Question

In Friedel-Crafts reaction, benzene reacts with isopropyl bromide in the presence of aluminium trichloride to give:

• A. $ \text{Benzophenone} $
• B. $ \text{Acetophenone} $
• C. $ \text{Isopropyl benzene} $
• D. $ \text{n-Propyl benzene} $

Hint:

In Friedel–Crafts alkylation, secondary carbocations like isopropyl are stable and commonly formed. Always consider carbocation stability and rearrangement.

Answer 1

The Correct Option is C

- The Friedel–Crafts alkylation reaction involves the alkylation of an aromatic ring with an alkyl halide in the presence of a Lewis acid such as {AlCl}_3.
- In this case, isopropyl bromide is the alkyl halide and benzene is the aromatic compound.
- The AlCl(\_3) catalyst helps generate a carbocation (or a carbocation-like species) from isopropyl bromide:
$$ (\text{CH}_3)_2\text{CH–Br} + \text{AlCl}_3 \rightarrow (\text{CH}_3)_2\text{C}^+ + \text{AlCl}_3\text{Br}^- $$ - This isopropyl carbocation then reacts with benzene to form isopropyl benzene (cumene) via electrophilic aromatic substitution. - Other options:
- (A) Benzophenone and (B) Acetophenone are products of Friedel–Crafts acylation, not alkylation.
- (D) n-Propyl benzene is not formed here because n-propyl carbocation is unstable and rearranges to isopropyl carbocation.
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