Chapter 11, Problem 11.38P

Draw the organic products formed in each reaction.

a. f.

b. h.

d. i.

e. j.

Interpretation Introduction

(a)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: The addition of an electrophile to an alkyne is followed by Markovnikoff’s rule and anti-stereoselectivity.

The addition of a halogen to an alkyne chain leads to the formation of corresponding alkene or alkane. The reaction which includes the addition of bromine atoms to the alkyne chain is known as bromination. Bromination can be done by using the reagents like $\text{HBr}$ or ${\text{Br}}_{\text{2}}$. The reaction which includes the addition of chlorine atoms to the alkyne chain is known as chlorination. Chlorination can be done by using the reagents like $\text{HCl}$ or ${\text{Cl}}_{\text{2}}$.

Answer to Problem 11.38P

The product that is formed by the reaction is $2,2-\text{dibromoheptane}$.

Explanation of Solution

The reaction of $\text{hept}-1-\text{yne}$ with two equivalents of $\text{HBr}$ is shown as,

Figure 1

The reaction of $\text{hept}-1-\text{yne}$ with two equivalents of $\text{HBr}$ results in the formation of $2,2-\text{dibromoheptane}$. In this reaction, the negatively charged bromide ions from two equivalents of $\text{HBr}$ get substituted on the highly substituted carbon atom of $\text{hept}-1-\text{yne}$ to form the desired dihalide product.

Conclusion

The product that is formed by the reaction is $2,2-\text{dibromoheptane}$.

Interpretation Introduction

(b)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: The addition of an electrophile to an alkyne is followed by Markovnikoff’s rule and anti-stereoselectivity.

The addition of a halogen to an alkyne chain leads to the formation of corresponding alkene or alkane. The reaction which includes the addition of bromine atoms to the alkyne chain is known as bromination. Bromination can be done by using the reagents like $\text{HBr}$ or ${\text{Br}}_{\text{2}}$. The reaction which includes the addition of chlorine atoms to the alkyne chain is known as chlorination. Chlorination can be done by using the reagents like $\text{HCl}$ or ${\text{Cl}}_{\text{2}}$.

Answer to Problem 11.38P

The product that is formed by the reaction is $1,1,2,2-\text{tetrachloro}-3,3-\text{dimethylbutane}$.

Explanation of Solution

The reaction of $3,3-\text{dimethylbut}-1-\text{yne}$ with two equivalents of ${\text{Cl}}_2$ is shown as,

Figure 2

The reaction of $3,3-\text{dimethylbut}-1-\text{yne}$ with two equivalents ${\text{Cl}}_2$ results in the formation of $1,1,2,2-\text{tetrachloro}-3,3-\text{dimethylbutane}$. In this reaction, the negatively charged chloride ions from two equivalents of ${\text{Cl}}_2$ get substituted on the highly substituted carbon atom of $3,3-\text{dimethylbut}-1-\text{yne}$ to form the desired tetrahalide product.

Conclusion

The product that is formed by the reaction is $1,1,2,2-\text{tetrachloro}-3,3-\text{dimethylbutane}$.

Interpretation Introduction

(c)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: The addition of a halogen to an alkyne chain leads to the formation of corresponding alkene or alkane. The reaction which includes the addition of bromine atoms to the alkyne chain is known as bromination.

The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Answer to Problem 11.38P

The product that is formed by the reaction is $1,2-\text{diphenylethyne}$.

Explanation of Solution

The reaction of $1,2-\text{diphenylethene}$ with one equivalent of ${\text{Cl}}_2$ is shown as,

Figure 3

The reaction of $1,2-\text{diphenylethene}$ with one equivalent ${\text{Cl}}_2$ results in the formation of $1,2-\text{dichloro}-1,2-\text{diphenylethane}$. In this reaction, the negatively charged chloride ions from one equivalents of ${\text{Cl}}_2$ get substituted on the highly substituted carbon atom of $3,3-\text{dimethylbut}-1-\text{yne}$ to form the desired dihalide product.

The reaction of $1,2-\text{dichloro}-1,2-\text{diphenylethane}$ with two equivalents of ${\text{NaNH}}_{\text{2}}$ is shown as,

Figure 4

The reaction of $1,2-\text{dichloro}-1,2-\text{diphenylethane}$ with two equivalents of ${\text{NaNH}}_{\text{2}}$ results in the formation of $1,2-\text{diphenylethyne}$. In this reaction, the strong base abstracts a proton from $1,2-\text{dichloro}-1,2-\text{diphenylethane}$ to form the desired alkyne product.

Conclusion

The product that is formed by the reaction is $1,2-\text{diphenylethyne}$.

Interpretation Introduction

(d)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: A stepwise procedure of transforming an alkyne into a carbonyl group is known hydroboration-oxidation reaction. In a hydroboration-oxidation reaction, a terminal alkyne reacts with ${\text{R}}_{\text{2}}\text{BH}$ followed by ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$, ${}^{-}\text{OH}$ to form an aldehyde.

Answer to Problem 11.38P

The product that is formed by the reaction is heptanal.

Explanation of Solution

The reaction of $\text{hept}-1-\text{yne}$ with ${\text{R}}_{\text{2}}\text{BH}$ followed by ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$, ${}^{-}\text{OH}$ is shown as,

Figure 5

The reaction of $\text{hept}-1-\text{yne}$ with ${\text{R}}_{\text{2}}\text{BH}$ followed by ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$, ${}^{-}\text{OH}$ results into the formation of alkenylborane of an aldehyde, heptanal. In this reaction, first of all, ${\text{R}}_{\text{2}}\text{BH}$ reacts with $\text{hept}-1-\text{yne}$ to form an alkenylborane. Then, alkenylborane reacts with ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$, ${}^{-}\text{OH}$ to form $\text{hept}-1-\text{en}-1-\text{ol}$. Then, $\text{hept}-1-\text{en}-1-\text{ol}$ undergoes tautomerization to form the desired product, heptanal.

Conclusion

The product that is formed by the reaction is heptanal.

Interpretation Introduction

(e)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Answer to Problem 11.38P

The product that is formed by the reaction is deuteroacetylene.

Explanation of Solution

The reaction of $\text{ethyn}-1-\text{ide}$ with ${\text{D}}_{\text{2}}\text{O}$ is shown as,

Figure 6

The reaction of $\text{ethyn}-1-\text{ide}$ with ${\text{D}}_{\text{2}}\text{O}$ results in the formation of deuteroacetylene. In this reaction, the deuterium from ${\text{D}}_{\text{2}}\text{O}$ gets substituted to $\text{ethyn}-1-\text{ide}$ that forms the desired alkyne product.

Conclusion

The product that is formed by the reaction is deuteroacetylene.

Interpretation Introduction

(f)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: A terminal alkyne reacts with ${\text{H}}_{\text{2}}\text{O}$ and ${\text{H}}_{\text{2}}{\text{SO}}_4$ (strong acid) to get hydrolyzed and forms an alcohol. In this case, hydrogen atom and $\text{OH}$ will get attach to the less substituted carbon atom to form a stable carbocation.

Answer to Problem 11.38P

The product that is formed by the reaction is $\left(E\right)-1-\text{cyclopentylprop}-1-\text{en}-\text{ol}$.

Explanation of Solution

The reaction of $\text{prop}-1-\text{yne}-1-\text{ylcyclopentane}$ with ${\text{H}}_{\text{2}}\text{O}$ and ${\text{H}}_{\text{2}}{\text{SO}}_4$ (strong acid) is shown as,

Figure 7

The reaction of $\text{prop}-1-\text{yne}-1-\text{ylcyclopentane}$ with ${\text{H}}_{\text{2}}\text{O}$ and ${\text{H}}_{\text{2}}{\text{SO}}_4$ (strong acid) results in the formation of $\left(E\right)-1-\text{cyclopentylprop}-1-\text{en}-\text{ol}$. In this reaction, $\text{prop}-1-\text{yne}-1-\text{ylcyclopentane}$ gets hydrolysed in the presence of strong acid that forms the desired product.

Conclusion

The product that is formed by the reaction is $\left(E\right)-1-\text{cyclopentylprop}-1-\text{en}-\text{ol}$.

Interpretation Introduction

(g)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Answer to Problem 11.38P

The product that is formed by the reaction is $\text{hept}-3-\text{yne}$.

Explanation of Solution

The reaction of $\text{but}-1-\text{yne}-1-\text{ide}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OTs}$ is shown as,

Figure 8

The reaction of $\text{but}-1-\text{yne}-1-\text{ide}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OTs}$ results in the formation of $\text{hept}-3-\text{yne}$. In this reaction, the alkyl group from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OTs}$ gets substituted to $\text{but}-1-\text{yne}-1-\text{ide}$ that forms the desired alkyne product.

Conclusion

The product that is formed by the reaction is $\text{hept}-3-\text{yne}$.

Interpretation Introduction

(h)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Answer to Problem 11.38P

The product that is formed by the reaction is $\left(1S,2R\right)-2-\text{ethynyl}-1-\text{methylcyclohexanol}$.

Explanation of Solution

The reaction of $\left(1S,6R\right)-1-\text{methyl}-7-\text{oxacyclo}\left[\mathrm{4.1.0}\right]\text{heptane}$ with acetylide ion is shown as,

Figure 9

The reaction of $\left(1S,6R\right)-1-\text{methyl}-7-\text{oxacyclo}\left[\mathrm{4.1.0}\right]\text{heptane}$ with acetylide ion results in the formation of $\left(1S,2R\right)-2-\text{ethynyl}-1-\text{methylcyclohexanol}$. In this reaction, epoxide ring gets open and the ethynyl group gets substituted to the intermediate that forms the desired alkyne product.

Conclusion

The product that is formed by the reaction is $\left(1S,2R\right)-2-\text{ethynyl}-1-\text{methylcyclohexanol}$.

Interpretation Introduction

(i)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Answer to Problem 11.38P

The product that is formed by the reaction is $\text{pent}-2-\text{yne}-1-\text{ylcyclohexane}$.

Explanation of Solution

The reaction of $\text{but}-1-\text{yne}$ with ${\text{NaNH}}_{\text{2}}$ followed by other reagent is shown as,

Figure 10

The reaction of $\text{but}-1-\text{yne}$ with ${\text{NaNH}}_{\text{2}}$ followed by $\left(\text{iodomethyl}\right)\text{cyclohexane}$ results in the formation of $\text{pent}-2-\text{yne}-1-\text{ylcyclohexane}$. In this reaction, the strong base abstracts a proton from $\text{but}-1-\text{yne}$ and cyclohexane ring gets substituted to the intermediate to form the desired alkyne product.

Conclusion

The product that is formed by the reaction is $\text{pent}-2-\text{yne}-1-\text{ylcyclohexane}$.

Interpretation Introduction

(j)

Interpretation: The product that is formed by the reaction is to be drawn.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Answer to Problem 11.38P

The product that is formed by the reaction is $4-\text{phenylbut}-3-\text{yn}-\text{ol}$.

Explanation of Solution

The reaction of ethynylbenzene with $\text{NaH}$ followed by other reagent is shown as,

Figure 11

The reaction of ethynylbenzene with $\text{NaH}$ results in the formation of $\text{sodium}\text{phenylethyn}-1-\text{ide}$. In this reaction, the strong base abstracts a proton from ethynyl that results in the formation of $\text{sodium}\text{phenylethyn}-1-\text{ide}$.

The reaction of $\text{sodium}\text{phenylethyn}-1-\text{ide}$ with oxirane followed by water molecule is shown as,

Figure 12

The reaction of $\text{sodium}\text{phenylethyn}-1-\text{ide}$ with oxirane followed by water molecule results in the formation of $4-\text{phenylbut}-3-\text{yn}-\text{ol}$. In this reaction, $\text{sodium}\text{phenylethyn}-1-\text{ide}$ behaves as a nucleophile that opens the oxirane ring to the more substituted carbon that results in the formation of $4-\text{phenylbut}-3-\text{yn}-1-\text{olate}$. Then, $4-\text{phenylbut}-3-\text{yn}-1-\text{olate}$ gets protonated by water molecule to form the desired product.

Conclusion

The product that is formed by the reaction is $4-\text{phenylbut}-3-\text{yn}-\text{ol}$.