Draw the major organic product of the following Friedel-Crafts alkylation. (An excess of benzene is present.) The drawn product would require the formation of a rather unstable primary carbocation intermediate. The aluminum chloride polarizes the alkyl halide’s chlorine-carbon bond. Consider the possibility of a rearrangement reaction.
The Correct Answer And Explanation is:
To draw the major organic product of a Friedel-Crafts alkylation reaction, we need to consider a few important concepts:
🔬 Key Concepts to Consider:
- Friedel-Crafts Alkylation Reaction Basics:
- This is an electrophilic aromatic substitution reaction.
- Benzene reacts with an alkyl halide (R–Cl) in the presence of a Lewis acid catalyst like AlCl₃.
- Carbocation Formation and Rearrangement:
- The Lewis acid AlCl₃ helps form a carbocation from the alkyl halide.
- If a primary carbocation would form (which is unstable), a hydride or methyl shift may occur to generate a more stable secondary or tertiary carbocation before the aromatic ring attacks.
- Excess Benzene:
- Excess benzene ensures monosubstitution rather than polysubstitution.
🧪 Example Case:
Let’s say the alkyl halide used is 1-chlorobutane (CH₃CH₂CH₂CH₂Cl).
- AlCl₃ polarizes the C–Cl bond → forms a primary carbocation (CH₃CH₂CH₂CH₂⁺).
- A 1,2-hydride shift occurs → forms a more stable secondary carbocation (CH₃CH₂CH⁺CH₃).
- Benzene acts as a nucleophile → attacks the secondary carbocation.
- The major product is sec-butylbenzene (C₆H₅CH(CH₃)CH₂CH₃).
🧑🎨 Structure of the Product:
Here’s the major product structure based on the above rearrangement:
CHCH3
|
Benzene–CH2CH3
Or in IUPAC naming: sec-butylbenzene
-chlorobutane.