Concept:Electrophilic aromatic substitution demands a powerful electrophile, normally generated via the interaction of the reagent with a Lewis acid catalyst.
Formula:$$ \text{Br}_2 + \text{FeBr}_3 \rightarrow \text{Br}^+ + \text{FeBr}_4^- $$
Solution:- When bromine (\( \text{Br}_2 \)) interacts with a Lewis acid (like \( \text{FeBr}_3 \) or \( \text{FeCl}_3 \)), the halogen bond is polarized and broken.
- This heterolytic cleavage produces the highly reactive bromonium ion, \( \text{Br}^+ \), which is electron-deficient.
- \( \text{Br}^+ \) is the actual electrophile that attacks the electron-rich pi cloud of the benzene ring.
Why other options are incorrect:- \( \text{FeCl}_4^- \) is a negatively charged counterion (a weak nucleophile), not an electrophile.
- \( \text{Fe}^{+3} \) and \( \text{Fe}^{+2} \) represent oxidation states of iron in the catalyst; they do not act as the carbon-attacking electrophile in the aromatic substitution step.
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