The curved arrow notation and products for each elementary step described by the given sequence are to be drawn. Concept introduction: The 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 H − C − C − L or H − C = C − 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 C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene ( C=C ) or alkyne ( C ≡ C ). An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

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Answer 1

Question

Chapter 7, Problem 7.60P

Interpretation Introduction

Interpretation:

The curved arrow notation and products for each elementary step described by the given sequence are to be drawn.

Concept introduction:

The 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 H−C−C−L or H−C=C−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 C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene (C=C) or alkyne (C≡C).

An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

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Answer 2

Question

Chapter 7, Problem 7.60P

Interpretation Introduction

Interpretation:

The curved arrow notation and products for each elementary step described by the given sequence are to be drawn.

Concept introduction:

The 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 H−C−C−L or H−C=C−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 C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene (C=C) or alkyne (C≡C).

An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

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Answer 3

Question

Chapter 7, Problem 7.60P

Interpretation Introduction

Interpretation:

The curved arrow notation and products for each elementary step described by the given sequence are to be drawn.

Concept introduction:

The 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 H−C−C−L or H−C=C−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 C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene (C=C) or alkyne (C≡C).

An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

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Answer 4

Question

Chapter 7, Problem 7.60P

Interpretation Introduction

Interpretation:

The curved arrow notation and products for each elementary step described by the given sequence are to be drawn.

Concept introduction:

The 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 H−C−C−L or H−C=C−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 C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene (C=C) or alkyne (C≡C).

An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

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Answer 5

Chapter 7, Problem 7.60P

Interpretation Introduction

Interpretation:

The curved arrow notation and products for each elementary step described by the given sequence are to be drawn.

Concept introduction:

The 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 H−C−C−L or H−C=C−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 C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene (C=C) or alkyne (C≡C).

An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

Answer 6

Chapter 7, Problem 7.60P

Interpretation Introduction

Interpretation:

The curved arrow notation and products for each elementary step described by the given sequence are to be drawn.

Concept introduction:

The 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 H−C−C−L or H−C=C−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 C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene (C=C) or alkyne (C≡C).

An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

Answer 7

Chapter 7, Problem 7.60P

Interpretation Introduction

Interpretation:

The curved arrow notation and products for each elementary step described by the given sequence are to be drawn.

Concept introduction:

The 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 H−C−C−L or H−C=C−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 C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene (C=C) or alkyne (C≡C).

An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

Answer 8

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Answer 9

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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 the curved arrow notation and products for each elementary step described by the given sequence.

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