Problem 37 - General Chemistry: Principles and Modern Applications

Question

Alkylation of benzene can be accomplished by treating benzene with haloalkane (RX) in the presence of \(\mathrm{AlCl}_{3}\) The reaction is known as a Friedel-Crafts alkylation reaction. (The reaction is named after Charles Friedel, a French chemist, and James M. Crafts, an American chemist, who discovered this method of making alkylbenzenes in \(1877 .\) ) An example of a Friedel-Crafts alkylation reaction is shown below: The mechanism for this reaction involves the following steps. First, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) and \(\mathrm{AlCl}_{3}\) react in a Lewis acid-base reaction to form an adduct, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{Cl}-\mathrm{AlCl}_{3},\) in which a chlorine atom is bonded to both carbon and aluminum. The adduct then dissociates to \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+},\) a carbocation, and \(\mathrm{AlCl}_{4}{^-}\). The carbocation acts as an electrophile in a reaction with benzene, forming an arenium ion. Finally, a proton is removed from the arenium ion by \(\mathrm{AlCl}_{4}{^-},\) yielding an alkylbenzene, \(\mathrm{HCl},\) and \(\mathrm{AlCl}_{3}\) Write chemical equations for the elementary processes involved in forming 1 -methyl-1-phenylethane and \(\mathrm{HCl}\) from \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) and benzene. Use curved arrows to show the movement of electrons.

Step-by-Step Solution

Answer

The chemical equations for the elementary processes involved in forming 1-methyl-1-phenylethane and HCl from \((\mathrm{CH}_{3})_{2} \mathrm{CHCl}\) and benzene involve four main steps: A Lewis acid-base reaction, dissociation of the formed adduct, reaction of the carbocation with benzene to form an arenium ion, and final proton abstraction yielding the products.

Step 1: Lewis Acid-Base Reaction

Firstly, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) and \(\mathrm{AlCl}_{3}\) react together. This is a Lewis acid-base reaction.\[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl} + \mathrm{AlCl}_{3} \rightarrow \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{Cl}-\mathrm{AlCl}_{3}\]In this reaction, a new compound, an adduct, is formed with a chlorine atom bonded to both carbon and aluminium.

Step 2: Dissociation of Adduct

Secondly, the adduct dissociates to form a carbocation and an ion.\[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{Cl}-\mathrm{AlCl}_{3} \rightarrow \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+} + \mathrm{AlCl}_{4}{^-}\]The carbocation which is formed will act as an electrophile in the next step.

Step 3: Reaction with Benzene

Thirdly, the carbocation reacts with benzene to form an arenium ion.\[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+} + \mathrm{C}_{6}\mathrm{H}_{6} \rightarrow \mathrm{C}_{6}\mathrm{H}_{5}-\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+} + \mathrm{H}^{+}\]In this step, benzene acts as a nucleophile.

Step 4: Protonation and Final Products

Lastly, a proton is abstracted from the arenium ion by \(\mathrm{AlCl}_{4}{^-}\) yielding the alkylbenzene, \(\mathrm{HCl}\), and \(\mathrm{AlCl}_{3}\).\[\mathrm{C}_{6}\mathrm{H}_{5}-\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+} + \mathrm{AlCl}_{4}{^-} \rightarrow 1\mathrm{-Methyl}-1\mathrm{-phenylethane} + \mathrm{HCl} + \mathrm{AlCl}_{3}\]

Key Concepts

Alkylation of Benzene

The process of alkylation of benzene involves the addition of an alkyl group to the benzene ring. This is typically done using a haloalkane and a Lewis acid catalyst such as aluminum chloride (abla AlCl_3 ). During the Friedel-Crafts alkylation reaction, the presence of the Lewis acid is crucial as it enables the formation of a highly reactive electrophile, which is necessary for the subsequent reaction with benzene. The product of this reaction is alkylbenzene, a compound where the benzene ring is substituted by an alkyl group.

It's important to ensure that the alkylating agent (the haloalkane) does not lead to carbocation rearrangements, as this can result in a mixture of products. Moreover, over-alkylation can occur if the newly formed alkylbenzene acts as the reactant in further alkylation steps. To mitigate these issues, the proper choice of haloalkane and experimental conditions is essential.

Lewis Acid-Base Reaction

In the context of the Friedel-Crafts alkylation, a Lewis acid-base reaction is the initial step where the haloalkane reacts with the Lewis acid catalyst. In this reaction, AlCl_3 acts as a Lewis acid by accepting a pair of electrons from the chloride ion of the haloalkane. This interaction forms a complex, which then leads to the generation of the carbocation, the species that will undergo subsequent reaction with benzene.

Understanding the Lewis acid-base concept is vital as it explains how the catalyst activates the haloalkane and enables the formation of the carbocation intermediate. Learners should remember that in Lewis acid-base reactions, acids are electron pair acceptors, and bases are electron pair donors. This principle is foundational not only for the Friedel-Crafts alkylation but also for many other organic transformations.

Carbocation Formation

Carbocation formation is a pivotal step in the Friedel-Crafts alkylation process. After the initial Lewis acid-base reaction, the complex dissociates to yield a positively charged carbocation. This species is essential for the subsequent electrophilic aromatic substitution, as it acts as the electrophile looking to accept electrons from the aromatic ring.

Carbocations are generally unstable and seek to attain a more stable state by reacting with electron-rich species. In the context of the reaction, the stability of the carbocation is a determinant for the overall reaction rate. Furthermore, students should be aware that the structure of the carbocation can influence the outcome of the reaction, as different carbocations can produce different isomers in the final product.

Electrophilic Aromatic Substitution

Electrophilic aromatic substitution is the mechanism by which the carbocation reacts with the benzene ring. Benzene, being rich in π-electrons, acts as a nucleophile and donates a pair of electrons to the carbocation. This results in the formation of an arenium ion intermediate, which then loses a proton to regenerate the aromatic system in the presence of the Lewis acid catalyst.

The beauty of electrophilic aromatic substitution lies in its ability to modify benzene rings by introducing a variety of functional groups, expanding benzene's utility in synthesizing complex molecules. Students should appreciate the significance of preserving the aromaticity of the benzene ring throughout the reaction, as this drives the release of the proton in the final step, culminating in the production of the substituted benzene derivative.