Chapter 14, Problem 14.27AP

Interpretation Introduction

(a)

Interpretation:

The product on reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}\text{, Pd/C}$ catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is ${\text{C}}_n{\text{H}}_{\text{2}n-\text{2}}$. The name of the alkyne compounds ends with suffix $-yne$. Hydrocarbons can also exist in ring like structures.

Answer to Problem 14.27AP

The product on reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}\text{, Pd/C}$ catalyst is octane, ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_3$.

Explanation of Solution

The reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}\text{, Pd/C}$ catalyst is shown below.

Figure 1

In the above reaction, octyne reacts with two moles of hydrogen to form octane. Octyne is an unsaturated molecule consisting of a triple bond. It reacts with hydrogen to form octane. Therefore, the product on reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}\text{, Pd/C}$ catalyst is octane, ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_3$.

Conclusion

The product on reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}\text{, Pd/C}$ catalyst is octane.

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}$, Lindlar’s catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is ${\text{C}}_n{\text{H}}_{\text{2}n-\text{2}}$. The name of the alkyne compounds ends with suffix $-yne$. Hydrocarbons can also exist in ring like structures.

Answer to Problem 14.27AP

The product on reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}$, Lindlar catalyst is shown below.

Explanation of Solution

The reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}$, Lindlar catalyst is shown below.

Figure 2

In the above reaction, octyne reacts with hydrogen to form octene. Octyne is an unsaturated molecule consisting of a triple bond. It reacts with hydrogen in presence of Lindlar’s catalyst to form $cis-$ octene. Therefore, the product on reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}$, Lindlar’s catalyst is shown below.

Figure 3

Conclusion

The product on reaction of $4-\text{octyne}$ with ${\text{H}}_{\text{2}}$, Lindlar’s catalyst is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The product on reaction of the product formed in part (b) with ${\text{O}}_3$, then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{/H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is ${\text{C}}_n{\text{H}}_{\text{2}n-\text{2}}$. The name of the alkyne compounds ends with suffix $-yne$. Hydrocarbons can also exist in ring like structures.

Answer to Problem 14.27AP

The product on reaction of the product formed in part (b) with ${\text{O}}_3$, then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{/H}}_{\text{2}}\text{O}$ is butanoic acid, ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_2\text{COOH}$.

Explanation of Solution

The product formed in part (b) is shown below.

Figure 4

The reaction of the product formed in part (b) with ${\text{O}}_3$, then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{/H}}_{\text{2}}\text{O}$ is shown below.

Figure 5

In the above reaction, octene a product from part (b) reacts with ozone to form ozonide. This ozonide formed hydrolysis to give two moles of butanoic acid as shown in figure 6. Therefore, the product formed on ozonolysis of octene is butanoic acid, ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_2\text{COOH}$.

Conclusion

The product on reaction of the product formed in part (b) with ${\text{O}}_3$, then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{/H}}_{\text{2}}\text{O}$ is butanoic acid.

Interpretation Introduction

(d)

Interpretation:

The product on reaction of the $4-\text{octyne}$ with $\text{Na}$ metal in liquid ${\text{NH}}_{\text{3}}$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is ${\text{C}}_n{\text{H}}_{\text{2}n-\text{2}}$. The name of the alkyne compounds ends with suffix $-yne$. Hydrocarbons can also exist in ring like structures.

Answer to Problem 14.27AP

The product on reaction of the $4-\text{octyne}$ with $\text{Na}$ metal in liquid ${\text{NH}}_{\text{3}}$ is shown below.

Explanation of Solution

The reaction of $4-\text{octyne}$ with $\text{Na}$ metal in liquid ${\text{NH}}_{\text{3}}$ is shown below.

Figure 6

In the above reaction, octyne reacts with sodium in liquid ammonia to form $trans-$ octene. Sodium in liquid ammonia is a reagent that specifically forms alkene in trans form. Therefore, the product formed on reaction of $4-\text{octyne}$ with $\text{Na}$ metal in liquid ${\text{NH}}_{\text{3}}$ is shown below.

Figure 7

Conclusion

The product on reaction of the $4-\text{octyne}$ with $\text{Na}$ metal in liquid ${\text{NH}}_{\text{3}}$ is shown in Figure 7.

Interpretation Introduction

(e)

Interpretation:

The product on reaction of $4-\text{octyne}$ with ${\text{Hg}}^{2+}{\text{, H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{, H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is ${\text{C}}_n{\text{H}}_{\text{2}n-\text{2}}$. The name of the alkyne compounds ends with suffix $-yne$. Hydrocarbons can also exist in ring like structures.

Answer to Problem 14.27AP

The product on reaction of the $4-\text{octyne}$ with ${\text{Hg}}^{2+}{\text{, H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{, H}}_{\text{2}}\text{O}$ is shown below.

Explanation of Solution

The reaction of $4-\text{octyne}$ with ${\text{Hg}}^{2+}{\text{, H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{, H}}_{\text{2}}\text{O}$ is shown below.

Figure 8

In the above reaction, octyne reacts with ${\text{Hg}}^{2+}{\text{, H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{, H}}_{\text{2}}\text{O}$ to form a ketone of eight carbon. This reaction is also known as oxymercuration-demercuration reaction. In the initial step an enol is formed which tautomerises to form a ketone. Therefore, the product formed on reaction of $4-\text{octyne}$ with ${\text{Hg}}^{2+}{\text{, H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{, H}}_{\text{2}}\text{O}$ is shown below.

Figure 9

Conclusion

The product on reaction of the $4-\text{octyne}$ with ${\text{Hg}}^{2+}{\text{, H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{, H}}_{\text{2}}\text{O}$ is shown in Figure 9.

Interpretation Introduction

(f)

Interpretation:

The product on reaction of the $4-\text{octyne}$ with ${\text{BH}}_{\text{3}}$, then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{/}^{-}\text{OH}$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is ${\text{C}}_n{\text{H}}_{\text{2}n-\text{2}}$. The name of the alkyne compounds ends with suffix $-yne$. Hydrocarbons can also exist in ring like structures.

Answer to Problem 14.27AP

The product formed on reaction of $4-\text{octyne}$ with ${\text{BH}}_{\text{3}}$ then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{/}^{-}\text{OH}$ is shown below.

Explanation of Solution

The reaction of $4-\text{octyne}$ with ${\text{BH}}_{\text{3}}$, then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{/}^{-}\text{OH}$ is shown below.

Figure 10

In the above reaction, octyne reacts with boron hydride to form organoborane. Organoborane reacts with peroxide and hydroxide ion to form an enol. This enol tautomerises to form $4-\text{octanone}$. Therefore, the product formed on reaction of $4-\text{octyne}$ with ${\text{BH}}_{\text{3}}$ then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{/}^{-}\text{OH}$ is shown below.

Figure 11

Conclusion

The product on reaction of the $4-\text{octyne}$ with ${\text{BH}}_{\text{3}}$ then ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{/}^{-}\text{OH}$ is shown in Figure 11.