Chapter 9, Problem 9.69P

Interpretation Introduction

(a)

Interpretation:

For the given substrate, when $\text{NaOH}$ is used as the base, the major E1 product is to be drawn.

Concept introduction:

As the number of alkyl groups on the carbon bonded to the leaving group increases, the rate of E1 reaction increases. Each additional alkyl group, which is electron donating, stabilizes the carbocation intermediate produced and thus helps the leaving group leave. The ${\text{S}}_{\text{N}}\text{1}$ and E1 reactions are more sensitive to the leaving group ability. Excellent leaving groups favor ${\text{S}}_{\text{N}}\text{1}$ and E1. Leaving group ability is determined largely by charge stability. The stronger the leaving group’s conjugate acid, the better the leaving group. After heterolysis, a carbocation is formed. To consider the stability of a carbocation, the rearrangements must be considered. Elimination usually takes place so as to produce the most stable, i.e., highly alkyl substituted, alkene.

Answer to Problem 9.69P

The major E1 product for the given reaction is shown below:

Explanation of Solution

Elimination usually takes place so as to produce the most stable, i.e., highly alkyl substituted, alkene.

The possible E1 products are obtained by eliminating the leaving group and a proton on carbon adjacent to the one bonded to the leaving group. Here the ${\text{Br}}^{\text{-}}$ is a good leaving group. After heterolysis, a secondary carbocation is formed. The tertiary carbocation is more stable than the secondary carbocation. Therefore, the 1, 2-hydride shift takes place and a tertiary carbocation is formed. The possible products for this elimination reaction are shown below:

The first and second alkene product is highly substituted than the third alkene product. Therefore, the first and second alkene products are more stable. The base in this reaction is OH, which is not bulky, so the first and second products are the major products.

Conclusion

From the stability of the product formed, the major E1 product is drawn.

Interpretation Introduction

(b)

Interpretation:

For the given substrate, when $\text{NaOH}$ is used as the base, the major E1 product is to be drawn.

Concept introduction:

As the number of alkyl groups on the carbon bonded to the leaving group increases, the rate of E1 reaction increases. Each additional alkyl group, which is electron donating, stabilizes the carbocation intermediate produced and thus helps the leaving group leave. The ${\text{S}}_{\text{N}}\text{1}$ and E1 reactions are more sensitive to the leaving group ability. Excellent leaving groups favor ${\text{S}}_{\text{N}}\text{1}$ and E1. Leaving group ability is determined largely by charge stability. The stronger the leaving group’s conjugate acid, the better the leaving group. After heterolysis, a carbocation is formed. To consider the stability of a carbocation, the rearrangements must be considered. Elimination usually takes place so as to produce the most stable, i.e., highly alkyl substituted, alkene.

Answer to Problem 9.69P

The major E1 product for the given reaction is shown below:

Explanation of Solution

Elimination usually takes place so as to produce the most stable, i.e., highly alkyl substituted, alkene.

The possible E1 products are obtained by eliminating the leaving group and a proton on carbon adjacent to the one bonded to the leaving group. Here the ${\text{Br}}^{\text{-}}$ is a good leaving group. After heterolysis, the secondary carbocation is formed. The tertiary carbocation is more stable than the secondary carbocation. Therefore, the 1, 2-hydride shift takes place, and a tertiary carbocation is formed. The major product for this elimination reaction is shown below:

Conclusion

From the stability of the product formed, the major E1 product is drawn.

Interpretation Introduction

(c)

Interpretation:

For the given substrate, when $\text{NaOH}$ is used as the base, the major E1 product is to be drawn.

Concept introduction:

As the number of alkyl groups on the carbon bonded to the leaving group increases, the rate of E1 reaction increases. Each additional alkyl group, which is electron donating, stabilizes the carbocation intermediate produced and thus helps the leaving group leave. The ${\text{S}}_{\text{N}}\text{1}$ and E1 reactions are more sensitive to the leaving group ability. Excellent leaving groups favor ${\text{S}}_{\text{N}}\text{1}$ and E1. Leaving group ability is determined largely by charge stability. The stronger the leaving group’s conjugate acid, the better the leaving group. After heterolysis, a carbocation is formed. To consider the stability of a carbocation, the rearrangements must be considered. Elimination usually takes place so as to produce the most stable, i.e., highly alkyl substituted, alkene.

Answer to Problem 9.69P

The major E1 product for the given reaction is shown below:

Explanation of Solution

Elimination usually takes place so as to produce the most stable, i.e., highly alkyl substituted, alkene.

The possible E1 products are obtained by eliminating the leaving group and a proton on carbon adjacent to the one bonded to the leaving group. Here the ${\text{Br}}^{\text{-}}$ is a good leaving group. After heterolysis, the secondary carbocation is formed. The tertiary carbocation is more stable than the secondary carbocation. Therefore, the 1, 2-hydride shift takes place, and a tertiary carbocation is formed. The major product for this elimination reaction is shown below:

Conclusion

From the stability of the product formed, the major E1 product is drawn.

Interpretation Introduction

(d)

Interpretation:

For the given substrate, when $\text{NaOH}$ is used as the base, the major E1 product is to be drawn.

Concept introduction:

As the number of alkyl groups on the carbon bonded to the leaving group increases, the rate of E1 reaction increases. Each additional alkyl group, which is electron donating, stabilizes the carbocation intermediate produced and thus helps the leaving group leave. The ${\text{S}}_{\text{N}}\text{1}$ and E1 reactions are more sensitive to the leaving group ability. Excellent leaving groups favor ${\text{S}}_{\text{N}}\text{1}$ and E1. Leaving group ability is determined largely by charge stability. The stronger the leaving group’s conjugate acid, the better the leaving group. After heterolysis, a carbocation is formed. To consider the stability of a carbocation, the rearrangements must be considered. Elimination usually takes place so as to produce the most stable, i.e., highly alkyl substituted, alkene.

Answer to Problem 9.69P

The major E1 product for the given reaction is shown below:

Explanation of Solution

Elimination usually takes place so as to produce the most stable that means highly alkyl substituted alkene.

The possible E1 products are obtained by eliminating the leaving group and a proton on the carbon adjacent to the one bonded to the leaving group. Here the ${\text{Br}}^{\text{-}}$ is a good leaving group. After heterolysis, the secondary carbocation is formed. The tertiary carbocation is more stable than the secondary carbocation. Therefore, the 1, 2-methyl shift takes place, and a tertiary carbocation is formed. The major product for this elimination reaction is shown below:

The first alkene product is highly substituted than the second alkene product. Therefore, the first alkene product is more stable. The base in this reaction is OH, which is not bulky, so the first and second products are the major products.

Conclusion

From the stability of the product formed, the major E1 product is drawn.

Interpretation Introduction

(e)

Interpretation:

For the given substrate, when $\text{NaOH}$ is used as the base, the major E1 product is to be drawn.

Concept introduction:

As the number of alkyl groups on the carbon bonded to the leaving group increases, the rate of E1 reaction increases. Each additional alkyl group, which is electron donating, stabilizes the carbocation intermediate produced and thus helps the leaving group leave. The ${\text{S}}_{\text{N}}\text{1}$ and E1 reactions are more sensitive to the leaving group ability. Excellent leaving groups favor ${\text{S}}_{\text{N}}\text{1}$ and E1. Leaving group ability is determined largely by charge stability. The stronger the leaving group’s conjugate acid, the better the leaving group. After heterolysis, a carbocation is formed. To consider the stability of a carbocation, the rearrangements must be considered. Elimination usually takes place so as to produce the most stable, i.e., highly alkyl substituted, alkene.

Answer to Problem 9.69P

The major E1 product for the given reaction is shown below:

Explanation of Solution

Elimination usually takes place so as to produce the most stable that means highly alkyl substituted alkene.

The possible E1 products are obtained by eliminating the leaving group and a proton on the carbon adjacent to the one bonded to the leaving group. Here the ${\text{Br}}^{\text{-}}$ is a good leaving group. After heterolysis, the secondary carbocation is formed. The tertiary carbocation is more stable than the secondary carbocation. Therefore, the 1, 2-methyl shift takes place, and a tertiary carbocation is formed. The major product for this elimination reaction is shown below:

Conclusion

From the stability of the product formed, the major E1 product is drawn.