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Chapter 18: Q 54. (page 724)

Question: Draw a stepwise mechanism for the following reaction.

Short Answer

Expert verified

The stepwise mechanism for the given reaction is as follows:

Formation of product C and product D

Formation of the resonance-stabilized carbocation

Carbocation A

Carbocation B

Formation of the product C

Formation of the product D

Step by step solution

01

Friedel-Craft's alkylation

In order to introduce an alkyl group to the benzene ring, Friedel-craft’s alkylation is used. It is a one-step reaction in which an alkyl halide is allowed to react with a benzene ring in the presence of aluminum chloride.

02

Mechanism for the given reaction

In the given reaction, benzene reacts with alkyl halide to produce the following products:

Formation of product C and product D

The step-wise mechanism for the given reaction is following:

  • Formation of the resonance-stabilized carbocation: The role of aluminum chloride is the abstraction of the halide ion from the given alkyl halide which results in the formation of the carbocations as shown in the following structure:

Formation of the resonance-stabilized carbocation

Carbocation A

Carbocation B

  • In the next step, carbocation A attacks the benzene molecule to result in the formation of the resonance hybrid and the product C as shown below:

Formation of the product C

  • The attack of the carbocation B also takes place in the same manner as carbocation A and results in the formation of the product D as shown below:

Formation of the product D

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Most popular questions from this chapter

Question: Rank the aryl halides in each group in order of increasing reactivity in nucleophilic aromatic substitution by an addition–elimination mechanism.

  1. chlorobenzene, p-fluoronitrobenzene, m-fluoronitrobenzene
  2. 1-fluoro-2,4-dinitrobenzene, 1-fluoro-3,5-dinitrobenzene, 1-fluoro-3,4-dinitrobenzene
  3. 1-fluoro-2,4-dinitrobenzene, 4-chloro-3-nitrotoluene, 4-fluoro-3-nitrotoluene

Synthesize each compound from benzene.

The 1 H NMR spectrum of phenol (\({C_6}{H_5}OH\)) shows three absorptions in the aromatic region: 6.70 (2 ortho H’s), 7.14 (2 meta H’s), and 6.80 (1 para H) ppm. Explain why the ortho and para absorptions occur at lower chemical shift than the meta absorption.

Benzyl bromide\({\bf{(}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br)}}\) reacts rapidly with \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\)to afford benzyl methyl ether\({\bf{(}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OC}}{{\bf{H}}_{\bf{3}}}{\bf{)}}\). Draw a stepwise mechanism for the reaction, and explain why this 1° alkyl halide reacts rapidly with a weak nucleophile under conditions that favor an\({{\bf{S}}_{\bf{N}}}{\bf{1}}\) mechanism. Would you expect the para-substituted benzylic halides \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{O}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br}}\) and \({{\bf{O}}_{\bf{2}}}{\bf{N}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br}}\) to each be more or less reactive than benzyl bromide in this reaction?Explain your reasoning.

Question: Draw a stepwise mechanism for the following reaction.
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