(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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Part V. Label ad match the carbons in compounds Jane and Diane w/ the corresponding peak no. in the Spectra (Note: use the given peak no. To label the carbons, other peak no are intentionally omitted) 7 4 2 -0.13 -0.12 -0.11 -0.10 -0.08 8 CI Jane 1 -0.09 5 210 200 190 180 170 160 150 140 130 120 110 100 -8 90 f1 (ppm) 11 8 172.4 172.0 f1 (ppr HO CI NH Diane 7 3 11 80 80 -80 -R 70 60 60 2 5 -8 50 40 8. 170 160 150 140 130 120 110 100 90 -0 80 70 20 f1 (ppm) 15 30 -20 20 -60 60 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 -0.01 10 -0.17 16 15 56 16 -0.16 -0.15 -0.14 -0.13 -0.12 -0.11 -0.10 -0.09 -0.08 -0.07 -0.06 -0.05 -0.04 17.8 17.6 17.4 17.2 17.0 f1 (ppm) -0.03 -0.02 550 106 40 30 20 20 -0.01 -0.00 F-0.01 10 0

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 11

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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