Chapter 11, Problem 11.60P

Devise a synthesis of each compound using ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material (a) ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_3$; (b) ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}\text{CH}\left(\text{OH}\right){\text{CH}}_3$. You may use any other needed reagents.

Interpretation Introduction

(a)

Interpretation: A synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is to be shown.

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 11.60P

A synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is,

Explanation of Solution

The first step of the synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is shown as,

Figure 1

In this step, the starting organic material $\text{propan}-1-\text{ol}$ undergoes dehydration reaction due to the presence of acid, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ that results in the formation of an alkene, $\text{prop}-1-\text{ene}$. Then, $\text{prop}-1-\text{ene}$ reacts with a halogen like bromine to form vicinal dihalide, which is an electrophilic addition reaction. This vicinal dihalide on reaction with the two equivalents of the strong base, sodium amide forms $\text{prop}-1-\text{yne}$. This reaction is called dehydrohalogenation.

The second step of the synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is shown as,

Figure 2

In this step, the starting organic material $\text{propan}-1-\text{ol}$ reacts with ${\text{PCl}}_{\text{3}}$ to form $1-\text{chloropropane}$ . In this reaction, ethanol undergoes nucleophilic substitution reaction.

The third step of the synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is shown as,

Figure 3

In this step, the intermediates, $\text{prop}-1-\text{yne}$ and $1-\text{chloropropane}$ reacts together in the presence of sodium amide to form the desired product, $\text{hex}-2-\text{yne}$.

Conclusion

A synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is shown in Figure 1, Figure 2 and Figure 3.

Interpretation Introduction

(b)

Interpretation: A synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}\text{CH}\left(\text{OH}\right){\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is to be shown.

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 11.60P

A synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}\text{CH}\left(\text{OH}\right){\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is,

Explanation of Solution

The first step of the synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}\text{CH}\left(\text{OH}\right){\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is shown as,

Figure 4

In this step, the starting organic material $\text{propan}-1-\text{ol}$ undergoes dehydration reaction due to the presence of acid, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ that results in the formation of an alkene, $\text{prop}-1-\text{ene}$. Then, $\text{prop}-1-\text{ene}$ reacts with a halogen like bromine to form vicinal dihalide, which is an electrophilic addition reaction. This vicinal dihalide on reaction with the two equivalents of the strong base, sodium amide forms $\text{prop}-1-\text{yne}$. This reaction is called dehydrohalogenation.

The second step of the synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}\text{CH}\left(\text{OH}\right){\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is shown as,

Figure 5

In this step, the starting organic material $\text{1}-\text{propene}$ formed in the first step reacts with ${\text{X}}_2={\text{Cl}}_{\text{2}}\text{or}{\text{Br}}_{\text{2}}$ and water to form halohydrin. Then, this halohydrin reacts with base, in which the base cause deprotonation of alcohol and the ring gets close by the displacement of halide through intramolecular ${\text{S}}_{\text{N}}\text{2}$ reaction. This ring closure forms $2-\text{methyloxirane}$.

The third step of the synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}\text{CH}\left(\text{OH}\right){\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is shown as,

Figure 6

In this step, the intermediates, $\text{prop}-1-\text{yne}$ and $2-\text{methyloxirane}$ reacts together in the presence of sodium amide to form the desired product, $\text{hex}-4-\text{yne}-\text{2}-\text{ol}$.

Conclusion

A synthesis of ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{CCH}}_{\text{2}}\text{CH}\left(\text{OH}\right){\text{CH}}_3$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$ as the only organic starting material is shown in Figure 4, Figure 5, and Figure 6.