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The following reaction is known as $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{Cl} \frac{\mathrm{AlCl}_{3}}{\text { (anhy) }} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}+\mathrm{HCl} $$ (a) Wurtz-Fittig reaction (b) Friedel-Crafts reaction (c) Rosenmund reaction (d) Sandmeyer reaction.

Short Answer

Expert verified
The correct answer is (b) Friedel-Crafts reaction.

Step by step solution

01

Identify the Type of Reaction

This reaction involves the conversion of an aromatic compound, benzene (C_6H_6), with an alkyl chloride (CH_3Cl) in the presence of a Lewis acid catalyst (AlCl_3), leading to the formation of an alkylated aromatic compound (C_6H_5CH_3) and hydrogen chloride (HCl). This process is characteristic of a specific type of aromatic substitution reaction.
02

Examine the Options

Review the provided reaction options and assess which type of reaction matches the process described in the reaction. Wurtz-Fittig Reaction involves reactions between alkyl halides and aryl halides. Friedel-Crafts Reactions are typically aromatic substitution reactions involving an alkyl halide and a Lewis acid catalyst. Rosenmund Reaction is a reduction process. Sandmeyer Reaction involves the conversion of an aromatic diazonium salt into a substituted benzene.
03

Determine the Correct Reaction

Based on the characteristics of the reaction, where an alkyl group from the alkyl chloride is substituted onto the aromatic ring in the presence of a Lewis acid catalyst, without reducing or converting diazonium salts, the reaction is identified as a Friedel-Crafts alkylation reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aromatic Substitution Reaction
In an aromatic substitution reaction, one atom or group of atoms in an aromatic compound is replaced by another atom or group of atoms. The uniqueness of these reactions lies in the preservation of the aromatic system's stability throughout the process.

An aromatic compound like benzene has a ring of alternating double bonds, known as the aromatic ring, which provides high stability due to delocalized π electrons. In the context of our exercise, benzene undergoes a reaction where one of its hydrogen atoms is substituted by a methyl group (-CH3), resulting in toluene. This type of reaction is facilitated by the presence of a catalyst, which in the given exercise is aluminum chloride (AlCl3).

It is important for students to understand the mechanism involved in this transformation, which proceeds through the formation of an intermediate complex where the aromaticity of the ring is temporarily lost, paving the way for the incoming group to attach. Afterwards, the aromaticity is restored, completing the substitution. Understanding these steps can clarify how these reactions work and why they are so important in organic synthesis.
Lewis Acid Catalyst
A Lewis acid acts as an electron pair acceptor, and in the case of aromatic substitution reactions such as Friedel-Crafts alkylation, it plays a pivotal role. When we look at our exercise, aluminum chloride (AlCl3) serves as the Lewis acid catalyst.

By accepting an electron pair, AlCl3 increases the reactivity of the alkyl chloride (CH3Cl) and forms a complex that is more electrophilic. This makes it easier for the alkyl group to attach itself to the aromatic ring. Simplifying this, imagine the Lewis acid catalyst as a tool that makes the nail (alkyl group) sharp and ready to be driven into the wood (aromatic ring).

For students, understanding the role of a Lewis acid in such reactions is key. It not only activates the halide, but it is also involved in the recovery of aromaticity in the final step by accepting the hydrogen chloride (HCl) released, which verifies its crucial position in the overall reaction mechanism.
Alkylated Aromatic Compound
Upon completion of a Friedel-Crafts alkylation reaction, the product formed is an alkylated aromatic compound. From our exercise, the result is toluene (C6H5CH3), where a methyl group has been added to the original benzene ring.

Alkylated aromatic compounds are significant in chemistry and widely used in the production of various materials, including pharmaceuticals, dyes, and polymers. The alkyl group provides different physical and chemical properties than the original aromatic compound. For instance, toluene is more reactive in certain types of reactions when compared to benzene and serves as a starting material for further chemical transformations.

It's essential for students to learn not just about the reaction that forms these compounds, but also about their practical applications and significance in real-world scenarios. Such understanding aids in appreciating why mastering the art of chemical synthesis, starting with reactions like Friedel-Crafts alkylation, is incredibly valuable.

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Most popular questions from this chapter

2-Bromopentane is treated with alcoholic KOH solution. What will be the major product formed in this reaction and what is the type of elimination) called? (a) Pent-1-ene, \(\beta\)-Elimination (b) Pent-2-ene, \(\beta\)-Elimination (c) Pent-1-ene, Nucleophilic substitution (d) Pent-2-ene, Nucleophilic substitution.

What happens when calcium carbide is treated with water? (a) Ethane is formed. (b) Methane and ethane are formed. (c) Ethyne is formed. (d) Ethene and ethyne are formed.

What happens when methane reacts with conc. \(\mathrm{HNO}_{3}\) at high temperature? (a) Nitromethane is formed. (b) Methanol is formed. (c) \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are formed. (d) \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are formed.

The ozonolysis product(s) of the following reaction is(are) $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{CH} \frac{\text { (i) } \mathrm{O}_{3}}{\text { (ii) } \mathrm{H}_{2} \mathrm{O}} \text { Product }(\mathrm{s}) \\ &\text { (a) } \mathrm{CH}_{3} \mathrm{COCH}_{3} \\ &\text { (b) } \mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{HCHO} \end{aligned} $$ (c) \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{HCOOH}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{HCOOH}\)

Identify the product for the following reaction: \(\mathrm{CH} \equiv \mathrm{CH}+\mathrm{HOCl} \longrightarrow\) (a) \(\mathrm{Cl}_{2} \mathrm{CHCHO}\) (b) \(\mathrm{CH}(\mathrm{OH})=\mathrm{CHCl}\) (c) \(\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{COCl}\)

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