(a) Interpretation: The C − H bonds in increasing order of bond strength are to be ranked with reference to the indicated C − H bonds in 2 − methylbutane . Concept introduction: Cleavage of 1 ∘ C − H bond form 1 ∘ radical, cleavage of 2 ∘ C − H bond forms 2 ∘ radical and cleavage of 3 ∘ C − H bond forms 3 ∘ radical.

Question

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Answer 1

Question

Chapter 15, Problem 15.30P

Interpretation Introduction

(a)

Interpretation: The C−H bonds in increasing order of bond strength are to be ranked with reference to the indicated C−H bonds in 2−methylbutane.

Concept introduction: Cleavage of 1∘ C−H bond form 1∘ radical, cleavage of 2∘ C−H bond forms 2∘ radical and cleavage of 3∘ C−H bond forms 3∘ radical.

Expert Solution

Answer to Problem 15.30P

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Explanation of Solution

The energy required to break the 1∘ C−H bond is more than the energy required to break the 2∘ C−H bond and energy required to break the 2∘C−H bond is more than the energy required to break the 3∘ C−H bond.

Thus, the increasing order of bond strength is,

3∘ C−H < 2∘ C−H < 1∘ C−H

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Interpretation Introduction

(b)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be drawn and is to be classified as 1°, 2°, or 3°.

Concept introduction: Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

Expert Solution

Answer to Problem 15.30P

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is,

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 1

Figure 1

Explanation of Solution

Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

The homolytic cleavage of the C−H bond generates primary, secondary, and tertiary radicals.

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 2

Figure 1

Conclusion

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is shown in Figure 2

Interpretation Introduction

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 3

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Conclusion

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Interpretation Introduction

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are 1° C−H bond<2° C−H bond<3° C−H bond.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 4

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Hydrogen atoms are less polarizable than alkyl groups. Therefore, alkyl group can easily donate electron density to the electron deficient carbon radical. Therefore, the increasing ease of H abstraction will be 1° C−H bond<2° C−H bond<3° C−H bond.

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

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Students have asked these similar questions

4. Draw all resonance structures for the carbocation formed by ortho attack of the electrophile +NO2 on each starting material. Label any resonance structures that are especially stable or unstable. C(CH3)3 LOH CHO a. b. С.

1. Free radicals are generated by subjecting a molecule such as chlorine or bromine gas to A. high temperature conditions. B. photolysis. C. ionizing conditions. D. choices A and B 2. Which of the following radical intermediates is the most stable? q •CF3 A. B. C. D. 3. In the reaction shown below the intermediate that is formed at the fastest rate is Cl₂ A. uv CH, CH₂CHCH₂CH₂CH₂ B. CH, CH.CCH₂CH₂CH, CH₂ C. CH.CHCHCH.CH, D. CH₂ CH₂CHCH₂CHCH

With reference to the indicated C-H bonds in the following compound: a. Rank the C-H bonds in order of increasing bond strength b. Draw the radical resulting from cleavage of each C-H bond, and classify it as 1º, 2º, or 3º c. Rank the radicals in order of increasing stability d. Rank the C-H bonds in order of increasing ease of H abstraction in a radical halogenation reaction.

Answer 2

Question

Chapter 15, Problem 15.30P

Interpretation Introduction

(a)

Interpretation: The C−H bonds in increasing order of bond strength are to be ranked with reference to the indicated C−H bonds in 2−methylbutane.

Concept introduction: Cleavage of 1∘ C−H bond form 1∘ radical, cleavage of 2∘ C−H bond forms 2∘ radical and cleavage of 3∘ C−H bond forms 3∘ radical.

Expert Solution

Answer to Problem 15.30P

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Explanation of Solution

The energy required to break the 1∘ C−H bond is more than the energy required to break the 2∘ C−H bond and energy required to break the 2∘C−H bond is more than the energy required to break the 3∘ C−H bond.

Thus, the increasing order of bond strength is,

3∘ C−H < 2∘ C−H < 1∘ C−H

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Interpretation Introduction

(b)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be drawn and is to be classified as 1°, 2°, or 3°.

Concept introduction: Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

Expert Solution

Answer to Problem 15.30P

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is,

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 1

Figure 1

Explanation of Solution

Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

The homolytic cleavage of the C−H bond generates primary, secondary, and tertiary radicals.

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 2

Figure 1

Conclusion

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is shown in Figure 2

Interpretation Introduction

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 3

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Conclusion

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Interpretation Introduction

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are 1° C−H bond<2° C−H bond<3° C−H bond.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 4

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Hydrogen atoms are less polarizable than alkyl groups. Therefore, alkyl group can easily donate electron density to the electron deficient carbon radical. Therefore, the increasing ease of H abstraction will be 1° C−H bond<2° C−H bond<3° C−H bond.

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

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Answer 3

Question

Chapter 15, Problem 15.30P

Interpretation Introduction

(a)

Interpretation: The C−H bonds in increasing order of bond strength are to be ranked with reference to the indicated C−H bonds in 2−methylbutane.

Concept introduction: Cleavage of 1∘ C−H bond form 1∘ radical, cleavage of 2∘ C−H bond forms 2∘ radical and cleavage of 3∘ C−H bond forms 3∘ radical.

Expert Solution

Answer to Problem 15.30P

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Explanation of Solution

The energy required to break the 1∘ C−H bond is more than the energy required to break the 2∘ C−H bond and energy required to break the 2∘C−H bond is more than the energy required to break the 3∘ C−H bond.

Thus, the increasing order of bond strength is,

3∘ C−H < 2∘ C−H < 1∘ C−H

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Interpretation Introduction

(b)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be drawn and is to be classified as 1°, 2°, or 3°.

Concept introduction: Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

Expert Solution

Answer to Problem 15.30P

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is,

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 1

Figure 1

Explanation of Solution

Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

The homolytic cleavage of the C−H bond generates primary, secondary, and tertiary radicals.

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 2

Figure 1

Conclusion

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is shown in Figure 2

Interpretation Introduction

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 3

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Conclusion

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Interpretation Introduction

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are 1° C−H bond<2° C−H bond<3° C−H bond.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 4

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Hydrogen atoms are less polarizable than alkyl groups. Therefore, alkyl group can easily donate electron density to the electron deficient carbon radical. Therefore, the increasing ease of H abstraction will be 1° C−H bond<2° C−H bond<3° C−H bond.

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

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Answer 4

Question

Chapter 15, Problem 15.30P

Interpretation Introduction

(a)

Interpretation: The C−H bonds in increasing order of bond strength are to be ranked with reference to the indicated C−H bonds in 2−methylbutane.

Concept introduction: Cleavage of 1∘ C−H bond form 1∘ radical, cleavage of 2∘ C−H bond forms 2∘ radical and cleavage of 3∘ C−H bond forms 3∘ radical.

Expert Solution

Answer to Problem 15.30P

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Explanation of Solution

The energy required to break the 1∘ C−H bond is more than the energy required to break the 2∘ C−H bond and energy required to break the 2∘C−H bond is more than the energy required to break the 3∘ C−H bond.

Thus, the increasing order of bond strength is,

3∘ C−H < 2∘ C−H < 1∘ C−H

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Interpretation Introduction

(b)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be drawn and is to be classified as 1°, 2°, or 3°.

Concept introduction: Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

Expert Solution

Answer to Problem 15.30P

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is,

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 1

Figure 1

Explanation of Solution

Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

The homolytic cleavage of the C−H bond generates primary, secondary, and tertiary radicals.

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 2

Figure 1

Conclusion

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is shown in Figure 2

Interpretation Introduction

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 3

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Conclusion

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Interpretation Introduction

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are 1° C−H bond<2° C−H bond<3° C−H bond.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 4

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Hydrogen atoms are less polarizable than alkyl groups. Therefore, alkyl group can easily donate electron density to the electron deficient carbon radical. Therefore, the increasing ease of H abstraction will be 1° C−H bond<2° C−H bond<3° C−H bond.

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

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Answer 5

Chapter 15, Problem 15.30P

Interpretation Introduction

(a)

Interpretation: The C−H bonds in increasing order of bond strength are to be ranked with reference to the indicated C−H bonds in 2−methylbutane.

Concept introduction: Cleavage of 1∘ C−H bond form 1∘ radical, cleavage of 2∘ C−H bond forms 2∘ radical and cleavage of 3∘ C−H bond forms 3∘ radical.

Expert Solution

Answer to Problem 15.30P

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Explanation of Solution

The energy required to break the 1∘ C−H bond is more than the energy required to break the 2∘ C−H bond and energy required to break the 2∘C−H bond is more than the energy required to break the 3∘ C−H bond.

Thus, the increasing order of bond strength is,

3∘ C−H < 2∘ C−H < 1∘ C−H

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Interpretation Introduction

(b)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be drawn and is to be classified as 1°, 2°, or 3°.

Concept introduction: Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

Expert Solution

Answer to Problem 15.30P

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is,

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 1

Figure 1

Explanation of Solution

Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

The homolytic cleavage of the C−H bond generates primary, secondary, and tertiary radicals.

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 2

Figure 1

Conclusion

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is shown in Figure 2

Interpretation Introduction

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 3

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Conclusion

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Interpretation Introduction

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are 1° C−H bond<2° C−H bond<3° C−H bond.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 4

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Hydrogen atoms are less polarizable than alkyl groups. Therefore, alkyl group can easily donate electron density to the electron deficient carbon radical. Therefore, the increasing ease of H abstraction will be 1° C−H bond<2° C−H bond<3° C−H bond.

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Answer 6

Chapter 15, Problem 15.30P

Interpretation Introduction

(a)

Interpretation: The C−H bonds in increasing order of bond strength are to be ranked with reference to the indicated C−H bonds in 2−methylbutane.

Concept introduction: Cleavage of 1∘ C−H bond form 1∘ radical, cleavage of 2∘ C−H bond forms 2∘ radical and cleavage of 3∘ C−H bond forms 3∘ radical.

Expert Solution

Answer to Problem 15.30P

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Explanation of Solution

The energy required to break the 1∘ C−H bond is more than the energy required to break the 2∘ C−H bond and energy required to break the 2∘C−H bond is more than the energy required to break the 3∘ C−H bond.

Thus, the increasing order of bond strength is,

3∘ C−H < 2∘ C−H < 1∘ C−H

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Answer 7

Chapter 15, Problem 15.30P

Interpretation Introduction

(a)

Interpretation: The C−H bonds in increasing order of bond strength are to be ranked with reference to the indicated C−H bonds in 2−methylbutane.

Concept introduction: Cleavage of 1∘ C−H bond form 1∘ radical, cleavage of 2∘ C−H bond forms 2∘ radical and cleavage of 3∘ C−H bond forms 3∘ radical.

Answer 8

Expert Solution

Answer to Problem 15.30P

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The energy required to break the 1∘ C−H bond is more than the energy required to break the 2∘ C−H bond and energy required to break the 2∘C−H bond is more than the energy required to break the 3∘ C−H bond.

Thus, the increasing order of bond strength is,

3∘ C−H < 2∘ C−H < 1∘ C−H

Answer 13

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Answer 14

Interpretation Introduction

(b)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be drawn and is to be classified as 1°, 2°, or 3°.

Concept introduction: Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

Expert Solution

Answer to Problem 15.30P

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is,

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 1

Figure 1

Explanation of Solution

Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

The homolytic cleavage of the C−H bond generates primary, secondary, and tertiary radicals.

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 2

Figure 1

Conclusion

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is shown in Figure 2

Answer 15

Interpretation Introduction

(b)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be drawn and is to be classified as 1°, 2°, or 3°.

Concept introduction: Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

Answer 16

Expert Solution

Answer to Problem 15.30P

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is,

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 1

Figure 1

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

Primary C−H are those in which the C−H bond is attached to one carbon atom. Secondary C−H are those in which the C−H bond is attached to two carbon atoms. Tertiary C−H are those in which the C−H bond is attached to three carbon atom.

The homolytic cleavage of the C−H bond generates primary, secondary, and tertiary radicals.

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 2

Figure 1

Answer 21

Conclusion

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is shown in Figure 2

Answer 22

Interpretation Introduction

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 3

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Conclusion

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Answer 23

Interpretation Introduction

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Answer 24

Expert Solution

Answer to Problem 15.30P

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 3

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Answer 29

Conclusion

The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Answer 30

Interpretation Introduction

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Expert Solution

Answer to Problem 15.30P

The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are 1° C−H bond<2° C−H bond<3° C−H bond.

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 4

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Hydrogen atoms are less polarizable than alkyl groups. Therefore, alkyl group can easily donate electron density to the electron deficient carbon radical. Therefore, the increasing ease of H abstraction will be 1° C−H bond<2° C−H bond<3° C−H bond.

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Answer 31

Interpretation Introduction

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Answer 32

Expert Solution

Answer to Problem 15.30P

The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are 1° C−H bond<2° C−H bond<3° C−H bond.

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

The radicals resulting from the cleavage of each C−H bond in 2−methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.30P , additional homework tip 4

Figure 1

The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are 1° radical<2° radical<3° radical.

Hydrogen atoms are less polarizable than alkyl groups. Therefore, alkyl group can easily donate electron density to the electron deficient carbon radical. Therefore, the increasing ease of H abstraction will be 1° C−H bond<2° C−H bond<3° C−H bond.

Answer 37

Conclusion

The C−H bonds in increasing order of bond strength are 3∘ C−H < 2∘ C−H < 1∘ C−H .

Answer 38

Ch. 15 - Prob. 15.1P Ch. 15 - Prob. 15.2P Ch. 15 - Prob. 15.3P Ch. 15 - Prob. 15.4P Ch. 15 - Prob. 15.5P Ch. 15 - Problem 15.6 Using mechanism 15.1 as guide, write...Ch. 15 - Prob. 15.7P Ch. 15 - Problem 15.8 Which bond in the each compound is...Ch. 15 - Prob. 15.9P Ch. 15 - Prob. 15.10P

Concept explainers

Answer to Problem 15.30P

Explanation of Solution

Answer to Problem 15.30P

Explanation of Solution

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Answer to Problem 15.30P

Explanation of Solution

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Answer to Problem 15.30P

Explanation of Solution

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Chapter 15 Solutions

Concept explainers

Answer to Problem 15.30P

Explanation of Solution

Answer to Problem 15.30P

Explanation of Solution

(c) Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability. Concept introduction:

Answer to Problem 15.30P

Explanation of Solution

(d) Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked. Concept introduction:

Answer to Problem 15.30P

Explanation of Solution

Want to see more full solutions like this?

Chapter 15 Solutions

(c)

Interpretation: The radicals resulting from the cleavage of each C−H bond in 2−methylbutane is to be ranked in increasing order of stability.

Concept introduction:

(d)

Interpretation: The C−H bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction: