Question

The alkylation of benzene with \(n\) -propyl chloride in the presence of anhydrous \(\Lambda l \mathrm{Cl}_{3}\) produces (1) \(n\) -propyl benzene (2) isopropyl benzene (3) 0 -dipropyl benzenc (4) a mixture of all these

Short answer

The major product is isopropyl benzene (2).

Step-by-step solution

Understand the reaction conditions

The given reaction is an alkylation of benzene using an alkyl halide (-propyl chloride) in the presence of a Lewis acid catalyst (anhydrous \(\text{AlCl}_3\)). This is known as the Friedel-Crafts alkylation reaction.

Identify the electrophile

-Propyl chloride in the presence of anhydrous \(\text{AlCl}_3\) will form a -propyl carbocation (-\text{C\(_3\)H\(_7\)}^{+}). This is the electrophile that will attack the benzene.

Consider carbocation rearrangement

The -propyl carbocation (-\text{C\(_3\)H\(_7\)}^{+}) formed can undergo rearrangement to a more stable secondary carbocation. Therefore, the -propyl carbocation will rearrange to an isopropyl carbocation (-\text{(CH\(_3\))\(_2\)CH}^{+}).

Analyze the products

The isopropyl carbocation (-\text{(CH\(_3\))\(_2\)CH}^{+}) will react with benzene to form isopropyl benzene (cumene). Therefore, the major product will be isopropyl benzene. -propyl benzene is not formed due to carbocation rearrangement, and no evidence suggests multiple substitutions to form o-dipropyl benzene.

Confirm the outcome

Since the major product will be isopropyl benzene, the correct answer is (2) isopropyl benzene.

Key concepts

alkylation of benzene

Let's dive into the alkylation of benzene, a key reaction in organic chemistry. Alkylation essentially means adding an alkyl group to a molecule. In this case, our molecule is benzene.
The reaction we are referring to is the Friedel-Crafts alkylation. This reaction uses an alkyl halide (like propyl chloride) and a Lewis acid catalyst to attach an alkyl group to benzene.
Benzene, being an electron-rich aromatic ring, reacts with an electrophile (an electron-poor species). By using a Lewis acid catalyst such as anhydrous \(\text{AlCl}_3\), the alkyl halide is activated, forming a carbocation (a positively charged species).
This carbocation then reacts with benzene to form the alkylated product. The overall process adds the alkyl group to the benzene ring, resulting in a new benzene derivative.

carbocation rearrangement

A crucial concept in Friedel-Crafts alkylation is carbocation rearrangement. When the alkyl halide reacts with \(\text{AlCl}_3\), it forms a primary carbocation.
However, primary carbocations ( \frac{C3H7^{+}} ) are unstable. They tend to rearrange to more stable forms.
In this case, the primary \(\text{propyl carbocation}\) rearranges to a secondary \(\text{isopropyl carbocation}\) because secondary carbocations are generally more stable than primary ones.
Thermodynamics drive this rearrangement, a process where the carbon structure rearranges to balance electronic and structural stability.
The result is that the initially formed primary carbocation often doesn't form the final product, influencing which alkylated benzene you end up with.

Lewis acid catalyst

Lewis acids are substances that accept an electron pair. In Friedel-Crafts reactions, \(\text{AlCl}_3\) serves as the Lewis acid catalyst.
This catalyst is essential because it helps generate the carbocation from the alkyl halide. \(\text{AlCl}_3\) interacts with the halide (chlorine in propyl chloride), creating an environment conducive to carbocation formation.
Here's how it works: The \(\text{AlCl}_3\) accepts an electron pair from the chlorine atom of the alkyl halide, weakening the carbon-chlorine bond and facilitating the formation of the carbocation.
This carbocation then becomes the key reactive species that benzene reacts with during alkylation. Without this catalyst, forming the necessary carbocation would be much more difficult, inhibiting the reaction from proceeding efficiently.