Question

What is formed when benzene reacts with ethyl bromide in the presence of anhydrous AlCl$_3$? Give equation.

Hint:

Friedel–Crafts alkylation = Benzene + Alkyl halide (in presence of AlCl$_3$) → Alkylbenzene.

Answer 1

Step 1: Type of reaction.
This is an example of a Friedel–Crafts alkylation reaction. It involves the introduction of an alkyl group into an aromatic ring using an alkyl halide and a Lewis acid catalyst like anhydrous AlCl$_3$.
Step 2: Reaction.
\[ C_6H_6 + C_2H_5Br \xrightarrow{AlCl_3} C_6H_5C_2H_5 + HBr \] Step 3: Product formed.
The main product is ethylbenzene.
Step 4: Mechanism overview.
- AlCl$_3$ reacts with ethyl bromide to generate the ethyl carbocation (C$_2$H$_5^+$).
- This carbocation then attacks the benzene ring, forming ethylbenzene after deprotonation.
Step 5: Importance.
This reaction is an important synthetic route for alkyl-substituted aromatic compounds.