An explanation for the formation of the same anion being formed by the loss of the most acidic proton from both 1-methyl-1,3-cyclopentadiene and 5-methyl-1,3-cyclopentadiene is to be provided. Concept introduction: The most acidic proton in cyclopentadiene and its derivatives (except C5 disubstituted) is the one bonded to C5 . A tertiary carbanion is unstable because of the three inductive electron donating groups bonded to the negatively charged carbon.

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Answer 1

Question

Chapter 12, Problem 54P

Interpretation Introduction

Interpretation:

An explanation for the formation of the same anion being formed by the loss of the most acidic proton from both 1-methyl-1,3-cyclopentadiene and 5-methyl-1,3-cyclopentadiene is to be provided.

Concept introduction:

The most acidic proton in cyclopentadiene and its derivatives (except C5 disubstituted) is the one bonded to C5.

A tertiary carbanion is unstable because of the three inductive electron donating groups bonded to the negatively charged carbon.

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Answer 2

Question

Chapter 12, Problem 54P

Interpretation Introduction

Interpretation:

An explanation for the formation of the same anion being formed by the loss of the most acidic proton from both 1-methyl-1,3-cyclopentadiene and 5-methyl-1,3-cyclopentadiene is to be provided.

Concept introduction:

The most acidic proton in cyclopentadiene and its derivatives (except C5 disubstituted) is the one bonded to C5.

A tertiary carbanion is unstable because of the three inductive electron donating groups bonded to the negatively charged carbon.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 12, Problem 54P

Interpretation Introduction

Interpretation:

An explanation for the formation of the same anion being formed by the loss of the most acidic proton from both 1-methyl-1,3-cyclopentadiene and 5-methyl-1,3-cyclopentadiene is to be provided.

Concept introduction:

The most acidic proton in cyclopentadiene and its derivatives (except C5 disubstituted) is the one bonded to C5.

A tertiary carbanion is unstable because of the three inductive electron donating groups bonded to the negatively charged carbon.

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Answer 4

Question

Chapter 12, Problem 54P

Interpretation Introduction

Interpretation:

An explanation for the formation of the same anion being formed by the loss of the most acidic proton from both 1-methyl-1,3-cyclopentadiene and 5-methyl-1,3-cyclopentadiene is to be provided.

Concept introduction:

The most acidic proton in cyclopentadiene and its derivatives (except C5 disubstituted) is the one bonded to C5.

A tertiary carbanion is unstable because of the three inductive electron donating groups bonded to the negatively charged carbon.

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Answer 5

Chapter 12, Problem 54P

Interpretation Introduction

Interpretation:

An explanation for the formation of the same anion being formed by the loss of the most acidic proton from both 1-methyl-1,3-cyclopentadiene and 5-methyl-1,3-cyclopentadiene is to be provided.

Concept introduction:

The most acidic proton in cyclopentadiene and its derivatives (except C5 disubstituted) is the one bonded to C5.

A tertiary carbanion is unstable because of the three inductive electron donating groups bonded to the negatively charged carbon.

Answer 6

Chapter 12, Problem 54P

Interpretation Introduction

Interpretation:

An explanation for the formation of the same anion being formed by the loss of the most acidic proton from both 1-methyl-1,3-cyclopentadiene and 5-methyl-1,3-cyclopentadiene is to be provided.

Concept introduction:

The most acidic proton in cyclopentadiene and its derivatives (except C5 disubstituted) is the one bonded to C5.

A tertiary carbanion is unstable because of the three inductive electron donating groups bonded to the negatively charged carbon.

Answer 7

Chapter 12, Problem 54P

Interpretation Introduction

Interpretation:

An explanation for the formation of the same anion being formed by the loss of the most acidic proton from both 1-methyl-1,3-cyclopentadiene and 5-methyl-1,3-cyclopentadiene is to be provided.

Concept introduction:

The most acidic proton in cyclopentadiene and its derivatives (except C5 disubstituted) is the one bonded to C5.

A tertiary carbanion is unstable because of the three inductive electron donating groups bonded to the negatively charged carbon.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 12.2 - Write structural formulas for toluene (C6H5CH3)...Ch. 12.3 - Prob. 2P Ch. 12.5 - Prob. 3P Ch. 12.5 - Prob. 4P Ch. 12.6 - Prob. 5P Ch. 12.6 - Chrysene is an aromatic hydrocarbon found in coal...Ch. 12.8 - Prob. 7P Ch. 12.9 - As measured by their first-order rate constants,...Ch. 12.9 - Give the structure of the principal organic...Ch. 12.9 - Prob. 10P

Solution Summary: The author explains that the most acidic proton in cyclopentadiene and its derivatives is the one bonded to C5.

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Chapter 12 Solutions