The complete and detailed mechanism leading to the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16 but not shown explicitly in Equation 12-18 is to be drawn. Concept introduction: When an alkene reacts with bromine in an aqueous solution, bromohydrin is produced instead of a vicinal dibromide. The first step of the reaction is the electrophilic addition of bromine to form a cyclic bromonium intermediate. In the second step, instead of a bromide ion, a molecule of water attacks one of the electron-poor carbons from the cyclic intermediate. An attack by the water molecule is far more likely because as a solvent, its concentration is far higher than that of bromide ions. This results in the formation of bromohydrin. Because the intermediate formed in the first step is cyclic, the water molecule must attack from the direction opposite to bromine. As a result, the overall addition is anti-addition. This determines the stereochemistry of the product. Except when substituents present on nearby carbons prevent an attack from a particular direction, the reaction produces a racemic mixture of enantiomers if the product is chiral.

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Chapter 12, Problem 12.8P

Interpretation Introduction

Interpretation:

The complete and detailed mechanism leading to the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16 but not shown explicitly in Equation 12-18 is to be drawn.

Concept introduction:

When an alkene reacts with bromine in an aqueous solution, bromohydrin is produced instead of a vicinal dibromide.

The first step of the reaction is the electrophilic addition of bromine to form a cyclic bromonium intermediate. In the second step, instead of a bromide ion, a molecule of water attacks one of the electron-poor carbons from the cyclic intermediate. An attack by the water molecule is far more likely because as a solvent, its concentration is far higher than that of bromide ions. This results in the formation of bromohydrin.

Because the intermediate formed in the first step is cyclic, the water molecule must attack from the direction opposite to bromine. As a result, the overall addition is anti-addition. This determines the stereochemistry of the product.

Except when substituents present on nearby carbons prevent an attack from a particular direction, the reaction produces a racemic mixture of enantiomers if the product is chiral.

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Answer 2

Question

Chapter 12, Problem 12.8P

Interpretation Introduction

Interpretation:

The complete and detailed mechanism leading to the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16 but not shown explicitly in Equation 12-18 is to be drawn.

Concept introduction:

When an alkene reacts with bromine in an aqueous solution, bromohydrin is produced instead of a vicinal dibromide.

The first step of the reaction is the electrophilic addition of bromine to form a cyclic bromonium intermediate. In the second step, instead of a bromide ion, a molecule of water attacks one of the electron-poor carbons from the cyclic intermediate. An attack by the water molecule is far more likely because as a solvent, its concentration is far higher than that of bromide ions. This results in the formation of bromohydrin.

Because the intermediate formed in the first step is cyclic, the water molecule must attack from the direction opposite to bromine. As a result, the overall addition is anti-addition. This determines the stereochemistry of the product.

Except when substituents present on nearby carbons prevent an attack from a particular direction, the reaction produces a racemic mixture of enantiomers if the product is chiral.

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Answer 3

Question

Chapter 12, Problem 12.8P

Interpretation Introduction

Interpretation:

The complete and detailed mechanism leading to the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16 but not shown explicitly in Equation 12-18 is to be drawn.

Concept introduction:

When an alkene reacts with bromine in an aqueous solution, bromohydrin is produced instead of a vicinal dibromide.

The first step of the reaction is the electrophilic addition of bromine to form a cyclic bromonium intermediate. In the second step, instead of a bromide ion, a molecule of water attacks one of the electron-poor carbons from the cyclic intermediate. An attack by the water molecule is far more likely because as a solvent, its concentration is far higher than that of bromide ions. This results in the formation of bromohydrin.

Because the intermediate formed in the first step is cyclic, the water molecule must attack from the direction opposite to bromine. As a result, the overall addition is anti-addition. This determines the stereochemistry of the product.

Except when substituents present on nearby carbons prevent an attack from a particular direction, the reaction produces a racemic mixture of enantiomers if the product is chiral.

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Answer 4

Question

Chapter 12, Problem 12.8P

Interpretation Introduction

Interpretation:

The complete and detailed mechanism leading to the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16 but not shown explicitly in Equation 12-18 is to be drawn.

Concept introduction:

When an alkene reacts with bromine in an aqueous solution, bromohydrin is produced instead of a vicinal dibromide.

The first step of the reaction is the electrophilic addition of bromine to form a cyclic bromonium intermediate. In the second step, instead of a bromide ion, a molecule of water attacks one of the electron-poor carbons from the cyclic intermediate. An attack by the water molecule is far more likely because as a solvent, its concentration is far higher than that of bromide ions. This results in the formation of bromohydrin.

Because the intermediate formed in the first step is cyclic, the water molecule must attack from the direction opposite to bromine. As a result, the overall addition is anti-addition. This determines the stereochemistry of the product.

Except when substituents present on nearby carbons prevent an attack from a particular direction, the reaction produces a racemic mixture of enantiomers if the product is chiral.

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Answer 5

Chapter 12, Problem 12.8P

Interpretation Introduction

Interpretation:

The complete and detailed mechanism leading to the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16 but not shown explicitly in Equation 12-18 is to be drawn.

Concept introduction:

When an alkene reacts with bromine in an aqueous solution, bromohydrin is produced instead of a vicinal dibromide.

The first step of the reaction is the electrophilic addition of bromine to form a cyclic bromonium intermediate. In the second step, instead of a bromide ion, a molecule of water attacks one of the electron-poor carbons from the cyclic intermediate. An attack by the water molecule is far more likely because as a solvent, its concentration is far higher than that of bromide ions. This results in the formation of bromohydrin.

Because the intermediate formed in the first step is cyclic, the water molecule must attack from the direction opposite to bromine. As a result, the overall addition is anti-addition. This determines the stereochemistry of the product.

Except when substituents present on nearby carbons prevent an attack from a particular direction, the reaction produces a racemic mixture of enantiomers if the product is chiral.

Answer 6

Chapter 12, Problem 12.8P

Interpretation Introduction

Interpretation:

The complete and detailed mechanism leading to the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16 but not shown explicitly in Equation 12-18 is to be drawn.

Concept introduction:

When an alkene reacts with bromine in an aqueous solution, bromohydrin is produced instead of a vicinal dibromide.

The first step of the reaction is the electrophilic addition of bromine to form a cyclic bromonium intermediate. In the second step, instead of a bromide ion, a molecule of water attacks one of the electron-poor carbons from the cyclic intermediate. An attack by the water molecule is far more likely because as a solvent, its concentration is far higher than that of bromide ions. This results in the formation of bromohydrin.

Because the intermediate formed in the first step is cyclic, the water molecule must attack from the direction opposite to bromine. As a result, the overall addition is anti-addition. This determines the stereochemistry of the product.

Except when substituents present on nearby carbons prevent an attack from a particular direction, the reaction produces a racemic mixture of enantiomers if the product is chiral.

Answer 7

Chapter 12, Problem 12.8P

Interpretation Introduction

Interpretation:

The complete and detailed mechanism leading to the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16 but not shown explicitly in Equation 12-18 is to be drawn.

Concept introduction:

When an alkene reacts with bromine in an aqueous solution, bromohydrin is produced instead of a vicinal dibromide.

The first step of the reaction is the electrophilic addition of bromine to form a cyclic bromonium intermediate. In the second step, instead of a bromide ion, a molecule of water attacks one of the electron-poor carbons from the cyclic intermediate. An attack by the water molecule is far more likely because as a solvent, its concentration is far higher than that of bromide ions. This results in the formation of bromohydrin.

Because the intermediate formed in the first step is cyclic, the water molecule must attack from the direction opposite to bromine. As a result, the overall addition is anti-addition. This determines the stereochemistry of the product.

Except when substituents present on nearby carbons prevent an attack from a particular direction, the reaction produces a racemic mixture of enantiomers if the product is chiral.

Answer 8

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Answer 11

Ch. 12 - Prob. 12.1P Ch. 12 - Prob. 12.2P Ch. 12 - Prob. 12.3P Ch. 12 - Prob. 12.4P Ch. 12 - Prob. 12.5P Ch. 12 - Prob. 12.6P Ch. 12 - Prob. 12.7P Ch. 12 - Prob. 12.8P Ch. 12 - Prob. 12.9P Ch. 12 - Prob. 12.10P

Solution Summary: The author explains the mechanism for the formation of the second enantiomer of 2-bromocyclohexanol from Equation 12-16.

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