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NCERT Exemplar · Q4

Q.Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is

(i) Electrophilic elimination reaction
(ii) Electrophilic substitution reaction
(iii) Free radical addition reaction
(iv) Nucleophilic substitution reaction
Meghalaya MboseMCQ· 1mImportance★★★★★
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Toluene undergoes electrophilic aromatic substitution with a halogen in the presence of FeCl₃. The halogen is activated by FeCl₃ to form an electrophile, which attacks the electron-rich benzene ring. The correct option is (ii).

Electrophilic halogenation of toluene
Electrophilic halogenation of toluene

The key here is to recognise what kind of reagent a halogen becomes when mixed with a Lewis acid like FeCl₃. On its own, Cl₂ or Br₂ is not strongly electrophilic — it needs a "push" to become reactive toward an aromatic ring. FeCl₃ does exactly that: it polarises the halogen molecule, making one end strongly positive (the electrophile). This is a classic setup for electrophilic aromatic substitution, not addition or free-radical chemistry.

Let’s walk through the reasoning step by step.

  1. Identify the reaction type from the reagents.

    Toluene (methylbenzene) is an aromatic compound. Halogens (Cl₂, Br₂) do not normally add to aromatic rings under these conditions — they substitute. The presence of FeCl₃ (a Lewis acid) is a dead giveaway: it’s the catalyst used in the halogenation of aromatic rings. This is not a free-radical or nucleophilic process.

  2. Understand the role of FeCl₃.

    FeCl₃ coordinates with the halogen, say Cl₂, forming a complex like FeCl₄⁻ and Cl⁺ (or a highly polarised Cl–Cl bond). This generates a strong electrophile — a chlorine atom with a partial or full positive charge.

Cl2+FeCl3→Clδ+⋯FeCl4−\text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^\delta+ \cdots \text{FeCl}_4^-

  1. Why electrophilic substitution, not addition?

    The aromatic ring is rich in π electrons. It attacks the electrophile (Cl⁺) to form a carbocation intermediate (the arenium ion). If addition occurred, the aromaticity would be permanently lost — but here, the intermediate loses a proton (H⁺) to restore the aromatic ring. That’s the hallmark of substitution: one H is replaced by Cl, and the ring remains aromatic.

  2. Why ortho/para products?

    Toluene has a methyl group, which is an activating, ortho/para-directing group. It donates electron density into the ring (hyperconjugation + inductive effect), making the ortho and para positions more nucleophilic. So the incoming halogen preferentially attacks those positions, giving a mixture of ortho- and para-halotoluene.

  3. Eliminate the wrong options. …

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