2019 H2 Chem P1 Q19 — Arene Solution

The answer is A.

This is Friedel-Crafts acylation, an electrophilic substitution (see figure).
1. \(\text{AlCl}_3\) (a Lewis acid) removes Cl from ethanoyl chloride to form the electrophile, the acylium cation: \(\text{CH}_3\text{COCl} + \text{AlCl}_3 \rightarrow \text{CH}_3\text{CO}^+ + \text{AlCl}_4^-\). This is the first intermediate.
2. The \(\pi\) electrons of the ring attack \(\text{CH}_3\text{CO}^+\), forming the arenium cation. The ring carbon that gains the \(\text{COCH}_3\) group becomes sp\(^3\) (it carries H and \(\text{COCH}_3\)), so the positive charge is delocalised over the other five ring carbons. This is the second intermediate. Loss of \(\text{H}^+\) then restores the aromatic ring.
In option A the arc drawn inside the ring leaves its gap at the sp\(^3\) carbon bearing H and \(\text{COCH}_3\), showing delocalisation over the five sp\(^2\) carbons.

• B: \([\text{CH}_3\text{CO}]^-\) is an anion. An anion cannot act as the electrophile that attacks the electron-rich ring.
• C: correct acylium cation, but the arc leaves its gap at the carbon bearing \(\text{CH}_3\) and runs through the sp\(^3\) carbon, which has no p orbital to take part in the delocalisation.
• D: an anion as the first intermediate, and the arc leaves its gap at the wrong carbon, as in C.