Why do we get isopropyl benzene ontreating benzene with 1-chloropropane insteadof n-propyl benzene?

The reaction of benzene with 1-chloropropane in the presence of a Lewis acid like AlCl3 follows the Friedel-Crafts alkylation mechanism. Here are the steps:

1. The Lewis acid AlCl3 coordinates with the chlorine atom in 1-chloropropane, making the carbon atom more electrophilic.

2. The benzene ring, acting as a nucleophile, attacks the electrophilic carbon, leading to the formation of a carbocation intermediate.

3. The carbocation intermediate can rearrange if a more stable carbocation is possible. In this case, a 1° carbocation can rearrange to a 2° carbocation through a hydride shift. This is because secondary carbocations are more stable than primary carbocations due to hyperconjugation.

4. The rearranged carbocation then loses a proton to the AlCl4- ion, restoring the aromaticity of the benzene ring.

5. The final product is isopropyl benzene, not n-propyl benzene, due to the rearrangement of the carbocation intermediate to a more stable form.

This is a classic example of the Friedel-Crafts alkylation reaction, which often leads to rearrangement products due to the stability of secondary and tertiary carbocations over primary carbocations.