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Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne because
- a)NaNH2 is a very strong base
- b)NaNH2 abstract H from less hindered carbon
- c)terminal bromo alkene is formed in first step of elimination reaction
- d)terminal alkyne forms salt with NaNH2, phased out of equilibrium
Correct answer is option 'D'. Can you explain this answer?
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Terminal alkyne being therm odynamically less stable than an internal ...
Answer:
Introduction:
In the given question, we are asked to explain why 1-butyne is formed as the major product instead of 2-butyne when 2,2-dibromobutane is treated with NaNH2. The correct answer is option 'D', which states that the terminal alkyne forms a salt with NaNH2 and is phased out of equilibrium. Let's understand this in detail.
Explanation:
When 2,2-dibromobutane is treated with NaNH2, an elimination reaction occurs, leading to the formation of an alkyne. Let's analyze the factors that contribute to the formation of 1-butyne as the major product.
1. Strong Base:
NaNH2 is a very strong base. It has a high affinity for protons and can abstract a hydrogen atom from a carbon atom. This property of NaNH2 plays a crucial role in determining the product formed.
2. H Abstraction from Less Hindered Carbon:
When NaNH2 acts as a base, it abstracts a hydrogen atom from a carbon atom to form an alkene. In this case, the less hindered carbon is more accessible for hydrogen abstraction. The less hindered carbon is the one that is directly attached to the bromine atom at one end of the molecule.
3. Terminal Bromoalkene Formation:
The first step of the elimination reaction involves the formation of a terminal bromoalkene. This is the result of one of the bromine atoms being abstracted by NaNH2, leaving a double bond between the adjacent carbon atoms.
4. Terminal Alkyne Salt Formation:
Now, when NaNH2 abstracts a hydrogen atom from the less hindered carbon, it forms a salt with the terminal alkyne. This salt is phased out of equilibrium, which means it is removed from the reaction mixture and does not contribute to the final product.
Conclusion:
Based on the above explanations, we can conclude that the formation of 1-butyne as the major product instead of 2-butyne is due to the terminal alkyne forming a salt with NaNH2, which is phased out of equilibrium.
Introduction:
In the given question, we are asked to explain why 1-butyne is formed as the major product instead of 2-butyne when 2,2-dibromobutane is treated with NaNH2. The correct answer is option 'D', which states that the terminal alkyne forms a salt with NaNH2 and is phased out of equilibrium. Let's understand this in detail.
Explanation:
When 2,2-dibromobutane is treated with NaNH2, an elimination reaction occurs, leading to the formation of an alkyne. Let's analyze the factors that contribute to the formation of 1-butyne as the major product.
1. Strong Base:
NaNH2 is a very strong base. It has a high affinity for protons and can abstract a hydrogen atom from a carbon atom. This property of NaNH2 plays a crucial role in determining the product formed.
2. H Abstraction from Less Hindered Carbon:
When NaNH2 acts as a base, it abstracts a hydrogen atom from a carbon atom to form an alkene. In this case, the less hindered carbon is more accessible for hydrogen abstraction. The less hindered carbon is the one that is directly attached to the bromine atom at one end of the molecule.
3. Terminal Bromoalkene Formation:
The first step of the elimination reaction involves the formation of a terminal bromoalkene. This is the result of one of the bromine atoms being abstracted by NaNH2, leaving a double bond between the adjacent carbon atoms.
4. Terminal Alkyne Salt Formation:
Now, when NaNH2 abstracts a hydrogen atom from the less hindered carbon, it forms a salt with the terminal alkyne. This salt is phased out of equilibrium, which means it is removed from the reaction mixture and does not contribute to the final product.
Conclusion:
Based on the above explanations, we can conclude that the formation of 1-butyne as the major product instead of 2-butyne is due to the terminal alkyne forming a salt with NaNH2, which is phased out of equilibrium.
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Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer?
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Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer?.
Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer?.
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Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne becausea)NaNH2 is a very strong baseb)NaNH2 abstract H from less hindered carbonc)terminal bromo alkene is formed in first step of elimination reactiond)terminal alkyne forms salt with NaNH2, phased out of equilibriumCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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