Chapter 7, Problem 7.27P

Interpretation Introduction

(a)

Interpretation:

Whether the product of the given step can eliminate a leaving group to form a different compound than the reactant is to be predicted. The product for the given nucleophilic elimination step with appropriate curved arrows is to be drawn.

Concept introduction:

In the nucleophilic elimination step, the more electronegative atom bears full or partial negative charge. This is an electron rich atom, and the less electronegative atom is relatively electron poor. The curved arrow drawn from the lone pair of electron rich atom points to the bonding region between the more electronegative atom and less electronegative atom representing the electron flow from the electron rich site to the electron poor site. The second curved arrow is drawn to represent the breaking of the bond between the less electronegative atom and leaving group to avoid exceeding an octet on the less electronegative atom.

Answer to Problem 7.27P

Products formed after the elimination of the leaving group are not the same as the reactant. Product formed in the nucleophilic elimination step with appropriate curved arrows is drawn as:

Explanation of Solution

Product for the given nucleophilic addition step is:

In the given product, there are two possible groups that can leave to form two different products.

In the first nucleophilic elimination step, the oxygen atom with negative charge is an electron rich site, and the carbon bonded to it is an electron poor site. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curved arrow is drawn to show the breaking of the bond between the carbon and the leaving group ${\text{R}}^{\text{-}}$ on the left side.

The respective product formed is different from the reactant in the given nucleophilic addition step. The X sign on the arrow represents that this nucleophilic elimination is unfeasible as ${\text{R}}^{\text{-}}$ group eliminates from the reactant.

In the second nucleophilic elimination step, the oxygen atom with negative charge is the electron rich site, and the carbon bonded to it is the electron poor site. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curve arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curve arrow is drawn to show the breaking of the bond between the carbon and the leaving group ${\text{R}}^{\text{-}}$ on the right side.

The respective product formed is different from the reactant in the given nucleophilic addition step. The X sign on the arrow represents that this nucleophilic elimination is unfeasible as ${\text{R}}^{\text{-}}$ group eliminates from the reactant.

Conclusion

Products formed in the elimination steps are different from the reactant in the given nucleophilic addition step.

Interpretation Introduction

(b)

Interpretation:

Whether the product of the given step can eliminate a leaving group to form a different compound than the reactant is to be predicted. The product for the given nucleophilic elimination step with appropriate curved arrows is to be drawn.

Concept introduction:

In the nucleophilic elimination step, the more electronegative atom bears full negative charge or partial negative charge. This is the electron rich atom and the less electronegative atom is relatively electron poor. The curved arrow drawn from the lone pair of electron rich atom points to the bonding region between the more electronegative atom and less electronegative atom representing the electron flow from the electron rich site to the electron poor site. The second curved arrow is drawn to represent the breaking of bond between the less electronegative atom and leaving group to avoid exceeding an octet on the less electronegative atom.

Answer to Problem 7.27P

The product formed after the elimination of the leaving group is not the same as the reactant. Product formed in the nucleophilic elimination step with an appropriate curved arrow is drawn as:

Explanation of Solution

Product for the given nucleophilic addition step is:

In the nucleophilic elimination step, the oxygen atom with negative charge is an electron rich site, and the carbon bonded to it is an electron poor site. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curved arrow is drawn to show the breaking of bond between the carbon and the leaving group ${\text{R}}^{\text{-}}$.

The respective product formed is different from the reactant in the given nucleophilic addition step. The X sign on the arrow represents this nucleophilic elimination is unfeasible as ${\text{R}}^{\text{-}}$ group eliminates from the reactant.

Conclusion

Product formed in the elimination step is different from the reactant in the given nucleophilic addition step.

Interpretation Introduction

(c)

Interpretation:

The product of the given step can eliminate a leaving group to form different compound than reactant is to be predicted. The product for the given nucleophilic elimination step with appropriate curved arrows is to be drawn.

Concept introduction:

In nucleophilic elimination step, the more electronegative atom bears full negative charge or partial negative charge. This is the electron rich atom and the less electronegative atom is relatively electron poor. The curved arrow drawn from the lone pair of electron rich atom points to the bonding region between the more electronegative atom and less electronegative atom representing the electron flow from electron rich site to electron poor site. The second curved arrow drawn to represent the breaking of bond between the less electronegative atom and leaving group to avoid exceeding an octet on the less electronegative atom.

Answer to Problem 7.27P

The products formed after the elimination of the leaving group are not the same as the reactant. Product formed in the nucleophilic elimination step with appropriate curved arrow is drawn as:

Explanation of Solution

Product for the given nucleophilic addition step is:

In the nucleophilic elimination step, the nitrogen atom with negative charge is electron rich site, and the carbon bonded to it is electron poor site. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged nitrogen to the mid of $\text{C = N}$ bond, and the second curved arrow is drawn to show the breaking of bond between the carbon and the leaving group ${\text{R}}^{\text{-}}$ on the right side.

The respective product formed is different from the reactant in the given nucleophilic addition step. The X sign on the arrow represents this nucleophilic elimination is unfeasible as ${\text{R}}^{\text{-}}$ group eliminates from the reactant.

Conclusion

Product formed in the elimination step is different from the reactant in the given nucleophilic addition step.

Interpretation Introduction

(d)

Interpretation:

Whether the product of the given step can eliminate a leaving group to form different compound than reactant is to be predicted. The product for the given nucleophilic elimination step with appropriate curved arrows is to be drawn.

Concept introduction:

In nucleophilic elimination step, the more electronegative atom bears full negative charge or partial negative charge. This is the electron rich atom and the less electronegative atom is relatively electron poor. The curved arrow drawn from the lone pair of electron rich atom points to the bonding region between the more electronegative atom and less electronegative atom representing the electron flow from electron rich site to electron poor site. The second curved arrow is drawn to represent the breaking of bond between the less electronegative atom and leaving group to avoid exceeding an octet on the less electronegative atom.

Answer to Problem 7.27P

Products formed after the elimination of the leaving group are not the same as the reactant. Product formed in the nucleophilic elimination step with appropriate curved arrow is drawn as:

Explanation of Solution

Product for the given nucleophilic addition step is:

In the first nucleophilic elimination step, the oxygen atom with negative charge is electron rich site, and the chlorine atom is a good leaving group. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curved arrow is drawn to show the breaking of bond between the carbon and the leaving group ${\text{Cl}}^{\text{-}}$. This nucleophilic elimination step involves elimination of the good leaving group, and hence, this step is feasible.

The respective product formed is different from the reactant in the given nucleophilic addition step.

In the second nucleophilic elimination step, the oxygen atom with negative charge is electron rich site and the carbon bonded to it is electron poor site. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curved arrow is drawn to show the breaking of bond between the carbon and the leaving group ${\text{R}}^{\text{-}}$.

The respective product formed is different from the reactant in the given nucleophilic addition step. The X sign on the arrow represents this nucleophilic elimination is unfeasible as ${\text{R}}^{\text{-}}$ group eliminates from the reactant.

Conclusion

Products formed in the elimination steps are different from the reactant in the given nucleophilic addition step.

Interpretation Introduction

(e)

Interpretation:

Whether the product of the given step can eliminate a leaving group to form different compound than reactant is to be predicted. The product for the given nucleophilic elimination step with appropriate curved arrows is to be drawn.

Concept introduction:

In nucleophilic elimination step, the more electronegative atom bears full negative charge or partial negative charge. This is the electron rich atom and the less electronegative atom is relatively electron poor. The curved arrow is drawn from the lone pair of electron rich atom points to the bonding region between the more electronegative atom and less electronegative atom representing the electron flow from electron rich site to electron poor site. The second curved arrow is drawn to represent the breaking of bond between the less electronegative atom and leaving group to avoid exceeding an octet on the less electronegative atom.

Answer to Problem 7.27P

Products formed after the elimination of the leaving group are not same as the reactant. Product formed in the nucleophilic elimination step with appropriate curved arrow is drawn as:

Explanation of Solution

Product for the given nucleophilic addition step is:

In the given product, there are two possible groups that can leave to form two different products.

In the first nucleophilic elimination step, the oxygen atom with negative charge is electron rich site, and the carbon bonded to it is electron poor site. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curved arrow is drawn to show the breaking of bond between carbon and the leaving group ${\text{R}}^{\text{-}}$ on the left side.

The respective product formed is different from the reactant in the given nucleophilic addition step. The X sign on the arrow represents this nucleophilic elimination is unfeasible as ${\text{R}}^{\text{-}}$ group is eliminated from the reactant.

In the second nucleophilic elimination step, the oxygen atom with negative charge is electron rich site and ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$ group is the leaving group. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curved arrow is drawn to show the breaking of bond between the carbon and the leaving group ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$. This nucleophilic elimination step involves elimination of the good leaving group, hence, this step is feasible.

The respective product formed is different from the reactant in the given nucleophilic addition step.

Conclusion

Products formed in the elimination steps are different from the reactant in the given nucleophilic addition step.

Interpretation Introduction

(f)

Interpretation:

Whether the product of the given step can eliminate a leaving group to form different compound than reactant is to be predicted. The product for the given nucleophilic elimination step with appropriate curved arrows is to be drawn.

Concept introduction:

In nucleophilic elimination step, the more electronegative atom bears full negative charge or partial negative charge. This is the electron rich atom and the less electronegative atom is relatively electron poor. The curved arrow drawn from the lone pair of electron rich atom points to bonding region between the more electronegative atom and less electronegative atom representing the electron flow from electron rich site to electron poor site. The second curved arrow is drawn to represent the breaking of bond between the less electronegative atom and leaving group to avoid exceeding an octet on the less electronegative atom.

Answer to Problem 7.27P

Products formed after the elimination of the leaving group are not same as the reactant. Product formed in the nucleophilic elimination step with appropriate curved arrow is drawn as:

Explanation of Solution

Product for the given nucleophilic addition step is:

In the given product, there are two possible groups that can leave to form two different products.

In the first nucleophilic elimination step, the oxygen atom with negative charge is electron rich site and the carbon bonded to it is electron poor site. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curved arrow is drawn to show the breaking of bond between the carbon and the leaving group ${\text{R}}^{\text{-}}$ on the left side.

The respective product formed is different from the reactant in the given nucleophilic addition step. The X sign on the arrow represents this nucleophilic elimination is unfeasible as ${\text{R}}^{\text{-}}$ group eliminates from the reactant.

The second nucleophilic elimination step, the oxygen atom with negative charge is electron rich site and the carbon bonded to it is electron poor site. The curved arrow mechanism for this given nucleophilic elimination step forming the respective product is:

The first curved arrow is drawn from the lone pair of negatively charged oxygen to the mid of $\text{C-O}$ bond, and the second curved arrow is drawn to show the breaking of bond between the carbon and the leaving group ${\text{R}}^{\text{-}}$ on the right side.

The respective product formed is different from the reactant in the given nucleophilic addition step. The X sign on the arrow represents this nucleophilic elimination is unfeasible as ${\text{R}}^{\text{-}}$ group eliminates from the reactant.

Conclusion

Products formed in the elimination steps are different from the reactant in the given nucleophilic addition step.