Chapter 15, Problem 15.24P

Draw the product(s) formed when each alkene is treated with either $\left[\text{1}\right]\text{HBr}$ alone; or $\left[2\right]\text{HBr}$ in the presence of peroxides.

Interpretation Introduction

(a)

Interpretation:

The product(s) formed by the reaction of given alkene with $\left[1\right]\text{HBr}$; or $\left[\text{2}\right]\text{HBr}$ in the presence of peroxides is to be drawn.

Concept introduction:

The reaction of hydrogen halide with alkene results in the formation of alkyl halide. This type of reaction is an electrophilic addition of hydrogen halide. Electrophilic addition reactions are those in which breaking of pi bond take place to form new sigma bond.

Answer to Problem 15.24P

The product formed by the reaction of given alkene with $\left[1\right]\text{HBr}$ is shown below.

The product formed by the reaction of given alkene with $\text{HBr}$ in the presence of peroxide is shown below.

Explanation of Solution

Electrophilic addition reaction follows Markovnikov rule. According to Markovnikov rule, the positive part of halogen acid attached to that carbon atom in $\text{C=C}$ bond which carries higher number of hydrogen atoms and the negative part of halogen acid will attach to that carbon atom in $\text{C=C}$ bond which has lesser number of hydrogen atoms.

The given alkene is shown below.

Figure 1

The steps followed by electrophilic addition reaction are stated below:

➢ First protonation of the alkene take place to generate the carbocation.

➢ The halide ion will attack on the carbocation to give the final product.

The product formed by the reaction of given alkene with $\text{HBr}$ is shown below.

Figure 2

The product formed by the reaction of given alkene with $\text{HBr}$ is $2-\text{bromohexane}$.

The addition of $\text{HBr}$ in absence of light or peroxide occurs via carbocation intermediate, whereas addition of $\text{HBr}$ in presence of light or peroxide occurs via radical intermediate.

The addition of $\text{HBr}$ in presence of light or peroxide follows anti- markovnikov rule.

The product formed by the reaction of given alkene with $\text{HBr}$ in presence of peroxide is shown below.

Figure 3

The product formed by the reaction of given alkene with $\text{HBr}$ in the presence of peroxide is $1-\text{bromohexane}$.

Conclusion

The product(s) formed by the reaction of given alkene with $\left[1\right]\text{HBr}$; or $\left[\text{2}\right]\text{HBr}$ in the presence of peroxides is shown in Figure 2 and 3.

Interpretation Introduction

(b)

Interpretation:

The product(s) formed by the reaction of given alkene with $\left[1\right]\text{HBr}$; or $\left[\text{2}\right]\text{HBr}$ in the presence of peroxides is to be drawn.

Concept introduction:

The reaction of hydrogen halide with alkene results in the formation of alkyl halide. This type of reaction is an electrophilic addition of hydrogen halide. Electrophilic addition reactions are those in which breaking of pi bond take place to form new sigma bond.

Answer to Problem 15.24P

The product formed by the reaction of given alkene with $\left[1\right]\text{HBr}$ is shown below.

The product formed by the reaction of given alkene with $\text{HBr}$ in presence of peroxide is shown below.

Explanation of Solution

Electrophilic addition reaction follows Markovnikov rule. According to Markovnikov rule, the positive part of halogen acid attached to that carbon atom in $\text{C=C}$ bond which carries higher number of hydrogen atoms and the negative part of halogen acid will attach to that carbon atom in $\text{C=C}$ bond which has lesser number of hydrogen atoms.

The given alkene is shown below.

Figure 4

The steps followed by electrophilic addition reaction are stated below:

➢ First protonation of the alkene take place to generate the carbocation.

➢ The halide ion will attack on the carbocation to give the final product.

The product formed by the reaction of given alkene with $\text{HBr}$ is shown below.

Figure 5

The product formed by the reaction of given alkene with $\text{HBr}$ is $1-\text{bromo}-1-\text{methylcyclohexane}$.

The addition of $\text{HBr}$ in absence of light or peroxide occurs via carbocation intermediate, whereas addition of $\text{HBr}$ in presence of light or peroxide occurs via radical intermediate.

The addition of $\text{HBr}$ in presence of light or peroxide follows anti- markovnikov rule.

The product formed by the reaction of given alkene with $\text{HBr}$ in presence of peroxide is shown below.

Figure 6

The product formed by the reaction of given alkene with $\text{HBr}$ in presence of peroxide is $1-\text{bromo}-2-\text{methylcyclohexane}$.

Conclusion

The product(s) formed by the reaction of given alkene with $\left[1\right]\text{HBr}$; or $\left[\text{2}\right]\text{HBr}$ in the presence of peroxides is shown in Figure 5 and 6.

Interpretation Introduction

(c)

Interpretation:

The product(s) formed by the reaction of given alkene with $\left[1\right]\text{HBr}$; or $\left[\text{2}\right]\text{HBr}$ in the presence of peroxides is to be drawn.

Concept introduction:

The reaction of hydrogen halide with alkene results in the formation of alkyl halide. This type of reaction is an electrophilic addition of hydrogen halide. Electrophilic addition reactions are those in which breaking of pi bond take place to form new sigma bond.

Answer to Problem 15.24P

The product formed by the reaction of given alkene with $\left[1\right]\text{HBr}$ is shown below.

The product formed by the reaction of given alkene with $\left[1\right]\text{HBr}$ in the presence of peroxide is shown below.

Explanation of Solution

Electrophilic addition reaction follows Markovnikov rule. According to Markovnikov rule, the positive part of halogen acid attached to that carbon atom in $\text{C=C}$ bond which carries higher number of hydrogen atoms and the negative part of halogen acid will attach to that carbon atom in $\text{C=C}$ bond which has lesser number of hydrogen atoms.

The given alkene is shown below.

Figure 7

The steps followed by electrophilic addition reaction are stated below:

➢ First protonation of the alkene take place to generate the carbocation.

➢ The halide ion will attack on the carbocation to give the final product.

The product formed by the reaction of given alkene with $\text{HBr}$ is shown below.

Figure 8

The reaction of given alkene with $\text{HBr}$ forms two carbocation. Thus, the product formed by the reaction of given alkene with $\text{HBr}$ are $2-\text{bromohexane}$ and $3-\text{bromohexane}$.

The addition of $\text{HBr}$ in absence of light or peroxide occurs via carbocation intermediate, whereas addition of $\text{HBr}$ in presence of light or peroxide occurs via radical intermediate.

The addition of $\text{HBr}$ in presence of light or peroxide follows anti- markovnikov rule.

The product formed by the reaction of given alkene with $\text{HBr}$ in presence of peroxide is shown below.

Figure 9

The reaction of given alkene with $\text{HBr}$ in presence of peroxide forms two carbocation. Thus, the product formed by the reaction of given alkene with $\text{HBr}$ are $2-\text{bromohexane}$ and $3-\text{bromohexane}$.

Conclusion

The product(s) formed by the reaction of given alkene with $\left[1\right]\text{HBr}$; or $\left[\text{2}\right]\text{HBr}$ in the presence of peroxides is shown in Figure 8 and 9.