Organic Chemistry
Organic Chemistry
4th Edition
ISBN: 9780073402772
Author: Janice G. Smith
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 15, Problem 15.55P

Consider the following bromination: ( CH 3 ) 3 CH + Br 2 Δ ( CH 3 ) 3 CBr + HBr .

a. Calculate Δ H ο for this reaction by using the bond dissociation energies in Table 6.2.

b. Draw out a stepwise mechanism for the reaction, including the initiation, propagation, and termination steps.

c. Calculate Δ H ο for each propagation step.

d. Draw an energy diagram for the propagation steps.

e. Draw the structure of the transition state of each propagation step.

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation: The value of ΔHο for the given reaction by the use of bond dissociation energies is to be calculated.

Concept introduction: The value of ΔHο is calculated by the formula,

ΔHο=BDE(B.B.)BDE(B.F.)

Answer to Problem 15.55P

The value of ΔHο for the given reaction by the use of bond dissociation energies is 16kcal/mol.

Explanation of Solution

The given bromination reaction is,

(CH3)3CH+Br2Δ(CH3)3CBr+HBr

In the given reaction, BrBr and CH bond is broken and CBr and HBr bond is formed.

The value of ΔHο is calculated by the formula,

ΔHο=BDE(B.B.)BDE(B.F.)

Where,

BDE(B.B.) is the sum of bond dissociation energy of bond broken.

BDE(B.F.) is the sum of bond dissociation energy of bond formed.

The bond dissociation energy of (CH3)3CH bond is 91kcal/mol.

The bond dissociation energy of (CH3)3CBr bond is 65kcal/mol.

The bond dissociation energy of BrBr bond is 46kcal/mol.

The bond dissociation energy of HBr bond is 88kcal/mol.

Substitute the bond dissociation energy of (CH3)3CH, (CH3)3CBr, BrBr and HBr in the above formula.

ΔHο=(91kcal/mol+46kcal/mol)(65kcal/mol+88kcal/mol)=16kcal/mol

Therefore, the value of ΔHο for the given reaction by the use of bond dissociation energies is 16kcal/mol.

Conclusion

The value of ΔHο for the given reaction by the use of bond dissociation energies is 16kcal/mol.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation: The stepwise mechanism for the given reaction which includes initiation, propagation and termination steps is to be drawn.

Concept introduction: The general steps followed by free-radical reaction are stated below:

1. First step is initiation that involves formation of radical.

2. Second step is propagation.

3. Third step is the termination that involves the formation of stable bond.

Answer to Problem 15.55P

The stepwise mechanism for the given reaction which includes initiation, propagation and termination steps is,

Organic Chemistry, Chapter 15, Problem 15.55P , additional homework tip  1

Figure 1

Explanation of Solution

The initiation step involves the generation of bromine radical by hemolytic cleavage of Br2. In the propagation step, bromine radical on reaction with (CH3)3CH generates tertiary radical which on reaction with Br2 converts into (CH3)3CBr. In the termination step, all free radicals combine to form stable molecules. The stepwise mechanism for the given reaction which includes initiation, propagation and termination steps is shown below.

Organic Chemistry, Chapter 15, Problem 15.55P , additional homework tip  2

Figure 1

Conclusion

The stepwise mechanism for the given reaction which includes initiation, propagation and termination steps is shown in Figure 1.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation: The value of ΔHο for each propagation step is to be calculated.

Concept introduction: The general steps followed by free-radical reaction are stated below:

1. First step is initiation that involves formation of radical.

2. Second step is propagation.

3. Third step is the termination that involves the formation of stable bond.

Answer to Problem 15.55P

The value of ΔHο for each propagation step is 3kcal/mol and 19kcal/mol.

Explanation of Solution

The propagation step of given reaction involves two steps. In the first step (CH3)3CH bond breaks and HBr bond is formed. In the second step, BrBr bond breaks and (CH3)3CBr bond is formed.

The value of ΔHο is calculated by the formula,

ΔHο=BDE(B.B.)BDE(B.F.)

Where,

BDE(B.B.) is the sum of bond dissociation energy of bond broken.

BDE(B.F.) is the sum of bond dissociation energy of bond formed.

The bond dissociation energy of (CH3)3CH bond is 91kcal/mol.

The bond dissociation energy of (CH3)3CBr bond is 65kcal/mol.

The bond dissociation energy of BrBr bond is 46kcal/mol.

The bond dissociation energy of HBr bond is 88kcal/mol.

The value of ΔHο for first propagation step is,

ΔHο=91kcal/mol88kcal/mol=3kcal/mol

The value of ΔHο for second propagation step is,

ΔHο=46kcal/mol65kcal/mol=19kcal/mol

Conclusion

The value of ΔHο for each propagation step is 3kcal/mol and 19kcal/mol.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation: The energy level diagram for the propagation steps is to be drawn.

Concept introduction: The graphical representation of chemical reaction in which x-axis represents energy of the reaction and y-axis represents the reaction coordinate is called energy profile diagram.

Answer to Problem 15.55P

The energy level diagram for the propagation steps is,

Organic Chemistry, Chapter 15, Problem 15.55P , additional homework tip  3

Figure 2

Explanation of Solution

The energy level diagram for the propagation steps is shown in Figure 2.

Organic Chemistry, Chapter 15, Problem 15.55P , additional homework tip  4

Figure 2

Conclusion

The energy level diagram for the propagation steps is shown in Figure 2.

Expert Solution
Check Mark
Interpretation Introduction

(e)

Interpretation: The structure of transition state of each propagation step is to be drawn.

Concept introduction: The general steps followed by free-radical reaction are stated below:

1. First step is initiation that involves formation of radical.

2. Second step is propagation.

3. Third step is the termination that involves the formation of stable bond.

Answer to Problem 15.55P

The structure of transition state of first and second propagation step is,

Organic Chemistry, Chapter 15, Problem 15.55P , additional homework tip  5

Figure 3

Explanation of Solution

The propagation step of given reaction involves two steps. In the first step (CH3)3CH bond breaks and HBr bond is formed. In the second step, BrBr bond breaks and (CH3)3CBr bond is formed.

The transition state of the first propagation step is,

Organic Chemistry, Chapter 15, Problem 15.55P , additional homework tip  6

Figure 4

The transition state of the first propagation step is,

Organic Chemistry, Chapter 15, Problem 15.55P , additional homework tip  7

Figure 5

Conclusion

The structure of transition state of first and second propagation step is shown in Figure 4 and Figure 5, respectively.

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
3. Why did breaking the P-C single bond lead to the formation of 2 species containing a free radical (ie. where did those 2 electrons come from)? 4. Do you think the photoinitiator 2-hydroxy-2-methylpropiophenone would require a shorter or longer wavelength of light to create a free radical? Explain your answer using the table provided in question #2 (be specific). The structure of 2-hydroxy-2- methylpropiophenone is shown below. CH3 Figure 5: Structure of 2-hydroxy-2- methylpropiophenone. The wavy line shows OH the bond that is photo-sensitive (ie. the bond that will be broken by light). (Taken from Tehfe et al. 2013, http://www.mdpi.com/2076- 3417/3/2/490/htm) ČH3 5. Why is the PEGDA monomer soluble in water, but the polymer is not? In other words, why do you think forming a large molecule would lead to the formation of a solid? (Hint: think about the movement of molecules in a liquid vs solid phase). 6. Thinking about the octet rule, why do you think many free radicals are very…
#9
Draw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. • • You do not have to consider stereochemistry. • Include all valence lone pairs in your answer. Br2 Br In cases where there is more than one answer, just draw one. + √n [ ? N ZI

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
Knowledge Booster
Background pattern image
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
Text book image
Organic Chemistry: A Guided Inquiry
Chemistry
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Cengage Learning
Enzymes - Effect of cofactors on enzyme; Author: Tutorials Point (India) Ltd;https://www.youtube.com/watch?v=AkAbIwxyUs4;License: Standard YouTube License, CC-BY
Enzyme Catalysis Part-I; Author: NPTEL-NOC IITM;https://www.youtube.com/watch?v=aZE740JWZuQ;License: Standard Youtube License