Chapter 11, Problem 11.20P

Interpretation Introduction

(a)

Interpretation: The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ 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 11.20P

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ is,

Explanation of Solution

The reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ is shown as,

Figure 1

Thus, the product that is formed by the reaction of $\text{but}-1-\text{yn}-1-\text{ylsodium}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with propyl bromide, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ is $\text{hept}-3-\text{yne}$. In the above reaction, butylate anion behaves as a strong nucleophile which undergoes nucleophilic substitution reaction with propyl bromide or $1-\text{bromopropane}$ (primary alkyl halide). Thus, in this reaction, $\text{NaBr}$ gets eliminated that result in the formation of the desired product, $\text{hept}-3-\text{yne}$.

Conclusion

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ is $\text{hept}-3-\text{yne}$.

Interpretation Introduction

(b)

Interpretation: The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}$ 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 11.20P

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}$ is,

Explanation of Solution

The reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}$ is shown as,

Figure 2

Thus, the product that is formed by the reaction of $\text{but}-1-\text{yn}-1-\text{ylsodium}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with $\text{1}-\text{chloro}-\text{3}-\text{methylbutane}$, ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}$ is $\text{7}-\text{methyloct}-3-\text{yne}$. In the above reaction, butylate anion behaves as a strong nucleophile which undergoes nucleophilic substitution reaction with $\text{1}-\text{chloro}-\text{3}-\text{methylbutane}$ (primary alkyl halide). Thus, in this reaction, $\text{NaCl}$ gets eliminated that result in the formation of the desired product, $\text{7}-\text{methyloct}-3-\text{yne}$.

Conclusion

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}$ is $\text{7}-\text{methyloct}-3-\text{yne}$.

Interpretation Introduction

(c)

Interpretation: The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_3\text{CCl}$ is to be drawn.

Concept introduction: The bimolecular elimination reaction that favours the removal of a HX substituent from an alkyl halide that needs a base to remove $\text{H}$ from carbon adjacent to the leaving group that in turn forms a double bond is termed ${\text{E}}^2$ elimination reaction.

Answer to Problem 11.20P

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_3\text{CCl}$ is,

Explanation of Solution

The reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_3\text{CCl}$ is shown as,

Figure 3

Thus, the products that are formed by the reaction of $\text{but}-1-\text{yn}-1-\text{ylsodium}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with $\text{3}-\text{chloro}-\text{3}-\text{ethylpentane}$, ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_3\text{CCl}$ are $\text{3}-\text{ethylpent}-2-\text{ene}$ and $\text{but}-1-\text{yne}$. Butylate anion behaves as a strong nucleophile, but in case of secondary and tertiary alkyl halide, it undergoes $\text{E2}$ elimination reaction due to steric hindrance. Thus, the butylate anion undergoes ${\text{E}}^2$ elimination reaction with $\text{3}-\text{chloro}-\text{3}-\text{ethylpentane}$ (tertiary alkyl halide). Thus, this results in the formation of two products, $\text{3}-\text{ethylpent}-2-\text{ene}$ and $\text{but}-1-\text{yne}$.

Conclusion

The products that are formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_3\text{CCl}$ are $\text{3}-\text{ethylpent}-2-\text{ene}$ and $\text{but}-1-\text{yne}$.

Interpretation Introduction

(d)

Interpretation: The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\text{BrCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ 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 11.20P

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\text{BrCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ is,

Explanation of Solution

The reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\text{BrCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ is shown as,

Figure 4

Thus, the major product that is formed by the reaction of $\text{but}-1-\text{yn}-1-\text{ylsodium}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with $\text{4}-\text{bromobutan}-\text{1}-\text{ol}$, ${\text{BrCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ is $\text{but}-1-\text{yne}$. In this reaction, butylate anion behaves as a strong nucleophile. Thus, it abstracts an acidic hydrogen atom from the alcoholic group.. Thus, this results in the formation of two products, $\text{4}-\text{bromobutan}-1-\text{olate}$ (minor) and $\text{but}-1-\text{yne}$ (major).

Conclusion

The major product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ${\text{BrCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ is $\text{but}-1-\text{yne}$.

Interpretation Introduction

(e)

Interpretation: The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ethylene oxide followed by ${\text{H}}_{\text{2}}\text{O}$ 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 11.20P

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ethylene oxide followed by ${\text{H}}_{\text{2}}\text{O}$ is,

Explanation of Solution

The reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ethylene oxide or oxirane followed by ${\text{H}}_{\text{2}}\text{O}$ is shown as,

Figure 5

Thus, the major product that is formed by the reaction of $\text{but}-1-\text{yn}-1-\text{ylsodium}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ethylene oxide or oxirane followed by ${\text{H}}_{\text{2}}\text{O}$ is $\text{hex}-3-\text{yne}-1-\text{ol}$. In this reaction, butylate anion behaves as a strong nucleophile which on reaction with oxirane forms $\text{hex}-3-\text{yne}-1-\text{ol}$.

Conclusion

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with ethylene oxide followed by ${\text{H}}_{\text{2}}\text{O}$ is $\text{hex}-3-\text{yne}-1-\text{ol}$.

Interpretation Introduction

(f)

Interpretation: The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with propene oxide followed by ${\text{H}}_{\text{2}}\text{O}$ 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 11.20P

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with propene oxide followed by ${\text{H}}_{\text{2}}\text{O}$ is,

Explanation of Solution

The reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with propene oxide or $2-\text{methyloxirane}$ followed by ${\text{H}}_{\text{2}}\text{O}$ is shown as,

Figure 6

Thus, the major product that is formed by the reaction of $\text{but}-1-\text{yn}-1-\text{ylsodium}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with propene oxide or $2-\text{methyloxirane}$ followed by ${\text{H}}_{\text{2}}\text{O}$ is $\text{hept}-4-\text{yne}-2-\text{ol}$. In this reaction, butylate anion behaves as a strong nucleophile which on reaction with $2-\text{methyloxirane}$ forms $\text{hept}-4-\text{yne}-2-\text{ol}$.

Conclusion

The product that is formed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ with propene oxide followed by ${\text{H}}_{\text{2}}\text{O}$ is $\text{hept}-4-\text{yne}-2-\text{ol}$.