Organic Chemistry
Organic Chemistry
5th Edition
ISBN: 9780078021558
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
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Chapter 15, Problem 15.2P
Interpretation Introduction

(a)

Interpretation: The most stable radical that can result from cleavage of CH bond in the given molecule is to be drawn.

Concept introduction: A free radical is an atom or ion with unpaired electrons. They are reactive intermediates formed by the homolysis of covalent bond. Free radicals are classified as 1°, 2° or 3° depending upon the number of alkyl groups attached to it. As number of alkyl substitutents increases, the stability of radical also increases. This is due to the fact alkyl substituents donate their electrons to electron deficient radicals.

Expert Solution
Check Mark

Answer to Problem 15.2P

The most stable radical that can result from cleavage of CH bond in the given molecule is shown below.

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  1

Explanation of Solution

The given species is,

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  2

Figure 1

Three types of radicals can be formed by the cleavage of CH bond in the given molecule. The cleavage of CH bond in the given molecule is shown below.

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  3

Figure 2

The number of alkyl substitutents increases, the stability of radical increases. The order of stability is 1°<2°<3°. The most stable radical is 3° radical.

Conclusion

The most stable radical that can result from cleavage of CH bond in the given molecule is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation: The most stable radical that can result from cleavage of CH bond in the given molecule is to be drawn.

Concept introduction: A free radical is an atom or ion with unpaired electrons. They are reactive intermediates formed by the homolysis of covalent bond. Free radicals are classified as 1°, 2° or 3° depending upon the number of alkyl groups attached to it. The number of alkyl substitutents increases, the stability of radical increases. This is due to the fact alkyl substituents donate their electrons to electron deficient radicals.

Expert Solution
Check Mark

Answer to Problem 15.2P

The most stable radical that can result from cleavage of CH bond in the given molecule is shown below.

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  4

Explanation of Solution

The given species is,

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  5

Figure 3

Two types of radicals can be formed from cleavage of CH bond in the given molecule. The cleavage of CH bond in the given molecule is shown below.

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  6

Figure 4

The number of alkyl substitutents increases, the stability of radical increases. The order of stability is 1°<2°<3°. The most stable radical is 2° radical.

Conclusion

The most stable radical that can result from cleavage of CH bond in the given molecule is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation: The most stable radical that can result from cleavage of CH bond in the given molecule is to be drawn.

Concept introduction: A free radical is an atom or ion with unpaired electrons. They are reactive intermediates formed by the homolysis of covalent bond. Free radicals are classified as 1°, 2° or 3° depending upon the number of alkyl groups attached to it. The number of alkyl substitutents increases, the stability of radical increases. This is due to the fact alkyl substituents donate their electrons to electron deficient radicals.

Expert Solution
Check Mark

Answer to Problem 15.2P

The most stable radical that can results from the cleavage of CH bond in the given molecule is shown below.

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  7

Explanation of Solution

The given species is,

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  8

Figure 5

Only 1° radical can be formed from cleavage of CH bond in the given molecule. The cleavage of CH bond in the given molecule is shown below.

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  9

Figure 6

Conclusion

The most stable radical that can result from cleavage of CH bond in the given molecule is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation: The most stable radical that can result from cleavage of CH bond in the given molecule is to be drawn.

Concept introduction: A free radical is an atom or ion with unpaired electrons. They are reactive intermediates formed by the homolysis of covalent bond. Free radicals are classified as 1°, 2° or 3° depending upon the number of alkyl groups attached to it. The number of alkyl substitutents increases, the stability of radical increases. This is due to the fact alkyl substituents donate their electrons to electron deficient radicals.

Expert Solution
Check Mark

Answer to Problem 15.2P

The most stable radical that can result from cleavage of CH bond in the given molecule is shown below.

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  10

Explanation of Solution

The given species is,

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  11

Figure 7

Two types of radicals can be formed from cleavage of CH bond in the given molecule. The cleavage of CH bond in the given molecule is shown below.

Organic Chemistry, Chapter 15, Problem 15.2P , additional homework tip  12

Figure 8

The number of alkyl substitutents increases, the stability of radical increases. The order of stability is 1°<2°<3°. The most stable radical is 3° radical.

Conclusion

The most stable radical that can result from cleavage of CH bond in the given molecule is shown in Figure 8.

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Chapter 15 Solutions

Organic Chemistry

Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Draw the products of each reaction. a. b. c. Ch. 15 - Draw all constitutional isomers formed when each...Ch. 15 - Draw the structure of the four allylic halides...Ch. 15 - Problem 15.20 Which compounds can be prepared in...Ch. 15 - Which CH bond is most readily cleaved in linolenic...Ch. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Problem 15.24 When adds to under radical...Ch. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Problem 15.27 Draw the steps of the mechanism that...Ch. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - 15.35 What is the major monobromination product...Ch. 15 - Prob. 15.36PCh. 15 - 15.37 What alkane is needed to make each alkyl...Ch. 15 - 15.38 Which alkyl halides can be prepared in good...Ch. 15 - Prob. 15.39PCh. 15 - 15.40 Explain why radical bromination of p-xylene...Ch. 15 - a. What product(s) (excluding stereoisomers) are...Ch. 15 - Prob. 15.42PCh. 15 - 15.43 Draw the products formed when each alkene is...Ch. 15 - 15.44 Draw all constitutional isomers formed when...Ch. 15 - 15.45 Draw the organic products formed in each...Ch. 15 - Prob. 15.46PCh. 15 - 15.47 Treatment of a hydrocarbon A (molecular...Ch. 15 - 15.48 Draw the products formed in each reaction...Ch. 15 - Prob. 15.49PCh. 15 - 15.50 Draw all the monochlorination products that...Ch. 15 - Prob. 15.51PCh. 15 - 15.52 (a) Draw the products (including...Ch. 15 - 15.53 Consider the following bromination: . a....Ch. 15 - 15.54 Draw a stepwise mechanism for the following...Ch. 15 - Prob. 15.55PCh. 15 - Prob. 15.56PCh. 15 - 15.57 Devise a synthesis of each compound from...Ch. 15 - Prob. 15.58PCh. 15 - Prob. 15.59PCh. 15 - 15.60 Devise a synthesis of each compound using ...Ch. 15 - Prob. 15.61PCh. 15 - Prob. 15.62PCh. 15 - 15.63 As described in Section 9.16, the...Ch. 15 - 15.64 Ethers are oxidized with to form...Ch. 15 - Prob. 15.65PCh. 15 - Prob. 15.66PCh. 15 - 15.67 In cells, vitamin C exists largely as its...Ch. 15 - What monomer is needed to form each...Ch. 15 - Prob. 15.69PCh. 15 - Prob. 15.70PCh. 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. 15.73PCh. 15 - 15.74 A and B, isomers of molecular formula , are...Ch. 15 - Prob. 15.75PCh. 15 - 15.76 Draw a stepwise mechanism for the...Ch. 15 - Prob. 15.77PCh. 15 - Prob. 15.78PCh. 15 - Prob. 15.79P
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