(a) Interpretation: Whether the carbocation formed by protonation of isoprene at carbon − 1 or the carbocation formed by protonation of isoprene at carbon − 4 is more stable is to be stated. Concept introduction: The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Question

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Answer 1

Question

Chapter 15, Problem 15.65AP

Interpretation Introduction

(a)

Interpretation:

Whether the carbocation formed by protonation of isoprene at carbon−1 or the carbocation formed by protonation of isoprene at carbon−4 is more stable is to be stated.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The carbocation which is formed by protonation of isoprene at carbon−1 is the more stable one.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 1

Figure 1

At the time of protonation at carbon −1, a tertiary carbocation is formed. This tertiary carbocation possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 2

Figure 2

At the time of protonation at carbon −4, a secondary carbocation is formed. This secondary carbocation also possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 3

Figure 3

The more stable carbocation is selected on the basis of inductive effect as both carbocation at carbon −1 and carbon −4 possess two resonance structures. In case of carbon −1, tertiary carbocation is stabilized by two methyl groups whereas, in case of carbon −4, secondary carbocation is stabilized by only one methyl group.

Therefore, tertiary carbocation formed at carbon −1 is more stable than the secondary carbocation formed at carbon −4.

Conclusion

The carbocation which is formed by protonation of isoprene at carbon−1 as compared to the position at carbon−4 is more stable.

Interpretation Introduction

(b)

Interpretation:

The products that are formed by the addition of one equivalent of HBr to isoprene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed by the addition of one equivalent of HBr to isoprene are

3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 4

Figure 1

An isoprene always forms the 1,2− addition and 1,4− addition products because isoprenes are conjugated dienes. The protonation of the given isoprene at carbon −1 position leads to the formation of more stable tertiary carbocation. So, the reaction protonation of the given isoprene followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 5

Figure 4

Therefore, the products formed by the conjugated diene, isoprene are 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Conclusion

The products, 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene are formed by by the addition of one equivalent of HBr to the given isoprene.

Interpretation Introduction

(c)

Interpretation:

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Explanation of Solution

In case of trans−1,3,5−hexatriene, three resonance structures are formed. These three resonance structures form the three different products on reaction with HBr.

The protonation of trans−1,3,5−hexatriene at carbon −2, carbon −4 and carbon −6 followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 6

Figure 5

The reaction of trans−1,3,5−hexatriene with HBr forms three different products due to 1,2− addition, 1,4− addition, and 1,6− addition. Therefore, the products formed by the conjugated diene, trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Conclusion

The products, (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene are formed by the addition of one equivalent of HBr to trans−1,3,5−hexatriene.

Interpretation Introduction

(d)

Interpretation:

The products which are formed in part (b) and (c) whether kinetically controlled products or thermodynamically controlled products are to be identified.

Concept introduction:

In the given conditions of the reaction, if the products of any reaction do not attain the equilibrium then the reaction is known as kinetically controlled reaction.

In the given conditions of the reaction, if the products of any reaction attain the equilibrium then the reaction is known as thermodynamically controlled reaction.

Expert Solution

Answer to Problem 15.65AP

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Explanation of Solution

The kinetically controlled products are formed much faster than thermodynamically controlled products. So, the 1,2− addition reaction is faster than 1,4− addition reaction and 1,6− addition reaction. Thus, 1,2− addition products are kinetically controlled product and 1,4− addition products are thermodynamically controlled products. The thermodynamically controlled products form slowly.

In part (b), the reaction of the given isoprene with HBr forms 3−bromo−3−methylbut−1−ene as a 1,2− addition product and 1−bromo−3−methylbut−2−ene as a 1,4− addition product. Therefore, 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 7

Figure 6

In part (c), the slowest addition is the 1,6− addition reaction. So, the reaction of one equivalent of HBr with trans−1,3,5−hexatriene forms (E)−5−bromohexa−1,3−diene as a 1,2− addition product, 3−bromohexa−1,5−diene as a 1,4− addition product and (2E,4E)−1−bromohexa−2,4−diene as a 1,6− addition product.

Therefore, (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 8

Figure 7

Conclusion

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

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Answer 2

Question

Chapter 15, Problem 15.65AP

Interpretation Introduction

(a)

Interpretation:

Whether the carbocation formed by protonation of isoprene at carbon−1 or the carbocation formed by protonation of isoprene at carbon−4 is more stable is to be stated.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The carbocation which is formed by protonation of isoprene at carbon−1 is the more stable one.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 1

Figure 1

At the time of protonation at carbon −1, a tertiary carbocation is formed. This tertiary carbocation possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 2

Figure 2

At the time of protonation at carbon −4, a secondary carbocation is formed. This secondary carbocation also possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 3

Figure 3

The more stable carbocation is selected on the basis of inductive effect as both carbocation at carbon −1 and carbon −4 possess two resonance structures. In case of carbon −1, tertiary carbocation is stabilized by two methyl groups whereas, in case of carbon −4, secondary carbocation is stabilized by only one methyl group.

Therefore, tertiary carbocation formed at carbon −1 is more stable than the secondary carbocation formed at carbon −4.

Conclusion

The carbocation which is formed by protonation of isoprene at carbon−1 as compared to the position at carbon−4 is more stable.

Interpretation Introduction

(b)

Interpretation:

The products that are formed by the addition of one equivalent of HBr to isoprene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed by the addition of one equivalent of HBr to isoprene are

3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 4

Figure 1

An isoprene always forms the 1,2− addition and 1,4− addition products because isoprenes are conjugated dienes. The protonation of the given isoprene at carbon −1 position leads to the formation of more stable tertiary carbocation. So, the reaction protonation of the given isoprene followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 5

Figure 4

Therefore, the products formed by the conjugated diene, isoprene are 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Conclusion

The products, 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene are formed by by the addition of one equivalent of HBr to the given isoprene.

Interpretation Introduction

(c)

Interpretation:

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Explanation of Solution

In case of trans−1,3,5−hexatriene, three resonance structures are formed. These three resonance structures form the three different products on reaction with HBr.

The protonation of trans−1,3,5−hexatriene at carbon −2, carbon −4 and carbon −6 followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 6

Figure 5

The reaction of trans−1,3,5−hexatriene with HBr forms three different products due to 1,2− addition, 1,4− addition, and 1,6− addition. Therefore, the products formed by the conjugated diene, trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Conclusion

The products, (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene are formed by the addition of one equivalent of HBr to trans−1,3,5−hexatriene.

Interpretation Introduction

(d)

Interpretation:

The products which are formed in part (b) and (c) whether kinetically controlled products or thermodynamically controlled products are to be identified.

Concept introduction:

In the given conditions of the reaction, if the products of any reaction do not attain the equilibrium then the reaction is known as kinetically controlled reaction.

In the given conditions of the reaction, if the products of any reaction attain the equilibrium then the reaction is known as thermodynamically controlled reaction.

Expert Solution

Answer to Problem 15.65AP

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Explanation of Solution

The kinetically controlled products are formed much faster than thermodynamically controlled products. So, the 1,2− addition reaction is faster than 1,4− addition reaction and 1,6− addition reaction. Thus, 1,2− addition products are kinetically controlled product and 1,4− addition products are thermodynamically controlled products. The thermodynamically controlled products form slowly.

In part (b), the reaction of the given isoprene with HBr forms 3−bromo−3−methylbut−1−ene as a 1,2− addition product and 1−bromo−3−methylbut−2−ene as a 1,4− addition product. Therefore, 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 7

Figure 6

In part (c), the slowest addition is the 1,6− addition reaction. So, the reaction of one equivalent of HBr with trans−1,3,5−hexatriene forms (E)−5−bromohexa−1,3−diene as a 1,2− addition product, 3−bromohexa−1,5−diene as a 1,4− addition product and (2E,4E)−1−bromohexa−2,4−diene as a 1,6− addition product.

Therefore, (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 8

Figure 7

Conclusion

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

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Answer 3

Question

Chapter 15, Problem 15.65AP

Interpretation Introduction

(a)

Interpretation:

Whether the carbocation formed by protonation of isoprene at carbon−1 or the carbocation formed by protonation of isoprene at carbon−4 is more stable is to be stated.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The carbocation which is formed by protonation of isoprene at carbon−1 is the more stable one.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 1

Figure 1

At the time of protonation at carbon −1, a tertiary carbocation is formed. This tertiary carbocation possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 2

Figure 2

At the time of protonation at carbon −4, a secondary carbocation is formed. This secondary carbocation also possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 3

Figure 3

The more stable carbocation is selected on the basis of inductive effect as both carbocation at carbon −1 and carbon −4 possess two resonance structures. In case of carbon −1, tertiary carbocation is stabilized by two methyl groups whereas, in case of carbon −4, secondary carbocation is stabilized by only one methyl group.

Therefore, tertiary carbocation formed at carbon −1 is more stable than the secondary carbocation formed at carbon −4.

Conclusion

The carbocation which is formed by protonation of isoprene at carbon−1 as compared to the position at carbon−4 is more stable.

Interpretation Introduction

(b)

Interpretation:

The products that are formed by the addition of one equivalent of HBr to isoprene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed by the addition of one equivalent of HBr to isoprene are

3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 4

Figure 1

An isoprene always forms the 1,2− addition and 1,4− addition products because isoprenes are conjugated dienes. The protonation of the given isoprene at carbon −1 position leads to the formation of more stable tertiary carbocation. So, the reaction protonation of the given isoprene followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 5

Figure 4

Therefore, the products formed by the conjugated diene, isoprene are 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Conclusion

The products, 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene are formed by by the addition of one equivalent of HBr to the given isoprene.

Interpretation Introduction

(c)

Interpretation:

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Explanation of Solution

In case of trans−1,3,5−hexatriene, three resonance structures are formed. These three resonance structures form the three different products on reaction with HBr.

The protonation of trans−1,3,5−hexatriene at carbon −2, carbon −4 and carbon −6 followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 6

Figure 5

The reaction of trans−1,3,5−hexatriene with HBr forms three different products due to 1,2− addition, 1,4− addition, and 1,6− addition. Therefore, the products formed by the conjugated diene, trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Conclusion

The products, (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene are formed by the addition of one equivalent of HBr to trans−1,3,5−hexatriene.

Interpretation Introduction

(d)

Interpretation:

The products which are formed in part (b) and (c) whether kinetically controlled products or thermodynamically controlled products are to be identified.

Concept introduction:

In the given conditions of the reaction, if the products of any reaction do not attain the equilibrium then the reaction is known as kinetically controlled reaction.

In the given conditions of the reaction, if the products of any reaction attain the equilibrium then the reaction is known as thermodynamically controlled reaction.

Expert Solution

Answer to Problem 15.65AP

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Explanation of Solution

The kinetically controlled products are formed much faster than thermodynamically controlled products. So, the 1,2− addition reaction is faster than 1,4− addition reaction and 1,6− addition reaction. Thus, 1,2− addition products are kinetically controlled product and 1,4− addition products are thermodynamically controlled products. The thermodynamically controlled products form slowly.

In part (b), the reaction of the given isoprene with HBr forms 3−bromo−3−methylbut−1−ene as a 1,2− addition product and 1−bromo−3−methylbut−2−ene as a 1,4− addition product. Therefore, 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 7

Figure 6

In part (c), the slowest addition is the 1,6− addition reaction. So, the reaction of one equivalent of HBr with trans−1,3,5−hexatriene forms (E)−5−bromohexa−1,3−diene as a 1,2− addition product, 3−bromohexa−1,5−diene as a 1,4− addition product and (2E,4E)−1−bromohexa−2,4−diene as a 1,6− addition product.

Therefore, (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 8

Figure 7

Conclusion

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

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Answer 4

Question

Chapter 15, Problem 15.65AP

Interpretation Introduction

(a)

Interpretation:

Whether the carbocation formed by protonation of isoprene at carbon−1 or the carbocation formed by protonation of isoprene at carbon−4 is more stable is to be stated.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The carbocation which is formed by protonation of isoprene at carbon−1 is the more stable one.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 1

Figure 1

At the time of protonation at carbon −1, a tertiary carbocation is formed. This tertiary carbocation possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 2

Figure 2

At the time of protonation at carbon −4, a secondary carbocation is formed. This secondary carbocation also possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 3

Figure 3

The more stable carbocation is selected on the basis of inductive effect as both carbocation at carbon −1 and carbon −4 possess two resonance structures. In case of carbon −1, tertiary carbocation is stabilized by two methyl groups whereas, in case of carbon −4, secondary carbocation is stabilized by only one methyl group.

Therefore, tertiary carbocation formed at carbon −1 is more stable than the secondary carbocation formed at carbon −4.

Conclusion

The carbocation which is formed by protonation of isoprene at carbon−1 as compared to the position at carbon−4 is more stable.

Interpretation Introduction

(b)

Interpretation:

The products that are formed by the addition of one equivalent of HBr to isoprene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed by the addition of one equivalent of HBr to isoprene are

3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 4

Figure 1

An isoprene always forms the 1,2− addition and 1,4− addition products because isoprenes are conjugated dienes. The protonation of the given isoprene at carbon −1 position leads to the formation of more stable tertiary carbocation. So, the reaction protonation of the given isoprene followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 5

Figure 4

Therefore, the products formed by the conjugated diene, isoprene are 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Conclusion

The products, 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene are formed by by the addition of one equivalent of HBr to the given isoprene.

Interpretation Introduction

(c)

Interpretation:

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Explanation of Solution

In case of trans−1,3,5−hexatriene, three resonance structures are formed. These three resonance structures form the three different products on reaction with HBr.

The protonation of trans−1,3,5−hexatriene at carbon −2, carbon −4 and carbon −6 followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 6

Figure 5

The reaction of trans−1,3,5−hexatriene with HBr forms three different products due to 1,2− addition, 1,4− addition, and 1,6− addition. Therefore, the products formed by the conjugated diene, trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Conclusion

The products, (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene are formed by the addition of one equivalent of HBr to trans−1,3,5−hexatriene.

Interpretation Introduction

(d)

Interpretation:

The products which are formed in part (b) and (c) whether kinetically controlled products or thermodynamically controlled products are to be identified.

Concept introduction:

In the given conditions of the reaction, if the products of any reaction do not attain the equilibrium then the reaction is known as kinetically controlled reaction.

In the given conditions of the reaction, if the products of any reaction attain the equilibrium then the reaction is known as thermodynamically controlled reaction.

Expert Solution

Answer to Problem 15.65AP

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Explanation of Solution

The kinetically controlled products are formed much faster than thermodynamically controlled products. So, the 1,2− addition reaction is faster than 1,4− addition reaction and 1,6− addition reaction. Thus, 1,2− addition products are kinetically controlled product and 1,4− addition products are thermodynamically controlled products. The thermodynamically controlled products form slowly.

In part (b), the reaction of the given isoprene with HBr forms 3−bromo−3−methylbut−1−ene as a 1,2− addition product and 1−bromo−3−methylbut−2−ene as a 1,4− addition product. Therefore, 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 7

Figure 6

In part (c), the slowest addition is the 1,6− addition reaction. So, the reaction of one equivalent of HBr with trans−1,3,5−hexatriene forms (E)−5−bromohexa−1,3−diene as a 1,2− addition product, 3−bromohexa−1,5−diene as a 1,4− addition product and (2E,4E)−1−bromohexa−2,4−diene as a 1,6− addition product.

Therefore, (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 8

Figure 7

Conclusion

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

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Answer 5

Chapter 15, Problem 15.65AP

Interpretation Introduction

(a)

Interpretation:

Whether the carbocation formed by protonation of isoprene at carbon−1 or the carbocation formed by protonation of isoprene at carbon−4 is more stable is to be stated.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The carbocation which is formed by protonation of isoprene at carbon−1 is the more stable one.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 1

Figure 1

At the time of protonation at carbon −1, a tertiary carbocation is formed. This tertiary carbocation possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 2

Figure 2

At the time of protonation at carbon −4, a secondary carbocation is formed. This secondary carbocation also possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 3

Figure 3

The more stable carbocation is selected on the basis of inductive effect as both carbocation at carbon −1 and carbon −4 possess two resonance structures. In case of carbon −1, tertiary carbocation is stabilized by two methyl groups whereas, in case of carbon −4, secondary carbocation is stabilized by only one methyl group.

Therefore, tertiary carbocation formed at carbon −1 is more stable than the secondary carbocation formed at carbon −4.

Conclusion

The carbocation which is formed by protonation of isoprene at carbon−1 as compared to the position at carbon−4 is more stable.

Interpretation Introduction

(b)

Interpretation:

The products that are formed by the addition of one equivalent of HBr to isoprene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed by the addition of one equivalent of HBr to isoprene are

3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 4

Figure 1

An isoprene always forms the 1,2− addition and 1,4− addition products because isoprenes are conjugated dienes. The protonation of the given isoprene at carbon −1 position leads to the formation of more stable tertiary carbocation. So, the reaction protonation of the given isoprene followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 5

Figure 4

Therefore, the products formed by the conjugated diene, isoprene are 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Conclusion

The products, 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene are formed by by the addition of one equivalent of HBr to the given isoprene.

Interpretation Introduction

(c)

Interpretation:

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Explanation of Solution

In case of trans−1,3,5−hexatriene, three resonance structures are formed. These three resonance structures form the three different products on reaction with HBr.

The protonation of trans−1,3,5−hexatriene at carbon −2, carbon −4 and carbon −6 followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 6

Figure 5

The reaction of trans−1,3,5−hexatriene with HBr forms three different products due to 1,2− addition, 1,4− addition, and 1,6− addition. Therefore, the products formed by the conjugated diene, trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Conclusion

The products, (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene are formed by the addition of one equivalent of HBr to trans−1,3,5−hexatriene.

Interpretation Introduction

(d)

Interpretation:

The products which are formed in part (b) and (c) whether kinetically controlled products or thermodynamically controlled products are to be identified.

Concept introduction:

In the given conditions of the reaction, if the products of any reaction do not attain the equilibrium then the reaction is known as kinetically controlled reaction.

In the given conditions of the reaction, if the products of any reaction attain the equilibrium then the reaction is known as thermodynamically controlled reaction.

Expert Solution

Answer to Problem 15.65AP

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Explanation of Solution

The kinetically controlled products are formed much faster than thermodynamically controlled products. So, the 1,2− addition reaction is faster than 1,4− addition reaction and 1,6− addition reaction. Thus, 1,2− addition products are kinetically controlled product and 1,4− addition products are thermodynamically controlled products. The thermodynamically controlled products form slowly.

In part (b), the reaction of the given isoprene with HBr forms 3−bromo−3−methylbut−1−ene as a 1,2− addition product and 1−bromo−3−methylbut−2−ene as a 1,4− addition product. Therefore, 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 7

Figure 6

In part (c), the slowest addition is the 1,6− addition reaction. So, the reaction of one equivalent of HBr with trans−1,3,5−hexatriene forms (E)−5−bromohexa−1,3−diene as a 1,2− addition product, 3−bromohexa−1,5−diene as a 1,4− addition product and (2E,4E)−1−bromohexa−2,4−diene as a 1,6− addition product.

Therefore, (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 8

Figure 7

Conclusion

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Answer 6

Chapter 15, Problem 15.65AP

Interpretation Introduction

(a)

Interpretation:

Whether the carbocation formed by protonation of isoprene at carbon−1 or the carbocation formed by protonation of isoprene at carbon−4 is more stable is to be stated.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The carbocation which is formed by protonation of isoprene at carbon−1 is the more stable one.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 1

Figure 1

At the time of protonation at carbon −1, a tertiary carbocation is formed. This tertiary carbocation possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 2

Figure 2

At the time of protonation at carbon −4, a secondary carbocation is formed. This secondary carbocation also possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 3

Figure 3

The more stable carbocation is selected on the basis of inductive effect as both carbocation at carbon −1 and carbon −4 possess two resonance structures. In case of carbon −1, tertiary carbocation is stabilized by two methyl groups whereas, in case of carbon −4, secondary carbocation is stabilized by only one methyl group.

Therefore, tertiary carbocation formed at carbon −1 is more stable than the secondary carbocation formed at carbon −4.

Conclusion

The carbocation which is formed by protonation of isoprene at carbon−1 as compared to the position at carbon−4 is more stable.

Answer 7

Chapter 15, Problem 15.65AP

Interpretation Introduction

(a)

Interpretation:

Whether the carbocation formed by protonation of isoprene at carbon−1 or the carbocation formed by protonation of isoprene at carbon−4 is more stable is to be stated.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Answer 8

Expert Solution

Answer to Problem 15.65AP

The carbocation which is formed by protonation of isoprene at carbon−1 is the more stable one.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 1

Figure 1

At the time of protonation at carbon −1, a tertiary carbocation is formed. This tertiary carbocation possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 2

Figure 2

At the time of protonation at carbon −4, a secondary carbocation is formed. This secondary carbocation also possesses two resonance structures due which it is the resonance stabilized as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 3

Figure 3

The more stable carbocation is selected on the basis of inductive effect as both carbocation at carbon −1 and carbon −4 possess two resonance structures. In case of carbon −1, tertiary carbocation is stabilized by two methyl groups whereas, in case of carbon −4, secondary carbocation is stabilized by only one methyl group.

Therefore, tertiary carbocation formed at carbon −1 is more stable than the secondary carbocation formed at carbon −4.

Answer 13

Conclusion

The carbocation which is formed by protonation of isoprene at carbon−1 as compared to the position at carbon−4 is more stable.

Answer 14

Interpretation Introduction

(b)

Interpretation:

The products that are formed by the addition of one equivalent of HBr to isoprene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed by the addition of one equivalent of HBr to isoprene are

3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 4

Figure 1

An isoprene always forms the 1,2− addition and 1,4− addition products because isoprenes are conjugated dienes. The protonation of the given isoprene at carbon −1 position leads to the formation of more stable tertiary carbocation. So, the reaction protonation of the given isoprene followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 5

Figure 4

Therefore, the products formed by the conjugated diene, isoprene are 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Conclusion

The products, 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene are formed by by the addition of one equivalent of HBr to the given isoprene.

Answer 15

Interpretation Introduction

(b)

Interpretation:

The products that are formed by the addition of one equivalent of HBr to isoprene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Answer 16

Expert Solution

Answer to Problem 15.65AP

The products that are formed by the addition of one equivalent of HBr to isoprene are

3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The given structure of isoprene is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 4

Figure 1

An isoprene always forms the 1,2− addition and 1,4− addition products because isoprenes are conjugated dienes. The protonation of the given isoprene at carbon −1 position leads to the formation of more stable tertiary carbocation. So, the reaction protonation of the given isoprene followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 5

Figure 4

Therefore, the products formed by the conjugated diene, isoprene are 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene.

Answer 21

Conclusion

The products, 3−bromo−3−methylbut−1−ene and 1−bromo−3−methylbut−2−ene are formed by by the addition of one equivalent of HBr to the given isoprene.

Answer 22

Interpretation Introduction

(c)

Interpretation:

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Expert Solution

Answer to Problem 15.65AP

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Explanation of Solution

In case of trans−1,3,5−hexatriene, three resonance structures are formed. These three resonance structures form the three different products on reaction with HBr.

The protonation of trans−1,3,5−hexatriene at carbon −2, carbon −4 and carbon −6 followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 6

Figure 5

The reaction of trans−1,3,5−hexatriene with HBr forms three different products due to 1,2− addition, 1,4− addition, and 1,6− addition. Therefore, the products formed by the conjugated diene, trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Conclusion

The products, (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene are formed by the addition of one equivalent of HBr to trans−1,3,5−hexatriene.

Answer 23

Interpretation Introduction

(c)

Interpretation:

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are to be stated. The reason for the formation of the corresponding products is to be explained.

Concept introduction:

The delocalization of electrons results in the formation of resonance structure. The curved-arrow notation traces the flow of the electrons in a compound. This notation is used to derive the resonance structure.

Answer 24

Expert Solution

Answer to Problem 15.65AP

The products that are formed the addition of one equivalent of HBr to trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

In case of trans−1,3,5−hexatriene, three resonance structures are formed. These three resonance structures form the three different products on reaction with HBr.

The protonation of trans−1,3,5−hexatriene at carbon −2, carbon −4 and carbon −6 followed by its reaction with HBr is shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 6

Figure 5

The reaction of trans−1,3,5−hexatriene with HBr forms three different products due to 1,2− addition, 1,4− addition, and 1,6− addition. Therefore, the products formed by the conjugated diene, trans−1,3,5−hexatriene are (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene.

Answer 29

Conclusion

The products, (E)−5−bromohexa−1,3−diene, 3−bromohexa−1,5−diene and (2E,4E)−1−bromohexa−2,4−diene are formed by the addition of one equivalent of HBr to trans−1,3,5−hexatriene.

Answer 30

Interpretation Introduction

(d)

Interpretation:

The products which are formed in part (b) and (c) whether kinetically controlled products or thermodynamically controlled products are to be identified.

Concept introduction:

In the given conditions of the reaction, if the products of any reaction do not attain the equilibrium then the reaction is known as kinetically controlled reaction.

In the given conditions of the reaction, if the products of any reaction attain the equilibrium then the reaction is known as thermodynamically controlled reaction.

Expert Solution

Answer to Problem 15.65AP

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Explanation of Solution

The kinetically controlled products are formed much faster than thermodynamically controlled products. So, the 1,2− addition reaction is faster than 1,4− addition reaction and 1,6− addition reaction. Thus, 1,2− addition products are kinetically controlled product and 1,4− addition products are thermodynamically controlled products. The thermodynamically controlled products form slowly.

In part (b), the reaction of the given isoprene with HBr forms 3−bromo−3−methylbut−1−ene as a 1,2− addition product and 1−bromo−3−methylbut−2−ene as a 1,4− addition product. Therefore, 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 7

Figure 6

In part (c), the slowest addition is the 1,6− addition reaction. So, the reaction of one equivalent of HBr with trans−1,3,5−hexatriene forms (E)−5−bromohexa−1,3−diene as a 1,2− addition product, 3−bromohexa−1,5−diene as a 1,4− addition product and (2E,4E)−1−bromohexa−2,4−diene as a 1,6− addition product.

Therefore, (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 8

Figure 7

Conclusion

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Answer 31

Interpretation Introduction

(d)

Interpretation:

The products which are formed in part (b) and (c) whether kinetically controlled products or thermodynamically controlled products are to be identified.

Concept introduction:

In the given conditions of the reaction, if the products of any reaction do not attain the equilibrium then the reaction is known as kinetically controlled reaction.

In the given conditions of the reaction, if the products of any reaction attain the equilibrium then the reaction is known as thermodynamically controlled reaction.

Answer 32

Expert Solution

Answer to Problem 15.65AP

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

The kinetically controlled products are formed much faster than thermodynamically controlled products. So, the 1,2− addition reaction is faster than 1,4− addition reaction and 1,6− addition reaction. Thus, 1,2− addition products are kinetically controlled product and 1,4− addition products are thermodynamically controlled products. The thermodynamically controlled products form slowly.

In part (b), the reaction of the given isoprene with HBr forms 3−bromo−3−methylbut−1−ene as a 1,2− addition product and 1−bromo−3−methylbut−2−ene as a 1,4− addition product. Therefore, 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 7

Figure 6

In part (c), the slowest addition is the 1,6− addition reaction. So, the reaction of one equivalent of HBr with trans−1,3,5−hexatriene forms (E)−5−bromohexa−1,3−diene as a 1,2− addition product, 3−bromohexa−1,5−diene as a 1,4− addition product and (2E,4E)−1−bromohexa−2,4−diene as a 1,6− addition product.

Therefore, (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product as shown below.

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 15, Problem 15.65AP , additional homework tip 8

Figure 7

Answer 37

Conclusion

In part (b), 3−bromo−3−methylbut−1−ene is a kinetically controlled product and 1−bromo−3−methylbut−2−ene is a thermodynamically controlled product.

In part (c), (E)−5−bromohexa−1,3−diene is a kinetically controlled product and (2E,4E)−1−bromohexa−2,4−diene is a thermodynamically controlled product.

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 - Prob. 15.6P Ch. 15 - Prob. 15.7P Ch. 15 - Prob. 15.8P Ch. 15 - Prob. 15.9P Ch. 15 - Prob. 15.10P

Answer to Problem 15.65AP

Explanation of Solution

Answer to Problem 15.65AP

Explanation of Solution

Answer to Problem 15.65AP

Explanation of Solution

Answer to Problem 15.65AP

Explanation of Solution

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