Chapter 14, Problem 14.33AP

Interpretation Introduction

(a)

Interpretation:

The synthesis of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CD}}_{\text{2}}{\text{CD}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CD}}_{\text{2}}{\text{CD}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ it forms sodium ethylide that further reacts with ethyl iodide to form $\text{1-butyne}$. Then $\text{1-butyne}$ reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ and forms sodium butylide that further reacts with ethyl iodide to form $\text{3-hexyne}$. Alkyne, $\text{3-hexyne}$ reacts with two equivalents of ${\text{D}}_{\text{2}}$ in presence of $\text{Pd}/\text{C}$ which gives the final product. Therefore, the synthesis of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CD}}_{\text{2}}{\text{CD}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ from acetylene is written as shown below.

Figure 1

Conclusion

The synthesis of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CD}}_{\text{2}}{\text{CD}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ from acetylene is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The synthesis of $1-\text{hexene}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of $1-\text{hexene}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ it forms sodium ethylide that further reacts with butyl iodide to form $1-\text{hexyne}$. Then $1-\text{hexyne}$ reacts with one equivalent of ${\text{H}}_{\text{2}}$ in presence of Lindlar catalyst which gives the final product $1-\text{hexene}$. Therefore, the synthesis of $1-\text{hexene}$ from acetylene is written as shown below.

Figure 2

Conclusion

The synthesis of $1-\text{hexene}$ from acetylene is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The synthesis of $3-\text{hexanol}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of $3-\text{hexanol}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ it forms sodium ethylide that further reacts with ethyl iodide to form $1-\text{butyne}$. Then $1-\text{butyne}$ reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ and forms sodium butylide that further reacts ethyl iodide to form $3-\text{hexyne}$. Then $3-\text{hexyne}$ reacts with one equivalent of ${\text{H}}_{\text{2}}$ in the presence of Lindlar catalyst that gives $3-\text{hexene}$. It reacts further under hydroboration oxidation reaction to give the final product that is $3-\text{hexanol}$. Therefore, the synthesis is written as shown below.

Figure 3

Conclusion

The synthesis of $3-\text{hexanol}$ from acetylene is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The synthesis of $\text{1-hexyne}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of $\text{1-hexyne}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ it forms sodium ethylide that further reacts with butyl iodide to form $\text{1-hexyne}$. Therefore, the synthesis is written as shown below.

Figure 4

Conclusion

The synthesis of $\text{1-hexyne}$ from acetylene is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The synthesis of ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}\text{COOH}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_{\text{n}}{\text{H}}_{\text{2n}-\text{2}}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of $\text{1-nonanoic}\text{acid}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ it forms sodium ethylide that further reacts with pentyl iodide to form $\text{1-heptyne}$. Then $\text{1-heptyne}$ reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ and forms sodium heptylide that further reacts with ethylene oxide which on hydrolysis gives $\text{3-nonyne-1-ol}$. It undergoes hydrogenation reaction with two equivalents of ${\text{H}}_{\text{2}}$ in presence of $\text{Pd}/\text{C}$ to give $\text{1-nonanol}$. It undergoes oxidation to gives the final product that is $\text{1-nonanoic}\text{acid}$. Therefore, the synthesis is written as shown below.

Figure 5

Conclusion

The synthesis of $\text{1-nonanoic}\text{acid}$ from acetylene is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The synthesis of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH=O}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH=O}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$, it forms sodium ethylide that further reacts with isopentyl iodide to form $\text{5-methyl-1-hexyne}$. Then $\text{5-methyl-1-hexyne}$ reacts with disimyl borane and forms organoborane. The organoborane undergoes hydroboration oxidation reaction to form enol form which tautomerizes to give final product that is heptanal. Therefore, the synthesis is written as shown below.

Figure 6

Conclusion

The synthesis of heptanal from acetylene is written as shown in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The synthesis of $cis\text{-2-pentene}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthsis of $cis\text{-2-pentene}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$, it forms sodium ethylide that further reacts with ethyl iodide to form $\text{1-butyne}$. Then $\text{1-butyne}$ reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ and forms sodium butylide that further reacts with methyl iodide to form $\text{3-pentyne}$. Then $\text{3-pentyne}$ reacts with one equivalents of ${\text{H}}_{\text{2}}$ in presence of Lindlar catalyst which gives the final product that is $cis\text{-2-pentene}$. Therefore, the synthesis is written as shown below.

Figure 7

Conclusion

The synthesis of $cis\text{-2-pentene}$ from acetylene is shown in Figure 7.

Interpretation Introduction

(h)

Interpretation:

The synthesis of $trans\text{-3-decene}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of $trans\text{-3-decene}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$, it forms sodium ethylide that further reacts ethyl iodide to form $\text{1-butyne}$. Then $\text{1-butyne}$ reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ and forms sodium butylide that further reacts hexyl iodide to form $\text{3-decyne}$. Then $\text{3-decyne}$ reacts with one equivalents of ${\text{H}}_{\text{2}}$ in presence of $\text{Na}/{\text{NH}}_{\text{3}}\left(\text{liq}\right)$ which gives the final product that is $trans\text{-3-decene}$. Therefore, the synthesis is written as shown below.

Figure 8

Conclusion

The synthesis of $trans\text{-3-decene}$ from acetylene is shown in Figure 8.

Interpretation Introduction

(i)

Interpretation:

The synthesis of $meso\text{-4, 5-octanediol}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of $meso\text{-4, 5-octanediol}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$, it forms sodium ethylide that further reacts propyl iodide to form $\text{1-pentyne}$. Then $\text{1-pentyne}$ reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ and forms sodium pentylide that further reacts with propyl iodide to form $\text{3-octyne}$. Then $\text{3-octyne}$ reacts with one equivalents of ${\text{H}}_{\text{2}}$ in presence of Lindlar catalyst which gives $cis\text{-4-octene}$. It undergoes oxidation reaction to give the final product that is $meso\text{-4, 5-octanediol}$. Therefore, the synthesis is written as shown below.

Figure 9

Conclusion

The synthesis of $meso\text{-4, 5-octanediol}$ from acetylene is shown in Figure 9.

Interpretation Introduction

(j)

Interpretation:

The synthesis of $\left(Z\right)\text{-3-hexen-1-ol}$ from acetylene is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon is of two types, alkenes and alkynes. The alkene contains a double bond between two carbon atoms. The alkynes contain a triple bond between two carbon atoms and follow a general formula ${\text{C}}_n{\text{H}}_{2n-2}$. The name of the alkyne compounds ends with the suffix $-yne$.

Answer to Problem 14.33AP

The synthesis of $\left(Z\right)\text{-3-hexen-1-ol}$ is written as shown below.

Explanation of Solution

When acetylene reacts with sodamide in $liq{\text{NH}}_{\text{3}}$, it forms sodium ethylide that further reacts with ethyl iodide to form $\text{1-butyne}$. Then $\text{1-butyne}$ reacts with sodamide in $liq{\text{NH}}_{\text{3}}$ and forms sodium butylide that further reacts with ethylene oxide followed by hydrolysis to form $\text{3-hexyne-1-ol}$. Then $\text{3-hexyne-1-ol}$ reacts with one equivalent of ${\text{H}}_{\text{2}}$ in presence of Lindlar catalyst which gives $\left(Z\right)\text{-3-hexen-1-ol}$. Therefore, the synthesis is written as shown below.

Figure 10

Conclusion

The synthesis of $\left(Z\right)\text{-3-hexen-1-ol}$ from acetylene is shown in Figure 10.