Chapter 7, Problem 7.29P

Interpretation Introduction

(a)

Interpretation:

Whether the given substrate undergoes E2 elimination step with base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$ is to be predicted. The product for $\text{E2}$ step with appropriate curved arrows is to be drawn.

Concept introduction:

$\text{E2}$ or bimolecular elimination step can take place when a strong base is in the presence of a substrate in which a leaving group (L) and a hydrogen atom are on adjacent carbon atoms.

$\text{E2}$ is the bimolecular elimination step. The step is bimolecular because it involves two reactants, a strong base, and the substrate. The strong base reacts with the substrate having the leaving group, and the hydrogen atoms are on adjacent carbons like $\text{H}-\text{C}-\text{C}-\text{L}$ or $\text{H}-\text{C}=\text{C}-\text{L}$. The first curved arrow is drawn from the negatively charged atom of a strong base to the hydrogen adjacent to the leaving group in the substrate. The second curved arrow is drawn from the $\text{C-H}$ bond to the region between the adjacent carbon atoms bearing hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid the exceeding octet of the carbon atom. The main product formed in this step is an alkene ($\text{C=C}$) or alkyne ($\text{C}\equiv \text{C}$), double bond between two carbon atoms.

Answer to Problem 7.29P

The given substrate cannot undergo E2 elimination step with base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$.

Explanation of Solution

The given substrate is

$\text{E2}$ or bimolecular elimination step can take place when a strong base is in the presence of a substrate in which a leaving group (L) and a hydrogen atom are on adjacent carbon atoms, but there is no H atom on the adjacent carbon atom to the leaving group.

Thus, the given substrate cannot undergo $\text{E2}$ elimination step.

Conclusion

No H atom on adjacent carbon atom.

Interpretation Introduction

(b)

Interpretation:

Whether the given substrate undergoes E2 elimination step with base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$, is to be predicted. The product for the $\text{E2}$ step with appropriate curved arrows is to be drawn.

Concept introduction:

$\text{E2}$ or bimolecular elimination step can take place when a strong base is in the presence of a substrate in which a leaving group (L) and a hydrogen atom are on adjacent carbon atoms.

$\text{E2}$ is the bimolecular elimination step. The step is bimolecular because it involves two reactants, a strong base, and the substrate. The strong base reacts with the substrate having the leaving group and the hydrogen atoms are on adjacent carbons like $\text{H}-\text{C}-\text{C}-\text{L}$ or $\text{H}-\text{C}=\text{C}-\text{L}$. The first curved arrow is drawn from the negatively charged atom of a strong base to the hydrogen adjacent to the leaving group in the substrate. The second curved arrow is drawn from the $\text{C-H}$ bond to the region between the adjacent carbon atoms bearing hydrogen and leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding the octet of carbon atom. The main product formed in this step is an alkene ($\text{C=C}$) or alkyne ($\text{C}\equiv \text{C}$), double bond between the carbon atoms.

Answer to Problem 7.29P

The given substrate undergoes E2 elimination step.

The E2 elimination step for the given substrate is drawn as:

Explanation of Solution

The given substrate is:

In the above given substrate, Cl acts as a leaving group with one H atom on each side. Any one of the hydrogen atom is taken away forming a C=C bond. The first curved arrow is drawn from the lone pair of nitrogen atom of a base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$. The second curved arrow is drawn from the $\text{C-H}$ bond to the region between two adjacent carbons having $\text{H}$ and $\text{Cl}$. The third curved arrow is drawn representing the breaking of $\text{C-Cl}$ bond to avoid exceeding the octet of carbon.

Conclusion

The product of $\text{E2}$ step, respective alkene, is drawn with appropriate curved arrows.

Interpretation Introduction

(c)

Interpretation:

Whether the given substrate undergo E2 elimination step with base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$ is to be predicted. The product for the $\text{E2}$ step with appropriate curved arrows is to be drawn.

Concept introduction:

$\text{E2}$ or bimolecular elimination step can take place when a strong base is in the presence of a substrate in which a leaving group (L) and a hydrogen atom are on adjacent carbon atoms.

$\text{E2}$ is the bimolecular elimination step. The step is bimolecular because it involves two reactants, a strong base, and the substrate. The strong base reacts with substrate having leaving group and the hydrogen atoms are on adjacent carbons like $\text{H}-\text{C}-\text{C}-\text{L}$ or $\text{H}-\text{C}=\text{C}-\text{L}$. The first curved arrow is drawn from the negatively charged atom of a strong base to the hydrogen adjacent to the leaving group in the substrate. The second curved arrow is drawn from the $\text{C-H}$ bond to the region between the adjacent carbon atoms bearing hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding the octet of carbon atom. The main product formed in this step is alkene ($\text{C=C}$) or alkyne ($\text{C}\equiv \text{C}$), double bond between the carbon atoms.

Answer to Problem 7.29P

The given substrate undergoes E2 elimination step.

The E2 elimination step for the given substrate is drawn as:

Explanation of Solution

The given substrate is:

In the above given substrate, $\text{I}$ acts as a leaving group. Any one of the hydrogen atom is taken away forming a C=C bond. The first curved arrow is drawn from the lone pair of nitrogen atom of base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$. The second curved arrow is drawn from the $\text{C-H}$ bond to the region between two adjacent carbons having $\text{H}$ and $\text{I}$. The third curved arrow is drawn representing the breaking of $\text{C-I}$ bond to avoid exceeding the octet of carbon.

Conclusion

The product of $\text{E2}$ step, respective alkene, is drawn with appropriate curved arrows.

Interpretation Introduction

(d)

Interpretation:

Whether the given substrate undergoes E2 elimination step with base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$ is to be predicted. The product for the $\text{E2}$ step with appropriate curved arrows is to be drawn.

Concept introduction:

$\text{E2}$ or bimolecular elimination step can take place when a strong base is in the presence of a substrate in which a leaving group (L) and a hydrogen atom are on adjacent carbon atoms.

$\text{E2}$ is the bimolecular elimination step. The step is bimolecular because it involves two reactants, a strong base, and the substrate. The strong base reacts with substrate having leaving group and the hydrogen atoms are on adjacent carbons like $\text{H}-\text{C}-\text{C}-\text{L}$ or $\text{H}-\text{C}=\text{C}-\text{L}$. The first curved arrow is drawn from the negatively charged atom of a strong base to the hydrogen adjacent to the leaving group in the substrate. The second curved arrow is drawn from the $\text{C-H}$ bond to the region between the adjacent carbon atoms bearing hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding the octet of carbon atom. The main product formed in this step is alkene ($\text{C=C}$) or alkyne ($\text{C}\equiv \text{C}$), double bond between the carbon atoms.

Answer to Problem 7.29P

The given substrate cannot undergoes E2 elimination step with base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$.

Explanation of Solution

The given substrate is

$\text{E2}$ or bimolecular elimination step can take place when a strong base is in the presence of a substrate in which a leaving group (L) and a hydrogen atom are on adjacent carbon atoms, but there is no H atom on the adjacent carbon atom to the leaving group.

Thus, the given substrate cannot undergo $\text{E2}$ elimination step.

Conclusion

There is no H atom on the adjacent carbon atom.

Interpretation Introduction

(e)

Interpretation:

Whether the given substrate undergo E2 elimination step with base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$ is to be predicted. The product for the $\text{E2}$ step with appropriate curved arrows is to be drawn.

Concept introduction:

$\text{E2}$ or bimolecular elimination step can take place when a strong base is in the presence of a substrate in which a leaving group (L) and a hydrogen atom are on adjacent carbon atoms.

The $\text{E2}$ is the bimolecular elimination step. The step is bimolecular because it involves two reactants, a strong base, and the substrate. The strong base reacts with substrate having leaving group and the hydrogen atoms are on adjacent carbons like $\text{H}-\text{C}-\text{C}-\text{L}$ or $\text{H}-\text{C}=\text{C}-\text{L}$. The first curved arrow is drawn from the negatively charged atom of a strong base to the hydrogen adjacent to the leaving group in the substrate. The second curved arrow is drawn from the $\text{C-H}$ bond to the region between the adjacent carbon atoms bearing hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of leaving group to avoid exceeding the octet of carbon atom. The main product form in such step is alkene ($\text{C=C}$) or alkyne ($\text{C}\equiv \text{C}$), double bond between two carbon atoms.

Answer to Problem 7.29P

The given substrate undergoes E2 elimination step.

The E2 elimination step for the given substrate is drawn as:

Explanation of Solution

The given substrate is

In the above given substrate, $\text{Br}$ acts as a leaving group. The hydrogen atom is taken away forming a $\text{C}\equiv \text{C}$ bond. The first curved arrow is drawn from the lone pair of nitrogen atom of a base ${\text{H}}_{\text{2}}{\text{N}}^{\text{-}}$. The second curved arrow is drawn from the $\text{C-H}$ bond to the region between two adjacent carbons having $\text{H}$ and $\text{Br}$. The third curved arrow is drawn representing the breaking of $\text{C-Br}$ bond to avoid exceeding the octet of carbon.

Conclusion

The product of $\text{E2}$ step and respective alkene is drawn with appropriate curved arrows.