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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Answer 2

This question is about what benzene forms when it reacts with isopropyl bromide and aluminium trichloride. Isopropyl bromide is an alkyl halide, not an acyl halide, so this is a Friedel-Crafts alkylation, and the aluminium trichloride generates a carbocation from the alkyl halide that then attacks the benzene ring. Checking each option against this alkylation pathway helps identify the product.

1. Benzophenone: This is a diphenyl ketone, formed when benzene reacts with an acyl chloride such as benzoyl chloride in a Friedel-Crafts acylation. Since isopropyl bromide is an alkyl halide and not an acyl halide, this ketone product cannot form here.
2. Acetophenone: This is a methyl phenyl ketone, formed when benzene reacts with acetyl chloride in a Friedel-Crafts acylation. Again, this needs an acyl halide as the starting reagent, so isopropyl bromide cannot give this product.
3. Isopropyl benzene: Aluminium trichloride pulls the bromide off isopropyl bromide to generate a secondary isopropyl carbocation, which is reasonably stable. This carbocation then attacks the benzene ring in an electrophilic aromatic substitution, attaching the isopropyl group directly to the ring and giving isopropyl benzene, also called cumene.
4. n-Propyl benzene: Forming this product would need a straight chain n-propyl carbocation to attack the ring. A primary carbocation like this is far less stable than the secondary isopropyl carbocation actually generated, so the reaction does not shift into this less stable form, and n-propyl benzene is not the product.

Since isopropyl bromide reacts through alkylation and forms the more stable secondary carbocation, the ring becomes substituted with an isopropyl group rather than an acyl group or a straight chain propyl group.

So the correct answer is Isopropyl benzene.