In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom. Concept introduction: Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms. According to this rule stable and unstable molecule is mentioned below, The first compound is stable. Although double bond is at bridgehead, the system is not bridged But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

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Answer 1

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Chapter 8, Problem 108P

Interpretation Introduction

Interpretation: In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom.

Concept introduction:

Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms.

According to this rule stable and unstable molecule is mentioned below,

The first compound is stable. Although double bond is at bridgehead, the system is not bridged

But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

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Answer 2

Question

Chapter 8, Problem 108P

Interpretation Introduction

Interpretation: In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom.

Concept introduction:

Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms.

According to this rule stable and unstable molecule is mentioned below,

The first compound is stable. Although double bond is at bridgehead, the system is not bridged

But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

Expert Solution & Answer

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Check out a sample textbook solution

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Answer 3

Question

Chapter 8, Problem 108P

Interpretation Introduction

Interpretation: In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom.

Concept introduction:

Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms.

According to this rule stable and unstable molecule is mentioned below,

The first compound is stable. Although double bond is at bridgehead, the system is not bridged

But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

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Answer 4

Question

Chapter 8, Problem 108P

Interpretation Introduction

Interpretation: In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom.

Concept introduction:

Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms.

According to this rule stable and unstable molecule is mentioned below,

The first compound is stable. Although double bond is at bridgehead, the system is not bridged

But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

Expert Solution & Answer

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Answer 5

Chapter 8, Problem 108P

Interpretation Introduction

Interpretation: In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom.

Concept introduction:

Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms.

According to this rule stable and unstable molecule is mentioned below,

The first compound is stable. Although double bond is at bridgehead, the system is not bridged

But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

Answer 6

Chapter 8, Problem 108P

Interpretation Introduction

Interpretation: In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom.

Concept introduction:

Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms.

According to this rule stable and unstable molecule is mentioned below,

The first compound is stable. Although double bond is at bridgehead, the system is not bridged

But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

Answer 7

Chapter 8, Problem 108P

Interpretation Introduction

Interpretation: In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom.

Concept introduction:

Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms.

According to this rule stable and unstable molecule is mentioned below,

The first compound is stable. Although double bond is at bridgehead, the system is not bridged

But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

Answer 8

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Answer 11

Ch. 8.1 - Prob. 1P Ch. 8.1 - Prob. 2P Ch. 8.4 - Prob. 3P Ch. 8.5 - Prob. 4P Ch. 8.5 - Prob. 6P Ch. 8.6 - a. Predict the relative bond lengths of the three...Ch. 8.6 - Prob. 8P Ch. 8.6 - Prob. 9P Ch. 8.6 - Prob. 10P Ch. 8.7 - Prob. 11P

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