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On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH−CH2CH3 gives:
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
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On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH&mi...
Generally, alkenes can easily be oxidized by potassium permanganate and other oxidizing agents. The products formed depend on the reaction conditions.
At low temperatures, with low concentrations of oxidizing reagents, alkenes form diols.
When more concentrated solutions of potassium permanganate and higher temperatures are employed (vigorous conditions), the reaction would not stop at this point.
The product of the reaction depends on the arrangement of groups around the carbon−carbon double bond. The potassium permanganate (VII) solution oxidizes the alkene by breaking the carbon - carbon double bond and replacing it with two carbon - oxgyen double bonds forming carbonyl compounds.
At low temperatures, with low concentrations of oxidizing reagents, alkenes form diols.
When more concentrated solutions of potassium permanganate and higher temperatures are employed (vigorous conditions), the reaction would not stop at this point.
The product of the reaction depends on the arrangement of groups around the carbon−carbon double bond. The potassium permanganate (VII) solution oxidizes the alkene by breaking the carbon - carbon double bond and replacing it with two carbon - oxgyen double bonds forming carbonyl compounds.
- If there are two alkyl groups at one end of the bond, that part of the molecule will give a ketone.
- If there is one alkyl group and one hydrogen at one end of the bond, that part of the molecule will give a carboxylic acid.
- If there are two hydrogens at one end of the bond, that part of the molecule will give carbon dioxide and water.
In 2- methyl 2−pentene, one carbon contains two methyl groups on one side of the double bond and the other carbon has one alkyl chain and one hydrogen. Hence, the product will be one ketone and one carboxylic acid.
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On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH−CH2CH3 gives:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?
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On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH−CH2CH3 gives:a)b)c)d)Correct 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 On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH−CH2CH3 gives:a)b)c)d)Correct 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 On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH−CH2CH3 gives:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?.
On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH−CH2CH3 gives:a)b)c)d)Correct 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 On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH−CH2CH3 gives:a)b)c)d)Correct 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 On vigorous oxidation by alkaline permanganate solution, (CH3)2C=CH−CH2CH3 gives:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?.
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