C_7H_8}$ is **toluene** (methylbenzene).
- **Key Insight:** If you don't identify the starting material correctly, every subsequent product will be wrong.

2. Mistake: Confusing the Reagent System ($\mathrm{Cl_2/FeCl_3}$)

• The Error: Students treat $\mathrm{Cl_2/FeCl_3}$ as a free-radical halogenation reagent (like $\mathrm{Cl_2}$/light or heat) and attack the side chain (methyl group) instead of the ring.
• Why It Happens: They memorize "chlorine + catalyst = substitution" but forget where the substitution occurs. $\mathrm{FeCl_3}$ is a Lewis acid that polarizes $\mathrm{Cl_2}$, making it an electrophilic aromatic substitution reagent.
• How to Avoid:
  • Remember the rule: $\mathrm{Cl_2/FeCl_3}$ (or $\mathrm{Br_2/FeBr_3}$) always substitutes on the aromatic ring (electrophilic substitution).
  • Side-chain substitution (free-radical) requires $\mathrm{Cl_2}$/light or heat (no $\mathrm{FeCl_3}$).
  • Mnemonic: "Catalyst = Ring; Light = Side-chain."

3. Mistake: Ignoring the Directing Effect of the Methyl Group

• The Error: Students write only one product (e.g., just ortho or just para) or write all three possible monochlorinated products (ortho, meta, para) without realizing that the methyl group is an ortho/para director.
• Why It Happens: They forget that substituents already on the ring influence where the next substituent goes.
• How to Avoid:
  • Recall the activating/directing groups: Alkyl groups (like $\mathrm{-CH_3}$) are activating and ortho/para directing.
  • Therefore, the major products are:
    1. ortho-chlorotoluene (1-chloro-2-methylbenzene)
    2. para-chlorotoluene (1-chloro-4-methylbenzene)
  • Meta product is a minor (negligible) product and is usually not listed in exam answers unless specifically asked.

4. Mistake: Writing Incorrect IUPAC Names or Structures

• The Error: Students write the correct skeleton but misnumber the carbons or use wrong locants (e.g., writing 2-chloro-1-methylbenzene instead of 1-chloro-2-methylbenzene).
• Why It Happens: Confusion about which substituent gets the lower number in IUPAC naming.
• How to Avoid: …