a) Interpretation: How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown. 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) S N 2 attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation. To show How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Question

Want to see more full solutions like this?

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 46AP

Interpretation Introduction

a)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 2

Explanation of Solution

The compound shown is a derivative of carboxylic acid. Hence it can be prepared using malonic ester synthesis. The acid has two methyl groups attached to the carbon adjacent to ester groups. It can be prepared by replacing the two hydrogens on the active methylene group of malonic ester by two methyl groups. This is achieved by treating the ester with two equivalents of sodium ethoxide and two equivalents of methyl bromide.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 5

Explanation of Solution

: The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,6- dibromohexane displaces the bromide ion to produce a Î±- substituted acetoacetic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield methyl cycloheptyl ketone.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 8

Explanation of Solution

The compound shown is a carboxylic acid. Hence it can be prepared using malonic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of malonic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,3- dibromopropane displaces the bromide ion to produce a Î±- substituted malonic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic diester. Upon treating with aqueous acids the ester groups get hydrolyzed to give a dicarboxylic acid. The dicarboxylic acid eliminates a CO₂ molecule on heating to yield cyclobutylcarboxylic acid.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 9

Interpretation Introduction

d)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 11

Explanation of Solution

The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on allyl bromide displaces the bromide ion to produce Î±- allylsubstituted acetoacetic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield hex-5-ene-2-one.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 12

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

1) Suppose 0.1 kg ice at 0°C (273K) is in 0.5kg water at 20°C (293K). What is the change in entropy of the ice as it melts at 0°? To produce the original "water gas" mixture, carbon (in a combustible form known as coke) is reacted with steam: 131.4 kJ + H20(g) + C(s) → CO(g) + H2(g) From this information and the equations in the previous problem, calculate the enthalpy for the combustion or carbon to form carbon dioxide. kindly show me how to solve both parts of the same long problem. Thanks

we were assigned to dilute 900ppm in to 18ppm by using only 250ml vol flask. firstly we did calc and convert 900ppm to 0.9 ppm to dilute in 1 liter. to begin the experiment we took 0,225g of kmno4 and dissolved in to 250 vol flask. then further we took 10 ml sample sol and dissolved in to 100 ml vol flask and put it in to a spectrometer and got value of 0.145A . upon further calc we got v2 as 50ml . need to find DF, % error (expval and accptVal), molarity, molality. please write the whole report. thank you The format, tables, introduction, procedure and observation, result, calculations, discussion and conclusion

Q5. Predict the organic product(s) for the following transformations. If no reaction will take place (or the reaction is not synthetically useful), write "N.R.". Determine what type of transition state is present for each reaction (think Hammond Postulate). I Br₂ CH3 F2, light CH3 Heat CH3 F₂ Heat Br2, light 12, light CH3 Cl2, light No

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 46AP

Interpretation Introduction

a)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 2

Explanation of Solution

The compound shown is a derivative of carboxylic acid. Hence it can be prepared using malonic ester synthesis. The acid has two methyl groups attached to the carbon adjacent to ester groups. It can be prepared by replacing the two hydrogens on the active methylene group of malonic ester by two methyl groups. This is achieved by treating the ester with two equivalents of sodium ethoxide and two equivalents of methyl bromide.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 5

Explanation of Solution

: The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,6- dibromohexane displaces the bromide ion to produce a Î±- substituted acetoacetic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield methyl cycloheptyl ketone.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 8

Explanation of Solution

The compound shown is a carboxylic acid. Hence it can be prepared using malonic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of malonic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,3- dibromopropane displaces the bromide ion to produce a Î±- substituted malonic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic diester. Upon treating with aqueous acids the ester groups get hydrolyzed to give a dicarboxylic acid. The dicarboxylic acid eliminates a CO₂ molecule on heating to yield cyclobutylcarboxylic acid.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 9

Interpretation Introduction

d)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 11

Explanation of Solution

The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on allyl bromide displaces the bromide ion to produce Î±- allylsubstituted acetoacetic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield hex-5-ene-2-one.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 12

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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 46AP

Interpretation Introduction

a)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 2

Explanation of Solution

The compound shown is a derivative of carboxylic acid. Hence it can be prepared using malonic ester synthesis. The acid has two methyl groups attached to the carbon adjacent to ester groups. It can be prepared by replacing the two hydrogens on the active methylene group of malonic ester by two methyl groups. This is achieved by treating the ester with two equivalents of sodium ethoxide and two equivalents of methyl bromide.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 5

Explanation of Solution

: The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,6- dibromohexane displaces the bromide ion to produce a Î±- substituted acetoacetic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield methyl cycloheptyl ketone.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 8

Explanation of Solution

The compound shown is a carboxylic acid. Hence it can be prepared using malonic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of malonic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,3- dibromopropane displaces the bromide ion to produce a Î±- substituted malonic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic diester. Upon treating with aqueous acids the ester groups get hydrolyzed to give a dicarboxylic acid. The dicarboxylic acid eliminates a CO₂ molecule on heating to yield cyclobutylcarboxylic acid.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 9

Interpretation Introduction

d)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 11

Explanation of Solution

The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on allyl bromide displaces the bromide ion to produce Î±- allylsubstituted acetoacetic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield hex-5-ene-2-one.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 12

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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 46AP

Interpretation Introduction

a)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 2

Explanation of Solution

The compound shown is a derivative of carboxylic acid. Hence it can be prepared using malonic ester synthesis. The acid has two methyl groups attached to the carbon adjacent to ester groups. It can be prepared by replacing the two hydrogens on the active methylene group of malonic ester by two methyl groups. This is achieved by treating the ester with two equivalents of sodium ethoxide and two equivalents of methyl bromide.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 5

Explanation of Solution

: The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,6- dibromohexane displaces the bromide ion to produce a Î±- substituted acetoacetic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield methyl cycloheptyl ketone.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 8

Explanation of Solution

The compound shown is a carboxylic acid. Hence it can be prepared using malonic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of malonic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,3- dibromopropane displaces the bromide ion to produce a Î±- substituted malonic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic diester. Upon treating with aqueous acids the ester groups get hydrolyzed to give a dicarboxylic acid. The dicarboxylic acid eliminates a CO₂ molecule on heating to yield cyclobutylcarboxylic acid.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 9

Interpretation Introduction

d)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 11

Explanation of Solution

The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on allyl bromide displaces the bromide ion to produce Î±- allylsubstituted acetoacetic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield hex-5-ene-2-one.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 12

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 22.SE, Problem 46AP

Interpretation Introduction

a)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 2

Explanation of Solution

The compound shown is a derivative of carboxylic acid. Hence it can be prepared using malonic ester synthesis. The acid has two methyl groups attached to the carbon adjacent to ester groups. It can be prepared by replacing the two hydrogens on the active methylene group of malonic ester by two methyl groups. This is achieved by treating the ester with two equivalents of sodium ethoxide and two equivalents of methyl bromide.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 5

Explanation of Solution

: The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,6- dibromohexane displaces the bromide ion to produce a Î±- substituted acetoacetic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield methyl cycloheptyl ketone.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 8

Explanation of Solution

The compound shown is a carboxylic acid. Hence it can be prepared using malonic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of malonic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,3- dibromopropane displaces the bromide ion to produce a Î±- substituted malonic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic diester. Upon treating with aqueous acids the ester groups get hydrolyzed to give a dicarboxylic acid. The dicarboxylic acid eliminates a CO₂ molecule on heating to yield cyclobutylcarboxylic acid.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 9

Interpretation Introduction

d)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 11

Explanation of Solution

The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on allyl bromide displaces the bromide ion to produce Î±- allylsubstituted acetoacetic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield hex-5-ene-2-one.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 12

Answer 6

Chapter 22.SE, Problem 46AP

Interpretation Introduction

a)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 2

Explanation of Solution

The compound shown is a derivative of carboxylic acid. Hence it can be prepared using malonic ester synthesis. The acid has two methyl groups attached to the carbon adjacent to ester groups. It can be prepared by replacing the two hydrogens on the active methylene group of malonic ester by two methyl groups. This is achieved by treating the ester with two equivalents of sodium ethoxide and two equivalents of methyl bromide.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 3

Answer 7

Chapter 22.SE, Problem 46AP

Interpretation Introduction

a)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Answer 8

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 2

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The compound shown is a derivative of carboxylic acid. Hence it can be prepared using malonic ester synthesis. The acid has two methyl groups attached to the carbon adjacent to ester groups. It can be prepared by replacing the two hydrogens on the active methylene group of malonic ester by two methyl groups. This is achieved by treating the ester with two equivalents of sodium ethoxide and two equivalents of methyl bromide.

Answer 13

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 3

Answer 14

Interpretation Introduction

b)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 5

Explanation of Solution

: The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,6- dibromohexane displaces the bromide ion to produce a Î±- substituted acetoacetic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield methyl cycloheptyl ketone.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 6

Answer 15

Interpretation Introduction

b)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Answer 16

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 5

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

: The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,6- dibromohexane displaces the bromide ion to produce a Î±- substituted acetoacetic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield methyl cycloheptyl ketone.

Answer 21

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 6

Answer 22

Interpretation Introduction

c)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 8

Explanation of Solution

The compound shown is a carboxylic acid. Hence it can be prepared using malonic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of malonic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,3- dibromopropane displaces the bromide ion to produce a Î±- substituted malonic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic diester. Upon treating with aqueous acids the ester groups get hydrolyzed to give a dicarboxylic acid. The dicarboxylic acid eliminates a CO₂ molecule on heating to yield cyclobutylcarboxylic acid.

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 9

Answer 23

Interpretation Introduction

c)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Answer 24

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 8

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The compound shown is a carboxylic acid. Hence it can be prepared using malonic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of malonic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,3- dibromopropane displaces the bromide ion to produce a Î±- substituted malonic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic diester. Upon treating with aqueous acids the ester groups get hydrolyzed to give a dicarboxylic acid. The dicarboxylic acid eliminates a CO₂ molecule on heating to yield cyclobutylcarboxylic acid.

Answer 29

Conclusion

The compound shown can be prepared by using malonic ester synthesis.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 9

Answer 30

Interpretation Introduction

d)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 11

Explanation of Solution

The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on allyl bromide displaces the bromide ion to produce Î±- allylsubstituted acetoacetic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield hex-5-ene-2-one.

Conclusion

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 12

Answer 31

Interpretation Introduction

d)

Interpretation:

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.

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) S_N² attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

To show

How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.

Answer 32

Expert Solution

Answer to Problem 46AP

The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.

ORGANIC CHEMISTRY W/OWL, Chapter 22.SE, Problem 46AP , additional homework tip 11

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on allyl bromide displaces the bromide ion to produce Î±- allylsubstituted acetoacetic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a Î²- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO₂ molecule on heating to yield hex-5-ene-2-one.