Organic Chemistry
Organic Chemistry
5th Edition
ISBN: 9780078021558
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
bartleby

Concept explainers

Question
Chapter 15, Problem 15.30P
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 15.30P

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

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 C

Conclusion

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

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 15.30P

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

Organic Chemistry, Chapter 15, Problem 15.30P , 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 15.30P , 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 15.30P

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 CH bond in 2methylbutane 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 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 15.30P

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 15.30P , 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°CHbond<2°CHbond<3°CHbond.

Conclusion

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
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 1°, 2°, or 3°. 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.
a.What carbon radical is formed by homolysis of the C–Ha bond in propylbenzene? Draw all reasonable resonance structures for this radical. b.What carbon radical is formed by homolysis of the C–Hb bond in propylbenzene? Draw all reasonable resonance structures for this radical. c. The bond dissociation energy of one of the C–H bonds is considerably less than the bond dissociation energy of the other. Which C–H bond is weaker? Offer an explanation.
Which C–H bond in each compound is most readily broken during radical halogenation?

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
Knowledge Booster
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
    • SEE MORE QUESTIONS
    Recommended textbooks for you
  • Organic Chemistry: A Guided Inquiry
    Chemistry
    ISBN:9780618974122
    Author:Andrei Straumanis
    Publisher:Cengage Learning
    Pushing Electrons
    Chemistry
    ISBN:9781133951889
    Author:Weeks, Daniel P.
    Publisher:Cengage Learning
  • Organic Chemistry: A Guided Inquiry
    Chemistry
    ISBN:9780618974122
    Author:Andrei Straumanis
    Publisher:Cengage Learning
    Pushing Electrons
    Chemistry
    ISBN:9781133951889
    Author:Weeks, Daniel P.
    Publisher:Cengage Learning