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A compound ‘A’
- a)Decolourizes B aeyer’s reagent
- b)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetone
- c)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylene
- d)Isobutylene
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A compound ‘A’a)Decolourizes B aeyer’s reagentb)On h...
The compound A, which is not mentioned in the question, exhibits certain reactions that allow us to determine its structure. Let's analyze the given reactions step by step.
a) Decolourization of B by A:
This implies that A undergoes a reaction with B, leading to the removal of color from B. This suggests that A is an oxidizing agent, as it is able to oxidize B and remove the colored impurities.
b) Hydroxylation and further oxidation of A by KMnO4:
The reaction of A with KMnO4 leads to the formation of formic acid and acetone. This indicates that A must contain a carbon-carbon double bond (C=C) since KMnO4 is known for its oxidative cleavage of C=C bonds. The oxidation of the C=C bond in A results in the formation of two carbonyl compounds, formic acid and acetone.
c) Ozonolysis of A:
Ozonolysis of A produces one molecule of acetone and one molecule of formaldehyde. This reaction is characteristic of compounds containing a double bond (C=C) or a triple bond (C≡C).
Based on the information obtained from the reactions, we can conclude that A must have a carbon-carbon double bond (C=C). Among the given options, only isobutylene (option D) contains a C=C bond. Therefore, the correct answer is option D, isobutylene.
In summary:
- Compound A is an oxidizing agent, as it decolorizes B.
- A contains a carbon-carbon double bond (C=C), as seen from its reaction with KMnO4 and ozonolysis.
- The reactions suggest that A is isobutylene, which is the correct answer (option D).
a) Decolourization of B by A:
This implies that A undergoes a reaction with B, leading to the removal of color from B. This suggests that A is an oxidizing agent, as it is able to oxidize B and remove the colored impurities.
b) Hydroxylation and further oxidation of A by KMnO4:
The reaction of A with KMnO4 leads to the formation of formic acid and acetone. This indicates that A must contain a carbon-carbon double bond (C=C) since KMnO4 is known for its oxidative cleavage of C=C bonds. The oxidation of the C=C bond in A results in the formation of two carbonyl compounds, formic acid and acetone.
c) Ozonolysis of A:
Ozonolysis of A produces one molecule of acetone and one molecule of formaldehyde. This reaction is characteristic of compounds containing a double bond (C=C) or a triple bond (C≡C).
Based on the information obtained from the reactions, we can conclude that A must have a carbon-carbon double bond (C=C). Among the given options, only isobutylene (option D) contains a C=C bond. Therefore, the correct answer is option D, isobutylene.
In summary:
- Compound A is an oxidizing agent, as it decolorizes B.
- A contains a carbon-carbon double bond (C=C), as seen from its reaction with KMnO4 and ozonolysis.
- The reactions suggest that A is isobutylene, which is the correct answer (option D).
Community Answer
A compound ‘A’a)Decolourizes B aeyer’s reagentb)On h...
Isobutylene ~. H2C=CH-(CH3)2
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A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer?
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A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer?.
A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer?.
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A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A compound ‘A’a)Decolourizes B aeyer’s reagentb)On hydroxylation and further oxidation with KMnO4 gave formic acid and acetonec)On ozonolysis gave one molecule of acetone and one of formaldehyde ‘A’ is (a) Propylene (b) Ethylene (c) n-butylened)IsobutyleneCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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