Chapter 9, Problem 9.45AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $\text{1-bromo-3-methylbutane}$ react 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 9.45AP

The products formed when $\text{1-bromo-3-methylbutane}$ reacting with $\text{KI}$ in aqueous acetone are shown below.

Explanation of Solution

The type of reactions those occurs when $\text{1-bromo-3-methylbutane}$ reacting with $\text{KI}$ in aqueous acetone are ${\text{S}}_{\text{N}}2$ and $\text{E2}$.

The products that are obtained via ${\text{S}}_{\text{N}}2$ and $\text{E2}$ type of reactions is shown below.

Figure 1

The ${\text{S}}_{\text{N}}\text{2}$ reaction is the concerted mechanism reaction in which a nucleophile substitutes a leaving group present in the reactant. The incoming nucleophile iodide attacks from the back side of reactant and leaving group bromide leaves from the front to give $\text{1-iodo-3-methylbutane}$ as a product. Only one ${\text{S}}_{\text{N}}\text{2}$ product is formed in a ${\text{S}}_{\text{N}}\text{2}$ reaction.

An $\text{E2}$ reaction is a base-promoted $\beta -\text{elimination}$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $\beta \text{-}$ proton is taken by the iodide ion and simultaneously bromide ion is removed. Only one alkene product ($\text{3-methyl-1-butene}$) is obtained as only one kind of $\beta \text{-}$ proton is present.

Conclusion

The products formed when $\text{1-bromo-3-methylbutane}$ reacting with $\text{KI}$ in aqueous acetone are shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The products expected when $\text{1-bromo-3-methylbutane}$ react with $\text{KOH}$ in aqueous 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 9.45AP

The products expected when $\text{1-bromo-3-methylbutane}$ react with $\text{KOH}$ in aqueous ethanol are shown below.

Explanation of Solution

The type of reactions which $\text{1-bromo-3-methylbutane}$ undergoes when reacted with $\text{KOH}$ in aqueous ethanol are ${\text{S}}_{\text{N}}2$ and $\text{E2}$.

The products that are obtained via ${\text{S}}_{\text{N}}2$ and $\text{E2}$ type of reactions are shown below.

Figure 2

The ${\text{S}}_{\text{N}}\text{2}$ reaction is the concerted mechanism reaction in which a nucleophile substitutes a leaving group present in the reactant. The incoming nucleophile hydroxide ion attacks from the back side of reactant and leaving group bromide leaves from the front to give $\text{3-methylbutan-1-ol}$ as a product. Only one ${\text{S}}_{\text{N}}\text{2}$ product is formed in a ${\text{S}}_{\text{N}}\text{2}$ reaction.

An $\text{E2}$ reaction is a base-promoted $\beta -\text{elimination}$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $\beta \text{-}$ proton is taken by the hydroxide ion and simultaneously bromide ion is removed. Only one alkene product ($\text{3-methyl-1-butene}$) is obtained as only one kind of $\beta \text{-}$ proton is present.

Conclusion

The products expected when $\text{1-bromo-3-methylbutane}$ react with $\text{KOH}$ in aqueous ethanol are shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The products expected when $\text{1-bromo-3-methylbutane}$ react with ${\text{K}}^{\text{+}}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{-}$ in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{OH}$ 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 9.45AP

The product expected when $\text{1-bromo-3-methylbutane}$ react with ${\text{K}}^{\text{+}}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{-}$ in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{OH}$ is shown below.

Explanation of Solution

The type of reaction which $\text{1-bromo-3-methylbutane}$ undergoes when react with ${\text{K}}^{\text{+}}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{-}$ in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{OH}$ is $\text{E2}$.

The product that is obtained via $\text{E2}$ type of reaction is shown below.

Figure 3

An $\text{E2}$ reaction is a base-promoted $\beta -\text{elimination}$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $\beta \text{-}$ proton is taken by the ${\text{K}}^{\text{+}}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{-}$ and simultaneously bromide ion is removed. Only one alkene product ($\text{3-methyl-1-butene}$) is obtained as only one kind of $\beta \text{-}$ proton is present.

Only $\text{E2}$ reaction takes place here because the base is strong and sterically hindered one.

Conclusion

The product expected when $\text{1-bromo-3-methylbutane}$ react with ${\text{K}}^{\text{+}}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{-}$ in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{OH}$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the product of part (c) and $\text{HBr}$ 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 9.45AP

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

Explanation of Solution

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 an alkene $\left(\text{3-methyl-1-butene}\right)$ with $\text{HBr}$ occurs by the formation of carbocation intermediate in the first step which then may rearrange leading to more stable carbocation by $1,2-$ hydride shift or methyl shift. The second step is the addition of bromide ion at carbocation position.

In the reaction between the product of part (c) and $\text{HBr}$ rearrangement of the carbocation formed which is a secondary carbocation takes place by $1,2-$ hydride shift to give tertiary carbocation and giving a corresponding product.

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

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The products expected when $\text{1-bromo-3-methylbutane}$ react with $\text{CsF}$ in $N,N\text{-dimethylformamide}$ 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 9.45AP

The products expected when $\text{1-bromo-3-methylbutane}$ react with $\text{CsF}$ in $N,N\text{-dimethylformamide}$ are shown below.

Explanation of Solution

The type of reaction which $\text{1-bromo-3-methylbutane}$ undergoes when react with $\text{CsF}$ in $N,N\text{-dimethylformamide}$ is ${\text{S}}_{\text{N}}\text{2}$.

The product for the ${\text{S}}_{\text{N}}\text{2}$ type of reaction is shown below.

Figure 5

The ${\text{S}}_{\text{N}}\text{2}$ reaction is the concerted mechanism reaction in which a nucleophile substitutes a leaving group present in the reactant. The incoming nucleophile fluoride ion attacks from the back side of reactant and bromide ion leave from the front to give $\text{1-fluoro-3-methylbutane}$ as a product. Only one ${\text{S}}_{\text{N}}\text{2}$ product is formed in a ${\text{S}}_{\text{N}}\text{2}$ reaction.

The fluoride ion only undergoes ${\text{S}}_{\text{N}}\text{2}$ reaction not elimination reaction because it is weak base but a good nucleophile.

Conclusion

The products expected when $\text{1-bromo-3-methylbutane}$ react with $\text{CsF}$ in $N,N\text{-dimethylformamide}$ are shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The products expected from the reaction of the product of part (c), chloroform and potassium $tert\text{-butoxide}$ are to be stated.

Concept introduction:

$\alpha -$ elimination is a type of reaction in which proton from the $\alpha -$ carbon is taken by the base being acidic leading to the formation of carbanion. This type of reaction is shown by chloroform in which carbanion formed is stabilized by the polar effects of three chlorine groups. After the formation of carbanion one of the chlorine leaves as chloride ion giving rise to the formation of dichloromethylene which is a carbene. Carbene undergoes addition reactions with an alkene.

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c), chloroform, and potassium $tert\text{-butoxide}$ are shown below.

Explanation of Solution

$\alpha -$ elimination reaction of chloroform in the presence of base leads to the formation of carbene dichloromethylene. The reaction mechanism for that is shown below.

Step-1: Take up of proton to give carbanion.

Figure 6

Step-2: Elimination of chloride ion to give carbene.

Figure 7

The carbene thus formed adds on the alkene in the leading to no change in the respective stereochemistry of substituents on alkene.

The same reaction is happening when the product of part (c) plus chloroform plus potassium $tert\text{-butoxide}$ are reacted.

Therefore, the products expected from the reaction of the product of part (c), chloroform and potassium $tert\text{-butoxide}$ are shown below.

Figure 8

Conclusion

The products expected from the reaction of the product of part (c) chloroform and potassium $tert\text{-butoxide}$ are shown in Figure 8.

Interpretation Introduction

(g)

Interpretation:

The products expected from the reaction of the product of part (c) and ${\text{CH}}_{\text{2}}{\text{I}}_{\text{2}}$ in the presence of a $\text{Zn}-\text{Cu}$ couple are to be stated.

Concept introduction:

The reaction of an alkene with the diiodomethane in the presence of $\text{Zn}-\text{Cu}$ couple is a name reaction known as Simons-Smith reaction. In this reaction formation of carbene takes place which adds on the alkene on the same side without changing the stereochemistry of the substituents on alkene respective to each other.

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c) and ${\text{CH}}_{\text{2}}{\text{I}}_{\text{2}}$ in the presence of a $\text{Zn}-\text{Cu}$ couple are shown below.

Explanation of Solution

The reaction of the product of part (c) and ${\text{CH}}_{\text{2}}{\text{I}}_{\text{2}}$ in the presence of a $\text{Zn}-\text{Cu}$ couple is a name reaction known as Simons-Smith reaction. In this reaction, a methylene carbene is formed which add on the double bond of the alkene $\left(\text{3-methyl-1-butene}\right)$ to give back isopropylcyclopropane.

The products expected from the reaction of the product of part (c) and ${\text{CH}}_{\text{2}}{\text{I}}_{\text{2}}$ in the presence of a $\text{Zn}-\text{Cu}$ couple are shown below.

Figure 9

Conclusion

The products expected from the reaction of the product of part (c) and ${\text{CH}}_{\text{2}}{\text{I}}_{\text{2}}$ in the presence of a $\text{Zn}-\text{Cu}$ couple are shown in Figure 9.

Interpretation Introduction

(h)

Interpretation:

The products expected from the reaction of $\text{1-bromo-3-methylbutane}$ with $\text{Li}$ in hexane, then ethanol is to be stated.

Concept introduction:

The reaction of an alkyl halide with a metal like lithium leads to the formation of organolithium compounds (alkyllithium). These compounds are very sensitive to moisture or polar hydrogens reacts immediately leading to the formation of alkane of the alkyl group.

Answer to Problem 9.45AP

The products expected from the reaction of $\text{1-bromo-3-methylbutane}$ with $\text{Li}$ in hexane, then ethanol is shown below.

Explanation of Solution

The reaction of an alkyl halide with a metal like lithium leads to the formation of organolithium compounds (alkyllithium), which are highly susceptible to humidity or react to polar hydrogen instantly leading to the alkyl group’s formation. The same reaction is happening in this case.

The products expected from the reaction of $\text{1-bromo-3-methylbutane}$ with $\text{Li}$ in hexane, then ethanol is shown below.

Figure 10

Conclusion

The products expected from the reaction of $\text{1-bromo-3-methylbutane}$ with $\text{Li}$ in hexane, then ethanol is shown in Figure 10.

Interpretation Introduction

(i)

Interpretation:

The products expected when $\text{1-bromo-3-methylbutane}$ react with sodium methoxide in methanol 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 9.45AP

The products expected when $\text{1-bromo-3-methylbutane}$ reacting with sodium methoxide in methanol are shown below.

Explanation of Solution

The type of reactions that occurs when $\text{1-bromo-3-methylbutane}$ reacting with sodium methoxide in methanol is ${\text{S}}_{\text{N}}2$ and $\text{E2}$.

The products that are obtained via ${\text{S}}_{\text{N}}2$ and $\text{E2}$ type of reaction are shown below.

Figure 11

The ${\text{S}}_{\text{N}}\text{2}$ reaction is the concerted mechanism reaction in which a nucleophile substitutes leaving group present in the reactant. In this reaction, the incoming nucleophile methoxide ion attacks from the back side of reactant and bromide ion leave from the front to give $\text{1-methoxy-3-methylbutane}$ as a product. Only one ${\text{S}}_{\text{N}}\text{2}$ product is formed in a ${\text{S}}_{\text{N}}\text{2}$ reaction.

An $\text{E2}$ reaction is a base-promoted $\beta -\text{elimination}$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $\beta \text{-}$ proton is taken by methoxide ion and simultaneously a bromide ion is removed. Only one alkene product ($\text{3-methyl-1-butene}$) is obtained as only one kind of $\beta \text{-}$ proton is present.

Conclusion

The products expected when $\text{1-bromo-3-methylbutane}$ reacting with sodium methoxide in methanol are shown in Figure 11.

Interpretation Introduction

(j)

Interpretation:

The products expected from the reaction of $\text{1-bromo-3-methylbutane}$ with magnesium and anhydrous ether, then ${\text{D}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

The reaction of an alkyl halide with a metal like magnesium in the presence of dry ether leads to the formation of $\text{RMgX}$ (organometallic compounds) also known as a Grignard reagent. These compounds are very sensitive to moisture or polar hydrogens reacts immediately leading to the formation of alkane of the alkyl group.

Answer to Problem 9.45AP

The products expected from the reaction of $\text{1-bromo-3-methylbutane}$ with magnesium and anhydrous ether then ${\text{D}}_{\text{2}}\text{O}$ is shown below.

Explanation of Solution

The reaction of an alkyl halide with a metal like magnesium in the presence of dry ether results in the formation of (organometallic compounds) also known as a Grignard reagent. These compounds are very susceptible to reactions of moisture or polar hydrogen leading to the creation of alkane of the alkyl group instantly. In this case, the same reaction is occurring.

The Grignard reagent obtained in this reaction is isopentylmagnesium bromide and product obtained after treatement with heavy water is deuterated isopentane.

The products expected from the reaction of $\text{1-bromo-3-methylbutane}$ with magnesium and anhydrous ether then ${\text{D}}_{\text{2}}\text{O}$ is shown below.

Figure 12

Conclusion

The products expected from the reaction of $\text{1-bromo-3-methylbutane}$ with magnesium and anhydrous ether then ${\text{D}}_{\text{2}}\text{O}$ is shown in Figure 12.