A detailed mechanism for the reaction between the given compound and potassium tert-butoxide in N, N-dimethylformamide is to be drawn and why it gives a single product is to be explained. A detailed mechanism for the reaction of the compound with dilute potassium ethoxide in ethanol is to be drawn, and why a mixture of two products is obtained is to be explained. Concept introduction: Alkyl halides undergo elimination when treated with a strong base, particularly at high temperatures. The halogen is eliminated as halide along with a proton from an adjacent carbon. This results in the formation of a double bond between the carbon atoms. The reaction can occur via two mechanisms, E1 or E2. A strong base at high concentration in a polar aprotic solvent favors the E2 mechanism. A weak base or a strong base at low concentrations in a polar protic solvent favors the E1 mechanism. The E2 mechanism is a single-step, bimolecular mechanism. The base extracts a proton from a carbon adjacent to the carbon with the halogen attached to it. The C − H bond pair moves toward the carbon with the attached halogen to form a π bond. At the same time, the carbon-halogen bond breaks heterolytically to give the product. Since this is a single-step, concerted action, the proton and the halogen must be in anti to each other. This also results in the formation of just one elimination product. The E1 mechanism is a two-step, unimolecular mechanism. The leaving group breaks off in the first, rate-determining step, leading to the formation of a carbocation. The base then extracts a proton from an adjacent carbon to give the elimination product. The carbocation formed in the first step can rearrange if it results in the formation of a more stable carbocation. This results in the formation of a mixture of products.

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Chapter 9, Problem 9.48P

Interpretation Introduction

Interpretation:

A detailed mechanism for the reaction between the given compound and potassium tert-butoxide in N, N-dimethylformamide is to be drawn and why it gives a single product is to be explained. A detailed mechanism for the reaction of the compound with dilute potassium ethoxide in ethanol is to be drawn, and why a mixture of two products is obtained is to be explained.

Concept introduction:

Alkyl halides undergo elimination when treated with a strong base, particularly at high temperatures. The halogen is eliminated as halide along with a proton from an adjacent carbon. This results in the formation of a double bond between the carbon atoms.

The reaction can occur via two mechanisms, E1 or E2. A strong base at high concentration in a polar aprotic solvent favors the E2 mechanism. A weak base or a strong base at low concentrations in a polar protic solvent favors the E1 mechanism.

The E2 mechanism is a single-step, bimolecular mechanism. The base extracts a proton from a carbon adjacent to the carbon with the halogen attached to it. The C−H bond pair moves toward the carbon with the attached halogen to form a π bond. At the same time, the carbon-halogen bond breaks heterolytically to give the product. Since this is a single-step, concerted action, the proton and the halogen must be in anti to each other. This also results in the formation of just one elimination product.

The E1 mechanism is a two-step, unimolecular mechanism. The leaving group breaks off in the first, rate-determining step, leading to the formation of a carbocation. The base then extracts a proton from an adjacent carbon to give the elimination product. The carbocation formed in the first step can rearrange if it results in the formation of a more stable carbocation. This results in the formation of a mixture of products.

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Answer 2

Question

Chapter 9, Problem 9.48P

Interpretation Introduction

Interpretation:

A detailed mechanism for the reaction between the given compound and potassium tert-butoxide in N, N-dimethylformamide is to be drawn and why it gives a single product is to be explained. A detailed mechanism for the reaction of the compound with dilute potassium ethoxide in ethanol is to be drawn, and why a mixture of two products is obtained is to be explained.

Concept introduction:

Alkyl halides undergo elimination when treated with a strong base, particularly at high temperatures. The halogen is eliminated as halide along with a proton from an adjacent carbon. This results in the formation of a double bond between the carbon atoms.

The reaction can occur via two mechanisms, E1 or E2. A strong base at high concentration in a polar aprotic solvent favors the E2 mechanism. A weak base or a strong base at low concentrations in a polar protic solvent favors the E1 mechanism.

The E2 mechanism is a single-step, bimolecular mechanism. The base extracts a proton from a carbon adjacent to the carbon with the halogen attached to it. The C−H bond pair moves toward the carbon with the attached halogen to form a π bond. At the same time, the carbon-halogen bond breaks heterolytically to give the product. Since this is a single-step, concerted action, the proton and the halogen must be in anti to each other. This also results in the formation of just one elimination product.

The E1 mechanism is a two-step, unimolecular mechanism. The leaving group breaks off in the first, rate-determining step, leading to the formation of a carbocation. The base then extracts a proton from an adjacent carbon to give the elimination product. The carbocation formed in the first step can rearrange if it results in the formation of a more stable carbocation. This results in the formation of a mixture of products.

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Answer 3

Question

Chapter 9, Problem 9.48P

Interpretation Introduction

Interpretation:

A detailed mechanism for the reaction between the given compound and potassium tert-butoxide in N, N-dimethylformamide is to be drawn and why it gives a single product is to be explained. A detailed mechanism for the reaction of the compound with dilute potassium ethoxide in ethanol is to be drawn, and why a mixture of two products is obtained is to be explained.

Concept introduction:

Alkyl halides undergo elimination when treated with a strong base, particularly at high temperatures. The halogen is eliminated as halide along with a proton from an adjacent carbon. This results in the formation of a double bond between the carbon atoms.

The reaction can occur via two mechanisms, E1 or E2. A strong base at high concentration in a polar aprotic solvent favors the E2 mechanism. A weak base or a strong base at low concentrations in a polar protic solvent favors the E1 mechanism.

The E2 mechanism is a single-step, bimolecular mechanism. The base extracts a proton from a carbon adjacent to the carbon with the halogen attached to it. The C−H bond pair moves toward the carbon with the attached halogen to form a π bond. At the same time, the carbon-halogen bond breaks heterolytically to give the product. Since this is a single-step, concerted action, the proton and the halogen must be in anti to each other. This also results in the formation of just one elimination product.

The E1 mechanism is a two-step, unimolecular mechanism. The leaving group breaks off in the first, rate-determining step, leading to the formation of a carbocation. The base then extracts a proton from an adjacent carbon to give the elimination product. The carbocation formed in the first step can rearrange if it results in the formation of a more stable carbocation. This results in the formation of a mixture of products.

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Answer 4

Question

Chapter 9, Problem 9.48P

Interpretation Introduction

Interpretation:

A detailed mechanism for the reaction between the given compound and potassium tert-butoxide in N, N-dimethylformamide is to be drawn and why it gives a single product is to be explained. A detailed mechanism for the reaction of the compound with dilute potassium ethoxide in ethanol is to be drawn, and why a mixture of two products is obtained is to be explained.

Concept introduction:

Alkyl halides undergo elimination when treated with a strong base, particularly at high temperatures. The halogen is eliminated as halide along with a proton from an adjacent carbon. This results in the formation of a double bond between the carbon atoms.

The reaction can occur via two mechanisms, E1 or E2. A strong base at high concentration in a polar aprotic solvent favors the E2 mechanism. A weak base or a strong base at low concentrations in a polar protic solvent favors the E1 mechanism.

The E2 mechanism is a single-step, bimolecular mechanism. The base extracts a proton from a carbon adjacent to the carbon with the halogen attached to it. The C−H bond pair moves toward the carbon with the attached halogen to form a π bond. At the same time, the carbon-halogen bond breaks heterolytically to give the product. Since this is a single-step, concerted action, the proton and the halogen must be in anti to each other. This also results in the formation of just one elimination product.

The E1 mechanism is a two-step, unimolecular mechanism. The leaving group breaks off in the first, rate-determining step, leading to the formation of a carbocation. The base then extracts a proton from an adjacent carbon to give the elimination product. The carbocation formed in the first step can rearrange if it results in the formation of a more stable carbocation. This results in the formation of a mixture of products.

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Answer 5

Chapter 9, Problem 9.48P

Interpretation Introduction

Interpretation:

A detailed mechanism for the reaction between the given compound and potassium tert-butoxide in N, N-dimethylformamide is to be drawn and why it gives a single product is to be explained. A detailed mechanism for the reaction of the compound with dilute potassium ethoxide in ethanol is to be drawn, and why a mixture of two products is obtained is to be explained.

Concept introduction:

Alkyl halides undergo elimination when treated with a strong base, particularly at high temperatures. The halogen is eliminated as halide along with a proton from an adjacent carbon. This results in the formation of a double bond between the carbon atoms.

The reaction can occur via two mechanisms, E1 or E2. A strong base at high concentration in a polar aprotic solvent favors the E2 mechanism. A weak base or a strong base at low concentrations in a polar protic solvent favors the E1 mechanism.

The E2 mechanism is a single-step, bimolecular mechanism. The base extracts a proton from a carbon adjacent to the carbon with the halogen attached to it. The C−H bond pair moves toward the carbon with the attached halogen to form a π bond. At the same time, the carbon-halogen bond breaks heterolytically to give the product. Since this is a single-step, concerted action, the proton and the halogen must be in anti to each other. This also results in the formation of just one elimination product.

The E1 mechanism is a two-step, unimolecular mechanism. The leaving group breaks off in the first, rate-determining step, leading to the formation of a carbocation. The base then extracts a proton from an adjacent carbon to give the elimination product. The carbocation formed in the first step can rearrange if it results in the formation of a more stable carbocation. This results in the formation of a mixture of products.

Answer 6

Chapter 9, Problem 9.48P

Interpretation Introduction

Interpretation:

A detailed mechanism for the reaction between the given compound and potassium tert-butoxide in N, N-dimethylformamide is to be drawn and why it gives a single product is to be explained. A detailed mechanism for the reaction of the compound with dilute potassium ethoxide in ethanol is to be drawn, and why a mixture of two products is obtained is to be explained.

Concept introduction:

Alkyl halides undergo elimination when treated with a strong base, particularly at high temperatures. The halogen is eliminated as halide along with a proton from an adjacent carbon. This results in the formation of a double bond between the carbon atoms.

The reaction can occur via two mechanisms, E1 or E2. A strong base at high concentration in a polar aprotic solvent favors the E2 mechanism. A weak base or a strong base at low concentrations in a polar protic solvent favors the E1 mechanism.

The E2 mechanism is a single-step, bimolecular mechanism. The base extracts a proton from a carbon adjacent to the carbon with the halogen attached to it. The C−H bond pair moves toward the carbon with the attached halogen to form a π bond. At the same time, the carbon-halogen bond breaks heterolytically to give the product. Since this is a single-step, concerted action, the proton and the halogen must be in anti to each other. This also results in the formation of just one elimination product.

The E1 mechanism is a two-step, unimolecular mechanism. The leaving group breaks off in the first, rate-determining step, leading to the formation of a carbocation. The base then extracts a proton from an adjacent carbon to give the elimination product. The carbocation formed in the first step can rearrange if it results in the formation of a more stable carbocation. This results in the formation of a mixture of products.

Answer 7

Chapter 9, Problem 9.48P

Interpretation Introduction

Interpretation:

A detailed mechanism for the reaction between the given compound and potassium tert-butoxide in N, N-dimethylformamide is to be drawn and why it gives a single product is to be explained. A detailed mechanism for the reaction of the compound with dilute potassium ethoxide in ethanol is to be drawn, and why a mixture of two products is obtained is to be explained.

Concept introduction:

Alkyl halides undergo elimination when treated with a strong base, particularly at high temperatures. The halogen is eliminated as halide along with a proton from an adjacent carbon. This results in the formation of a double bond between the carbon atoms.

The reaction can occur via two mechanisms, E1 or E2. A strong base at high concentration in a polar aprotic solvent favors the E2 mechanism. A weak base or a strong base at low concentrations in a polar protic solvent favors the E1 mechanism.

The E2 mechanism is a single-step, bimolecular mechanism. The base extracts a proton from a carbon adjacent to the carbon with the halogen attached to it. The C−H bond pair moves toward the carbon with the attached halogen to form a π bond. At the same time, the carbon-halogen bond breaks heterolytically to give the product. Since this is a single-step, concerted action, the proton and the halogen must be in anti to each other. This also results in the formation of just one elimination product.

The E1 mechanism is a two-step, unimolecular mechanism. The leaving group breaks off in the first, rate-determining step, leading to the formation of a carbocation. The base then extracts a proton from an adjacent carbon to give the elimination product. The carbocation formed in the first step can rearrange if it results in the formation of a more stable carbocation. This results in the formation of a mixture of products.

Answer 8

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Answer 11

Ch. 9 - Prob. 9.1P Ch. 9 - Prob. 9.2P Ch. 9 - Prob. 9.3P Ch. 9 - Prob. 9.4P Ch. 9 - Prob. 9.5P Ch. 9 - Prob. 9.6P Ch. 9 - Prob. 9.7P Ch. 9 - Prob. 9.8P Ch. 9 - Prob. 9.9P Ch. 9 - Prob. 9.10P

Solution Summary: The author explains the mechanism for the reaction of the compound with potassium tert-butoxide in N, N-dimethylformamide and why it gives a single product.

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