Chapter 10, Problem 10.46P

Draw the products formed when ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ is treated with each reagent.

a. $\text{HBr}$

b. ${\text{H}}_{\text{2}}\text{O},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

c. ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

d. ${\text{Cl}}_2$

e. ${\text{Br}}_2,{\text{H}}_{\text{2}}\text{O}$

f. $\text{NBS}\left(\text{aqueous}\text{DMSO}\right)$

g. $\left[1\right]{\text{BH}}_{\text{3}}$; $\left[2\right]{\text{H}}_{\text{2}}{\text{O}}_{\text{2}},{\text{HO}}^{-}$

Interpretation Introduction

(a)

Interpretation: The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\text{HBr}$ 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 reaction follows Markovnikov’s rule.

According to Markovnikov’s rule, the positive part of halogen acid attached to that carbon atom in $\text{C=C}$ bond which carries larger 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.

Answer to Problem 10.46P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\text{HBr}$ is $2-\text{bromo}-2-\text{methylpropane}$.

Explanation of Solution

A reaction of hydrogen halide with alkene results in the formation of alkyl halide. This type of reaction is an electrophilic addition of hydrogen halide.

The steps involved in the 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 ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\text{HBr}$ is shown below.

Figure 1

Therefore, the product formed is $2-\text{bromo}-2-\text{methylpropane}$.

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\text{HBr}$ is $2-\text{bromo}-2-\text{methylpropane}$.

Interpretation Introduction

(b)

Interpretation: The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\text{O},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be drawn.

Concept introduction: Hydration of alkenes is one the method used for the formation of alcohol.

The general steps followed by hydration reaction are stated below:

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

• Formation of protonated alcohol.

• Deprotonation.

Answer to Problem 10.46P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\text{O},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $2-\text{methylpropan}-2-\text{ol}$.

Explanation of Solution

Hydration of alkenes is one the method used for the formation of alcohol.

The general steps followed by hydration reaction are stated below:

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

• Formation of protonated alcohol.

• Deprotonation.

In this type of reaction, hydroxyl group attacks on the more substituted carbon.

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\text{O},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

Figure 2

The first step in the given reaction is the abstraction of proton from sulfuric acid. This results in the formation of carbocation. In the next step, attack of water molecule takes place on the electrophilic carbon. The last step is the removal of ${\text{H}}^{+}$ ion to give the final product. Therefore, the product formed is $2-\text{methylpropan}-2-\text{ol}$.

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\text{O},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $2-\text{methylpropan}-2-\text{ol}$.

Interpretation Introduction

(c)

Interpretation: The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be drawn.

Concept introduction: Reaction of alkene with alcohol is one the method used for the formation of ether.

The general steps followed by the reaction are stated below:

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

• Deprotonation to give the final product.

Answer to Problem 10.46P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $2-\text{ethoxy}-2-\text{methylpropane}$.

Explanation of Solution

Reaction of alkene with alcohol in presence of sulfuric acid is one the method used for the formation of ether.

The general steps followed by the reaction are stated below:

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

• Deprotonation to give the final product.

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

Figure 3

Conclusion

Therefore, the product formed is $2-\text{ethoxy}-2-\text{methylpropane}$.

Interpretation Introduction

(d)

Interpretation: The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{Cl}}_2$ is to be drawn.

Concept introduction: The general steps followed by the addition of Halogens to alkenes are stated below:

• The first step is the electrophilic attack of halide ion to form a halonium ion.

• The second step is the attack of halide ion from back side to opens the halonium ion.

Answer to Problem 10.46P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{Cl}}_2$ is $1,2-\text{dichloro}-2-\text{methylpropane}$.

Explanation of Solution

The general steps followed by the addition of Halogens to alkenes are stated below:

• The first step is the electrophilic attack of halide ion to form a halonium ion.

• The second step is the attack of halide ion from back side to opens the halonium ion to give the final product.

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{Cl}}_2$ is shown below.

Figure 4

Therefore, the product formed is $2-\text{bromo}-2-\text{methylpropane}$.

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{Cl}}_2$ is $1,2-\text{dichloro}-2-\text{methylpropane}$.

Interpretation Introduction

(e)

Interpretation: The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{Br}}_2,{\text{H}}_{\text{2}}\text{O}$ is to be drawn.

Concept introduction: The reaction of alkene with halogen and water results in the formation of halohydrin product.

The general steps for the formation of halohydrin are stated below:

• The first step is the attack of halide ion to form a halonium ion.

• The second step is the attack of water from back side to opens the halonium ion.

• The last step is deprotonation to give the halohydrin product.

Answer to Problem 10.46P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{Br}}_2,{\text{H}}_{\text{2}}\text{O}$ is $1-\text{bromo}-2-\text{methylpropan}-2-\text{ol}$.

Explanation of Solution

The reaction of alkene with halogen and water results in the formation of halohydrin product.

The general steps for the formation of halohydrin are stated below:

• The first step is the attack of halide ion to form a halonium ion.

• The second step is the attack of water from back side to opens the halonium ion.

• The last step is deprotonation to give the halohydrin product.

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{Br}}_2,{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 5

Therefore, the product formed is $1-\text{bromo}-2-\text{methylpropan}-2-\text{ol}$.

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with ${\text{Br}}_2,{\text{H}}_{\text{2}}\text{O}$ is $1-\text{bromo}-2-\text{methylpropan}-2-\text{ol}$.

Interpretation Introduction

(f)

Interpretation: The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\text{NBS}\left(\text{aqueous}\text{DMSO}\right)$ is to be drawn.

Concept introduction: The reaction of $N-\text{bromosuccinimide}$ in aqueous DMSO results in the formation of bromohydrins. $\text{NBS}$ is the source of ${\text{Br}}_{\text{2}}$.

Answer to Problem 10.46P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\text{NBS}\left(\text{aqueous}\text{DMSO}\right)$ is $1-\text{bromo}-2-\text{methylpropan}-2-\text{ol}$.

Explanation of Solution

The reaction of $N-\text{bromosuccinimide}$ in aqueous DMSO results in the formation of bromohydrins. $\text{NBS}$ is the source of ${\text{Br}}_{\text{2}}$.

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\text{NBS}\left(\text{aqueous}\text{DMSO}\right)$ is shown below.

Figure 6

Therefore, the product formed is $1-\text{bromo}-2-\text{methylpropan}-2-\text{ol}$.

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\text{NBS}\left(\text{aqueous}\text{DMSO}\right)$ is $1-\text{bromo}-2-\text{methylpropan}-2-\text{ol}$.

Interpretation Introduction

(g)

Interpretation: The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\left[1\right]{\text{BH}}_{\text{3}}$; $\left[2\right]{\text{H}}_{\text{2}}{\text{O}}_{\text{2}},{\text{OH}}^{-}$ is to be drawn.

Concept introduction: Hydroboration reaction is a two step reaction, which involves conversion of alkene into alcohol. This type of reaction follows anti-markovinokov’s rule.

Anti markovinokov’s rule states that the positive part of acid attached to that carbon atom in $\text{C=C}$ bond which carries minimum number of hydrogen atoms and the negative part of acid will attach to that carbon atom in $\text{C=C}$ bond which has maximum number of hydrogen atoms.

Answer to Problem 10.46P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\left[1\right]{\text{BH}}_{\text{3}}$; $\left[2\right]{\text{H}}_{\text{2}}{\text{O}}_{\text{2}},{\text{OH}}^{-}$ is $2-\text{methylpropan}-1-\text{ol}$.

Explanation of Solution

Hydroboration reaction is a two step reaction, which involves conversion of alkene into alcohol. This type of reaction follows anti-markovinokov’s rule.

Anti markovinokov’s rule states that the positive part of acid attached to that carbon atom in $\text{C=C}$ bond which carries minimum number of hydrogen atoms and the negative part of acid will attach to that carbon atom in $\text{C=C}$ bond which has maximum number of hydrogen atoms.

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\left[1\right]{\text{BH}}_{\text{3}}$; $\left[2\right]{\text{H}}_{\text{2}}{\text{O}}_{\text{2}},{\text{OH}}^{-}$ is shown below.

Figure 7

During hydroboration of $2-\text{methylprop}-1-\text{ene}$, addition of boron and hydrogen is from same side. Therefore, it leads to syn addtion from above or belwo the double bond to form alkylborane. This alkylborane gets oxidised and replaces boron with a $-\text{OH}$ group.

Therefore, the product formed is $2-\text{methylpropan}-1-\text{ol}$.

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}$ with $\left[1\right]{\text{BH}}_{\text{3}}$; $\left[2\right]{\text{H}}_{\text{2}}{\text{O}}_{\text{2}},{\text{OH}}^{-}$ is $2-\text{methylpropan}-1-\text{ol}$.