Organic Chemistry
Organic Chemistry
4th Edition
ISBN: 9780073402772
Author: Janice G. Smith
Publisher: MCG
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Chapter 15, Problem 15.32P
Interpretation Introduction

(a)

Interpretation:

The CH bonds in increasing order of bond strength are to be ranked with reference to the indicated CH bonds in 2methylbutane.

Concept introduction:

Cleavage of 1 CH bond form 1 radical, cleavage of 2 CH bond forms 2 radical and cleavage of 3 CH bond forms 3 radical.

Expert Solution
Check Mark

Answer to Problem 15.32P

The CH bonds in increasing order of bond strength are 3°CH<2°CH<1°CH.

Explanation of Solution

The energy required to break the 1 CH bond is more than the energy required to break the 2 CH bond and energy required to break the 2CH bond is more than the energy required to break the 3 CH bond.

Thus, the increasing order of bond strength is,

3°CH<2°CH<1°CH

Conclusion

The CH bonds in increasing order of bond strength are 3°CH<2°CH<1°CH.

Interpretation Introduction

(b)

Interpretation:

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

Concept introduction:

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

Expert Solution
Check Mark

Answer to Problem 15.32P

The radicals resulting from the cleavage of each CH bond in 2methylbutane is,

Organic Chemistry, Chapter 15, Problem 15.32P , additional homework tip  1

Figure 1

Explanation of Solution

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

The homolytic cleavage of the CH bond generates primary, secondary, and tertiary radicals.

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

Organic Chemistry, Chapter 15, Problem 15.32P , additional homework tip  2

Figure 1

Conclusion

The radicals resulting from the cleavage of each CH bond in 2methylbutane is shown in Figure 2

Interpretation Introduction

(c)

Interpretation:

The radicals resulting from the cleavage of each CH bond in 2methylbutane is to be ranked in increasing order of stability.

Concept introduction:

Expert Solution
Check Mark

Answer to Problem 15.32P

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 CH bond in 2methylbutane and its classification as 1°, 2°, or 3° is shown in Figure 2

Organic Chemistry, Chapter 15, Problem 15.32P , 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 CH bonds in order of increasing ease of H abstraction in a radical halogenation reaction are to be ranked.

Concept introduction:

Expert Solution
Check Mark

Answer to Problem 15.32P

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

Explanation of Solution

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

Organic Chemistry, Chapter 15, Problem 15.32P , 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°CHbond<2°CHbond<3°CHbond.

Conclusion

The CH bonds in increasing order of bond strength are 3°CH<2°CH<1°CH.

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Chapter 15 Solutions

Organic Chemistry

Ch. 15 - Prob. 15.11PCh. 15 - Synthesize each compound from (CH3)3CH. a....Ch. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Draw all constitutional isomers formed when each...Ch. 15 - Draw the structure of the four allylic halides...Ch. 15 - Which compounds can be prepared in good yield by...Ch. 15 - Which CH bond is most readily cleaved in linolenic...Ch. 15 - Prob. 15.23PCh. 15 - Draw the products formed when each alkene is...Ch. 15 - Problem 15.24 When adds to under radical...Ch. 15 - Prob. 15.26PCh. 15 - Draw an energy diagram for the two propagation...Ch. 15 - Prob. 15.28PCh. 15 - Problem 15.27 Draw the steps of the mechanism that...Ch. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Why is a benzylic CH bond labeled in red unusually...Ch. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - What alkane is needed to make each alkyl halide by...Ch. 15 - Which alkyl halides can be prepared in good yield...Ch. 15 - Prob. 15.41PCh. 15 - 15.40 Explain why radical bromination of p-xylene...Ch. 15 - a. What product(s) (excluding stereoisomers) are...Ch. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Prob. 15.46PCh. 15 - 15.44 Draw all constitutional isomers formed when...Ch. 15 - Draw the organic products formed in each reaction....Ch. 15 - Prob. 15.49PCh. 15 - 15.47 Treatment of a hydrocarbon A (molecular...Ch. 15 - Prob. 15.51PCh. 15 - Prob. 15.52PCh. 15 - Prob. 15.53PCh. 15 - Prob. 15.54PCh. 15 - 15.53 Consider the following bromination: . a....Ch. 15 - 15.54 Draw a stepwise mechanism for the following...Ch. 15 - Prob. 15.57PCh. 15 - An alternative mechanism for the propagation steps...Ch. 15 - Prob. 15.59PCh. 15 - Prob. 15.60PCh. 15 - Devise a synthesis of each compound from...Ch. 15 - Devise a synthesis of each target compound from...Ch. 15 - Devisea synthesis of each target compound from the...Ch. 15 - Devise a synthesis of each compound using CH3CH3...Ch. 15 - Prob. 15.65PCh. 15 - 15.63 As described in Section 9.16, the...Ch. 15 - 15.64 Ethers are oxidized with to form...Ch. 15 - Prob. 15.68PCh. 15 - Prob. 15.69PCh. 15 - 15.67 In cells, vitamin C exists largely as its...Ch. 15 - What monomer is needed to form each...Ch. 15 - Prob. 15.72PCh. 15 - Prob. 15.73PCh. 15 - 15.71 Draw a stepwise mechanism for the following...Ch. 15 - 15.72 As we will learn in Chapter 30, styrene...Ch. 15 - Prob. 15.76PCh. 15 - 15.74 A and B, isomers of molecular formula , are...Ch. 15 - Prob. 15.78PCh. 15 - Radical chlorination of CH3CH3 forms two minor...Ch. 15 - 15.76 Draw a stepwise mechanism for the...Ch. 15 - Prob. 15.81PCh. 15 - Prob. 15.82PCh. 15 - Prob. 15.83P
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