(a) Interpretation: The product on reaction of 1-hexyne with HBr is to be stated. Concept introduction: The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is C n H 2 n − 2 . The name of the alkyne compounds ends with suffix − y n e . Hydrocarbons can also exist in ring like structures.

Question

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Answer 1

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 14, Problem 14.26AP

Interpretation Introduction

(a)

Interpretation:

The product on reaction of $1-hexyne$ with $HBr$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 1

Explanation of Solution

The reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 2

Figure 1

In the above reaction, hexyne reacts with $HBr$ to form $2-bromohex-1-ene$ . Hexyne is an unsaturated molecule consisting of a triple bond. It reacts with to form a haloalkene. Therefore, the product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 4

Figure 2

Conclusion

The product on reaction of $1-hexyne$ with $HBr$ is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 5

Explanation of Solution

The reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 6

Figure 3

In the above reaction, hexyne reacts with hydrogen to form hexene. Hexyne is an unsaturated molecule consisting of a triple bond. It reacts hydrogen in presence of $Pd/C$ to give alkene. Therefore, the product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 7

Figure 4

Conclusion

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , 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 $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 8

Explanation of Solution

The reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 9

Figure 5

In the above reaction, hexyne reacts with $H2,Pd/C$ in presence of Lindlar’s catalyst to form an alkene. Due to the presence of Lindlar’s catalyst the hydrogenation of alkyne stops at alkene and does not proceed further to give alkane. Therefore, the product formed on reaction of $1-hexyne$ with the $H2,Pd/C$ in presence of Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 10

Figure 6

Conclusion

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation:

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 11

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 12

Figure 7

The reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 13

Figure 8

The above reaction is known as ozonlysis reaction. In the above reaction, a double bond is cleaved and oxidized to give two products. Reaction of hexene with $O3$ followed by $(CH3)2S$ forms pentanal and formaldehyde. Therefore, the products formed on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 14

Figure 9

Conclusion

The products on reaction of reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown in Figure 9.

Interpretation Introduction

(e)

Interpretation:

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 15

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 16

Figure 7

The reaction of above compound with $BH3$ in $THF$ , then $H2O2/−OH$

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 17

Figure 10

In the above reaction, hexene reacts with boron hydride to form organoborane which further reacts with peroxide to form an alcohol. The alcohol thus formed is by anti-markovnikov addition.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 18

Figure 11

Conclusion

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown in Figure 11.

Interpretation Introduction

(f)

Interpretation:

The product on reaction of the product formed in part (c) with $Br2$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 19

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 20

Figure 7

The reaction of above compound with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 21

Figure 12

In the above reaction, hexene being unsaturated reacts with bromine molecule to form a dibromoalkane, that is $1, 2-dibromohexane$ . In the above reaction, unsaturated molecule is forming a saturated dihaloalkane on reaction with bromine molecule. Therefore, the product formed on reaction of product formed in (c) with bromine molecule is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 22

Figure 13

Conclusion

The product on reaction of the product formed in part (c) with $Br2$ is shown in Figure 13.

Interpretation Introduction

(g)

Interpretation:

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 23

Explanation of Solution

The reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 24

Figure 14

Alkynes react with sodamide $(NaNH2)$ in presence of liquid ammonia to release an acidic proton and to form an anion of alkyne. Therefore, on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia releases the acidic proton to form an anion of hexyne. The product therefore formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 25

Figure 15

Conclusion

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown in Figure 15.

Interpretation Introduction

(h)

Interpretation:

The product on reaction of the product formed in part (g) with $CH3CH2I$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 26

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 27

Figure 15

The reaction of above compound with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 28

Figure 16

In the above reaction, the anion of alkyne obtained reacts with iodoethane to form an alkyne of eight carbons. It is non-terminal alkyne, that is, triple bond is not situated at the terminal end. The product formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 29

Figure 17

Conclusion

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown in Figure 17.

Interpretation Introduction

(i)

Interpretation:

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 30

Explanation of Solution

The reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 31

Figure 18

In the above reaction, hexyne reacts with $Hg2+, H2SO4, H2O$ to form a ketone of six carbons. This reaction is also known as oxymercuration-demercuration reaction. In the initial step an enol is formed which tautomerism to form a ketone. Therefore, the product formed on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 32

Figure 19

Conclusion

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown in Figure 19.

Interpretation Introduction

(j)

Interpretation:

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 33

Explanation of Solution

The reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 34

Figure 20

In the above reaction, hexyne reacts with disiamyl borane to form an organoborane. This reaction is also known as hydroboration-oxidation reaction. The organo-borane gives an enol which tautomerism to form an aldehyde of six carbons. Therefore, the product formed on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 35

Figure 21

Conclusion

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown in Figure 21.

Interpretation Introduction

(k)

Interpretation:

The product on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 36

Explanation of Solution

The reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 37

Figure 22

In the above reaction, $CH3CH2MgBr$ is Grignard reagent that reacts with alkyne to form an alkane, that is ethane and another Grignard reagent. Hexyne on reaction with Grignard reagent releases the cationic terminal hydrogen which goes on to react with the negative part of Grignard reagent and forms an alkane. The anionic part of alkyne and positive part of Grignard reagent react to form another Grignard reagent. The products therefore obtained are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 38

Figure 23

Conclusion

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ are shown in Figure 23.

Interpretation Introduction

(l)

Interpretation:

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 39

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 40

Figure 23

The reaction of above compound with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 41

Figure 24

In the above reaction, the Grignard reagent obtained in part (k) reacts with ethylene oxide followed by hydrolysis to form an alcohol with a triple bond in between the chain. The product formed on reaction of product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 42

Figure 25

Conclusion

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is in Figure 25.

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Answer 2

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 14, Problem 14.26AP

Interpretation Introduction

(a)

Interpretation:

The product on reaction of $1-hexyne$ with $HBr$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 1

Explanation of Solution

The reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 2

Figure 1

In the above reaction, hexyne reacts with $HBr$ to form $2-bromohex-1-ene$ . Hexyne is an unsaturated molecule consisting of a triple bond. It reacts with to form a haloalkene. Therefore, the product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 4

Figure 2

Conclusion

The product on reaction of $1-hexyne$ with $HBr$ is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 5

Explanation of Solution

The reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 6

Figure 3

In the above reaction, hexyne reacts with hydrogen to form hexene. Hexyne is an unsaturated molecule consisting of a triple bond. It reacts hydrogen in presence of $Pd/C$ to give alkene. Therefore, the product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 7

Figure 4

Conclusion

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , 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 $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 8

Explanation of Solution

The reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 9

Figure 5

In the above reaction, hexyne reacts with $H2,Pd/C$ in presence of Lindlar’s catalyst to form an alkene. Due to the presence of Lindlar’s catalyst the hydrogenation of alkyne stops at alkene and does not proceed further to give alkane. Therefore, the product formed on reaction of $1-hexyne$ with the $H2,Pd/C$ in presence of Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 10

Figure 6

Conclusion

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation:

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 11

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 12

Figure 7

The reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 13

Figure 8

The above reaction is known as ozonlysis reaction. In the above reaction, a double bond is cleaved and oxidized to give two products. Reaction of hexene with $O3$ followed by $(CH3)2S$ forms pentanal and formaldehyde. Therefore, the products formed on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 14

Figure 9

Conclusion

The products on reaction of reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown in Figure 9.

Interpretation Introduction

(e)

Interpretation:

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 15

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 16

Figure 7

The reaction of above compound with $BH3$ in $THF$ , then $H2O2/−OH$

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 17

Figure 10

In the above reaction, hexene reacts with boron hydride to form organoborane which further reacts with peroxide to form an alcohol. The alcohol thus formed is by anti-markovnikov addition.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 18

Figure 11

Conclusion

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown in Figure 11.

Interpretation Introduction

(f)

Interpretation:

The product on reaction of the product formed in part (c) with $Br2$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 19

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 20

Figure 7

The reaction of above compound with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 21

Figure 12

In the above reaction, hexene being unsaturated reacts with bromine molecule to form a dibromoalkane, that is $1, 2-dibromohexane$ . In the above reaction, unsaturated molecule is forming a saturated dihaloalkane on reaction with bromine molecule. Therefore, the product formed on reaction of product formed in (c) with bromine molecule is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 22

Figure 13

Conclusion

The product on reaction of the product formed in part (c) with $Br2$ is shown in Figure 13.

Interpretation Introduction

(g)

Interpretation:

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 23

Explanation of Solution

The reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 24

Figure 14

Alkynes react with sodamide $(NaNH2)$ in presence of liquid ammonia to release an acidic proton and to form an anion of alkyne. Therefore, on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia releases the acidic proton to form an anion of hexyne. The product therefore formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 25

Figure 15

Conclusion

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown in Figure 15.

Interpretation Introduction

(h)

Interpretation:

The product on reaction of the product formed in part (g) with $CH3CH2I$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 26

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 27

Figure 15

The reaction of above compound with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 28

Figure 16

In the above reaction, the anion of alkyne obtained reacts with iodoethane to form an alkyne of eight carbons. It is non-terminal alkyne, that is, triple bond is not situated at the terminal end. The product formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 29

Figure 17

Conclusion

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown in Figure 17.

Interpretation Introduction

(i)

Interpretation:

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 30

Explanation of Solution

The reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 31

Figure 18

In the above reaction, hexyne reacts with $Hg2+, H2SO4, H2O$ to form a ketone of six carbons. This reaction is also known as oxymercuration-demercuration reaction. In the initial step an enol is formed which tautomerism to form a ketone. Therefore, the product formed on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 32

Figure 19

Conclusion

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown in Figure 19.

Interpretation Introduction

(j)

Interpretation:

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 33

Explanation of Solution

The reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 34

Figure 20

In the above reaction, hexyne reacts with disiamyl borane to form an organoborane. This reaction is also known as hydroboration-oxidation reaction. The organo-borane gives an enol which tautomerism to form an aldehyde of six carbons. Therefore, the product formed on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 35

Figure 21

Conclusion

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown in Figure 21.

Interpretation Introduction

(k)

Interpretation:

The product on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 36

Explanation of Solution

The reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 37

Figure 22

In the above reaction, $CH3CH2MgBr$ is Grignard reagent that reacts with alkyne to form an alkane, that is ethane and another Grignard reagent. Hexyne on reaction with Grignard reagent releases the cationic terminal hydrogen which goes on to react with the negative part of Grignard reagent and forms an alkane. The anionic part of alkyne and positive part of Grignard reagent react to form another Grignard reagent. The products therefore obtained are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 38

Figure 23

Conclusion

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ are shown in Figure 23.

Interpretation Introduction

(l)

Interpretation:

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 39

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 40

Figure 23

The reaction of above compound with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 41

Figure 24

In the above reaction, the Grignard reagent obtained in part (k) reacts with ethylene oxide followed by hydrolysis to form an alcohol with a triple bond in between the chain. The product formed on reaction of product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 42

Figure 25

Conclusion

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is in Figure 25.

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Answer 3

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 14, Problem 14.26AP

Interpretation Introduction

(a)

Interpretation:

The product on reaction of $1-hexyne$ with $HBr$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 1

Explanation of Solution

The reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 2

Figure 1

In the above reaction, hexyne reacts with $HBr$ to form $2-bromohex-1-ene$ . Hexyne is an unsaturated molecule consisting of a triple bond. It reacts with to form a haloalkene. Therefore, the product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 4

Figure 2

Conclusion

The product on reaction of $1-hexyne$ with $HBr$ is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 5

Explanation of Solution

The reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 6

Figure 3

In the above reaction, hexyne reacts with hydrogen to form hexene. Hexyne is an unsaturated molecule consisting of a triple bond. It reacts hydrogen in presence of $Pd/C$ to give alkene. Therefore, the product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 7

Figure 4

Conclusion

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , 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 $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 8

Explanation of Solution

The reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 9

Figure 5

In the above reaction, hexyne reacts with $H2,Pd/C$ in presence of Lindlar’s catalyst to form an alkene. Due to the presence of Lindlar’s catalyst the hydrogenation of alkyne stops at alkene and does not proceed further to give alkane. Therefore, the product formed on reaction of $1-hexyne$ with the $H2,Pd/C$ in presence of Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 10

Figure 6

Conclusion

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation:

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 11

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 12

Figure 7

The reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 13

Figure 8

The above reaction is known as ozonlysis reaction. In the above reaction, a double bond is cleaved and oxidized to give two products. Reaction of hexene with $O3$ followed by $(CH3)2S$ forms pentanal and formaldehyde. Therefore, the products formed on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 14

Figure 9

Conclusion

The products on reaction of reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown in Figure 9.

Interpretation Introduction

(e)

Interpretation:

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 15

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 16

Figure 7

The reaction of above compound with $BH3$ in $THF$ , then $H2O2/−OH$

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 17

Figure 10

In the above reaction, hexene reacts with boron hydride to form organoborane which further reacts with peroxide to form an alcohol. The alcohol thus formed is by anti-markovnikov addition.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 18

Figure 11

Conclusion

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown in Figure 11.

Interpretation Introduction

(f)

Interpretation:

The product on reaction of the product formed in part (c) with $Br2$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 19

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 20

Figure 7

The reaction of above compound with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 21

Figure 12

In the above reaction, hexene being unsaturated reacts with bromine molecule to form a dibromoalkane, that is $1, 2-dibromohexane$ . In the above reaction, unsaturated molecule is forming a saturated dihaloalkane on reaction with bromine molecule. Therefore, the product formed on reaction of product formed in (c) with bromine molecule is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 22

Figure 13

Conclusion

The product on reaction of the product formed in part (c) with $Br2$ is shown in Figure 13.

Interpretation Introduction

(g)

Interpretation:

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 23

Explanation of Solution

The reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 24

Figure 14

Alkynes react with sodamide $(NaNH2)$ in presence of liquid ammonia to release an acidic proton and to form an anion of alkyne. Therefore, on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia releases the acidic proton to form an anion of hexyne. The product therefore formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 25

Figure 15

Conclusion

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown in Figure 15.

Interpretation Introduction

(h)

Interpretation:

The product on reaction of the product formed in part (g) with $CH3CH2I$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 26

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 27

Figure 15

The reaction of above compound with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 28

Figure 16

In the above reaction, the anion of alkyne obtained reacts with iodoethane to form an alkyne of eight carbons. It is non-terminal alkyne, that is, triple bond is not situated at the terminal end. The product formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 29

Figure 17

Conclusion

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown in Figure 17.

Interpretation Introduction

(i)

Interpretation:

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 30

Explanation of Solution

The reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 31

Figure 18

In the above reaction, hexyne reacts with $Hg2+, H2SO4, H2O$ to form a ketone of six carbons. This reaction is also known as oxymercuration-demercuration reaction. In the initial step an enol is formed which tautomerism to form a ketone. Therefore, the product formed on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 32

Figure 19

Conclusion

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown in Figure 19.

Interpretation Introduction

(j)

Interpretation:

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 33

Explanation of Solution

The reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 34

Figure 20

In the above reaction, hexyne reacts with disiamyl borane to form an organoborane. This reaction is also known as hydroboration-oxidation reaction. The organo-borane gives an enol which tautomerism to form an aldehyde of six carbons. Therefore, the product formed on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 35

Figure 21

Conclusion

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown in Figure 21.

Interpretation Introduction

(k)

Interpretation:

The product on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 36

Explanation of Solution

The reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 37

Figure 22

In the above reaction, $CH3CH2MgBr$ is Grignard reagent that reacts with alkyne to form an alkane, that is ethane and another Grignard reagent. Hexyne on reaction with Grignard reagent releases the cationic terminal hydrogen which goes on to react with the negative part of Grignard reagent and forms an alkane. The anionic part of alkyne and positive part of Grignard reagent react to form another Grignard reagent. The products therefore obtained are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 38

Figure 23

Conclusion

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ are shown in Figure 23.

Interpretation Introduction

(l)

Interpretation:

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 39

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 40

Figure 23

The reaction of above compound with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 41

Figure 24

In the above reaction, the Grignard reagent obtained in part (k) reacts with ethylene oxide followed by hydrolysis to form an alcohol with a triple bond in between the chain. The product formed on reaction of product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 42

Figure 25

Conclusion

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is in Figure 25.

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Answer 4

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 14, Problem 14.26AP

Interpretation Introduction

(a)

Interpretation:

The product on reaction of $1-hexyne$ with $HBr$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 1

Explanation of Solution

The reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 2

Figure 1

In the above reaction, hexyne reacts with $HBr$ to form $2-bromohex-1-ene$ . Hexyne is an unsaturated molecule consisting of a triple bond. It reacts with to form a haloalkene. Therefore, the product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 4

Figure 2

Conclusion

The product on reaction of $1-hexyne$ with $HBr$ is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 5

Explanation of Solution

The reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 6

Figure 3

In the above reaction, hexyne reacts with hydrogen to form hexene. Hexyne is an unsaturated molecule consisting of a triple bond. It reacts hydrogen in presence of $Pd/C$ to give alkene. Therefore, the product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 7

Figure 4

Conclusion

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , 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 $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 8

Explanation of Solution

The reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 9

Figure 5

In the above reaction, hexyne reacts with $H2,Pd/C$ in presence of Lindlar’s catalyst to form an alkene. Due to the presence of Lindlar’s catalyst the hydrogenation of alkyne stops at alkene and does not proceed further to give alkane. Therefore, the product formed on reaction of $1-hexyne$ with the $H2,Pd/C$ in presence of Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 10

Figure 6

Conclusion

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation:

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 11

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 12

Figure 7

The reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 13

Figure 8

The above reaction is known as ozonlysis reaction. In the above reaction, a double bond is cleaved and oxidized to give two products. Reaction of hexene with $O3$ followed by $(CH3)2S$ forms pentanal and formaldehyde. Therefore, the products formed on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 14

Figure 9

Conclusion

The products on reaction of reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown in Figure 9.

Interpretation Introduction

(e)

Interpretation:

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 15

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 16

Figure 7

The reaction of above compound with $BH3$ in $THF$ , then $H2O2/−OH$

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 17

Figure 10

In the above reaction, hexene reacts with boron hydride to form organoborane which further reacts with peroxide to form an alcohol. The alcohol thus formed is by anti-markovnikov addition.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 18

Figure 11

Conclusion

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown in Figure 11.

Interpretation Introduction

(f)

Interpretation:

The product on reaction of the product formed in part (c) with $Br2$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 19

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 20

Figure 7

The reaction of above compound with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 21

Figure 12

In the above reaction, hexene being unsaturated reacts with bromine molecule to form a dibromoalkane, that is $1, 2-dibromohexane$ . In the above reaction, unsaturated molecule is forming a saturated dihaloalkane on reaction with bromine molecule. Therefore, the product formed on reaction of product formed in (c) with bromine molecule is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 22

Figure 13

Conclusion

The product on reaction of the product formed in part (c) with $Br2$ is shown in Figure 13.

Interpretation Introduction

(g)

Interpretation:

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 23

Explanation of Solution

The reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 24

Figure 14

Alkynes react with sodamide $(NaNH2)$ in presence of liquid ammonia to release an acidic proton and to form an anion of alkyne. Therefore, on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia releases the acidic proton to form an anion of hexyne. The product therefore formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 25

Figure 15

Conclusion

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown in Figure 15.

Interpretation Introduction

(h)

Interpretation:

The product on reaction of the product formed in part (g) with $CH3CH2I$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 26

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 27

Figure 15

The reaction of above compound with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 28

Figure 16

In the above reaction, the anion of alkyne obtained reacts with iodoethane to form an alkyne of eight carbons. It is non-terminal alkyne, that is, triple bond is not situated at the terminal end. The product formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 29

Figure 17

Conclusion

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown in Figure 17.

Interpretation Introduction

(i)

Interpretation:

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 30

Explanation of Solution

The reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 31

Figure 18

In the above reaction, hexyne reacts with $Hg2+, H2SO4, H2O$ to form a ketone of six carbons. This reaction is also known as oxymercuration-demercuration reaction. In the initial step an enol is formed which tautomerism to form a ketone. Therefore, the product formed on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 32

Figure 19

Conclusion

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown in Figure 19.

Interpretation Introduction

(j)

Interpretation:

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 33

Explanation of Solution

The reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 34

Figure 20

In the above reaction, hexyne reacts with disiamyl borane to form an organoborane. This reaction is also known as hydroboration-oxidation reaction. The organo-borane gives an enol which tautomerism to form an aldehyde of six carbons. Therefore, the product formed on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 35

Figure 21

Conclusion

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown in Figure 21.

Interpretation Introduction

(k)

Interpretation:

The product on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 36

Explanation of Solution

The reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 37

Figure 22

In the above reaction, $CH3CH2MgBr$ is Grignard reagent that reacts with alkyne to form an alkane, that is ethane and another Grignard reagent. Hexyne on reaction with Grignard reagent releases the cationic terminal hydrogen which goes on to react with the negative part of Grignard reagent and forms an alkane. The anionic part of alkyne and positive part of Grignard reagent react to form another Grignard reagent. The products therefore obtained are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 38

Figure 23

Conclusion

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ are shown in Figure 23.

Interpretation Introduction

(l)

Interpretation:

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 39

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 40

Figure 23

The reaction of above compound with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 41

Figure 24

In the above reaction, the Grignard reagent obtained in part (k) reacts with ethylene oxide followed by hydrolysis to form an alcohol with a triple bond in between the chain. The product formed on reaction of product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 42

Figure 25

Conclusion

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is in Figure 25.

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Answer 5

Chapter 14, Problem 14.26AP

Interpretation Introduction

(a)

Interpretation:

The product on reaction of $1-hexyne$ with $HBr$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 1

Explanation of Solution

The reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 2

Figure 1

In the above reaction, hexyne reacts with $HBr$ to form $2-bromohex-1-ene$ . Hexyne is an unsaturated molecule consisting of a triple bond. It reacts with to form a haloalkene. Therefore, the product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 4

Figure 2

Conclusion

The product on reaction of $1-hexyne$ with $HBr$ is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 5

Explanation of Solution

The reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 6

Figure 3

In the above reaction, hexyne reacts with hydrogen to form hexene. Hexyne is an unsaturated molecule consisting of a triple bond. It reacts hydrogen in presence of $Pd/C$ to give alkene. Therefore, the product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 7

Figure 4

Conclusion

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , 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 $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 8

Explanation of Solution

The reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 9

Figure 5

In the above reaction, hexyne reacts with $H2,Pd/C$ in presence of Lindlar’s catalyst to form an alkene. Due to the presence of Lindlar’s catalyst the hydrogenation of alkyne stops at alkene and does not proceed further to give alkane. Therefore, the product formed on reaction of $1-hexyne$ with the $H2,Pd/C$ in presence of Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 10

Figure 6

Conclusion

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation:

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 11

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 12

Figure 7

The reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 13

Figure 8

The above reaction is known as ozonlysis reaction. In the above reaction, a double bond is cleaved and oxidized to give two products. Reaction of hexene with $O3$ followed by $(CH3)2S$ forms pentanal and formaldehyde. Therefore, the products formed on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 14

Figure 9

Conclusion

The products on reaction of reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown in Figure 9.

Interpretation Introduction

(e)

Interpretation:

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 15

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 16

Figure 7

The reaction of above compound with $BH3$ in $THF$ , then $H2O2/−OH$

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 17

Figure 10

In the above reaction, hexene reacts with boron hydride to form organoborane which further reacts with peroxide to form an alcohol. The alcohol thus formed is by anti-markovnikov addition.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 18

Figure 11

Conclusion

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown in Figure 11.

Interpretation Introduction

(f)

Interpretation:

The product on reaction of the product formed in part (c) with $Br2$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 19

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 20

Figure 7

The reaction of above compound with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 21

Figure 12

In the above reaction, hexene being unsaturated reacts with bromine molecule to form a dibromoalkane, that is $1, 2-dibromohexane$ . In the above reaction, unsaturated molecule is forming a saturated dihaloalkane on reaction with bromine molecule. Therefore, the product formed on reaction of product formed in (c) with bromine molecule is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 22

Figure 13

Conclusion

The product on reaction of the product formed in part (c) with $Br2$ is shown in Figure 13.

Interpretation Introduction

(g)

Interpretation:

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 23

Explanation of Solution

The reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 24

Figure 14

Alkynes react with sodamide $(NaNH2)$ in presence of liquid ammonia to release an acidic proton and to form an anion of alkyne. Therefore, on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia releases the acidic proton to form an anion of hexyne. The product therefore formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 25

Figure 15

Conclusion

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown in Figure 15.

Interpretation Introduction

(h)

Interpretation:

The product on reaction of the product formed in part (g) with $CH3CH2I$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 26

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 27

Figure 15

The reaction of above compound with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 28

Figure 16

In the above reaction, the anion of alkyne obtained reacts with iodoethane to form an alkyne of eight carbons. It is non-terminal alkyne, that is, triple bond is not situated at the terminal end. The product formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 29

Figure 17

Conclusion

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown in Figure 17.

Interpretation Introduction

(i)

Interpretation:

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 30

Explanation of Solution

The reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 31

Figure 18

In the above reaction, hexyne reacts with $Hg2+, H2SO4, H2O$ to form a ketone of six carbons. This reaction is also known as oxymercuration-demercuration reaction. In the initial step an enol is formed which tautomerism to form a ketone. Therefore, the product formed on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 32

Figure 19

Conclusion

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown in Figure 19.

Interpretation Introduction

(j)

Interpretation:

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 33

Explanation of Solution

The reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 34

Figure 20

In the above reaction, hexyne reacts with disiamyl borane to form an organoborane. This reaction is also known as hydroboration-oxidation reaction. The organo-borane gives an enol which tautomerism to form an aldehyde of six carbons. Therefore, the product formed on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 35

Figure 21

Conclusion

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown in Figure 21.

Interpretation Introduction

(k)

Interpretation:

The product on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 36

Explanation of Solution

The reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 37

Figure 22

In the above reaction, $CH3CH2MgBr$ is Grignard reagent that reacts with alkyne to form an alkane, that is ethane and another Grignard reagent. Hexyne on reaction with Grignard reagent releases the cationic terminal hydrogen which goes on to react with the negative part of Grignard reagent and forms an alkane. The anionic part of alkyne and positive part of Grignard reagent react to form another Grignard reagent. The products therefore obtained are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 38

Figure 23

Conclusion

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ are shown in Figure 23.

Interpretation Introduction

(l)

Interpretation:

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 39

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 40

Figure 23

The reaction of above compound with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 41

Figure 24

In the above reaction, the Grignard reagent obtained in part (k) reacts with ethylene oxide followed by hydrolysis to form an alcohol with a triple bond in between the chain. The product formed on reaction of product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 42

Figure 25

Conclusion

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is in Figure 25.

Answer 6

Chapter 14, Problem 14.26AP

Interpretation Introduction

(a)

Interpretation:

The product on reaction of $1-hexyne$ with $HBr$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 1

Explanation of Solution

The reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 2

Figure 1

In the above reaction, hexyne reacts with $HBr$ to form $2-bromohex-1-ene$ . Hexyne is an unsaturated molecule consisting of a triple bond. It reacts with to form a haloalkene. Therefore, the product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 4

Figure 2

Conclusion

The product on reaction of $1-hexyne$ with $HBr$ is shown in Figure 2.

Answer 7

Chapter 14, Problem 14.26AP

Interpretation Introduction

(a)

Interpretation:

The product on reaction of $1-hexyne$ with $HBr$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 8

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 1

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 2

Figure 1

In the above reaction, hexyne reacts with $HBr$ to form $2-bromohex-1-ene$ . Hexyne is an unsaturated molecule consisting of a triple bond. It reacts with to form a haloalkene. Therefore, the product on reaction of $1-hexyne$ with $HBr$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 4

Figure 2

Answer 13

Conclusion

The product on reaction of $1-hexyne$ with $HBr$ is shown in Figure 2.

Answer 14

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 5

Explanation of Solution

The reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 6

Figure 3

In the above reaction, hexyne reacts with hydrogen to form hexene. Hexyne is an unsaturated molecule consisting of a triple bond. It reacts hydrogen in presence of $Pd/C$ to give alkene. Therefore, the product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 7

Figure 4

Conclusion

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown in Figure 4.

Answer 15

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 16

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 5

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 6

Figure 3

In the above reaction, hexyne reacts with hydrogen to form hexene. Hexyne is an unsaturated molecule consisting of a triple bond. It reacts hydrogen in presence of $Pd/C$ to give alkene. Therefore, the product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 7

Figure 4

Answer 21

Conclusion

The product on reaction of $1-hexyne$ with $H2,Pd/C$ catalyst is shown in Figure 4.

Answer 22

Interpretation Introduction

(c)

Interpretation:

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , 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 $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 8

Explanation of Solution

The reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 9

Figure 5

In the above reaction, hexyne reacts with $H2,Pd/C$ in presence of Lindlar’s catalyst to form an alkene. Due to the presence of Lindlar’s catalyst the hydrogenation of alkyne stops at alkene and does not proceed further to give alkane. Therefore, the product formed on reaction of $1-hexyne$ with the $H2,Pd/C$ in presence of Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 10

Figure 6

Conclusion

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown in Figure 6.

Answer 23

Interpretation Introduction

(c)

Interpretation:

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , 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 $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 24

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 8

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 9

Figure 5

In the above reaction, hexyne reacts with $H2,Pd/C$ in presence of Lindlar’s catalyst to form an alkene. Due to the presence of Lindlar’s catalyst the hydrogenation of alkyne stops at alkene and does not proceed further to give alkane. Therefore, the product formed on reaction of $1-hexyne$ with the $H2,Pd/C$ in presence of Lindlar’s catalyst is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 10

Figure 6

Answer 29

Conclusion

The product on reaction of $1-hexyne$ with the $H2,Pd/C$ , Lindlar’s catalyst is shown in Figure 6.

Answer 30

Interpretation Introduction

(d)

Interpretation:

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 11

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 12

Figure 7

The reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 13

Figure 8

The above reaction is known as ozonlysis reaction. In the above reaction, a double bond is cleaved and oxidized to give two products. Reaction of hexene with $O3$ followed by $(CH3)2S$ forms pentanal and formaldehyde. Therefore, the products formed on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 14

Figure 9

Conclusion

The products on reaction of reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown in Figure 9.

Answer 31

Interpretation Introduction

(d)

Interpretation:

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 32

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 11

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 12

Figure 7

The reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 13

Figure 8

The above reaction is known as ozonlysis reaction. In the above reaction, a double bond is cleaved and oxidized to give two products. Reaction of hexene with $O3$ followed by $(CH3)2S$ forms pentanal and formaldehyde. Therefore, the products formed on reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 14

Figure 9

Answer 37

Conclusion

The products on reaction of reaction of the product formed in part (c) with $O3$ and then with $(CH3)2S$ are shown in Figure 9.

Answer 38

Interpretation Introduction

(e)

Interpretation:

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 15

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 16

Figure 7

The reaction of above compound with $BH3$ in $THF$ , then $H2O2/−OH$

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 17

Figure 10

In the above reaction, hexene reacts with boron hydride to form organoborane which further reacts with peroxide to form an alcohol. The alcohol thus formed is by anti-markovnikov addition.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 18

Figure 11

Conclusion

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown in Figure 11.

Answer 39

Interpretation Introduction

(e)

Interpretation:

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 40

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 15

Answer 41

Expert Solution

Answer 42

Expert Solution

Answer 43

Expert Solution

Answer 44

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 16

Figure 7

The reaction of above compound with $BH3$ in $THF$ , then $H2O2/−OH$

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 17

Figure 10

In the above reaction, hexene reacts with boron hydride to form organoborane which further reacts with peroxide to form an alcohol. The alcohol thus formed is by anti-markovnikov addition.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 18

Figure 11

Answer 45

Conclusion

The product on reaction of the product formed in part (c) with $BH3$ in $THF$ , then $H2O2/−OH$ is shown in Figure 11.

Answer 46

Interpretation Introduction

(f)

Interpretation:

The product on reaction of the product formed in part (c) with $Br2$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 19

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 20

Figure 7

The reaction of above compound with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 21

Figure 12

In the above reaction, hexene being unsaturated reacts with bromine molecule to form a dibromoalkane, that is $1, 2-dibromohexane$ . In the above reaction, unsaturated molecule is forming a saturated dihaloalkane on reaction with bromine molecule. Therefore, the product formed on reaction of product formed in (c) with bromine molecule is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 22

Figure 13

Conclusion

The product on reaction of the product formed in part (c) with $Br2$ is shown in Figure 13.

Answer 47

Interpretation Introduction

(f)

Interpretation:

The product on reaction of the product formed in part (c) with $Br2$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 48

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (c) with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 19

Answer 49

Expert Solution

Answer 50

Expert Solution

Answer 51

Expert Solution

Answer 52

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 20

Figure 7

The reaction of above compound with $Br2$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 21

Figure 12

In the above reaction, hexene being unsaturated reacts with bromine molecule to form a dibromoalkane, that is $1, 2-dibromohexane$ . In the above reaction, unsaturated molecule is forming a saturated dihaloalkane on reaction with bromine molecule. Therefore, the product formed on reaction of product formed in (c) with bromine molecule is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 22

Figure 13

Answer 53

Conclusion

The product on reaction of the product formed in part (c) with $Br2$ is shown in Figure 13.

Answer 54

Interpretation Introduction

(g)

Interpretation:

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 23

Explanation of Solution

The reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 24

Figure 14

Alkynes react with sodamide $(NaNH2)$ in presence of liquid ammonia to release an acidic proton and to form an anion of alkyne. Therefore, on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia releases the acidic proton to form an anion of hexyne. The product therefore formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 25

Figure 15

Conclusion

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown in Figure 15.

Answer 55

Interpretation Introduction

(g)

Interpretation:

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 56

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 23

Answer 57

Expert Solution

Answer 58

Expert Solution

Answer 59

Expert Solution

Answer 60

Explanation of Solution

The reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 24

Figure 14

Alkynes react with sodamide $(NaNH2)$ in presence of liquid ammonia to release an acidic proton and to form an anion of alkyne. Therefore, on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia releases the acidic proton to form an anion of hexyne. The product therefore formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 25

Figure 15

Answer 61

Conclusion

The product on reaction of $1-hexyne$ with $NaNH2$ in liquid ammonia is shown in Figure 15.

Answer 62

Interpretation Introduction

(h)

Interpretation:

The product on reaction of the product formed in part (g) with $CH3CH2I$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 26

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 27

Figure 15

The reaction of above compound with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 28

Figure 16

In the above reaction, the anion of alkyne obtained reacts with iodoethane to form an alkyne of eight carbons. It is non-terminal alkyne, that is, triple bond is not situated at the terminal end. The product formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 29

Figure 17

Conclusion

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown in Figure 17.

Answer 63

Interpretation Introduction

(h)

Interpretation:

The product on reaction of the product formed in part (g) with $CH3CH2I$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 64

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 26

Answer 65

Expert Solution

Answer 66

Expert Solution

Answer 67

Expert Solution

Answer 68

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 27

Figure 15

The reaction of above compound with $CH3CH2I$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 28

Figure 16

In the above reaction, the anion of alkyne obtained reacts with iodoethane to form an alkyne of eight carbons. It is non-terminal alkyne, that is, triple bond is not situated at the terminal end. The product formed is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 29

Figure 17

Answer 69

Conclusion

The product on reaction of the product formed in part (g) with $CH3CH2I$ is shown in Figure 17.

Answer 70

Interpretation Introduction

(i)

Interpretation:

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 30

Explanation of Solution

The reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 31

Figure 18

In the above reaction, hexyne reacts with $Hg2+, H2SO4, H2O$ to form a ketone of six carbons. This reaction is also known as oxymercuration-demercuration reaction. In the initial step an enol is formed which tautomerism to form a ketone. Therefore, the product formed on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 32

Figure 19

Conclusion

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown in Figure 19.

Answer 71

Interpretation Introduction

(i)

Interpretation:

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 72

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 30

Answer 73

Expert Solution

Answer 74

Expert Solution

Answer 75

Expert Solution

Answer 76

Explanation of Solution

The reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 31

Figure 18

In the above reaction, hexyne reacts with $Hg2+, H2SO4, H2O$ to form a ketone of six carbons. This reaction is also known as oxymercuration-demercuration reaction. In the initial step an enol is formed which tautomerism to form a ketone. Therefore, the product formed on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 32

Figure 19

Answer 77

Conclusion

The product on reaction of $1-hexyne$ with $Hg2+, H2SO4, H2O$ is shown in Figure 19.

Answer 78

Interpretation Introduction

(j)

Interpretation:

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 33

Explanation of Solution

The reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 34

Figure 20

In the above reaction, hexyne reacts with disiamyl borane to form an organoborane. This reaction is also known as hydroboration-oxidation reaction. The organo-borane gives an enol which tautomerism to form an aldehyde of six carbons. Therefore, the product formed on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 35

Figure 21

Conclusion

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown in Figure 21.

Answer 79

Interpretation Introduction

(j)

Interpretation:

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 80

Expert Solution

Answer to Problem 14.26AP

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 33

Answer 81

Expert Solution

Answer 82

Expert Solution

Answer 83

Expert Solution

Answer 84

Explanation of Solution

The reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 34

Figure 20

In the above reaction, hexyne reacts with disiamyl borane to form an organoborane. This reaction is also known as hydroboration-oxidation reaction. The organo-borane gives an enol which tautomerism to form an aldehyde of six carbons. Therefore, the product formed on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 35

Figure 21

Answer 85

Conclusion

The product on reaction of $1-hexyne$ with disiamyl borane, $H2O2/−OH$ is shown in Figure 21.

Answer 86

Interpretation Introduction

(k)

Interpretation:

The product on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 36

Explanation of Solution

The reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 37

Figure 22

In the above reaction, $CH3CH2MgBr$ is Grignard reagent that reacts with alkyne to form an alkane, that is ethane and another Grignard reagent. Hexyne on reaction with Grignard reagent releases the cationic terminal hydrogen which goes on to react with the negative part of Grignard reagent and forms an alkane. The anionic part of alkyne and positive part of Grignard reagent react to form another Grignard reagent. The products therefore obtained are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 38

Figure 23

Conclusion

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ are shown in Figure 23.

Answer 87

Interpretation Introduction

(k)

Interpretation:

The product on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 88

Expert Solution

Answer to Problem 14.26AP

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 36

Answer 89

Expert Solution

Answer 90

Expert Solution

Answer 91

Expert Solution

Answer 92

Explanation of Solution

The reaction of $1-hexyne$ with $CH3CH2MgBr$ in ether is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 37

Figure 22

In the above reaction, $CH3CH2MgBr$ is Grignard reagent that reacts with alkyne to form an alkane, that is ethane and another Grignard reagent. Hexyne on reaction with Grignard reagent releases the cationic terminal hydrogen which goes on to react with the negative part of Grignard reagent and forms an alkane. The anionic part of alkyne and positive part of Grignard reagent react to form another Grignard reagent. The products therefore obtained are shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 38

Figure 23

Answer 93

Conclusion

The products formed on reaction of $1-hexyne$ with $CH3CH2MgBr$ are shown in Figure 23.

Answer 94

Interpretation Introduction

(l)

Interpretation:

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 39

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 40

Figure 23

The reaction of above compound with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 41

Figure 24

In the above reaction, the Grignard reagent obtained in part (k) reacts with ethylene oxide followed by hydrolysis to form an alcohol with a triple bond in between the chain. The product formed on reaction of product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 42

Figure 25

Conclusion

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is in Figure 25.

Answer 95

Interpretation Introduction

(l)

Interpretation:

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is to be stated.

Concept introduction:

The alkynes consist of a triple bond between two carbon atoms. The general formula of alkynes is $CnH2n−2$ . The name of the alkyne compounds ends with suffix $−yne$ . Hydrocarbons can also exist in ring like structures.

Answer 96

Expert Solution

Answer to Problem 14.26AP

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 39

Answer 97

Expert Solution

Answer 98

Expert Solution

Answer 99

Expert Solution

Answer 100

Explanation of Solution

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

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 40

Figure 23

The reaction of above compound with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 41

Figure 24

In the above reaction, the Grignard reagent obtained in part (k) reacts with ethylene oxide followed by hydrolysis to form an alcohol with a triple bond in between the chain. The product formed on reaction of product formed in part (k) with ethylene oxide and $H3O+$ is shown below.

Organic Chemistry, Chapter 14, Problem 14.26AP , additional homework tip 42

Figure 25

Answer 101

Conclusion

The product on reaction of the product formed in part (k) with ethylene oxide and $H3O+$ is in Figure 25.

Answer 102

Ch. 14 - Prob. 14.1P Ch. 14 - Prob. 14.2P Ch. 14 - Prob. 14.3P Ch. 14 - Prob. 14.4P Ch. 14 - Prob. 14.5P Ch. 14 - Prob. 14.6P Ch. 14 - Prob. 14.7P Ch. 14 - Prob. 14.8P Ch. 14 - Prob. 14.9P Ch. 14 - Prob. 14.10P

Concept explainers

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

Answer to Problem 14.26AP

Explanation of Solution

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