Organic Chemistry
Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 14, Problem 14.33AP
Interpretation Introduction

(a)

Interpretation:

The synthesis of CH3CH2CD2CD2CH2CH3 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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

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Answer to Problem 14.33AP

The synthesis of CH3CH2CD2CD2CH2CH3 is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  1

Explanation of Solution

When acetylene reacts with sodamide in liqNH3 it forms sodium ethylide that further reacts with ethyl iodide to form 1-butyne. Then 1-butyne reacts with sodamide in liqNH3 and forms sodium butylide that further reacts with ethyl iodide to form 3-hexyne. Alkyne, 3-hexyne reacts with two equivalents of D2 in presence of Pd/C which gives the final product. Therefore, the synthesis of CH3CH2CD2CD2CH2CH3 from acetylene is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  2

Figure 1

Conclusion

The synthesis of CH3CH2CD2CD2CH2CH3 from acetylene is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The synthesis of 1hexene 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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

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Answer to Problem 14.33AP

The synthesis of 1hexene is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  3

Explanation of Solution

When acetylene reacts with sodamide in liqNH3 it forms sodium ethylide that further reacts with butyl iodide to form 1hexyne. Then 1hexyne reacts with one equivalent of H2 in presence of Lindlar catalyst which gives the final product 1hexene. Therefore, the synthesis of 1hexene from acetylene is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  4

Figure 2

Conclusion

The synthesis of 1hexene from acetylene is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The synthesis of 3hexanol 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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

Expert Solution
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Answer to Problem 14.33AP

The synthesis of 3hexanol is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  5

Explanation of Solution

When acetylene reacts with sodamide in liqNH3 it forms sodium ethylide that further reacts with ethyl iodide to form 1butyne. Then 1butyne reacts with sodamide in liqNH3 and forms sodium butylide that further reacts ethyl iodide to form 3hexyne. Then 3hexyne reacts with one equivalent of H2 in the presence of Lindlar catalyst that gives 3hexene. It reacts further under hydroboration oxidation reaction to give the final product that is 3hexanol. Therefore, the synthesis is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  6

Figure 3

Conclusion

The synthesis of 3hexanol from acetylene is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The synthesis of 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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

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Answer to Problem 14.33AP

The synthesis of 1-hexyne is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  7

Explanation of Solution

When acetylene reacts with sodamide in liqNH3 it forms sodium ethylide that further reacts with butyl iodide to form 1-hexyne. Therefore, the synthesis is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  8

Figure 4

Conclusion

The synthesis of 1-hexyne from acetylene is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The synthesis of CH3(CH2)7COOH 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 CnH2n-2. The name of the alkyne compounds ends with the suffix yne.

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Answer to Problem 14.33AP

The synthesis of 1-nonanoicacid is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  9

Explanation of Solution

When acetylene reacts with sodamide in liqNH3 it forms sodium ethylide that further reacts with pentyl iodide to form 1-heptyne. Then 1-heptyne reacts with sodamide in liqNH3 and forms sodium heptylide that further reacts with ethylene oxide which on hydrolysis gives 3-nonyne-1-ol. It undergoes hydrogenation reaction with two equivalents of H2 in presence of Pd/C to give 1-nonanol. It undergoes oxidation to gives the final product that is 1-nonanoicacid. Therefore, the synthesis is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  10

Figure 5

Conclusion

The synthesis of 1-nonanoicacid from acetylene is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The synthesis of (CH3)2CHCH2CH2CH2CH=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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

Expert Solution
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Answer to Problem 14.33AP

The synthesis of (CH3)2CHCH2CH2CH2CH=O is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  11

Explanation of Solution

When acetylene reacts with sodamide in liqNH3, it forms sodium ethylide that further reacts with isopentyl iodide to form 5-methyl-1-hexyne. Then 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.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  12

Figure 6

Conclusion

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

Interpretation Introduction

(g)

Interpretation:

The synthesis of cis-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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

Expert Solution
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Answer to Problem 14.33AP

The synthsis of cis-2-pentene is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  13

Explanation of Solution

When acetylene reacts with sodamide in liqNH3, it forms sodium ethylide that further reacts with ethyl iodide to form 1-butyne. Then 1-butyne reacts with sodamide in liqNH3 and forms sodium butylide that further reacts with methyl iodide to form 3-pentyne. Then 3-pentyne reacts with one equivalents of H2 in presence of Lindlar catalyst which gives the final product that is cis-2-pentene. Therefore, the synthesis is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  14

Figure 7

Conclusion

The synthesis of cis-2-pentene from acetylene is shown in Figure 7.

Interpretation Introduction

(h)

Interpretation:

The synthesis of trans-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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

Expert Solution
Check Mark

Answer to Problem 14.33AP

The synthesis of trans-3-decene is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  15

Explanation of Solution

When acetylene reacts with sodamide in liqNH3, it forms sodium ethylide that further reacts ethyl iodide to form 1-butyne. Then 1-butyne reacts with sodamide in liqNH3 and forms sodium butylide that further reacts hexyl iodide to form 3-decyne. Then 3-decyne reacts with one equivalents of H2 in presence of Na/NH3(liq) which gives the final product that is trans-3-decene. Therefore, the synthesis is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  16

Figure 8

Conclusion

The synthesis of trans-3-decene from acetylene is shown in Figure 8.

Interpretation Introduction

(i)

Interpretation:

The synthesis of meso-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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

Expert Solution
Check Mark

Answer to Problem 14.33AP

The synthesis of meso-4, 5-octanediol is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  17

Explanation of Solution

When acetylene reacts with sodamide in liqNH3, it forms sodium ethylide that further reacts propyl iodide to form 1-pentyne. Then 1-pentyne reacts with sodamide in liqNH3 and forms sodium pentylide that further reacts with propyl iodide to form 3-octyne. Then 3-octyne reacts with one equivalents of H2 in presence of Lindlar catalyst which gives cis-4-octene. It undergoes oxidation reaction to give the final product that is meso-4, 5-octanediol. Therefore, the synthesis is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  18

Figure 9

Conclusion

The synthesis of meso-4, 5-octanediol from acetylene is shown in Figure 9.

Interpretation Introduction

(j)

Interpretation:

The synthesis of (Z)-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 CnH2n2. The name of the alkyne compounds ends with the suffix yne.

Expert Solution
Check Mark

Answer to Problem 14.33AP

The synthesis of (Z)-3-hexen-1-ol is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  19

Explanation of Solution

When acetylene reacts with sodamide in liqNH3, it forms sodium ethylide that further reacts with ethyl iodide to form 1-butyne. Then 1-butyne reacts with sodamide in liqNH3 and forms sodium butylide that further reacts with ethylene oxide followed by hydrolysis to form 3-hexyne-1-ol. Then 3-hexyne-1-ol reacts with one equivalent of H2 in presence of Lindlar catalyst which gives (Z)-3-hexen-1-ol. Therefore, the synthesis is written as shown below.

Organic Chemistry, Chapter 14, Problem 14.33AP , additional homework tip  20

Figure 10

Conclusion

The synthesis of (Z)-3-hexen-1-ol from acetylene is shown in Figure 10.

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