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CH3=CH2CH3 + H – I → CH3CH2CH2I + CH3CHICH3 (major). This reaction is
- a)Sandmeyer’s reaction
- b)Markovnikov’s rule.
- c)free radical halogenations
- d)Finkelstein reaction
Correct answer is option 'B'. Can you explain this answer?
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CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This react...
This is electrophilic addition reaction following Markovnikoffs rule.
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CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This react...
Explanation:
The given reaction involves the conversion of an alkene (CH3CH=CH2) to an alkyl halide (CH3CH2CH2I) through the addition of a halogen (Iodine, I) in the presence of a Lewis acid catalyst (AlCl3). This reaction follows the Markovnikov's rule.
Markovnikov's Rule:
According to the Markovnikov's rule, when a hydrogen halide (HX) is added to an alkene, the hydrogen atom adds to the carbon atom with fewer alkyl substituents, while the halide adds to the carbon atom with more alkyl substituents. In other words, the hydrogen adds to the less substituted carbon, while the halide adds to the more substituted carbon.
Reaction Mechanism:
The reaction proceeds through the following steps:
1. Formation of a carbocation:
The double bond in the alkene attacks the electrophilic iodine atom, breaking the iodine-halogen bond and forming a carbocation intermediate.
CH3CH=CH2 + I2 → CH3CHI-CH2• + I-
2. Addition of iodide ion:
The iodide ion (I-) acts as a nucleophile and attacks the carbocation, resulting in the addition of iodide to the more substituted carbon atom.
CH3CHI-CH2• + I- → CH3CHI-CH2I
3. Rearrangement (optional):
In some cases, rearrangement of the carbocation may occur to form a more stable carbocation before the addition of the iodide ion.
4. Formation of the alkyl iodide:
The final product is an alkyl iodide (CH3CH2CH2I), where the iodide has added to the more substituted carbon atom.
Conclusion:
The given reaction follows the Markovnikov's rule, where the iodide ion adds to the more substituted carbon atom. Therefore, the correct answer is option 'B' - Markovnikov's rule.
The given reaction involves the conversion of an alkene (CH3CH=CH2) to an alkyl halide (CH3CH2CH2I) through the addition of a halogen (Iodine, I) in the presence of a Lewis acid catalyst (AlCl3). This reaction follows the Markovnikov's rule.
Markovnikov's Rule:
According to the Markovnikov's rule, when a hydrogen halide (HX) is added to an alkene, the hydrogen atom adds to the carbon atom with fewer alkyl substituents, while the halide adds to the carbon atom with more alkyl substituents. In other words, the hydrogen adds to the less substituted carbon, while the halide adds to the more substituted carbon.
Reaction Mechanism:
The reaction proceeds through the following steps:
1. Formation of a carbocation:
The double bond in the alkene attacks the electrophilic iodine atom, breaking the iodine-halogen bond and forming a carbocation intermediate.
CH3CH=CH2 + I2 → CH3CHI-CH2• + I-
2. Addition of iodide ion:
The iodide ion (I-) acts as a nucleophile and attacks the carbocation, resulting in the addition of iodide to the more substituted carbon atom.
CH3CHI-CH2• + I- → CH3CHI-CH2I
3. Rearrangement (optional):
In some cases, rearrangement of the carbocation may occur to form a more stable carbocation before the addition of the iodide ion.
4. Formation of the alkyl iodide:
The final product is an alkyl iodide (CH3CH2CH2I), where the iodide has added to the more substituted carbon atom.
Conclusion:
The given reaction follows the Markovnikov's rule, where the iodide ion adds to the more substituted carbon atom. Therefore, the correct answer is option 'B' - Markovnikov's rule.
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CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This reaction isa)Sandmeyer’s reactionb)Markovnikov’s rule.c)free radical halogenationsd)Finkelstein reactionCorrect answer is option 'B'. Can you explain this answer?
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CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This reaction isa)Sandmeyer’s reactionb)Markovnikov’s rule.c)free radical halogenationsd)Finkelstein reactionCorrect answer is option 'B'. 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 CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This reaction isa)Sandmeyer’s reactionb)Markovnikov’s rule.c)free radical halogenationsd)Finkelstein reactionCorrect answer is option 'B'. 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 CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This reaction isa)Sandmeyer’s reactionb)Markovnikov’s rule.c)free radical halogenationsd)Finkelstein reactionCorrect answer is option 'B'. Can you explain this answer?.
CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This reaction isa)Sandmeyer’s reactionb)Markovnikov’s rule.c)free radical halogenationsd)Finkelstein reactionCorrect answer is option 'B'. 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 CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This reaction isa)Sandmeyer’s reactionb)Markovnikov’s rule.c)free radical halogenationsd)Finkelstein reactionCorrect answer is option 'B'. 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 CH3=CH2CH3+ H – I →CH3CH2CH2I+CH3CHICH3(major). This reaction isa)Sandmeyer’s reactionb)Markovnikov’s rule.c)free radical halogenationsd)Finkelstein reactionCorrect answer is option 'B'. Can you explain this answer?.
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