Chapter 8, Problem 8.64P

Interpretation Introduction

(a)

Interpretation:

The detailed mechanism for the given reaction occurring via $\text{E2}$ reaction involving the $\text{D-}$ labeled substrate is to be drawn, and the major product(s) is to be predicted.

Concept introduction:

The deuterium $\text{D}$ is an isotope of hydrogen atom, which can also be eliminated along with the leaving group in an elimination reaction. In case of $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$ or $\text{D}$) must be anti to each other. If the substrate has both hydrogen atom and deuterium atom attached to the carbon adjacent to the leaving group, then elimination can occur in two ways. In one case, the hydrogen atom is eliminated along with the leaving group, and in another case, the $\text{D}$ atom is eliminated along with the leaving group. As a result, a mixture of diastereomers is formed.

Answer to Problem 8.64P

The $\text{E2}$ mechanism for the given reaction involving given $\text{D-}$ labeled substrate is

Explanation of Solution

The given reaction equation is

Here, the chlorine atom is the leaving group, which can be eliminated along with either $\text{H}$ or $\text{D}$, as indicated in the following substrate:

In the given substrate, the leaving group $\text{Cl}$ is above the plane, represented by a wedge bond. For the $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$ or $\text{D}$) must be anti to each other. In the given substrate, only $\text{D}$ is anti to the leaving group $\text{Cl}$. Thus the reaction proceeds by the elimination of $\text{D}$ along with the leaving group $\text{Cl}$. In the mechanism, the base abstracts the atom $\text{D}$ and displaces the leaving group $\text{Cl}$, forming a $\text{C=C}$ bond. The curved arrow mechanism of elimination of $\text{DCl}$ is as follows:

Conclusion

The product formed for the given reaction from $\text{E2}$ mechanism is determined based on substrate stereochemistry.

Interpretation Introduction

(b)

Interpretation:

The detailed mechanism for the given reaction occurring via $\text{E1}$ reaction involving the $\text{D-}$ labeled substrate is to be drawn, and the major product(s) is to be predicted.

Concept introduction:

The deuterium $\text{D}$ is an isotope of hydrogen atom, which can also be eliminated along with the leaving group in an elimination reaction. The $\text{E1}$ reaction proceeds with the formation of a carbocation intermediate by elimination of the leaving group in the first step. In the second step, either hydrogen or deuterium atom adjacent to the carbocation is removed by the base to form a $\text{C=C}$ bond.

Answer to Problem 8.64P

The $\text{E1}$ mechanism for the given reaction involving given $\text{D-}$ labeled substrate is as follows:

Elimination of $\text{HCl}$:

Elimination of $\text{DCl}$:

Explanation of Solution

The given reaction equation is

Here, the chlorine atom is the leaving group, which can be eliminated along with either $\text{H}$ or $\text{D}$, as indicated in the following substrate.

According to $\text{E1}$ mechanism, the first step is the formation of a carbocation by removing the leaving group. Thus, the tertiary carbocation formed by the elimination of the leaving group $\text{Cl}$ from the given substrate is shown below:

In the second step, both $\text{H}$ or $\text{D}$ adjacent to the carbocation can be removed by the base, and a $\text{C=C}$ bond is formed. The mechanisms for the elimination of $\text{H}$ and $\text{D}$ separately are shown below:

Conclusion

The product formed for the given reaction from $\text{E1}$ mechanism is determined based on the elimination of hydrogen and deuterium adjacent to the carbocation formed.

Interpretation Introduction

(c)

Interpretation:

The molar mass of each product from $\text{E2}$ and $\text{E1}$ mechanism for the given reaction is to be determined.

Concept introduction:

The molar mass is the sum of the atomic mass of each element in the molecular formula. The deuterium $\text{D}$ is an isotope of hydrogen atom, which can also be eliminated along with the leaving group in an elimination reaction. In case of $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$ or $\text{D}$) must be anti to each other. If the substrate has both hydrogen atom and deuterium atom attached to the carbon adjacent to the leaving group, then elimination can occur in two ways. In one case, the hydrogen atom is eliminated along with the leaving group, and in another case, the $\text{D}$ atom is eliminated along with the leaving group. As a result, a mixture of diastereomers is formed.

The $\text{E1}$ reaction proceeds with the formation of a carbocation intermediate by elimination of the leaving group in the first step. In the second step, either hydrogen or deuterium atom adjacent to the carbocation is removed by the base to form a $\text{C=C}$ bond.

Answer to Problem 8.64P

The molar masses of the products formed by $\text{E2}$ and $\text{E1}$ mechanism for the given reaction are ${\text{C}}_{\text{8}}{\text{H}}_{\text{14}}\text{=110}\text{u}$ and ${\text{C}}_{\text{8}}{\text{H}}_{\text{13}}\text{D = 111 u}$.

Explanation of Solution

The given reaction equation is

Here, the chlorine atom is the leaving group, which can be eliminated along with either $\text{H}$ or $\text{D}$, as indicated in the following substrate.

In the given substrate, the leaving group $\text{Cl}$ is above the plane, represented by a wedge bond. For the $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$ or $\text{D}$) must be anti to each other. In the given substrate, only $\text{D}$ is anti to the leaving group $\text{Cl}$. Thus, the reaction proceeds by elimination of $\text{D}$ along with the leaving group $\text{Cl}$. Therefore, the product formed in this mechanism is

The molecular formula for this product is ${\text{C}}_{\text{8}}{\text{H}}_{\text{14}}$.

According to $\text{E1}$ mechanism, the first step is the formation of a carbocation by removing the leaving group. Thus, the tertiary carbocation formed by elimination of the leaving group $\text{Cl}$ from the given substrate is shown below:

In the second step, both $\text{H}$ or $\text{D}$ adjacent to the carbocation can be removed by the base to form a $\text{C=C}$ bond. The mechanisms for the elimination of $\text{H}$ and $\text{D}$ separately are shown below:

The molecular formulae for both the products formed by $\text{E1}$ mechanism are ${\text{C}}_{\text{8}}{\text{H}}_{\text{14}}$ and ${\text{C}}_{\text{8}}{\text{H}}_{\text{13}}\text{D}$.

Hence the molar mass of the product ${\text{C}}_{\text{8}}{\text{H}}_{\text{14}}$ is $\text{110 u}$, and that of ${\text{C}}_{\text{8}}{\text{H}}_{\text{13}}\text{D}$ is $\text{111 u}$.

Conclusion

The molar masses of each product formed by $\text{E2}$ and $\text{E1}$ mechanism for the given reaction are determined by adding the atomic mass of each element multiplied by the number of each atom in the molecular formula.