Chapter 9, Problem 9.46AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $\text{2-}$ bromo $\text{-2-}$ methylhexane react with warm $1:1$ ethanol-water are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2,}$ $\text{E1}$ and $\text{E2}$.

All these reactions take place in the presence of basic compounds but in case of ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$ reactions, those basic compounds are termed as nucleophiles.

Answer to Problem 9.46AP

The products formed when $\text{2-}$ bromo $\text{-2-}$ methylhexane react with warm $1:1$ ethanol-water are shown below.

Explanation of Solution

The types of reactions which $\text{2-}$ bromo $\text{-2-}$ methylhexane undergoes when reacted with warm $1:1$ ethanol-water are ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$

The products that are obtained via ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ type of reactions are shown below.

Figure 1

The ${\text{S}}_{\text{N}}\text{1}$ reaction is a nucleophilic substitution reaction in which the substitution of nucleophile takes place. This reaction takes place in two steps. First is the elimination of bromide from the $\text{2-}$ bromo $\text{-2-}$ methylhexane to give carbocation intermediate. The second step is the addition of incoming nucleophiles (water and ethanol) to the carbocation intermediate to give back two products $\text{2-methylhexan-2-ol}$ and $\text{2-ethoxy-2-methylhexane}$.

An $\text{E1}$ reaction is the elimination reaction leading to the formation of the alkene. This reaction also takes place in two steps. The first step is the formation of carbocation by the leaving of bromide. The second step is the abstraction of a proton from the $\beta \text{-}$ carbons of $\text{2-}$ bromo $\text{-2-}$ methylhexane to give back alkenes ($\text{2-methyl-1-hexene}$ and $\text{2-methyl-2-hexene}$).

Conclusion

The products formed when $\text{2-}$ bromo $\text{-2-}$ methylhexane react with warm $1:1$ ethanol-water are shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The products expected when $\text{2-}$ bromo $\text{-2-}$ methylhexane react with sodium ethoxide in ethanol are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2,}$ $\text{E1}$ and $\text{E2}$.

All these reactions take place in the presence of basic compounds but in case of ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$ reactions, those basic compounds are termed as nucleophiles.

Answer to Problem 9.46AP

The products expected when $\text{2-}$ bromo $\text{-2-}$ methylhexane react with sodium ethoxide in ethanol are shown below.

Explanation of Solution

An $\text{E2}$ reaction takes place when $\text{2-}$ bromo $\text{-2-}$ methylhexane react with sodium ethoxide in ethanol because the alkyl halide is bulky and the base is strong base.

The products formed for this reaction are shown below.

Figure 2

An $\text{E2}$ reaction is a base-promoted $\beta -\text{elimination}$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, two $\beta \text{-}$ protons from the two different $\beta \text{-}$ carbons are taken by the base and simultaneously a bromide ion is removed. This leads to the formation of two products because of two different $\beta \text{-}$ protons abstraction to give back two alkenes ($\text{2-methyl-1-hexene}$ and $\text{2-methyl-2-hexene}$).

Conclusion

The products expected when $\text{2-}$ bromo $\text{-2-}$ methylhexane react with sodium ethoxide in ethanol are shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The products expected when $\text{2-}$ bromo $\text{-2-}$ methylhexane reacted with $\text{KI}$ in aqueous acetone are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2,}$ $\text{E1}$ and $\text{E2}$.

All these reactions take place in the presence of basic compounds but in case of ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$ reactions, those basic compounds are termed as nucleophiles.

Answer to Problem 9.46AP

The products expected when $\text{2-}$ bromo $\text{-2-}$ methylhexane reacted with KI in aqueous acetone are shown below.

Explanation of Solution

The type of reactions which $\text{2-}$ bromo $\text{-2-}$ methylhexane undergoeswhen reacted with KI in aqueous acetone are ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$.

The products that are obtained via ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ type of reaction are shown below.

Figure 3

The ${\text{S}}_{\text{N}}\text{1}$ reaction is a nucleophilic substitution reaction in which the substitution of nucleophile takes place. This reaction takes place in two steps. First is the elimination of bromide from the $\text{2-}$ bromo $\text{-2-}$ methylhexane to give carbocation intermediate. The second step is the addition of incoming nucleophile (water) to the carbocation intermediate to give back product $\text{2-methylhexan-2-ol}$.

An $\text{E1}$ reaction is the elimination reaction leading to the formation of the alkene. This reaction also takes place in two steps. The first step is the formation of carbocation by the leaving of bromide. The second step is the abstraction of a proton from the $\beta \text{-}$ carbons of $\text{2-}$ bromo $\text{-2-}$ methylhexane to give back alkenes ($\text{2-methyl-1-hexene}$ and $\text{2-methyl-2-hexene}$).

Conclusion

The products expected when $\text{2-}$ bromo $\text{-2-}$ methylhexane reacted with KI in aqueous acetone are shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the products of part (b) and $\text{HBr}$ in the presence of peroxides are to be stated.

Concept introduction:

An alkene undergoes addition reactions due to the presence of unsaturated bond in it. Markonikov’s gave a rule for the addition of protic acids or protic compounds takes place in such a way that proton goes to carbon having the highest number of hydrogen atoms and anion goes to other carbon.

Answer to Problem 9.46AP

The products expected from the reaction of the products of part (b) and $\text{HBr}$ in the presence of peroxides are shown below.

Explanation of Solution

Some reactions follow the rule of Markonikov’s addition but some do not. Markonikov’s gave the rule purely on the basis of the observation of products received by him when performed the addition reaction on alkene.

Some reactions do not follow the rule because they follow the stability of the intermediate formed in the mechanism of that particular reaction.

The reaction of alkenes ($\text{2-methyl-1-hexene}$ and $\text{2-methyl-2-hexene}$) with $\text{HBr}$ in the presence of peroxides leads to the formation of the the anti-Markonikov product ($\text{1-bromo-2-methylhexane}$ and $\text{3-bromo-2-methylhexane}$) due to the stability of the free radical intermediate formed during the reaction.

The products expected from the reaction of the products of part (b) and $\text{HBr}$ in the presence of peroxides are shown below.

Figure 4

Conclusion

The products expected from the reaction of the products of part (b) and $\text{HBr}$ in the presence of peroxides are shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The products expected from the reaction of the products of part (b) and ${\text{Hg(OAc)}}_{\text{2}}$ in THF-water, followed by ${\text{NaBH}}_{\text{4}}$ are to be stated.

Concept introduction:

An alkene undergoes addition reactions due to the presence of unsaturated bond in it. Markonikov’s gave a rule for the addition of protic acids or protic compounds takes place in such a way that proton goes to carbon having the highest number of hydrogen and anion goes to other carbon.

Answer to Problem 9.46AP

The products expected from the reaction of the products of part (b) and ${\text{Hg(OAc)}}_{\text{2}}$ in THF-water, followed by ${\text{NaBH}}_{\text{4}}$ are shown below.

Explanation of Solution

Some reactions follow the rule of Markonikov’s addition but some do not. Markonikov’s gave the rule purely on the basis of the observation of products received by him when performed the addition reaction on alkene.

Some reactions do not follow the rule because they follow the stability of the intermediate formed in the mechanism of that particular reaction.

The reaction of alkenes ($\text{2-methyl-1-hexene}$ and $\text{2-methyl-2-hexene}$) with ${\text{Hg(OAc)}}_{\text{2}}$ in THF-water, followed by ${\text{NaBH}}_{\text{4}}$ is a name reaction known as an oxymercuration-reduction reaction. This reaction leads to the formation of Markonikov product ($\text{2-methyl-2-hexanol}$).

The products expected from the reaction of the products of part (b) and ${\text{Hg(OAc)}}_{\text{2}}$ in THF-water, followed by ${\text{NaBH}}_{\text{4}}$ are shown below.

Figure 5

Conclusion

The products expected from the reaction of the products of part (b) and ${\text{Hg(OAc)}}_{\text{2}}$ in THF-water, followed by ${\text{NaBH}}_{\text{4}}$ are shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The products expected from the reaction of the products of part (b) and ${\text{BH}}_{\text{3}}$ in THF, followed by alkaline ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ are to be stated.

Concept introduction:

An alkene undergoes addition reactions due to the presence of unsaturated bond in it. Markonikov’s gave a rule for the addition of protic acids or protic compounds takes place in such a way that proton goes to carbon having the highest number of hydrogen and anion goes to other carbon.

Answer to Problem 9.46AP

The products expected from the reaction of the products of part (b) and ${\text{BH}}_{\text{3}}$ in THF, followed by alkaline ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ are shown below.

Explanation of Solution

Some reactions follow the rule of Markonikov’s addition but some do not.

Markonikov’s gave the rule purely on the basis of the observation of products received by him when performed the addition reaction on alkene.

Some reactions do not follow the rule because they follow the stability of the intermediate formed in the mechanism of that particular reaction.

The reaction of alkenes ($\text{2-methyl-1-hexene}$ and $\text{2-methyl-2-hexene}$) with ${\text{BH}}_{\text{3}}$ in THF, followed by alkaline ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is a name reaction known as Hydroboration-oxidation Reaction. This reaction leads to the formation of anti-Markonikov products ($\text{2-methyl-1-hexanol}$ and $\text{2-methyl-3-hexanol}$).

The products expected from the reaction of the products of part (b) and ${\text{BH}}_{\text{3}}$ in THF, followed by alkaline ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ are shown below.

Figure 6

Conclusion

The products expected from the reaction of the products of part (b) and ${\text{BH}}_{\text{3}}$ in THF, followed by alkaline ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ are shown in Figure 6.