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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19.78 Write the products of the following sequences of reactions. Refer to your reaction road- maps to see how the combined reactions allow you to "navigate" between the different functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18 roadmaps along with your new Chapter 19 roadmap for these. (a) 1. BHS 2. H₂O₂ 3. H₂CrO4 4. SOCI₂ (b) 1. Cl₂/hv 2. KOLBU 3. H₂O, catalytic H₂SO4 4. H₂CrO4 Reaction Roadmap An alkene 5. EtOH 6.0.5 Equiv. NaOEt/EtOH 7. Mild H₂O An alkane 1.0 2. (CH3)₂S 3. H₂CrO (d) (c) 4. Excess EtOH, catalytic H₂SO OH 4. Mild H₂O* 5.0.5 Equiv. NaOEt/EtOH An alkene 6. Mild H₂O* A carboxylic acid 7. Mild H₂O* 1. SOC₁₂ 2. EtOH 3.0.5 Equiv. NaOEt/E:OH 5.1.0 Equiv. NaOEt 6. NH₂ (e) 1. 0.5 Equiv. NaOEt/EtOH 2. Mild H₂O* Br (f) i H An aldehyde 1. Catalytic NaOE/EtOH 2. H₂O*, heat 3. (CH,CH₂)₂Culi 4. Mild H₂O* 5.1.0 Equiv. LDA Br An ester 4. NaOH, H₂O 5. Mild H₂O* 6. Heat 7. MgBr 8. Mild H₂O* 7. Mild H₂O+

Li+ is a hard acid. With this in mind, which if the following compounds should be most soluble in water? Group of answer choices LiBr LiI LiF LiCl

Q4: Write organic product(s) of the following reactions and show the curved-arrow mechanism of the reactions. Br MeOH OSO2CH3 MeOH

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 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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