(a) Interpretation: Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. The resonance structure for methyl acetate (CH 3 CO 2 CH 3 ) is to be drawn. Concept introduction: An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with C=O .

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Answer 1

Question

Chapter 15, Problem 15.44P

Interpretation Introduction

(a)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. The resonance structure for methyl acetate (CH3CO2CH3) is to be drawn.

Concept introduction:

An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with C=O.

Interpretation Introduction

(b)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. An analogous resonance structure can be drawn for an ester. Based on the fact that an ester’s C=O stretch frequency is higher than an amide’s, the resonance structure that has a larger contribution toward the resonance hybrid of an ester or an amide is to be determined.

Concept introduction:

The C-O single bond is weaker than C=O bond. The C-O single bond character decreases the stiffness of C=O bond. The stretching frequency of C=O band lowers by the Weakening of C=O bond. Oxygen is higher electronegative than nitrogen. The positive charge is less stable on O atom than N atom. In the resonance of amide, the C-O single bond character is more significant than that in ester.

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Answer 2

Question

Chapter 15, Problem 15.44P

Interpretation Introduction

(a)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. The resonance structure for methyl acetate (CH3CO2CH3) is to be drawn.

Concept introduction:

An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with C=O.

Interpretation Introduction

(b)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. An analogous resonance structure can be drawn for an ester. Based on the fact that an ester’s C=O stretch frequency is higher than an amide’s, the resonance structure that has a larger contribution toward the resonance hybrid of an ester or an amide is to be determined.

Concept introduction:

The C-O single bond is weaker than C=O bond. The C-O single bond character decreases the stiffness of C=O bond. The stretching frequency of C=O band lowers by the Weakening of C=O bond. Oxygen is higher electronegative than nitrogen. The positive charge is less stable on O atom than N atom. In the resonance of amide, the C-O single bond character is more significant than that in ester.

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Answer 3

Question

Chapter 15, Problem 15.44P

Interpretation Introduction

(a)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. The resonance structure for methyl acetate (CH3CO2CH3) is to be drawn.

Concept introduction:

An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with C=O.

Interpretation Introduction

(b)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. An analogous resonance structure can be drawn for an ester. Based on the fact that an ester’s C=O stretch frequency is higher than an amide’s, the resonance structure that has a larger contribution toward the resonance hybrid of an ester or an amide is to be determined.

Concept introduction:

The C-O single bond is weaker than C=O bond. The C-O single bond character decreases the stiffness of C=O bond. The stretching frequency of C=O band lowers by the Weakening of C=O bond. Oxygen is higher electronegative than nitrogen. The positive charge is less stable on O atom than N atom. In the resonance of amide, the C-O single bond character is more significant than that in ester.

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Answer 4

Question

Chapter 15, Problem 15.44P

Interpretation Introduction

(a)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. The resonance structure for methyl acetate (CH3CO2CH3) is to be drawn.

Concept introduction:

An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with C=O.

Interpretation Introduction

(b)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. An analogous resonance structure can be drawn for an ester. Based on the fact that an ester’s C=O stretch frequency is higher than an amide’s, the resonance structure that has a larger contribution toward the resonance hybrid of an ester or an amide is to be determined.

Concept introduction:

The C-O single bond is weaker than C=O bond. The C-O single bond character decreases the stiffness of C=O bond. The stretching frequency of C=O band lowers by the Weakening of C=O bond. Oxygen is higher electronegative than nitrogen. The positive charge is less stable on O atom than N atom. In the resonance of amide, the C-O single bond character is more significant than that in ester.

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Answer 5

Chapter 15, Problem 15.44P

Interpretation Introduction

(a)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. The resonance structure for methyl acetate (CH3CO2CH3) is to be drawn.

Concept introduction:

An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with C=O.

Interpretation Introduction

(b)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. An analogous resonance structure can be drawn for an ester. Based on the fact that an ester’s C=O stretch frequency is higher than an amide’s, the resonance structure that has a larger contribution toward the resonance hybrid of an ester or an amide is to be determined.

Concept introduction:

The C-O single bond is weaker than C=O bond. The C-O single bond character decreases the stiffness of C=O bond. The stretching frequency of C=O band lowers by the Weakening of C=O bond. Oxygen is higher electronegative than nitrogen. The positive charge is less stable on O atom than N atom. In the resonance of amide, the C-O single bond character is more significant than that in ester.

Answer 6

Chapter 15, Problem 15.44P

Interpretation Introduction

(a)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. The resonance structure for methyl acetate (CH3CO2CH3) is to be drawn.

Concept introduction:

An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with C=O.

Answer 7

Chapter 15, Problem 15.44P

Interpretation Introduction

(a)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. The resonance structure for methyl acetate (CH3CO2CH3) is to be drawn.

Concept introduction:

An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with C=O.

Answer 8

Interpretation Introduction

(b)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. An analogous resonance structure can be drawn for an ester. Based on the fact that an ester’s C=O stretch frequency is higher than an amide’s, the resonance structure that has a larger contribution toward the resonance hybrid of an ester or an amide is to be determined.

Concept introduction:

The C-O single bond is weaker than C=O bond. The C-O single bond character decreases the stiffness of C=O bond. The stretching frequency of C=O band lowers by the Weakening of C=O bond. Oxygen is higher electronegative than nitrogen. The positive charge is less stable on O atom than N atom. In the resonance of amide, the C-O single bond character is more significant than that in ester.

Answer 9

Interpretation Introduction

(b)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the C=O stretch frequency. An analogous resonance structure can be drawn for an ester. Based on the fact that an ester’s C=O stretch frequency is higher than an amide’s, the resonance structure that has a larger contribution toward the resonance hybrid of an ester or an amide is to be determined.

Concept introduction:

The C-O single bond is weaker than C=O bond. The C-O single bond character decreases the stiffness of C=O bond. The stretching frequency of C=O band lowers by the Weakening of C=O bond. Oxygen is higher electronegative than nitrogen. The positive charge is less stable on O atom than N atom. In the resonance of amide, the C-O single bond character is more significant than that in ester.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 15 - Prob. 15.1P Ch. 15 - Prob. 15.2P Ch. 15 - Prob. 15.3P Ch. 15 - Prob. 15.4P Ch. 15 - Prob. 15.5P Ch. 15 - Prob. 15.6P Ch. 15 - Prob. 15.7P Ch. 15 - Prob. 15.8P Ch. 15 - Prob. 15.9P Ch. 15 - Prob. 15.10P

Solution Summary: The author illustrates the resonance structure of an amide that accounts for the lowering of the C=O stretch frequency.

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Chapter 15 Solutions