Chapter 11, Problem 11.36P

Draw the products formed when $\text{hex}-1-\text{yne}$ is treated with each reagent.

a. $\text{HCl}\left(\text{2}\text{equiv}\right)$

b. $\text{HBr}\left(\text{2}\text{equiv}\right)$

c. ${\text{Cl}}_2\left(\text{2}\text{equiv}\right)$

d. ${\text{H}}_2\text{O}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{HgSO}}_{\text{4}}$

e. $\left[1\right]{\text{R}}_2\text{BH;}\left[2\right]{\text{H}}_2{\text{O}}_2,{\text{HO}}^{-}$

f. $\text{NaH}$

g. $\left[1\right]{}^{-}{\text{NH}}_2;\left[2\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$

h. $\left[1\right]{}^{-}{\text{NH}}_2;$[2] ; $\left[3\right]{\text{H}}_{\text{2}}\text{O}$

Interpretation Introduction

(a)

Interpretation: The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{HCl}\left(\text{2}\text{equiv}\right)$ 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 11.36P

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{HCl}\left(\text{2}\text{equiv}\right)$ is $2,2-\text{dichlorohexane}$.

Explanation of Solution

The reaction of $\text{hex}-1-\text{yne}$ with $\text{HCl}\left(\text{2}\text{equiv}\right)$ is shown as,

Figure 1

The reaction of $\text{hex}-1-\text{yne}$ with $\text{HCl}\left(\text{2}\text{equiv}\right)$ results in the formation of $2,2-\text{dichlorohexane}$. In this reaction, the negatively charged chloride ion from two equivalents of $\text{HCl}$ gets substituted on the highly substituted carbon atom of $\text{hex}-1-\text{yne}$ to form the desired dihalide product.

Conclusion

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{HCl}\left(\text{2}\text{equiv}\right)$ is $2,2-\text{dichlorohexane}$.

Interpretation Introduction

(b)

Interpretation: The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{HBr}\left(\text{2}\text{equiv}\right)$ 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 11.36P

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{HBr}\left(\text{2}\text{equiv}\right)$ is $2,2-\text{dibromohexane}$.

Explanation of Solution

The reaction of $\text{hex}-1-\text{yne}$ with $\text{HBr}\left(\text{2}\text{equiv}\right)$ is shown as,

Figure 2

The reaction of $\text{hex}-1-\text{yne}$ with $\text{HBr}\left(\text{2}\text{equiv}\right)$ results in the formation of $2,2-\text{dibromohexane}$. In this reaction, the negatively charged bromide ion from two equivalents of $\text{HBr}$ gets substituted on the highly substituted carbon atom of $\text{hex}-1-\text{yne}$ to form the desired dihalide product.

Conclusion

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{HBr}\left(\text{2}\text{equiv}\right)$ is $2,2-\text{dibromohexane}$.

Interpretation Introduction

(c)

Interpretation: The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with ${\text{Cl}}_2\left(\text{2}\text{equiv}\right)$ 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 11.36P

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with ${\text{Cl}}_2\left(\text{2}\text{equiv}\right)$ is $1,1,2,2-\text{tetrachlorohexane}$.

Explanation of Solution

The reaction of $\text{hex}-1-\text{yne}$ with ${\text{Cl}}_2\left(\text{2}\text{equiv}\right)$ is shown as,

Figure 3

The reaction of $\text{hex}-1-\text{yne}$ with ${\text{Cl}}_2\left(\text{2}\text{equiv}\right)$ results in the formation of $1,1,2,2-\text{tetrachlorohexane}$. In this reaction, the negatively charged chloride ion from two equivalents of ${\text{Cl}}_2$ gets substituted on the highly substituted carbon atom of $\text{hex}-1-\text{yne}$ to form the desired tetrahalide product.

Conclusion

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with ${\text{Cl}}_2\left(\text{2}\text{equiv}\right)$ is $1,1,2,2-\text{tetrachlorohexane}$.

Interpretation Introduction

(d)

Interpretation: The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with ${\text{H}}_2\text{O}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{HgSO}}_{\text{4}}$ is to be drawn.

Concept introduction: A terminal alkyne reacts with ${\text{H}}_{\text{2}}\text{O}$, ${\text{H}}_{\text{2}}{\text{SO}}_4$ (strong acid) and ${\text{HgSO}}_{\text{4}}$ to form a ketone. In this case, hydrogen atom and $\text{OH}$ will get attach to the less substituted carbon atom to form a stable carbocation.

The chemical equilibrium that exists between a keto form of a compound and an enol form of the same compound is known as keto-enol tautomerism. Tautomers refer to these keto and enol forms.

Answer to Problem 11.36P

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with ${\text{H}}_2\text{O}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{HgSO}}_{\text{4}}$ is $\text{hexan}-2-\text{one}$.

Explanation of Solution

The reaction of $\text{hex}-1-\text{yne}$ with ${\text{H}}_2\text{O}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{HgSO}}_{\text{4}}$ is shown as,

Figure 4

The reaction of $\text{hex}-1-\text{yne}$ with ${\text{H}}_2\text{O}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{HgSO}}_{\text{4}}$ results in the formation of $\text{hex}-1-\text{ene}-2-\text{ol}$. The terminal alkyne, $\text{hex}-1-\text{yne}$ reacts with the reagents ${\text{H}}_{\text{2}}\text{O}$, ${\text{HgSO}}_{\text{4}}$ in the presence of strong acid ${\text{H}}_{\text{2}}{\text{SO}}_4$ to form $\text{hex}-1-\text{ene}-2-\text{ol}$ as the desired product.

Then, the enol form, $\text{hex}-1-\text{ene}-2-\text{ol}$ forms a keto form that is shown below.

Figure 5

Thus, the keto-enol tautomerization of $\text{hex}-1-\text{ene}-2-\text{ol}$ forms the desired product, $\text{hexan}-2-\text{one}$.

Conclusion

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with ${\text{H}}_2\text{O}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{HgSO}}_{\text{4}}$ is $\text{hexan}-2-\text{one}$.

Interpretation Introduction

(e)

Interpretation: The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{\text{R}}_2\text{BH;}\left[2\right]{\text{H}}_2{\text{O}}_2,{\text{HO}}^{-}$ 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 11.36P

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{\text{R}}_2\text{BH;}\left[2\right]{\text{H}}_2{\text{O}}_2,{\text{HO}}^{-}$ is hexanal.

Explanation of Solution

The reaction of $\text{hex}-1-\text{yne}$ $\left[1\right]{\text{R}}_2\text{BH}$ is shown as,

Figure 6

The reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{\text{R}}_2\text{BH}$ results into the formation of alkenylborane.

Then, the reaction of alkenylborane with $\left[2\right]{\text{H}}_2{\text{O}}_2,{\text{HO}}^{-}$ is shown below.

Figure 7

Thus, the reaction of alkenylborane with $\left[2\right]{\text{H}}_2{\text{O}}_2,{\text{HO}}^{-}$ results in the formation of $\text{hex}-1-\text{en}-1-\text{ol}$.

In the last step, $\text{hex}-1-\text{en}-1-\text{ol}$ undergoes tautomerization as shown below.

Figure 8

Thus, the tautomerization of $\text{hex}-1-\text{en}-1-\text{ol}$ forms the desired product, hexanal.

Conclusion

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{\text{R}}_2\text{BH;}\left[2\right]{\text{H}}_2{\text{O}}_2,{\text{HO}}^{-}$ is hexanal.

Interpretation Introduction

(f)

Interpretation: The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{NaH}$ 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 11.36P

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{NaH}$ is $\text{hex}-1-\text{yn}-1-\text{ide}$.

Explanation of Solution

The reaction of $\text{hex}-1-\text{yne}$ with $\text{NaH}$ is shown as,

Figure 9

The reaction of $\text{hex}-1-\text{yne}$ with $\text{NaH}$ results in the formation of $\text{hex}-1-\text{yn}-1-\text{ide}$. In this reaction, the strong base abstracts a proton from $\text{hex}-1-\text{yne}$ to form the desired anionic product.

Conclusion

(a) The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\text{NaH}$ is $\text{hex}-1-\text{yn}-1-\text{ide}$.

Interpretation Introduction

(g)

Interpretation: The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2;\left[2\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ 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 11.36P

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2;\left[2\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ is $\text{oct}-3-\text{yne}$.

Explanation of Solution

The reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2$ is shown as,

Figure 10

The reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2$ results in the formation of $\text{hex}-1-\text{yn}-1-\text{ide}$. In this reaction, the strong base abstracts a proton from $\text{hex}-1-\text{yne}$ to form the anionic compound.

Then, the reaction of $\text{hex}-1-\text{yn}-1-\text{ide}$ with ethyl bromide is shown below.

Figure 11

Thus, in the above reaction, $\text{hex}-1-\text{yn}-1-\text{ide}$ behaves as a nucleophile that attach to the ethyl group and forms the desired product, $\text{oct}-3-\text{yne}$.

Conclusion

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2;\left[2\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ is $\text{oct}-3-\text{yne}$.

Interpretation Introduction

(h)

Interpretation: The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2;$;$\left[3\right]{\text{H}}_{\text{2}}\text{O}$ 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 11.36P

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2;$;$\left[3\right]{\text{H}}_{\text{2}}\text{O}$ is $\text{oct}-3-\text{yne}-1-\text{ol}$.

Explanation of Solution

The reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2$ is shown as,

Figure 12

The reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2$ results in the formation of $\text{hex}-1-\text{yn}-1-\text{ide}$. In this reaction, the strong base abstracts a proton from $\text{hex}-1-\text{yne}$ to form the anionic compound.

Then, the reaction of $\text{hex}-1-\text{yn}-1-\text{ide}$ with oxirane is shown below.

Figure 13

Thus, in the above reaction, $\text{hex}-1-\text{yn}-1-\text{ide}$ behaves as a nucleophile that opens the oxirane ring to the more substituted carbon that results in the formation of $\text{oct}-3-\text{yn}-1-\text{olate}$.

Then, the reaction of $\text{oct}-3-\text{yn}-1-\text{olate}$ with water molecule is shown below.

Figure 14

Thus, in the above reaction, $\text{oct}-3-\text{yn}-1-\text{olate}$ gets protonated to form the desired alcoholic product, $\text{oct}-3-\text{yne}-1-\text{ol}$.

Conclusion

The product that is formed by the reaction of $\text{hex}-1-\text{yne}$ with $\left[1\right]{}^{-}{\text{NH}}_2;$;$\left[3\right]{\text{H}}_{\text{2}}\text{O}$ is $\text{oct}-3-\text{yne}-1-\text{ol}$.