Question

Benzene reacts with n-propyl chloride in the presence of anhydrous $AlCl_3$ to give

• A. 3-propyl-1-chlorobenzene
• B. n-propyl benzene
• C. no reaction
• D. iso-propyl benzene

Answer 1

The Correct Option is D

 The given reaction involves benzene reacting with n-propyl chloride in the presence of anhydrous \(AlCl_3\), a classic example of a Friedel-Crafts alkylation. To solve this, let's go through the steps and logic behind the reaction:

1. The Friedel-Crafts alkylation is an electrophilic aromatic substitution reaction where an alkyl group is introduced into an aromatic ring. \(AlCl_3\) acts as a Lewis acid catalyst in this reaction, forming an electrophile.
2. In the presence of anhydrous \(AlCl_3\), n-propyl chloride (CH₃CH₂CH₂Cl) forms a carbocation. However, in this case, a 1^o propyl carbocation is initially formed, but being unstable, it can rearrange to a more stable 2^o carbocation (isopropyl carbocation).
3. The rearrangement happens due to the stability of carbocations: 3^o > 2^o > 1^o. The hydrogen shift leads to the formation of a more stable 2^o isopropyl carbocation (CH₃CH⁺CH₃).
4. The aromatic ring of benzene then attacks this stable isopropyl carbocation, resulting in the formation of isopropyl benzene (cumene) as the final product.

Hence, the correct outcome of the reaction is the formation of iso-propyl benzene.

Let's rule out other options:

• 3-propyl-1-chlorobenzene: This product is unlikely due to the rearrangement of the carbocation to form a more stable structure.
• n-propyl benzene: Direct formation of n-propyl benzene is not feasible due to the carbocation rearrangement.
• No reaction: A reaction does occur under these conditions as per the Friedel-Crafts alkylation process.

Therefore, the correct answer is: iso-propyl benzene.