Hydrocarbons: Question 4

Worked solution

Step 1: How the electrophile is generated

$\text{AlCl}_3$ is a strong Lewis acid (electron-pair acceptor). It accepts a lone pair from one chlorine atom of $\text{Cl}_2$, polarising and ultimately breaking the $\text{Cl}-\text{Cl}$ bond heterolytically:

$\text{Cl}_2 + \text{AlCl}_3 \rightarrow \text{Cl}^+ + \text{AlCl}_4^-$

This generates $\text{Cl}^+$, a powerful electrophile, together with the $\text{AlCl}_4^-$ counter-ion.

Step 2: Attack on the ring and formation of the arenium ion

The delocalised ring of $\pi$ electrons in benzene attacks the electrophilic $\text{Cl}^+$, forming a new $\text{C}-\text{Cl}$ bond. This uses two of the ring’s six delocalised electrons, so the carbon that bonds to chlorine becomes $\text{sp}^3$-hybridised (bonded to $\text{Cl}$, $\text{H}$, and two ring carbons) and the ring loses its full aromatic delocalisation at that point. The result is the arenium ion: a positively charged, non-aromatic intermediate in which the remaining positive charge is delocalised over the other five ring carbons.

Step 3: Restoring aromaticity

The arenium ion is high in energy (it has lost the extra stability of full aromatic delocalisation), so it does not persist. The $\text{AlCl}_4^-$ ion generated in Step 1 removes a proton ($\text{H}^+$) from the same carbon that now bears the chlorine substituent. This regenerates the fully delocalised aromatic ring (chlorobenzene) and reforms $\text{HCl}$ and the $\text{AlCl}_3$ catalyst:

$\text{AlCl}_4^- + \text{H}^+ \rightarrow \text{HCl} + \text{AlCl}_3$

Since $\text{AlCl}_3$ is regenerated at the end, it is acting as a true catalyst rather than being consumed overall.

Why the other options are wrong

• A: the electrophile is the positively charged $\text{Cl}^+$, not $\text{Cl}^-$; the arenium ion loses a proton (not gains one) to restore aromaticity.
• C: after losing $\text{H}^+$, the ring does regain its full aromatic delocalisation. It does not remain as a non-aromatic cyclohexadiene.
• D: $\text{Cl}_2$ as a whole molecule is not electrophilic enough to react with benzene without the Lewis acid catalyst polarising it first, and the arenium ion is a short-lived, highly reactive intermediate, not a stable isolable product.

Final answer

B. The electrophile is $\text{Cl}^+$; the arenium ion loses a proton to restore aromaticity, regenerating the $\text{AlCl}_3$ catalyst.