Name the following alkynes , and predict the products of their reaction with (1) H 2 in the presence of a Lindlar catalyst and (2) H 3 O + in the presence of HgSO 4 :

Question

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Answer 1

Textbook Question

Chapter 9.SE, Problem 14VC

Name the following alkynes, and predict the products of their reaction with (1) H₂ in the presence of a Lindlar catalyst and (2) H₃O₊ in the presence of HgSO₄:

Chapter 9.SE, Problem 14VC, Name the following alkynes, and predict the products of their reaction with (1) H2 in the presence

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.

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 1

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O₊ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 2

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 3

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 4

Explanation of Solution

The compound has a six carbon straight chain with two methyl groups on C₄ with a triple bond between C₁ & C₂. Hence it’s name is 4,4-dimethyl-1-hexyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, 4,4-dimethyl-1-hexene,is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond following Markovnikov regiochemistry. The OH group adds to more highly substituted carbon and H adds to the less highly substituted carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 5

Conclusion

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 6

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 7

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 8

b)

Expert Solution

Interpretation Introduction

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 9

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 10

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 11

Explanation of Solution

The compound has an eight carbon straight chain with two methyl groups on C₂ & C₇ with a triple bond between C₄ & C₅. Hence it’s name is 2,7-dimethyl-4-octyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, cis- 2,7-dimethyl-4-octene.is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond. As the alkene is symmetrical the Markovnikov regiochemistry cannot be applied. The OH group adds to one carbon and H adds to the other carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 12

Conclusion

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 13

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 14

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 15

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Answer 2

Textbook Question

Chapter 9.SE, Problem 14VC

Name the following alkynes, and predict the products of their reaction with (1) H₂ in the presence of a Lindlar catalyst and (2) H₃O₊ in the presence of HgSO₄:

Chapter 9.SE, Problem 14VC, Name the following alkynes, and predict the products of their reaction with (1) H2 in the presence

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.

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 1

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O₊ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 2

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 3

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 4

Explanation of Solution

The compound has a six carbon straight chain with two methyl groups on C₄ with a triple bond between C₁ & C₂. Hence it’s name is 4,4-dimethyl-1-hexyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, 4,4-dimethyl-1-hexene,is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond following Markovnikov regiochemistry. The OH group adds to more highly substituted carbon and H adds to the less highly substituted carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 5

Conclusion

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 6

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 7

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 8

b)

Expert Solution

Interpretation Introduction

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 9

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 10

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 11

Explanation of Solution

The compound has an eight carbon straight chain with two methyl groups on C₂ & C₇ with a triple bond between C₄ & C₅. Hence it’s name is 2,7-dimethyl-4-octyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, cis- 2,7-dimethyl-4-octene.is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond. As the alkene is symmetrical the Markovnikov regiochemistry cannot be applied. The OH group adds to one carbon and H adds to the other carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 12

Conclusion

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 13

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 14

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 15

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Answer 3

Textbook Question

Chapter 9.SE, Problem 14VC

Name the following alkynes, and predict the products of their reaction with (1) H₂ in the presence of a Lindlar catalyst and (2) H₃O₊ in the presence of HgSO₄:

Chapter 9.SE, Problem 14VC, Name the following alkynes, and predict the products of their reaction with (1) H2 in the presence

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.

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 1

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O₊ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 2

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 3

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 4

Explanation of Solution

The compound has a six carbon straight chain with two methyl groups on C₄ with a triple bond between C₁ & C₂. Hence it’s name is 4,4-dimethyl-1-hexyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, 4,4-dimethyl-1-hexene,is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond following Markovnikov regiochemistry. The OH group adds to more highly substituted carbon and H adds to the less highly substituted carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 5

Conclusion

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 6

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 7

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 8

b)

Expert Solution

Interpretation Introduction

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 9

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 10

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 11

Explanation of Solution

The compound has an eight carbon straight chain with two methyl groups on C₂ & C₇ with a triple bond between C₄ & C₅. Hence it’s name is 2,7-dimethyl-4-octyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, cis- 2,7-dimethyl-4-octene.is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond. As the alkene is symmetrical the Markovnikov regiochemistry cannot be applied. The OH group adds to one carbon and H adds to the other carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 12

Conclusion

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 13

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 14

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 15

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 9.SE, Problem 14VC

Name the following alkynes, and predict the products of their reaction with (1) H₂ in the presence of a Lindlar catalyst and (2) H₃O₊ in the presence of HgSO₄:

Chapter 9.SE, Problem 14VC, Name the following alkynes, and predict the products of their reaction with (1) H2 in the presence

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.

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 1

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O₊ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 2

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 3

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 4

Explanation of Solution

The compound has a six carbon straight chain with two methyl groups on C₄ with a triple bond between C₁ & C₂. Hence it’s name is 4,4-dimethyl-1-hexyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, 4,4-dimethyl-1-hexene,is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond following Markovnikov regiochemistry. The OH group adds to more highly substituted carbon and H adds to the less highly substituted carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 5

Conclusion

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 6

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 7

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 8

b)

Expert Solution

Interpretation Introduction

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 9

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 10

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 11

Explanation of Solution

The compound has an eight carbon straight chain with two methyl groups on C₂ & C₇ with a triple bond between C₄ & C₅. Hence it’s name is 2,7-dimethyl-4-octyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, cis- 2,7-dimethyl-4-octene.is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond. As the alkene is symmetrical the Markovnikov regiochemistry cannot be applied. The OH group adds to one carbon and H adds to the other carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 12

Conclusion

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 13

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 14

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 15

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 1

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O₊ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 2

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 3

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 4

Explanation of Solution

The compound has a six carbon straight chain with two methyl groups on C₄ with a triple bond between C₁ & C₂. Hence it’s name is 4,4-dimethyl-1-hexyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, 4,4-dimethyl-1-hexene,is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond following Markovnikov regiochemistry. The OH group adds to more highly substituted carbon and H adds to the less highly substituted carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 5

Conclusion

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 6

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 7

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 8

b)

Expert Solution

Interpretation Introduction

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 9

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 10

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 11

Explanation of Solution

The compound has an eight carbon straight chain with two methyl groups on C₂ & C₇ with a triple bond between C₄ & C₅. Hence it’s name is 2,7-dimethyl-4-octyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, cis- 2,7-dimethyl-4-octene.is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond. As the alkene is symmetrical the Markovnikov regiochemistry cannot be applied. The OH group adds to one carbon and H adds to the other carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 12

Conclusion

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 13

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 14

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 15

Answer 8

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 1

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O₊ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 2

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 3

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 4

Explanation of Solution

The compound has a six carbon straight chain with two methyl groups on C₄ with a triple bond between C₁ & C₂. Hence it’s name is 4,4-dimethyl-1-hexyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, 4,4-dimethyl-1-hexene,is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond following Markovnikov regiochemistry. The OH group adds to more highly substituted carbon and H adds to the less highly substituted carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 5

Conclusion

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 6

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 7

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 8

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

a)

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 1

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O₊ in the presence of HgSO₄.

Answer 13

Answer to Problem 14VC

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 2

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 3

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 4

Answer 14

Explanation of Solution

The compound has a six carbon straight chain with two methyl groups on C₄ with a triple bond between C₁ & C₂. Hence it’s name is 4,4-dimethyl-1-hexyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, 4,4-dimethyl-1-hexene,is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond following Markovnikov regiochemistry. The OH group adds to more highly substituted carbon and H adds to the less highly substituted carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 5

Answer 15

Conclusion

The name of the alkyne shown is 4,4-dimethyl-1-hexyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 6

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 7

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 4,4-dimethyl-2-hexanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 8

Answer 16

b)

Expert Solution

Interpretation Introduction

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄.

Answer to Problem 14VC

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 9

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 10

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 11

Explanation of Solution

The compound has an eight carbon straight chain with two methyl groups on C₂ & C₇ with a triple bond between C₄ & C₅. Hence it’s name is 2,7-dimethyl-4-octyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, cis- 2,7-dimethyl-4-octene.is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond. As the alkene is symmetrical the Markovnikov regiochemistry cannot be applied. The OH group adds to one carbon and H adds to the other carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 12

Conclusion

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 13

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 14

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 15

Answer 17

b)

Expert Solution

Answer 18

b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Interpretation Introduction

Interpretation:

The alkyne shown is to be named and the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄ is to be predicted.

Concept introduction:

The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.

When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H₃O⁺ in the presence of HgSO₄, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).

To give:

The name of the alkyne shown and to predict the products formed when it reacts with 1) H₂ in the presence of Lindlar catalyst and 2) H₃O⁺ in the presence of HgSO₄.

Answer 21

Answer to Problem 14VC

The name of the alkyne shown is 2,7-dimethyl-4-octyne.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 9

The product formed when it reacts with H₂ in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 10

The product formed when it reacts with H₃O⁺ in the presence of HgSO₄ is 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 11

Answer 22

Explanation of Solution

The compound has an eight carbon straight chain with two methyl groups on C₂ & C₇ with a triple bond between C₄ & C₅. Hence it’s name is 2,7-dimethyl-4-octyne.

When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, cis- 2,7-dimethyl-4-octene.is thus produced.

When treated with H₃O⁺ in the presence of HgSO₄, the addition of water takes place in the triple bond. As the alkene is symmetrical the Markovnikov regiochemistry cannot be applied. The OH group adds to one carbon and H adds to the other carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 2,7-dimethyl-4-octanone.

Organic Chemistry, Chapter 9.SE, Problem 14VC , additional homework tip 12