(a) Interpretation: The products expected when 1-bromo-3-methylbutane react with 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, S N 1, S N 2, E1 and E2 . All these reactions take place in the presence of basic compounds but in case of S N 1 and S N 2 reactions, those basic compounds are termed as nucleophiles.

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Answer 1

Question

Chapter 9, Problem 9.45AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KI$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 1

Explanation of Solution

The type of reactions those occurs when $1-bromo-3-methylbutane$ reacting with $KI$ in aqueous acetone are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 2

Figure 1

The $SN2$ 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 $1-iodo-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KOH$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 3

Explanation of Solution

The type of reactions which $1-bromo-3-methylbutane$ undergoes when reacted with $KOH$ in aqueous ethanol are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 4

Figure 2

The $SN2$ 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 $3-methylbutan-1-ol$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 5

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is $E2$ .

The product that is obtained via $E2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 6

Figure 3

An $E2$ reaction is a base-promoted $β−elimination$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $β-$ proton is taken by the $K+(CH3)3C−O−$ and simultaneously bromide ion is removed. Only one alkene product ( $3-methyl-1-butene$ ) is obtained as only one kind of $β-$ proton is present.

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

Conclusion

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the product of part (c) and $HBr$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 7

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 $(3-methyl-1-butene)$ with $HBr$ 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 $HBr$ 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 $HBr$ are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $CsF$ in $N,N-dimethylformamide$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 9

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $CsF$ in $N,N-dimethylformamide$ is $SN2$ .

The product for the $SN2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 10

Figure 5

The $SN2$ 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 $1-fluoro-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

The fluoride ion only undergoes $SN2$ reaction not elimination reaction because it is weak base but a good nucleophile.

Conclusion

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

Interpretation Introduction

(f)

Interpretation:

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

Concept introduction:

$α−$ elimination is a type of reaction in which proton from the $α−$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 11

Explanation of Solution

$α−$ 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.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 12

Figure 6

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 13

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 $tert-butoxide$ are reacted.

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 14

Figure 8

Conclusion

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

Interpretation Introduction

(g)

Interpretation:

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are to be stated.

Concept introduction:

The reaction of an alkene with the diiodomethane in the presence of $Zn−Cu$ 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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 15

Explanation of Solution

The reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ 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 $(3-methyl-1-butene)$ to give back isopropylcyclopropane.

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 16

Figure 9

Conclusion

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown in Figure 9.

Interpretation Introduction

(h)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with $Li$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 17

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 $1-bromo-3-methylbutane$ with $Li$ in hexane, then ethanol is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 18

Figure 10

Conclusion

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

Interpretation Introduction

(i)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 19

Explanation of Solution

The type of reactions that occurs when $1-bromo-3-methylbutane$ reacting with sodium methoxide in methanol is $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reaction are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 20

Figure 11

The $SN2$ 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 $1-methoxy-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(j)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether, then $D2O$ 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 $RMgX$ (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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 21

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 $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 22

Figure 12

Conclusion

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown in Figure 12.

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Answer 2

Question

Chapter 9, Problem 9.45AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KI$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 1

Explanation of Solution

The type of reactions those occurs when $1-bromo-3-methylbutane$ reacting with $KI$ in aqueous acetone are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 2

Figure 1

The $SN2$ 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 $1-iodo-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KOH$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 3

Explanation of Solution

The type of reactions which $1-bromo-3-methylbutane$ undergoes when reacted with $KOH$ in aqueous ethanol are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 4

Figure 2

The $SN2$ 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 $3-methylbutan-1-ol$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 5

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is $E2$ .

The product that is obtained via $E2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 6

Figure 3

An $E2$ reaction is a base-promoted $β−elimination$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $β-$ proton is taken by the $K+(CH3)3C−O−$ and simultaneously bromide ion is removed. Only one alkene product ( $3-methyl-1-butene$ ) is obtained as only one kind of $β-$ proton is present.

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

Conclusion

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the product of part (c) and $HBr$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 7

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 $(3-methyl-1-butene)$ with $HBr$ 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 $HBr$ 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 $HBr$ are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $CsF$ in $N,N-dimethylformamide$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 9

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $CsF$ in $N,N-dimethylformamide$ is $SN2$ .

The product for the $SN2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 10

Figure 5

The $SN2$ 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 $1-fluoro-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

The fluoride ion only undergoes $SN2$ reaction not elimination reaction because it is weak base but a good nucleophile.

Conclusion

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

Interpretation Introduction

(f)

Interpretation:

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

Concept introduction:

$α−$ elimination is a type of reaction in which proton from the $α−$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 11

Explanation of Solution

$α−$ 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.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 12

Figure 6

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 13

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 $tert-butoxide$ are reacted.

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 14

Figure 8

Conclusion

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

Interpretation Introduction

(g)

Interpretation:

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are to be stated.

Concept introduction:

The reaction of an alkene with the diiodomethane in the presence of $Zn−Cu$ 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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 15

Explanation of Solution

The reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ 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 $(3-methyl-1-butene)$ to give back isopropylcyclopropane.

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 16

Figure 9

Conclusion

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown in Figure 9.

Interpretation Introduction

(h)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with $Li$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 17

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 $1-bromo-3-methylbutane$ with $Li$ in hexane, then ethanol is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 18

Figure 10

Conclusion

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

Interpretation Introduction

(i)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 19

Explanation of Solution

The type of reactions that occurs when $1-bromo-3-methylbutane$ reacting with sodium methoxide in methanol is $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reaction are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 20

Figure 11

The $SN2$ 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 $1-methoxy-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(j)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether, then $D2O$ 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 $RMgX$ (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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 21

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 $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 22

Figure 12

Conclusion

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown in Figure 12.

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Answer 3

Question

Chapter 9, Problem 9.45AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KI$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 1

Explanation of Solution

The type of reactions those occurs when $1-bromo-3-methylbutane$ reacting with $KI$ in aqueous acetone are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 2

Figure 1

The $SN2$ 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 $1-iodo-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KOH$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 3

Explanation of Solution

The type of reactions which $1-bromo-3-methylbutane$ undergoes when reacted with $KOH$ in aqueous ethanol are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 4

Figure 2

The $SN2$ 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 $3-methylbutan-1-ol$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 5

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is $E2$ .

The product that is obtained via $E2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 6

Figure 3

An $E2$ reaction is a base-promoted $β−elimination$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $β-$ proton is taken by the $K+(CH3)3C−O−$ and simultaneously bromide ion is removed. Only one alkene product ( $3-methyl-1-butene$ ) is obtained as only one kind of $β-$ proton is present.

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

Conclusion

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the product of part (c) and $HBr$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 7

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 $(3-methyl-1-butene)$ with $HBr$ 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 $HBr$ 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 $HBr$ are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $CsF$ in $N,N-dimethylformamide$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 9

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $CsF$ in $N,N-dimethylformamide$ is $SN2$ .

The product for the $SN2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 10

Figure 5

The $SN2$ 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 $1-fluoro-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

The fluoride ion only undergoes $SN2$ reaction not elimination reaction because it is weak base but a good nucleophile.

Conclusion

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

Interpretation Introduction

(f)

Interpretation:

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

Concept introduction:

$α−$ elimination is a type of reaction in which proton from the $α−$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 11

Explanation of Solution

$α−$ 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.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 12

Figure 6

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 13

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 $tert-butoxide$ are reacted.

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 14

Figure 8

Conclusion

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

Interpretation Introduction

(g)

Interpretation:

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are to be stated.

Concept introduction:

The reaction of an alkene with the diiodomethane in the presence of $Zn−Cu$ 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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 15

Explanation of Solution

The reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ 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 $(3-methyl-1-butene)$ to give back isopropylcyclopropane.

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 16

Figure 9

Conclusion

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown in Figure 9.

Interpretation Introduction

(h)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with $Li$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 17

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 $1-bromo-3-methylbutane$ with $Li$ in hexane, then ethanol is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 18

Figure 10

Conclusion

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

Interpretation Introduction

(i)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 19

Explanation of Solution

The type of reactions that occurs when $1-bromo-3-methylbutane$ reacting with sodium methoxide in methanol is $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reaction are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 20

Figure 11

The $SN2$ 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 $1-methoxy-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(j)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether, then $D2O$ 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 $RMgX$ (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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 21

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 $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 22

Figure 12

Conclusion

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown in Figure 12.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 9, Problem 9.45AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KI$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 1

Explanation of Solution

The type of reactions those occurs when $1-bromo-3-methylbutane$ reacting with $KI$ in aqueous acetone are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 2

Figure 1

The $SN2$ 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 $1-iodo-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KOH$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 3

Explanation of Solution

The type of reactions which $1-bromo-3-methylbutane$ undergoes when reacted with $KOH$ in aqueous ethanol are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 4

Figure 2

The $SN2$ 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 $3-methylbutan-1-ol$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 5

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is $E2$ .

The product that is obtained via $E2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 6

Figure 3

An $E2$ reaction is a base-promoted $β−elimination$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $β-$ proton is taken by the $K+(CH3)3C−O−$ and simultaneously bromide ion is removed. Only one alkene product ( $3-methyl-1-butene$ ) is obtained as only one kind of $β-$ proton is present.

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

Conclusion

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the product of part (c) and $HBr$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 7

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 $(3-methyl-1-butene)$ with $HBr$ 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 $HBr$ 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 $HBr$ are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $CsF$ in $N,N-dimethylformamide$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 9

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $CsF$ in $N,N-dimethylformamide$ is $SN2$ .

The product for the $SN2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 10

Figure 5

The $SN2$ 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 $1-fluoro-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

The fluoride ion only undergoes $SN2$ reaction not elimination reaction because it is weak base but a good nucleophile.

Conclusion

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

Interpretation Introduction

(f)

Interpretation:

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

Concept introduction:

$α−$ elimination is a type of reaction in which proton from the $α−$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 11

Explanation of Solution

$α−$ 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.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 12

Figure 6

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 13

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 $tert-butoxide$ are reacted.

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 14

Figure 8

Conclusion

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

Interpretation Introduction

(g)

Interpretation:

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are to be stated.

Concept introduction:

The reaction of an alkene with the diiodomethane in the presence of $Zn−Cu$ 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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 15

Explanation of Solution

The reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ 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 $(3-methyl-1-butene)$ to give back isopropylcyclopropane.

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 16

Figure 9

Conclusion

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown in Figure 9.

Interpretation Introduction

(h)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with $Li$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 17

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 $1-bromo-3-methylbutane$ with $Li$ in hexane, then ethanol is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 18

Figure 10

Conclusion

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

Interpretation Introduction

(i)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 19

Explanation of Solution

The type of reactions that occurs when $1-bromo-3-methylbutane$ reacting with sodium methoxide in methanol is $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reaction are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 20

Figure 11

The $SN2$ 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 $1-methoxy-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(j)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether, then $D2O$ 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 $RMgX$ (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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 21

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 $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 22

Figure 12

Conclusion

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown in Figure 12.

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Answer 5

Chapter 9, Problem 9.45AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KI$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 1

Explanation of Solution

The type of reactions those occurs when $1-bromo-3-methylbutane$ reacting with $KI$ in aqueous acetone are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 2

Figure 1

The $SN2$ 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 $1-iodo-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KOH$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 3

Explanation of Solution

The type of reactions which $1-bromo-3-methylbutane$ undergoes when reacted with $KOH$ in aqueous ethanol are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 4

Figure 2

The $SN2$ 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 $3-methylbutan-1-ol$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 5

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is $E2$ .

The product that is obtained via $E2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 6

Figure 3

An $E2$ reaction is a base-promoted $β−elimination$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $β-$ proton is taken by the $K+(CH3)3C−O−$ and simultaneously bromide ion is removed. Only one alkene product ( $3-methyl-1-butene$ ) is obtained as only one kind of $β-$ proton is present.

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

Conclusion

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the product of part (c) and $HBr$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 7

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 $(3-methyl-1-butene)$ with $HBr$ 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 $HBr$ 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 $HBr$ are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $CsF$ in $N,N-dimethylformamide$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 9

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $CsF$ in $N,N-dimethylformamide$ is $SN2$ .

The product for the $SN2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 10

Figure 5

The $SN2$ 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 $1-fluoro-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

The fluoride ion only undergoes $SN2$ reaction not elimination reaction because it is weak base but a good nucleophile.

Conclusion

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

Interpretation Introduction

(f)

Interpretation:

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

Concept introduction:

$α−$ elimination is a type of reaction in which proton from the $α−$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 11

Explanation of Solution

$α−$ 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.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 12

Figure 6

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 13

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 $tert-butoxide$ are reacted.

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 14

Figure 8

Conclusion

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

Interpretation Introduction

(g)

Interpretation:

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are to be stated.

Concept introduction:

The reaction of an alkene with the diiodomethane in the presence of $Zn−Cu$ 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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 15

Explanation of Solution

The reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ 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 $(3-methyl-1-butene)$ to give back isopropylcyclopropane.

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 16

Figure 9

Conclusion

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown in Figure 9.

Interpretation Introduction

(h)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with $Li$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 17

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 $1-bromo-3-methylbutane$ with $Li$ in hexane, then ethanol is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 18

Figure 10

Conclusion

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

Interpretation Introduction

(i)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 19

Explanation of Solution

The type of reactions that occurs when $1-bromo-3-methylbutane$ reacting with sodium methoxide in methanol is $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reaction are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 20

Figure 11

The $SN2$ 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 $1-methoxy-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Interpretation Introduction

(j)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether, then $D2O$ 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 $RMgX$ (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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 21

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 $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 22

Figure 12

Conclusion

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown in Figure 12.

Answer 6

Chapter 9, Problem 9.45AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KI$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 1

Explanation of Solution

The type of reactions those occurs when $1-bromo-3-methylbutane$ reacting with $KI$ in aqueous acetone are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 2

Figure 1

The $SN2$ 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 $1-iodo-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Answer 7

Chapter 9, Problem 9.45AP

Interpretation Introduction

(a)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KI$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Answer 8

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 1

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The type of reactions those occurs when $1-bromo-3-methylbutane$ reacting with $KI$ in aqueous acetone are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 2

Figure 1

The $SN2$ 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 $1-iodo-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Answer 13

Conclusion

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

Answer 14

Interpretation Introduction

(b)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KOH$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 3

Explanation of Solution

The type of reactions which $1-bromo-3-methylbutane$ undergoes when reacted with $KOH$ in aqueous ethanol are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 4

Figure 2

The $SN2$ 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 $3-methylbutan-1-ol$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Answer 15

Interpretation Introduction

(b)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $KOH$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Answer 16

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 3

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The type of reactions which $1-bromo-3-methylbutane$ undergoes when reacted with $KOH$ in aqueous ethanol are $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reactions are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 4

Figure 2

The $SN2$ 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 $3-methylbutan-1-ol$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Answer 21

Conclusion

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

Answer 22

Interpretation Introduction

(c)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 5

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is $E2$ .

The product that is obtained via $E2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 6

Figure 3

An $E2$ reaction is a base-promoted $β−elimination$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $β-$ proton is taken by the $K+(CH3)3C−O−$ and simultaneously bromide ion is removed. Only one alkene product ( $3-methyl-1-butene$ ) is obtained as only one kind of $β-$ proton is present.

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

Conclusion

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown in Figure 3.

Answer 23

Interpretation Introduction

(c)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Answer 24

Expert Solution

Answer to Problem 9.45AP

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 5

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is $E2$ .

The product that is obtained via $E2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 6

Figure 3

An $E2$ reaction is a base-promoted $β−elimination$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $β-$ proton is taken by the $K+(CH3)3C−O−$ and simultaneously bromide ion is removed. Only one alkene product ( $3-methyl-1-butene$ ) is obtained as only one kind of $β-$ proton is present.

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

Answer 29

Conclusion

The product expected when $1-bromo-3-methylbutane$ react with $K+(CH3)3C−O−$ in $(CH3)3C−OH$ is shown in Figure 3.

Answer 30

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the product of part (c) and $HBr$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 7

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 $(3-methyl-1-butene)$ with $HBr$ 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 $HBr$ 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 $HBr$ are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 8

Figure 4

Conclusion

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

Answer 31

Interpretation Introduction

(d)

Interpretation:

The products expected from the reaction of the product of part (c) and $HBr$ 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 32

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 7

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

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 $(3-methyl-1-butene)$ with $HBr$ 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 $HBr$ 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 $HBr$ are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 8

Figure 4

Answer 37

Conclusion

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

Answer 38

Interpretation Introduction

(e)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $CsF$ in $N,N-dimethylformamide$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 9

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $CsF$ in $N,N-dimethylformamide$ is $SN2$ .

The product for the $SN2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 10

Figure 5

The $SN2$ 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 $1-fluoro-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

The fluoride ion only undergoes $SN2$ reaction not elimination reaction because it is weak base but a good nucleophile.

Conclusion

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

Answer 39

Interpretation Introduction

(e)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ react with $CsF$ in $N,N-dimethylformamide$ are to be stated.

Concept introduction:

An alkyl halide in the presence of basic compounds undergoes multiple kinds of reaction, for example, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Answer 40

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 9

Answer 41

Expert Solution

Answer 42

Expert Solution

Answer 43

Expert Solution

Answer 44

Explanation of Solution

The type of reaction which $1-bromo-3-methylbutane$ undergoes when react with $CsF$ in $N,N-dimethylformamide$ is $SN2$ .

The product for the $SN2$ type of reaction is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 10

Figure 5

The $SN2$ 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 $1-fluoro-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

The fluoride ion only undergoes $SN2$ reaction not elimination reaction because it is weak base but a good nucleophile.

Answer 45

Conclusion

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

Answer 46

Interpretation Introduction

(f)

Interpretation:

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

Concept introduction:

$α−$ elimination is a type of reaction in which proton from the $α−$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 11

Explanation of Solution

$α−$ 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.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 12

Figure 6

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 13

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 $tert-butoxide$ are reacted.

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 14

Figure 8

Conclusion

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

Answer 47

Interpretation Introduction

(f)

Interpretation:

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

Concept introduction:

$α−$ elimination is a type of reaction in which proton from the $α−$ 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 48

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 11

Answer 49

Expert Solution

Answer 50

Expert Solution

Answer 51

Expert Solution

Answer 52

Explanation of Solution

$α−$ 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.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 12

Figure 6

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 13

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 $tert-butoxide$ are reacted.

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 14

Figure 8

Answer 53

Conclusion

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

Answer 54

Interpretation Introduction

(g)

Interpretation:

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are to be stated.

Concept introduction:

The reaction of an alkene with the diiodomethane in the presence of $Zn−Cu$ 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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 15

Explanation of Solution

The reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ 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 $(3-methyl-1-butene)$ to give back isopropylcyclopropane.

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 16

Figure 9

Conclusion

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown in Figure 9.

Answer 55

Interpretation Introduction

(g)

Interpretation:

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are to be stated.

Concept introduction:

The reaction of an alkene with the diiodomethane in the presence of $Zn−Cu$ 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 56

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 15

Answer 57

Expert Solution

Answer 58

Expert Solution

Answer 59

Expert Solution

Answer 60

Explanation of Solution

The reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ 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 $(3-methyl-1-butene)$ to give back isopropylcyclopropane.

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 16

Figure 9

Answer 61

Conclusion

The products expected from the reaction of the product of part (c) and $CH2I2$ in the presence of a $Zn−Cu$ couple are shown in Figure 9.

Answer 62

Interpretation Introduction

(h)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with $Li$ 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.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 17

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 $1-bromo-3-methylbutane$ with $Li$ in hexane, then ethanol is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 18

Figure 10

Conclusion

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

Answer 63

Interpretation Introduction

(h)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with $Li$ 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 64

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 17

Answer 65

Expert Solution

Answer 66

Expert Solution

Answer 67

Expert Solution

Answer 68

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 $1-bromo-3-methylbutane$ with $Li$ in hexane, then ethanol is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 18

Figure 10

Answer 69

Conclusion

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

Answer 70

Interpretation Introduction

(i)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 19

Explanation of Solution

The type of reactions that occurs when $1-bromo-3-methylbutane$ reacting with sodium methoxide in methanol is $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reaction are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 20

Figure 11

The $SN2$ 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 $1-methoxy-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Conclusion

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

Answer 71

Interpretation Introduction

(i)

Interpretation:

The products expected when $1-bromo-3-methylbutane$ 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, $SN1, SN2,$ $E1$ and $E2$ .

All these reactions take place in the presence of basic compounds but in case of $SN1$ and $SN2$ reactions, those basic compounds are termed as nucleophiles.

Answer 72

Expert Solution

Answer to Problem 9.45AP

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

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 19

Answer 73

Expert Solution

Answer 74

Expert Solution

Answer 75

Expert Solution

Answer 76

Explanation of Solution

The type of reactions that occurs when $1-bromo-3-methylbutane$ reacting with sodium methoxide in methanol is $SN2$ and $E2$ .

The products that are obtained via $SN2$ and $E2$ type of reaction are shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 20

Figure 11

The $SN2$ 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 $1-methoxy-3-methylbutane$ as a product. Only one $SN2$ product is formed in a $SN2$ reaction.

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

Answer 77

Conclusion

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

Answer 78

Interpretation Introduction

(j)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether, then $D2O$ 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 $RMgX$ (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.

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 21

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 $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 22

Figure 12

Conclusion

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown in Figure 12.

Answer 79

Interpretation Introduction

(j)

Interpretation:

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether, then $D2O$ 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 $RMgX$ (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 80

Expert Solution

Answer to Problem 9.45AP

The products expected from the reaction of $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 21

Answer 81

Expert Solution

Answer 82

Expert Solution

Answer 83

Expert Solution

Answer 84

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 $1-bromo-3-methylbutane$ with magnesium and anhydrous ether then $D2O$ is shown below.

Organic Chemistry, Chapter 9, Problem 9.45AP , additional homework tip 22

Figure 12