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2-Bromopen tane is heated with potassium ethoxide in ethanol. The major product obtained is[1998]
- a)2-ethoxypentane
- b)pentene-1
- c)trans-2-pentene
- d)cis-pentene-2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
2-Bromopen tane is heated with potassium ethoxide in ethanol. The majo...
Potassium ethoxide is a strong base, and 2-bromopentane is a 2º bromide, so elimination raction predominates
Since trans- alkene is more stable than cis.thus trans-pentene -2 is the main product.
Most Upvoted Answer
2-Bromopen tane is heated with potassium ethoxide in ethanol. The majo...
Reaction: 2-Bromopentane + Potassium ethoxide → trans-2-pentene + Potassium bromide + Ethanol
Explanation:
The reaction involves the elimination of a bromide ion (Br-) from 2-bromopentane, resulting in the formation of an alkene. The reaction is carried out in the presence of potassium ethoxide (C2H5OK) in ethanol as the solvent.
Step 1: Formation of Potassium Bromide
- The reaction starts with the dissociation of potassium ethoxide in ethanol solvent:
C2H5OK → C2H5O- + K+
- The potassium ion (K+) acts as a catalyst in the reaction.
- The bromide ion (Br-) is displaced from 2-bromopentane and reacts with the potassium ion to form potassium bromide (KBr):
C5H11Br + K+ → C5H11 + KBr
Step 2: Elimination of Bromide Ion
- The alkoxide ion (C2H5O-) acts as a strong base and abstracts a proton from the carbon adjacent to the bromine atom in 2-bromopentane.
- This results in the formation of an alkene (trans-2-pentene) and the elimination of a bromide ion:
C5H11 + C2H5O- → C5H10 + C2H5OH
Step 3: Formation of trans-2-pentene
- The elimination of the bromide ion from 2-bromopentane leads to the formation of trans-2-pentene.
- In trans-2-pentene, the double bond is present between the second and third carbon atoms, and the two methyl groups are on opposite sides of the double bond.
- This is the major product obtained in the reaction.
Conclusion:
Upon heating 2-bromopentane with potassium ethoxide in ethanol, the major product obtained is trans-2-pentene. This is due to the elimination of a bromide ion from 2-bromopentane, resulting in the formation of the alkene trans-2-pentene.
Explanation:
The reaction involves the elimination of a bromide ion (Br-) from 2-bromopentane, resulting in the formation of an alkene. The reaction is carried out in the presence of potassium ethoxide (C2H5OK) in ethanol as the solvent.
Step 1: Formation of Potassium Bromide
- The reaction starts with the dissociation of potassium ethoxide in ethanol solvent:
C2H5OK → C2H5O- + K+
- The potassium ion (K+) acts as a catalyst in the reaction.
- The bromide ion (Br-) is displaced from 2-bromopentane and reacts with the potassium ion to form potassium bromide (KBr):
C5H11Br + K+ → C5H11 + KBr
Step 2: Elimination of Bromide Ion
- The alkoxide ion (C2H5O-) acts as a strong base and abstracts a proton from the carbon adjacent to the bromine atom in 2-bromopentane.
- This results in the formation of an alkene (trans-2-pentene) and the elimination of a bromide ion:
C5H11 + C2H5O- → C5H10 + C2H5OH
Step 3: Formation of trans-2-pentene
- The elimination of the bromide ion from 2-bromopentane leads to the formation of trans-2-pentene.
- In trans-2-pentene, the double bond is present between the second and third carbon atoms, and the two methyl groups are on opposite sides of the double bond.
- This is the major product obtained in the reaction.
Conclusion:
Upon heating 2-bromopentane with potassium ethoxide in ethanol, the major product obtained is trans-2-pentene. This is due to the elimination of a bromide ion from 2-bromopentane, resulting in the formation of the alkene trans-2-pentene.
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2-Bromopen tane is heated with potassium ethoxide in ethanol. The major product obtained is[1998]a)2-ethoxypentaneb)pentene-1c)trans-2-pentened)cis-pentene-2Correct answer is option 'C'. Can you explain this answer?
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2-Bromopen tane is heated with potassium ethoxide in ethanol. The major product obtained is[1998]a)2-ethoxypentaneb)pentene-1c)trans-2-pentened)cis-pentene-2Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 2-Bromopen tane is heated with potassium ethoxide in ethanol. The major product obtained is[1998]a)2-ethoxypentaneb)pentene-1c)trans-2-pentened)cis-pentene-2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2-Bromopen tane is heated with potassium ethoxide in ethanol. The major product obtained is[1998]a)2-ethoxypentaneb)pentene-1c)trans-2-pentened)cis-pentene-2Correct answer is option 'C'. Can you explain this answer?.
2-Bromopen tane is heated with potassium ethoxide in ethanol. The major product obtained is[1998]a)2-ethoxypentaneb)pentene-1c)trans-2-pentened)cis-pentene-2Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 2-Bromopen tane is heated with potassium ethoxide in ethanol. The major product obtained is[1998]a)2-ethoxypentaneb)pentene-1c)trans-2-pentened)cis-pentene-2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2-Bromopen tane is heated with potassium ethoxide in ethanol. The major product obtained is[1998]a)2-ethoxypentaneb)pentene-1c)trans-2-pentened)cis-pentene-2Correct answer is option 'C'. Can you explain this answer?.
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