It is to be explained why the first given heterolysis reaction takes place readily but the second does not. Concept introduction: Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species. An electronegative atom destabilizes the carbocation via electron-withdrawing inductive effect; however, the lone pairs on the electronegative atom can participate via resonance, and this stabilizes the positive charge. Moreover, in the resonance structures, the structure having complete octets for maximum number of atoms is more stable and thus, contributes more to the resonance hybrid. Resonance and inductive effects work in opposite directions, and out of the two, the resonance effect is dominant.

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Answer 1

Question

Chapter 7, Problem 7.39P

Interpretation Introduction

Interpretation:

It is to be explained why the first given heterolysis reaction takes place readily but the second does not.

Concept introduction:

Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.

An electronegative atom destabilizes the carbocation via electron-withdrawing inductive effect; however, the lone pairs on the electronegative atom can participate via resonance, and this stabilizes the positive charge. Moreover, in the resonance structures, the structure having complete octets for maximum number of atoms is more stable and thus, contributes more to the resonance hybrid. Resonance and inductive effects work in opposite directions, and out of the two, the resonance effect is dominant.

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Answer 2

Question

Chapter 7, Problem 7.39P

Interpretation Introduction

Interpretation:

It is to be explained why the first given heterolysis reaction takes place readily but the second does not.

Concept introduction:

Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.

An electronegative atom destabilizes the carbocation via electron-withdrawing inductive effect; however, the lone pairs on the electronegative atom can participate via resonance, and this stabilizes the positive charge. Moreover, in the resonance structures, the structure having complete octets for maximum number of atoms is more stable and thus, contributes more to the resonance hybrid. Resonance and inductive effects work in opposite directions, and out of the two, the resonance effect is dominant.

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Answer 3

Question

Chapter 7, Problem 7.39P

Interpretation Introduction

Interpretation:

It is to be explained why the first given heterolysis reaction takes place readily but the second does not.

Concept introduction:

Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.

An electronegative atom destabilizes the carbocation via electron-withdrawing inductive effect; however, the lone pairs on the electronegative atom can participate via resonance, and this stabilizes the positive charge. Moreover, in the resonance structures, the structure having complete octets for maximum number of atoms is more stable and thus, contributes more to the resonance hybrid. Resonance and inductive effects work in opposite directions, and out of the two, the resonance effect is dominant.

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Answer 4

Question

Chapter 7, Problem 7.39P

Interpretation Introduction

Interpretation:

It is to be explained why the first given heterolysis reaction takes place readily but the second does not.

Concept introduction:

Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.

An electronegative atom destabilizes the carbocation via electron-withdrawing inductive effect; however, the lone pairs on the electronegative atom can participate via resonance, and this stabilizes the positive charge. Moreover, in the resonance structures, the structure having complete octets for maximum number of atoms is more stable and thus, contributes more to the resonance hybrid. Resonance and inductive effects work in opposite directions, and out of the two, the resonance effect is dominant.

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Answer 5

Chapter 7, Problem 7.39P

Interpretation Introduction

Interpretation:

It is to be explained why the first given heterolysis reaction takes place readily but the second does not.

Concept introduction:

Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.

An electronegative atom destabilizes the carbocation via electron-withdrawing inductive effect; however, the lone pairs on the electronegative atom can participate via resonance, and this stabilizes the positive charge. Moreover, in the resonance structures, the structure having complete octets for maximum number of atoms is more stable and thus, contributes more to the resonance hybrid. Resonance and inductive effects work in opposite directions, and out of the two, the resonance effect is dominant.

Answer 6

Chapter 7, Problem 7.39P

Interpretation Introduction

Interpretation:

It is to be explained why the first given heterolysis reaction takes place readily but the second does not.

Concept introduction:

Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.

An electronegative atom destabilizes the carbocation via electron-withdrawing inductive effect; however, the lone pairs on the electronegative atom can participate via resonance, and this stabilizes the positive charge. Moreover, in the resonance structures, the structure having complete octets for maximum number of atoms is more stable and thus, contributes more to the resonance hybrid. Resonance and inductive effects work in opposite directions, and out of the two, the resonance effect is dominant.

Answer 7

Chapter 7, Problem 7.39P

Interpretation Introduction

Interpretation:

It is to be explained why the first given heterolysis reaction takes place readily but the second does not.

Concept introduction:

Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.

An electronegative atom destabilizes the carbocation via electron-withdrawing inductive effect; however, the lone pairs on the electronegative atom can participate via resonance, and this stabilizes the positive charge. Moreover, in the resonance structures, the structure having complete octets for maximum number of atoms is more stable and thus, contributes more to the resonance hybrid. Resonance and inductive effects work in opposite directions, and out of the two, the resonance effect is dominant.

Answer 8

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Answer 10

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Answer 11

Ch. 7 - Prob. 7.1P Ch. 7 - Prob. 7.2P Ch. 7 - Prob. 7.3P Ch. 7 - Prob. 7.4P Ch. 7 - Prob. 7.5P Ch. 7 - Prob. 7.6P Ch. 7 - Prob. 7.7P Ch. 7 - Prob. 7.8P Ch. 7 - Prob. 7.9P Ch. 7 - Prob. 7.10P

Solution Summary: The author explains why the first heterolysis reaction takes place readily but the second doesn't because the carbon atom in reaction 1 is more favorable and stable than reaction 2.

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