Chapter 23, Problem 23.15P

Interpretation Introduction

(a)

Interpretation: The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is to be identified.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

The bulky base like LDA always abstracts the proton from the side of less substituted $\text{C}-\text{atom}$.

Answer to Problem 23.15P

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown below.

Explanation of Solution

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown is shown as,

Figure 1

In this reaction, the given compound is treated with LDA in the presence of THF solution to form an enolate ion. The strong base, LDA abstracts a proton from the less substituted carbon atom of the compound. Then methyl iodide reacts with the enolte ion to form the desired product.

Conclusion

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is to be identified.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

The bulky base like LDA always abstracts the proton from the side of less substituted $\text{C}-\text{atom}$.

Answer to Problem 23.15P

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown below.

Explanation of Solution

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown is shown as,

Figure 2

In this reaction, the given compound is treated with LDA in the presence of THF solution to form an enolate ion. The strong base, LDA abstracts a proton from the less substituted carbon atom or $\alpha -\text{carbon}$ atom of the compound. Then methyl iodide reacts with the enolte ion to form the desired product.

Conclusion

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is to be identified.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

The bulky base like LDA always abstracts the proton from the side of less substituted $\text{C}-\text{atom}$.

Answer to Problem 23.15P

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown below.

Explanation of Solution

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown is shown as,

Figure 3

In this reaction, the given compound is treated with LDA in the presence of THF solution to form an enolate ion. The strong base, LDA abstracts a proton from the less substituted carbon atom or $\alpha -\text{carbon}$ atom of the compound. Then methyl iodide reacts with the enolte ion to form the desired product.

Conclusion

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is to be identified.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

The bulky base like LDA always abstracts the proton from the side of less substituted $\text{C}-\text{atom}$.

Answer to Problem 23.15P

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is,

Explanation of Solution

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown is shown as,

Figure 4

In this reaction, the given compound is treated with LDA in the presence of THF solution to form an enolate ion. The strong base, LDA abstracts a proton from the less substituted carbon atom or $\alpha -\text{carbon}$ atom of the compound. Then methyl iodide reacts with the enolte ion to form the desired product.

Conclusion

The product that is formed by the reaction of the given compound with LDA in the presence of THF solution at low temperature, followed by ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown in Figure 4.