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Chapter 17: Problem 6

(a) Write equations describing two different ways in which benzyl ethyl ether could be prepared by a Williamson ether synthesis. (b) Write an equation showing the most practical synthesis of allyl phenyl ether by the Williamson method

Short Answer

Expert verified
Benzyl ethyl ether can be synthesized using either benzyl alcohol with ethyl bromide or ethyl alcohol with benzyl chloride. Allyl phenyl ether is made using allyl alcohol with phenyl bromide.

Step by step solution

01

Understand Benzyl Ethyl Ether Synthesis

Benzyl ethyl ether can be prepared using two main pathways in the Williamson ether synthesis. In this method, an alkoxide ion acts as a nucleophile and attacks a primary alkyl halide.
02

Draw First Reaction Mechanism for Benzyl Ethyl Ether

Consider the first pathway: use benzyl alcohol to form benzyl alkoxide ion \((C_6H_5CH_2O^-)\) and ethyl bromide \((C_2H_5Br)\) as the alkyl halide.The reaction is: \[ C_6H_5CH_2OH + NaH \rightarrow C_6H_5CH_2O^-Na^+ + H_2 \]\[ C_6H_5CH_2O^-Na^+ + C_2H_5Br \rightarrow C_6H_5CH_2OC_2H_5 + NaBr \]
03

Draw Second Reaction Mechanism for Benzyl Ethyl Ether

Consider the second pathway: use ethyl alcohol to form ethoxide ion \((C_2H_5O^-)\) and benzyl chloride \((C_6H_5CH_2Cl)\) as the alkyl halide.The reaction is:\[ C_2H_5OH + NaH \rightarrow C_2H_5O^-Na^+ + H_2 \]\[ C_2H_5O^-Na^+ + C_6H_5CH_2Cl \rightarrow C_6H_5CH_2OC_2H_5 + NaCl \]
04

Analyze the Reactions for Practicality

Both reactions are valid, but the choice depends on the reactivity and availability of starting materials. Typically, using ethyl bromide or benzyl chloride as the alkyl halide is practical because they are good electrophiles.
05

Understand Allyl Phenyl Ether Synthesis

For preparing allyl phenyl ether, the procedure uses allyl alcohol to form an alkoxide ion (allyl alkoxide) and phenyl halide as the alkyl halide.
06

Write Reaction for Allyl Phenyl Ether

The Williamson synthesis for allyl phenyl ether uses an allyl alkoxide and phenyl bromide. The steps are:\[ CH_2=CHCH_2OH + NaH \rightarrow CH_2=CHCH_2O^-Na^+ + H_2 \]\[ CH_2=CHCH_2O^-Na^+ + C_6H_5Br \rightarrow C_6H_5OCH_2CH=CH_2 + NaBr \]
07

Final Reaction Practicality Consideration

Using a strong alkoxide ion like allyl alkoxide makes this reaction efficient. Phenyl bromide acts as a suitable substrate for attack due to good leaving group ability of bromide.

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Key Concepts

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

Benzyl Ethyl Ether
Benzyl ethyl ether is an organic compound formed when an alkoxide ion reacts with an alkyl halide. This is typically done through the Williamson Ether Synthesis, a method that allows the formation of ethers. The central idea is to use either benzyl alcohol or ethyl alcohol to form the necessary alkoxide ions. For instance:
  • Benzyl alcohol \(C_6H_5CH_2OH\) reacts with sodium hydride \(NaH\) to form benzyl alkoxide \(C_6H_5CH_2O^-Na^+\).
  • This alkoxide ion then attacks ethyl bromide, \(C_2H_5Br\), to form benzyl ethyl ether \(C_6H_5CH_2OC_2H_5\).
Alternatively, the reaction can start with ethyl alcohol to form ethoxide ions and then combine with benzyl chloride.
Allyl Phenyl Ether
Allyl phenyl ether is synthesized using the Williamson method by combining allylic and phenolic components. In this reaction sequence:
  • Allyl alcohol \(CH_2=CHCH_2OH\) is deprotonated by sodium hydride to form allyl alkoxide \(CH_2=CHCH_2O^-Na^+\).
  • This alkoxide ion subsequently reacts with phenyl bromide \(C_6H_5Br\).
  • The product of this reaction is allyl phenyl ether \(C_6H_5OCH_2CH=CH_2\).
This procedure efficiently produces allyl phenyl ether thanks to the excellent leaving group properties of the halide and the effective action of a strong alkoxide ion.
Alkoxide Ion
Alkoxide ions are the central players in Williamson ether synthesis. These ions are generated when an alcohol reacts with a strong base such as sodium hydride. For example:
  • The reaction between benzyl alcohol and sodium hydride yields benzyl alkoxide \(C_6H_5CH_2O^-Na^+\).
  • Similarly, ethyl alcohol will form ethoxide ion \(C_2H_5O^-Na^+\).
Alkoxide ions act as potent nucleophiles capable of attacking alkyl halides to form ether linkages. This step of the reaction is crucial for the formation of ethers in the Williamson synthesis.
Alkyl Halide
Alkyl halides play a key role in the Williamson ether synthesis as they serve as the electrophilic component that the alkoxide ion attacks. This reaction forms ethers through nucleophilic substitution.
  • Ethyl bromide \(C_2H_5Br\) and benzyl chloride \(C_6H_5CH_2Cl\) are typical examples used in these reactions.
  • The presence of a good leaving group, such as bromide or chloride, is crucial as it facilitates the substitution process.
Selection of the alkyl halide is important depending on its reactivity and availability, which can influence the yield and rate of the reaction.
Nucleophilic Substitution Reactions
Nucleophilic substitution reactions are central to the formation of ethers in the Williamson ether synthesis. These reactions involve:
  • A nucleophile, such as an alkoxide ion, attacking an electrophile like an alkyl halide.
  • The exchange of a leaving group from the electrophile, resulting in the formation of a new compound.
Understanding the mechanism of nucleophilic substitution helps in predicting the outcome of Williamson reactions, ensuring high efficiency and proper selection of substrates.

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Most popular questions from this chapter

The epoxide derived from benzene, 1,2 -epoxycyclohexa- 3,5 -diene, exists in equilibrium with a monocyclic isomer oxepine. Which statement is correct concerning the aromaticity of these two isomers? A. Both are aromatic. B. Neither is aromatic. C. 1,2-Epoxycyclohexa-3,5-diene is aromatic; oxepine is not aromatic. D. Oxepine is aromatic; 1,2 -epoxycyclohexa- 3,5 -diene is not aromatic.

Write the structures of all the constitutionally isomeric ethers of molecular formula \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\), and give an acceptable name for each.

Lithium aluminum hydride reduction of 1,2 -epoxy-2-methylpropane gives, as expected, predominantly tert-butyl alcohol. When the reduction is carried out with an LiAlH \(_{4} / \mathrm{AlCl}_{3}\) mixture, however, epoxide rearrangement precedes reduction and isobutyl alcohol becomes the major product. This rearrangement was confirmed by a deuterium-labeling experiment in which an LiAlD \(_{4} / \mathrm{AlCl}_{3}\) mixture was used. Where was the deuterium located in the isobutyl alcohol product?

A series of dialkyl ethers was allowed to react with excess hydrogen bromide, with the following results. Identify the ether in each case. (a) One ether gave a mixture of bromocyclopentane and 1 -bromobutane. (b) Another ether gave only benzyl bromide. (c) A third ether gave one mole of 1,5 -d bromopentane per mole of ether. Sample Solution (a) In the reaction of dialkyl ethers with excess hydrogen bromide, each alkyl group of the ether function is cleaved and forms an alkyl bromide. Because bromocyclopentane and 1 -bromobutane are the products, the starting ether must be butyl cyclopentyl ether.

The \({ }^{1}\) H NMR spectrum of compound \(\mathbf{A}\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}\right)\) consists of two resonances of equal area: \(\delta 5.1\) (singlet) and \(\delta 7.2\) (broad multiplet). Upon treatment with 2 equivalents of hydrogen bromide, compound \(\mathbf{A}\) is converted to a single dibromide \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{Br}_{2}\right)\) and water. The \({ }^{1} \mathrm{H}\) NMR spectrum of the dibromide consists of two resonances of equal area: \(\delta 4.7\) (singlet) and \(\delta 7.3\) (multiplet). Draw reasonable structures for compound \(\mathrm{A}\) and the dibromide derived from it and briefly explain your reasoning.
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