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Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How?
Most Upvoted Answer
Arrange the following alkyl halides in decreasing order of the rate of...
Explanation:
The rate of beta-elimination reaction with alcoholic KOH depends on the stability of the carbocation intermediate formed during the reaction. The stability of the carbocation can be determined by considering the following factors:
1. Hyperconjugation: Hyperconjugation is the stabilizing interaction between a filled sigma orbital (usually C-H or C-C bond) and an empty or partially filled orbital (usually an empty p orbital). The more hyperconjugative structures a carbocation can form, the more stable it is.
2. Inductive effect: The inductive effect is the electron-donating or electron-withdrawing effect of a substituent group. Electron donating groups stabilize carbocations, while electron-withdrawing groups destabilize them.
3. Resonance effect: Resonance stabilization occurs when a carbocation can delocalize its positive charge through pi bonds or lone pairs. The more resonance structures a carbocation can form, the more stable it is.
Based on these factors, we can determine the order of stability and therefore the rate of beta-elimination reaction for the given alkyl halides:
1. CH3CH2CH2Br:
- There are no hyperconjugative structures, as there are no adjacent hydrogen atoms to donate electrons.
- The inductive effect of the ethyl group (+CH2CH3) is weakly electron-donating, but it does not significantly stabilize the carbocation.
- There are no resonance structures, as the carbocation is primary and cannot delocalize its positive charge.
- Therefore, the carbocation formed from CH3CH2CH2Br is the least stable among the given alkyl halides, leading to the slowest rate of beta-elimination reaction.
2. CH3CH(CH3)CH2Br:
- There are three hyperconjugative structures, as there are three adjacent hydrogen atoms to donate electrons. This increases the stability of the carbocation.
- The inductive effect of the methyl groups (+CH3) is weakly electron-donating, further stabilizing the carbocation.
- There are no resonance structures, as the carbocation is secondary and cannot delocalize its positive charge.
- Therefore, the carbocation formed from CH3CH(CH3)CH2Br is more stable than the one formed from CH3CH2CH2Br, leading to a faster rate of beta-elimination reaction.
3. CH3CH2CH(Br)CH3:
- There are two hyperconjugative structures, as there are two adjacent hydrogen atoms to donate electrons. This provides some stability to the carbocation.
- The inductive effect of the ethyl group (+CH2CH3) is weakly electron-donating, further stabilizing the carbocation.
- There are no resonance structures, as the carbocation is primary and cannot delocalize its positive charge.
- Therefore, the carbocation formed from CH3CH2CH(Br)CH3 is less stable than the one formed from CH3CH(CH3)CH2Br but more stable than the one formed from CH3CH2CH2Br. As a result, the rate of beta-elimination reaction for CH3CH2CH(Br)CH3 falls between the other two alkyl halides.
Conclusion:
Based on the stability of the carbocation intermediates formed, the correct order of the rate of beta-
The rate of beta-elimination reaction with alcoholic KOH depends on the stability of the carbocation intermediate formed during the reaction. The stability of the carbocation can be determined by considering the following factors:
1. Hyperconjugation: Hyperconjugation is the stabilizing interaction between a filled sigma orbital (usually C-H or C-C bond) and an empty or partially filled orbital (usually an empty p orbital). The more hyperconjugative structures a carbocation can form, the more stable it is.
2. Inductive effect: The inductive effect is the electron-donating or electron-withdrawing effect of a substituent group. Electron donating groups stabilize carbocations, while electron-withdrawing groups destabilize them.
3. Resonance effect: Resonance stabilization occurs when a carbocation can delocalize its positive charge through pi bonds or lone pairs. The more resonance structures a carbocation can form, the more stable it is.
Based on these factors, we can determine the order of stability and therefore the rate of beta-elimination reaction for the given alkyl halides:
1. CH3CH2CH2Br:
- There are no hyperconjugative structures, as there are no adjacent hydrogen atoms to donate electrons.
- The inductive effect of the ethyl group (+CH2CH3) is weakly electron-donating, but it does not significantly stabilize the carbocation.
- There are no resonance structures, as the carbocation is primary and cannot delocalize its positive charge.
- Therefore, the carbocation formed from CH3CH2CH2Br is the least stable among the given alkyl halides, leading to the slowest rate of beta-elimination reaction.
2. CH3CH(CH3)CH2Br:
- There are three hyperconjugative structures, as there are three adjacent hydrogen atoms to donate electrons. This increases the stability of the carbocation.
- The inductive effect of the methyl groups (+CH3) is weakly electron-donating, further stabilizing the carbocation.
- There are no resonance structures, as the carbocation is secondary and cannot delocalize its positive charge.
- Therefore, the carbocation formed from CH3CH(CH3)CH2Br is more stable than the one formed from CH3CH2CH2Br, leading to a faster rate of beta-elimination reaction.
3. CH3CH2CH(Br)CH3:
- There are two hyperconjugative structures, as there are two adjacent hydrogen atoms to donate electrons. This provides some stability to the carbocation.
- The inductive effect of the ethyl group (+CH2CH3) is weakly electron-donating, further stabilizing the carbocation.
- There are no resonance structures, as the carbocation is primary and cannot delocalize its positive charge.
- Therefore, the carbocation formed from CH3CH2CH(Br)CH3 is less stable than the one formed from CH3CH(CH3)CH2Br but more stable than the one formed from CH3CH2CH2Br. As a result, the rate of beta-elimination reaction for CH3CH2CH(Br)CH3 falls between the other two alkyl halides.
Conclusion:
Based on the stability of the carbocation intermediates formed, the correct order of the rate of beta-
Community Answer
Arrange the following alkyl halides in decreasing order of the rate of...
Order tertiary greater than secondary greater than primary
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Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How?
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Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How?.
Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How?.
Solutions for Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How? in English & in Hindi are available as part of our courses for NEET.
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Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How?, a detailed solution for Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How? has been provided alongside types of Arrange the following alkyl halides in decreasing order of the rate of beta - elimination reaction with alcoholic KOH. (I) CH3CH2CH2Br (2) CH3CH(CH3) CH2Br (3) CH3CH2CH(Br)CH3 Option: (A) 2>3>1 (B) 3>2>1 (C) 1>2>3 (D) 2>1>3 But the correct option is (B). How? theory, EduRev gives you an
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