Chapter 10, Problem 10.16P

Interpretation Introduction

(a)

Interpretation:

It is to be determined whether and how[DK1] the given ether can be produced from a Williamson ether synthesis. And if there are two feasible syntheses for the given ether, it is to be determined which one is more preferable.

Concept introduction:

The Williamson ether synthesis is the most convenient method for an ether synthesis. In this synthesis, an alkyl halide $\left(\text{R-X}\right)$ is treated with a salt of an alkoxide ion $\left({\text{R-O}}^{\text{-}}\right)$ to form ether $\left(R-O-R\right)$ as a product via ${\text{ S}}_{\text{N}}\text{2}$ reaction. The alkyl halide $\left(\text{R-X}\right)$ is the substrate, and alkoxide ion $\left({\text{R-O}}^{\text{-}}\right)$ is the nucleophile in this reaction. The ${\text{ S}}_{\text{N}}\text{2}$ reaction cannot take place when the leaving group is on $\text{sp2}$ carbon of a substrate because the electron-rich sp2 carbon will repel the attack of the upcoming nucleophile. A leaving group must not be on a tertiary carbon because of the steric hindrance. Very poor yield for an ${\text{ S}}_{\text{N}}\text{2}$ reaction may occur, or instead of ${\text{ S}}_{\text{N}}\text{2}$ reaction, other reactions may take place. The general structure of ether is $\text{R-O-R}$, where R is an alkyl and/or aryl group.

Answer to Problem 10.16P

The given ether can be produced successfully from Williamson ether synthesis as below:

Explanation of Solution

The structure of the given ether is

In this ether, one of the R groups is a phenyl ring, and the other is an ethyl group.

So, there are two routes to produce the desired ether by Williamson ether synthesis. Route one is discussed below.

Route I:

This is a feasible synthesis because the phenoxide ion is a good nucleophile, and the halide group attached on the primary carbon atom (primary alkyl halide) is a good substrate for a ${\text{ S}}_{\text{N}}\text{2}$ reaction.

The second possible route is discussed below.

Route II:

In this method, the halide group is on sp2 hybridized carbon, which is not acceptable for an ${\text{ S}}_{\text{N}}\text{2}$ reaction because there will be a repulsion between a cloud of pi electrons of the phenyl ring and a lone pair of electrons on the oxygen atom in the nucleophile. This results in a low yield of the product. Thus, this route is not feasible for Williamson ether synthesis. Therefore, the only the first route is feasible for Williamson ether synthesis.

Conclusion

As Williamson ether synthesis is an ${\text{ S}}_{\text{N}}\text{2}$ reaction, the choice of substrate should be a primary or methyl alkyl halide.

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether and how[DK2] the given ether can be produced from a Williamson ether synthesis. And if there are two feasible syntheses for the given ether, it is to be determined which one is more preferable.

Concept introduction:

The Williamson ether synthesis is the most convenient method for an ether synthesis. In this synthesis, an alkyl halide $\left(\text{R-X}\right)$ is treated with a salt of an alkoxide ion $\left({\text{R-O}}^{\text{-}}\right)$ to form ether $\left(R-O-R\right)$ as a product via ${\text{ S}}_{\text{N}}\text{2}$ reaction. The alkyl halide $\left(\text{R-X}\right)$ is the substrate, and alkoxide ion $\left({\text{R-O}}^{\text{-}}\right)$ is the nucleophile in this reaction. The ${\text{ S}}_{\text{N}}\text{2}$ reaction cannot take place when the leaving group is on $\mathrm{s}\mathrm{p}2$ carbon of a substrate because the electron-rich sp2 carbon will repel the attack of the upcoming nucleophile. A leaving group must not be on a tertiary carbon because of the steric hindrance. Very poor yield for an ${\text{ S}}_{\text{N}}\text{2}$ reaction may occur, or instead of ${\text{ S}}_{\text{N}}\text{2}$ reaction, other reactions may take place. The general structure of ether is $\text{R-O-R}$, where R is an alkyl and/or aryl group.

Answer to Problem 10.16P

The given ether cannot be synthesized by Williamson ether synthesis.

Explanation of Solution

The structure of the given ether is

In this ether, one of the R groups is a phenyl ring, and the other is a tertiary butyl group. Those two groups would be the potential alkyl halides for a Williamson ether synthesis reaction. Route I is shown below:

The retrosynthesis suggests that the given ether can be synthesized from a tertiary butyl halide as a substrate and a phenoxide ion as a nucleophile. But the alkyl halide has a leaving group on the tertiary carbon, so it will not follow an ${\text{ S}}_{\text{N}}\text{2}$ reaction. Thus, this route is not a successful route to synthesize the given ether.

Route I:[DK3]

Instead, it shows an E1 reaction with the phenoxide ion because the phenoxide ion acts as a base instead of the nucleophile due to bulkiness.

The second route is not acceptable because the positive charge comes on [DK4] the carbon atom of a phenyl ring, which is already electron-rich and sp2, which is not good for the ${\text{ S}}_{\text{N}}\text{2}$ reaction. The reaction of this route is given below:

Route II:

Since both routes do not give the desired ether as a product via ${\text{ S}}_{\text{N}}\text{2}$ reaction, the given ether cannot be produced from Williamson ether synthesis.

Conclusion

As Williamson ether synthesis is an ${\text{ S}}_{\text{N}}\text{2}$ reaction, the choice of substrate should be a primary or methyl alkyl halide.

Interpretation Introduction

(c)

Interpretation:

It is to be determined whether and how[DK5] the given ether can be produced from a Williamson ether synthesis. And if there are two feasible syntheses for the given ether, it is to be determined which one is more preferable.

Concept introduction:

The Williamson ether synthesis is the most convenient method for an ether synthesis. In this synthesis, an alkyl halide $\left(\text{R-X}\right)$ is treated with a salt of an alkoxide ion $\left({\text{R-O}}^{\text{-}}\right)$ to form ether $\left(R-O-R\right)$ as a product via ${\text{ S}}_{\text{N}}\text{2}$ reaction. The alkyl halide $\left(\text{R-X}\right)$ is the substrate, and alkoxide ion $\left({\text{R-O}}^{\text{-}}\right)$ is the nucleophile in this reaction. The ${\text{ S}}_{\text{N}}\text{2}$ reaction cannot take place when the leaving group is on $\mathrm{s}\mathrm{p}2$ carbon of a substrate because the electron-rich sp2 carbon will repel the attack of the upcoming nucleophile. A leaving group must not be on a tertiary carbon because of the steric hindrance. Very poor yield for an ${\text{ S}}_{\text{N}}\text{2}$ reaction may occur, or instead of ${\text{ S}}_{\text{N}}\text{2}$ reaction, other reactions may take place. The general structure of ether is $\text{R-O-R}$, where R is an alkyl and/or aryl group.

Answer to Problem 10.16P

The given ether can be successfully produced from a Williamson ether synthesis via two routes as below:

Route I:

Route II:

The first route is more preferable as it makes use of a primary alkyl halide as a substrate.

Explanation of Solution

The given ether is

One R group in the given ether is cyclohexane while the other is an allyl group. Both of these R groups can be potentially used as substrates in the Williamson ether synthesis. Route I is shown below:

In this route, the leaving group (halogen atom, X) is on the primary carbon, and an alkoxide ion is also a good nucleophile, so the reaction can proceed through ${\text{ S}}_{\text{N}}\text{2}$ reaction to form the given ether. Thus, this route is a successful route for the synthesis of the given ether by Williamson ether synthesis.

The other route for the synthesis of the given ether is shown below:

Route II:

In this route, the leaving group (halogen atom, X) is on the secondary carbon, and an alkoxide ion is also a good nucleophile, so the reaction can proceed through ${\text{ S}}_{\text{N}}\text{2}$ reaction to form the desired ether as shown above. Thus, this route is also possible for the synthesis of the given ether by Williamson ether synthesis.

Note that both routes are feasible for the given ether synthesis, but the substrate of both routes is different. In the first route, the substrate (alkyl halide) has a leaving group on primary carbon while in the second route it is on the secondary carbon. Since an ${\text{ S}}_{\text{N}}\text{2}$ reaction takes place more readily when the leaving group is on the primary carbon, the first route is more preferable for the given ether synthesis via Williamson ether synthesis.

Conclusion

As Williamson ether synthesis is an ${\text{ S}}_{\text{N}}\text{2}$ reaction, the choice of substrate should be a primary or methyl alkyl halide.