Chapter 15, Problem 15.65AP

Interpretation Introduction

(a)

Interpretation:

Whether the carbocation formed by protonation of isoprene at $\text{carbon}-1$ or the carbocation formed by protonation of isoprene at $\text{carbon}-4$ 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 to Problem 15.65AP

The carbocation which is formed by protonation of isoprene at $\text{carbon}-1$ is the more stable one.

Explanation of Solution

The given structure of isoprene is shown below.

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.

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.

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 $\text{carbon}-1$ as compared to the position at $\text{carbon}-4$ is more stable.

Interpretation Introduction

(b)

Interpretation:

The products that are formed by the addition of one equivalent of $\text{HBr}$ 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 to Problem 15.65AP

The products that are formed by the addition of one equivalent of $\text{HBr}$ to isoprene are

$3-\text{bromo}-3-\text{methylbut}-1-\text{ene}$ and $1-\text{bromo}-3-\text{methylbut}-2-\text{ene}$.

Explanation of Solution

The given structure of isoprene is shown below.

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 $\text{HBr}$ is shown below.

Figure 4

Therefore, the products formed by the conjugated diene, isoprene are $3-\text{bromo}-3-\text{methylbut}-1-\text{ene}$ and $1-\text{bromo}-3-\text{methylbut}-2-\text{ene}$.

Conclusion

The products, $3-\text{bromo}-3-\text{methylbut}-1-\text{ene}$ and $1-\text{bromo}-3-\text{methylbut}-2-\text{ene}$ are formed by by the addition of one equivalent of $\text{HBr}$ to the given isoprene.

Interpretation Introduction

(c)

Interpretation:

The products that are formed the addition of one equivalent of $\text{HBr}$ to $trans-1,3,5-\text{hexatriene}$ 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 to Problem 15.65AP

The products that are formed the addition of one equivalent of $\text{HBr}$ to $trans-1,3,5-\text{hexatriene}$ are $\left(E\right)-5-\text{bromohexa}-1,3-\text{diene}$, $3-\text{bromohexa}-1,5-\text{diene}$ and $\left(2E,4E\right)-1-\text{bromohexa}-2,4-\text{diene}$.

Explanation of Solution

In case of $trans-1,3,5-\text{hexatriene}$, three resonance structures are formed. These three resonance structures form the three different products on reaction with $\text{HBr}$.

The protonation of $trans-1,3,5-\text{hexatriene}$ at carbon $-2$, carbon $-4$ and carbon $-6$ followed by its reaction with $\text{HBr}$ is shown below.

Figure 5

The reaction of $trans-1,3,5-\text{hexatriene}$ with $\text{HBr}$ forms three different products due to $1,2-$ addition, $1,4-$ addition, and $1,6-$ addition. Therefore, the products formed by the conjugated diene, $trans-1,3,5-\text{hexatriene}$ are $\left(E\right)-5-\text{bromohexa}-1,3-\text{diene}$, $3-\text{bromohexa}-1,5-\text{diene}$ and $\left(2E,4E\right)-1-\text{bromohexa}-2,4-\text{diene}$.

Conclusion

The products, $\left(E\right)-5-\text{bromohexa}-1,3-\text{diene}$, $3-\text{bromohexa}-1,5-\text{diene}$ and $\left(2E,4E\right)-1-\text{bromohexa}-2,4-\text{diene}$ are formed by the addition of one equivalent of $\text{HBr}$ to $trans-1,3,5-\text{hexatriene}$.

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.

Answer to Problem 15.65AP

In part (b), $3-\text{bromo}-3-\text{methylbut}-1-\text{ene}$ is a kinetically controlled product and $1-\text{bromo}-3-\text{methylbut}-2-\text{ene}$ is a thermodynamically controlled product.

In part (c), $\left(E\right)-5-\text{bromohexa}-1,3-\text{diene}$ is a kinetically controlled product and $\left(2E,4E\right)-1-\text{bromohexa}-2,4-\text{diene}$ 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 $\text{HBr}$ forms $3-\text{bromo}-3-\text{methylbut}-1-\text{ene}$ as a $1,2-$ addition product and $1-\text{bromo}-3-\text{methylbut}-2-\text{ene}$ as a $1,4-$ addition product. Therefore, $3-\text{bromo}-3-\text{methylbut}-1-\text{ene}$ is a kinetically controlled product and $1-\text{bromo}-3-\text{methylbut}-2-\text{ene}$ is a thermodynamically controlled product as shown below.

Figure 6

In part (c), the slowest addition is the $1,6-$ addition reaction. So, the reaction of one equivalent of $\text{HBr}$ with $trans-1,3,5-\text{hexatriene}$ forms $\left(E\right)-5-\text{bromohexa}-1,3-\text{diene}$ as a $1,2-$ addition product, $3-\text{bromohexa}-1,5-\text{diene}$ as a $1,4-$ addition product and $\left(2E,4E\right)-1-\text{bromohexa}-2,4-\text{diene}$ as a $1,6-$ addition product.

Therefore, $\left(E\right)-5-\text{bromohexa}-1,3-\text{diene}$ is a kinetically controlled product and $\left(2E,4E\right)-1-\text{bromohexa}-2,4-\text{diene}$ is a thermodynamically controlled product as shown below.

Figure 7

Conclusion

In part (b), $3-\text{bromo}-3-\text{methylbut}-1-\text{ene}$ is a kinetically controlled product and $1-\text{bromo}-3-\text{methylbut}-2-\text{ene}$ is a thermodynamically controlled product.

In part (c), $\left(E\right)-5-\text{bromohexa}-1,3-\text{diene}$ is a kinetically controlled product and $\left(2E,4E\right)-1-\text{bromohexa}-2,4-\text{diene}$ is a thermodynamically controlled product.