a) Interpretation: The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given. Concept introduction: Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms. Both reactions involve the same steps such as i) enolate ion formation ii) SN 2 attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Question

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Answer 1

Question

Definition Definition Class of organic compounds that contain a carboxyl group ( - COOH ) and has a general formula R - COOH or R - CO 2 H , where R refers to the alkyl, alkenyl, aryl, or other groups. They can undergo different chemical reactions, such as acid-base reactions, esterification, and oxidation. These are essential components of living organisms, playing important roles in metabolic processes, signaling, and as pharmaceuticals.

Chapter 22.SE, Problem 17VC

Interpretation Introduction

a)

Interpretation:

The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 2

Explanation of Solution

The compound represented by the model is 6-methylhept-5-ene-2-one. It is a methyl ketone and hence it can be prepared using acetoacetic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group of the ester to form the enolate ion. The enolate ion then attacks 1-bromo-3-methy-2-butene and displaces the bromine as bromide ion. The product obtained upon hydrolysis with dilute acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

The steps involved in preparing the compound represented by the model using either malonic ester synthesis or an acetoacetic ester synthesis are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 5

Explanation of Solution

The compound represented by the model is 2-methyl-3-phenylpropanoic acid and hence it can be prepared using malonic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group to form the enolate ion. The enolate ion then attacks benzyl bromide and displaces the bromine as bromide ion. The abstraction of another acidic hydrogen in the product by the base and the nucleophilic displacement of bromine from methyl bromide by enolate ion introduces a methyl group at α- position of the diester. The alkylated diester obtained upon hydrolysis with aqueous acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 6

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SYNTHESIS REACTIONS. For the following reactions, synthesize the given products from the given reactants. Multiple reactions/steps will be needed. For the one of the steps (ie reactions) in each synthesis, write out the mechanism for that reaction and draw an energy diagram showing the correct number of hills and valleys for that step's mechanism. CI b. a. Use acetylene (ethyne) and any alkyl halide as your starting materials Br C. d. "OH OH III. OH

Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely: (a) 0.200 M HCl

Answer 2

Question

Definition Definition Class of organic compounds that contain a carboxyl group ( - COOH ) and has a general formula R - COOH or R - CO 2 H , where R refers to the alkyl, alkenyl, aryl, or other groups. They can undergo different chemical reactions, such as acid-base reactions, esterification, and oxidation. These are essential components of living organisms, playing important roles in metabolic processes, signaling, and as pharmaceuticals.

Chapter 22.SE, Problem 17VC

Interpretation Introduction

a)

Interpretation:

The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 2

Explanation of Solution

The compound represented by the model is 6-methylhept-5-ene-2-one. It is a methyl ketone and hence it can be prepared using acetoacetic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group of the ester to form the enolate ion. The enolate ion then attacks 1-bromo-3-methy-2-butene and displaces the bromine as bromide ion. The product obtained upon hydrolysis with dilute acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

The steps involved in preparing the compound represented by the model using either malonic ester synthesis or an acetoacetic ester synthesis are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 5

Explanation of Solution

The compound represented by the model is 2-methyl-3-phenylpropanoic acid and hence it can be prepared using malonic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group to form the enolate ion. The enolate ion then attacks benzyl bromide and displaces the bromine as bromide ion. The abstraction of another acidic hydrogen in the product by the base and the nucleophilic displacement of bromine from methyl bromide by enolate ion introduces a methyl group at α- position of the diester. The alkylated diester obtained upon hydrolysis with aqueous acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 6

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Answer 3

Question

Definition Definition Class of organic compounds that contain a carboxyl group ( - COOH ) and has a general formula R - COOH or R - CO 2 H , where R refers to the alkyl, alkenyl, aryl, or other groups. They can undergo different chemical reactions, such as acid-base reactions, esterification, and oxidation. These are essential components of living organisms, playing important roles in metabolic processes, signaling, and as pharmaceuticals.

Chapter 22.SE, Problem 17VC

Interpretation Introduction

a)

Interpretation:

The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 2

Explanation of Solution

The compound represented by the model is 6-methylhept-5-ene-2-one. It is a methyl ketone and hence it can be prepared using acetoacetic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group of the ester to form the enolate ion. The enolate ion then attacks 1-bromo-3-methy-2-butene and displaces the bromine as bromide ion. The product obtained upon hydrolysis with dilute acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

The steps involved in preparing the compound represented by the model using either malonic ester synthesis or an acetoacetic ester synthesis are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 5

Explanation of Solution

The compound represented by the model is 2-methyl-3-phenylpropanoic acid and hence it can be prepared using malonic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group to form the enolate ion. The enolate ion then attacks benzyl bromide and displaces the bromine as bromide ion. The abstraction of another acidic hydrogen in the product by the base and the nucleophilic displacement of bromine from methyl bromide by enolate ion introduces a methyl group at α- position of the diester. The alkylated diester obtained upon hydrolysis with aqueous acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 6

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Answer 4

Question

Definition Definition Class of organic compounds that contain a carboxyl group ( - COOH ) and has a general formula R - COOH or R - CO 2 H , where R refers to the alkyl, alkenyl, aryl, or other groups. They can undergo different chemical reactions, such as acid-base reactions, esterification, and oxidation. These are essential components of living organisms, playing important roles in metabolic processes, signaling, and as pharmaceuticals.

Chapter 22.SE, Problem 17VC

Interpretation Introduction

a)

Interpretation:

The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 2

Explanation of Solution

The compound represented by the model is 6-methylhept-5-ene-2-one. It is a methyl ketone and hence it can be prepared using acetoacetic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group of the ester to form the enolate ion. The enolate ion then attacks 1-bromo-3-methy-2-butene and displaces the bromine as bromide ion. The product obtained upon hydrolysis with dilute acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

The steps involved in preparing the compound represented by the model using either malonic ester synthesis or an acetoacetic ester synthesis are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 5

Explanation of Solution

The compound represented by the model is 2-methyl-3-phenylpropanoic acid and hence it can be prepared using malonic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group to form the enolate ion. The enolate ion then attacks benzyl bromide and displaces the bromine as bromide ion. The abstraction of another acidic hydrogen in the product by the base and the nucleophilic displacement of bromine from methyl bromide by enolate ion introduces a methyl group at α- position of the diester. The alkylated diester obtained upon hydrolysis with aqueous acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 6

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Answer 5

Chapter 22.SE, Problem 17VC

Interpretation Introduction

a)

Interpretation:

The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 2

Explanation of Solution

The compound represented by the model is 6-methylhept-5-ene-2-one. It is a methyl ketone and hence it can be prepared using acetoacetic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group of the ester to form the enolate ion. The enolate ion then attacks 1-bromo-3-methy-2-butene and displaces the bromine as bromide ion. The product obtained upon hydrolysis with dilute acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

The steps involved in preparing the compound represented by the model using either malonic ester synthesis or an acetoacetic ester synthesis are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 5

Explanation of Solution

The compound represented by the model is 2-methyl-3-phenylpropanoic acid and hence it can be prepared using malonic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group to form the enolate ion. The enolate ion then attacks benzyl bromide and displaces the bromine as bromide ion. The abstraction of another acidic hydrogen in the product by the base and the nucleophilic displacement of bromine from methyl bromide by enolate ion introduces a methyl group at α- position of the diester. The alkylated diester obtained upon hydrolysis with aqueous acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 6

Answer 6

Chapter 22.SE, Problem 17VC

Interpretation Introduction

a)

Interpretation:

The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 2

Explanation of Solution

The compound represented by the model is 6-methylhept-5-ene-2-one. It is a methyl ketone and hence it can be prepared using acetoacetic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group of the ester to form the enolate ion. The enolate ion then attacks 1-bromo-3-methy-2-butene and displaces the bromine as bromide ion. The product obtained upon hydrolysis with dilute acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 3

Answer 7

Chapter 22.SE, Problem 17VC

Interpretation Introduction

a)

Interpretation:

The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Answer 8

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 2

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The compound represented by the model is 6-methylhept-5-ene-2-one. It is a methyl ketone and hence it can be prepared using acetoacetic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group of the ester to form the enolate ion. The enolate ion then attacks 1-bromo-3-methy-2-butene and displaces the bromine as bromide ion. The product obtained upon hydrolysis with dilute acids and decarboxylation by heating yields the product.

Answer 13

Conclusion

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 3

Answer 14

Interpretation Introduction

b)

Interpretation:

The steps involved in preparing the compound represented by the model using either malonic ester synthesis or an acetoacetic ester synthesis are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 5

Explanation of Solution

The compound represented by the model is 2-methyl-3-phenylpropanoic acid and hence it can be prepared using malonic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group to form the enolate ion. The enolate ion then attacks benzyl bromide and displaces the bromine as bromide ion. The abstraction of another acidic hydrogen in the product by the base and the nucleophilic displacement of bromine from methyl bromide by enolate ion introduces a methyl group at α- position of the diester. The alkylated diester obtained upon hydrolysis with aqueous acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 6

Answer 15

Interpretation Introduction

b)

Interpretation:

The steps involved in preparing the compound represented by the model using either malonic ester synthesis or an acetoacetic ester synthesis are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Answer 16

Expert Solution

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 5

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The compound represented by the model is 2-methyl-3-phenylpropanoic acid and hence it can be prepared using malonic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group to form the enolate ion. The enolate ion then attacks benzyl bromide and displaces the bromine as bromide ion. The abstraction of another acidic hydrogen in the product by the base and the nucleophilic displacement of bromine from methyl bromide by enolate ion introduces a methyl group at α- position of the diester. The alkylated diester obtained upon hydrolysis with aqueous acids and decarboxylation by heating yields the product.

Answer 21

Conclusion

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 17VC , additional homework tip 6

Answer 22

Ch. 22.1 - Prob. 1P Ch. 22.1 - How many acidic hydrogens does each of the...Ch. 22.1 - Prob. 3P Ch. 22.3 - Write the complete mechanism for the deuteration...Ch. 22.3 - Prob. 5P Ch. 22.4 - If methanol rather than water is added at the end...Ch. 22.5 - Prob. 7P Ch. 22.5 - Draw a resonance structure of the acetonitrile...Ch. 22.6 - If methanol rather than water is added at the end...Ch. 22.7 - Prob. 10P

Answer to Problem 17VC

Explanation of Solution

Answer to Problem 17VC

Explanation of Solution

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Chapter 22 Solutions