The complete, detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction. Concept introduction: The primary amine containing one or more beta hydrogen atom shows reaction with methyl iodide, followed by aqueous silver oxide, followed by heat. In this reaction, the primary amine is transformed to an alkene . Hoffman elimination reaction takes place in three stages. Stage one consists of methylation of the nitrogen atom in the given amine to form a quaternary ammonium salt. In stage two, aqueous silver oxide, when heated, produces silver hydroxide, which serves as a strong base. This silver hydroxide formed reacts with the quaternary ammonium salt to replace the iodine by hydroxyl group. Stage three consists of two steps. In step one, the strong base abstracts the alpha proton, forming an alkene, and at the same time, the primary amine leaves as a good leaving group. This creates the least substituted alkene as the base removes the proton from the least substituted carbon atom. This reaction is called as Hoffman elimination reaction. The major alkene formed in this reaction is the anti-Zaitsev product. In anti-Zaitsev product, a least substituted alkene is produced.

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Chapter 10, Problem 10.29P

Interpretation Introduction

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

Concept introduction:

The primary amine containing one or more beta hydrogen atom shows reaction with methyl iodide, followed by aqueous silver oxide, followed by heat. In this reaction, the primary amine is transformed to an alkene. Hoffman elimination reaction takes place in three stages. Stage one consists of methylation of the nitrogen atom in the given amine to form a quaternary ammonium salt. In stage two, aqueous silver oxide, when heated, produces silver hydroxide, which serves as a strong base. This silver hydroxide formed reacts with the quaternary ammonium salt to replace the iodine by hydroxyl group. Stage three consists of two steps. In step one, the strong base abstracts the alpha proton, forming an alkene, and at the same time, the primary amine leaves as a good leaving group. This creates the least substituted alkene as the base removes the proton from the least substituted carbon atom. This reaction is called as Hoffman elimination reaction. The major alkene formed in this reaction is the anti-Zaitsev product. In anti-Zaitsev product, a least substituted alkene is produced.

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Answer 2

Question

Chapter 10, Problem 10.29P

Interpretation Introduction

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

Concept introduction:

The primary amine containing one or more beta hydrogen atom shows reaction with methyl iodide, followed by aqueous silver oxide, followed by heat. In this reaction, the primary amine is transformed to an alkene. Hoffman elimination reaction takes place in three stages. Stage one consists of methylation of the nitrogen atom in the given amine to form a quaternary ammonium salt. In stage two, aqueous silver oxide, when heated, produces silver hydroxide, which serves as a strong base. This silver hydroxide formed reacts with the quaternary ammonium salt to replace the iodine by hydroxyl group. Stage three consists of two steps. In step one, the strong base abstracts the alpha proton, forming an alkene, and at the same time, the primary amine leaves as a good leaving group. This creates the least substituted alkene as the base removes the proton from the least substituted carbon atom. This reaction is called as Hoffman elimination reaction. The major alkene formed in this reaction is the anti-Zaitsev product. In anti-Zaitsev product, a least substituted alkene is produced.

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Answer 3

Question

Chapter 10, Problem 10.29P

Interpretation Introduction

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

Concept introduction:

The primary amine containing one or more beta hydrogen atom shows reaction with methyl iodide, followed by aqueous silver oxide, followed by heat. In this reaction, the primary amine is transformed to an alkene. Hoffman elimination reaction takes place in three stages. Stage one consists of methylation of the nitrogen atom in the given amine to form a quaternary ammonium salt. In stage two, aqueous silver oxide, when heated, produces silver hydroxide, which serves as a strong base. This silver hydroxide formed reacts with the quaternary ammonium salt to replace the iodine by hydroxyl group. Stage three consists of two steps. In step one, the strong base abstracts the alpha proton, forming an alkene, and at the same time, the primary amine leaves as a good leaving group. This creates the least substituted alkene as the base removes the proton from the least substituted carbon atom. This reaction is called as Hoffman elimination reaction. The major alkene formed in this reaction is the anti-Zaitsev product. In anti-Zaitsev product, a least substituted alkene is produced.

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Answer 4

Question

Chapter 10, Problem 10.29P

Interpretation Introduction

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

Concept introduction:

The primary amine containing one or more beta hydrogen atom shows reaction with methyl iodide, followed by aqueous silver oxide, followed by heat. In this reaction, the primary amine is transformed to an alkene. Hoffman elimination reaction takes place in three stages. Stage one consists of methylation of the nitrogen atom in the given amine to form a quaternary ammonium salt. In stage two, aqueous silver oxide, when heated, produces silver hydroxide, which serves as a strong base. This silver hydroxide formed reacts with the quaternary ammonium salt to replace the iodine by hydroxyl group. Stage three consists of two steps. In step one, the strong base abstracts the alpha proton, forming an alkene, and at the same time, the primary amine leaves as a good leaving group. This creates the least substituted alkene as the base removes the proton from the least substituted carbon atom. This reaction is called as Hoffman elimination reaction. The major alkene formed in this reaction is the anti-Zaitsev product. In anti-Zaitsev product, a least substituted alkene is produced.

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Answer 5

Chapter 10, Problem 10.29P

Interpretation Introduction

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

Concept introduction:

The primary amine containing one or more beta hydrogen atom shows reaction with methyl iodide, followed by aqueous silver oxide, followed by heat. In this reaction, the primary amine is transformed to an alkene. Hoffman elimination reaction takes place in three stages. Stage one consists of methylation of the nitrogen atom in the given amine to form a quaternary ammonium salt. In stage two, aqueous silver oxide, when heated, produces silver hydroxide, which serves as a strong base. This silver hydroxide formed reacts with the quaternary ammonium salt to replace the iodine by hydroxyl group. Stage three consists of two steps. In step one, the strong base abstracts the alpha proton, forming an alkene, and at the same time, the primary amine leaves as a good leaving group. This creates the least substituted alkene as the base removes the proton from the least substituted carbon atom. This reaction is called as Hoffman elimination reaction. The major alkene formed in this reaction is the anti-Zaitsev product. In anti-Zaitsev product, a least substituted alkene is produced.

Answer 6

Chapter 10, Problem 10.29P

Interpretation Introduction

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

Concept introduction:

The primary amine containing one or more beta hydrogen atom shows reaction with methyl iodide, followed by aqueous silver oxide, followed by heat. In this reaction, the primary amine is transformed to an alkene. Hoffman elimination reaction takes place in three stages. Stage one consists of methylation of the nitrogen atom in the given amine to form a quaternary ammonium salt. In stage two, aqueous silver oxide, when heated, produces silver hydroxide, which serves as a strong base. This silver hydroxide formed reacts with the quaternary ammonium salt to replace the iodine by hydroxyl group. Stage three consists of two steps. In step one, the strong base abstracts the alpha proton, forming an alkene, and at the same time, the primary amine leaves as a good leaving group. This creates the least substituted alkene as the base removes the proton from the least substituted carbon atom. This reaction is called as Hoffman elimination reaction. The major alkene formed in this reaction is the anti-Zaitsev product. In anti-Zaitsev product, a least substituted alkene is produced.

Answer 7

Chapter 10, Problem 10.29P

Interpretation Introduction

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

Concept introduction:

The primary amine containing one or more beta hydrogen atom shows reaction with methyl iodide, followed by aqueous silver oxide, followed by heat. In this reaction, the primary amine is transformed to an alkene. Hoffman elimination reaction takes place in three stages. Stage one consists of methylation of the nitrogen atom in the given amine to form a quaternary ammonium salt. In stage two, aqueous silver oxide, when heated, produces silver hydroxide, which serves as a strong base. This silver hydroxide formed reacts with the quaternary ammonium salt to replace the iodine by hydroxyl group. Stage three consists of two steps. In step one, the strong base abstracts the alpha proton, forming an alkene, and at the same time, the primary amine leaves as a good leaving group. This creates the least substituted alkene as the base removes the proton from the least substituted carbon atom. This reaction is called as Hoffman elimination reaction. The major alkene formed in this reaction is the anti-Zaitsev product. In anti-Zaitsev product, a least substituted alkene is produced.

Answer 8

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Ch. 10 - Prob. 10.1P Ch. 10 - Prob. 10.2P Ch. 10 - Prob. 10.3P Ch. 10 - Prob. 10.4P Ch. 10 - Prob. 10.5P Ch. 10 - Prob. 10.6P Ch. 10 - Prob. 10.7P Ch. 10 - Prob. 10.8P Ch. 10 - Prob. 10.9P Ch. 10 - Prob. 10.10P

Solution Summary: The author explains that the Hoffman elimination reaction takes place in three stages. The primary amine is transformed into an alkene.

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