(a) Interpretation: The complete mechanism and the major organic product of the given reaction are to be written. Concept introduction: In the reaction between a carboxylic acid and PBr 3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr 2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

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Answer 1

Question

Chapter 21, Problem 21.47P

Interpretation Introduction

(a)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Interpretation Introduction

(b)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

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Answer 2

Question

Chapter 21, Problem 21.47P

Interpretation Introduction

(a)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Interpretation Introduction

(b)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

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Answer 3

Question

Chapter 21, Problem 21.47P

Interpretation Introduction

(a)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Interpretation Introduction

(b)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

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Answer 4

Question

Chapter 21, Problem 21.47P

Interpretation Introduction

(a)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Interpretation Introduction

(b)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

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Answer 5

Chapter 21, Problem 21.47P

Interpretation Introduction

(a)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Interpretation Introduction

(b)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Answer 6

Chapter 21, Problem 21.47P

Interpretation Introduction

(a)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Answer 7

Chapter 21, Problem 21.47P

Interpretation Introduction

(a)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Answer 8

Interpretation Introduction

(b)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Answer 9

Interpretation Introduction

(b)

Interpretation:

The complete mechanism and the major organic product of the given reaction are to be written.

Concept introduction:

In the reaction between a carboxylic acid and PBr3 gives an acid bromide. In this reaction the bonds around the C=O and C-OH are broken and C=O and C-Br are formed. Therefore, the original carbon skeleton remains unchanged. Further HOPBr2 takes H from the alpha carbon atom to form a resonance stabilized carbonyl group. Enol is formed at the alpha carbon takes place after bromination. Finally, the hydrolysis of acyl bromide forms a tetrahedral intermediate which on the reformation of carbonyl forms corresponding carboxylic acid.

Answer 10

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Answer 12

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Answer 13

Ch. 21 - Prob. 21.1P Ch. 21 - Prob. 21.2P Ch. 21 - Prob. 21.3P Ch. 21 - Prob. 21.4P Ch. 21 - Prob. 21.5P Ch. 21 - Prob. 21.6P Ch. 21 - Prob. 21.7P Ch. 21 - Prob. 21.8P Ch. 21 - Prob. 21.9P Ch. 21 - Prob. 21.10P

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Chapter 21 Solutions