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You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?
- a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-one
- b)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanone
- c)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanone
- d)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanone
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
You have two C6H10O ketones, I and II. Both are optically active, but ...
Not racemises on treatment with base because its α -carbon is not chiral.
Most Upvoted Answer
You have two C6H10O ketones, I and II. Both are optically active, but ...
Explanation:
1. Racemization:
Racemization is the process of converting an enantiomerically pure compound into a racemic mixture (equal amounts of both enantiomers) by breaking and reforming bonds. In the given question, it is mentioned that ketone I is racemized by treatment with base, while ketone II is not. This information helps us in determining the structure of the ketones.
2. Wolff-Kishner Reduction:
Wolff-Kishner reduction is a chemical reaction that converts a ketone or aldehyde into an alkane by the reaction with hydrazine (NH2NH2) followed by treatment with a strong base, typically potassium hydroxide (KOH). In this reaction, the carbonyl group is converted into a methylene group (-CH2-). The given ketones I and II both undergo Wolff-Kishner reduction to form the same achiral hydrocarbon with the formula C6H12.
3. Analysis of the options:
Let's analyze the given options to determine the structures of ketones I and II.
a) I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-one
b) I is 2-methyl cyclopentanone; II is 3-methyl cyclopentanone
c) I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanone
d) I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanone
4. Reasoning:
From the given options, we need to find a pair of ketones where ketone I is racemized by treatment with base, while ketone II is not racemized. Additionally, both ketones should give the same achiral hydrocarbon (C6H12) upon Wolff-Kishner reduction.
In option b), ketone I is 2-methyl cyclopentanone and ketone II is 3-methyl cyclopentanone. In this case, ketone I can be racemized by treatment with base due to the presence of a chiral center at the carbon atom attached to the carbonyl group. On the other hand, ketone II does not have a chiral center and is not racemized by treatment with base.
Upon Wolff-Kishner reduction, both ketones I and II will form cyclopentane (C5H10), which is an achiral hydrocarbon with the formula C6H12.
5. Conclusion:
Based on the analysis, we can conclude that option b) is the correct answer. Ketone I is 2-methyl cyclopentanone and ketone II is 3-methyl cyclopentanone.
1. Racemization:
Racemization is the process of converting an enantiomerically pure compound into a racemic mixture (equal amounts of both enantiomers) by breaking and reforming bonds. In the given question, it is mentioned that ketone I is racemized by treatment with base, while ketone II is not. This information helps us in determining the structure of the ketones.
2. Wolff-Kishner Reduction:
Wolff-Kishner reduction is a chemical reaction that converts a ketone or aldehyde into an alkane by the reaction with hydrazine (NH2NH2) followed by treatment with a strong base, typically potassium hydroxide (KOH). In this reaction, the carbonyl group is converted into a methylene group (-CH2-). The given ketones I and II both undergo Wolff-Kishner reduction to form the same achiral hydrocarbon with the formula C6H12.
3. Analysis of the options:
Let's analyze the given options to determine the structures of ketones I and II.
a) I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-one
b) I is 2-methyl cyclopentanone; II is 3-methyl cyclopentanone
c) I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanone
d) I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanone
4. Reasoning:
From the given options, we need to find a pair of ketones where ketone I is racemized by treatment with base, while ketone II is not racemized. Additionally, both ketones should give the same achiral hydrocarbon (C6H12) upon Wolff-Kishner reduction.
In option b), ketone I is 2-methyl cyclopentanone and ketone II is 3-methyl cyclopentanone. In this case, ketone I can be racemized by treatment with base due to the presence of a chiral center at the carbon atom attached to the carbonyl group. On the other hand, ketone II does not have a chiral center and is not racemized by treatment with base.
Upon Wolff-Kishner reduction, both ketones I and II will form cyclopentane (C5H10), which is an achiral hydrocarbon with the formula C6H12.
5. Conclusion:
Based on the analysis, we can conclude that option b) is the correct answer. Ketone I is 2-methyl cyclopentanone and ketone II is 3-methyl cyclopentanone.
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You have two C6H10O ketones, I and II. Both are optically active, but ...
Not racemises on treatment with base because its α -carbon is not chiral.
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You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer?
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You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer?.
You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer?.
Solutions for You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice You have two C6H10O ketones, I and II. Both are optically active, but I is racemised by treatment with base and II is not. Wolff-Kishner reduction of both ketones give the same achiral hydrocarbon, formula C6H12. What reasonable structures may be assigned to I and II?a)I is 3-methyl-4-penten-2-one; II is 4-methyl-1-penten-3-oneb)I is 2-methyl cyclopentanone; II is 3-methyi cyclopentanonec)I is 3-methyl cyclopentanone; II is 2-methyl cyclopentanoned)I is 2-ethyl cyclobutanone; II is 3-ethyl cyclobutanoneCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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