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Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds through
- a)a more stable carbonium ion.
- b)a more stable carbanion.
- c)a more stable free radical.
- d)homolysis.
Correct answer is option 'A'. Can you explain this answer?
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Addition of HI to double bond of propene yields isopropyl iodide and ...
All are examples of electrophilic addition reactions and involve the formation of stable carbocation leading to the formation of addition product according to Markovnikov’s rule
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Addition of HI to double bond of propene yields isopropyl iodide and ...
Explanation:
When propene reacts with HI, the addition of HI to the double bond can occur in two ways: Markovnikov addition and anti-Markovnikov addition. In the case of propene, the major product formed is isopropyl iodide (2-iodopropane) rather than n-propyl iodide (1-iodopropane). This is because the addition proceeds through a more stable carbonium ion.
Markovnikov Addition:
- In Markovnikov addition, the hydrogen atom of HI adds to the carbon atom with fewer hydrogen atoms attached (the less substituted carbon atom) in the double bond.
- As a result, the iodine atom adds to the carbon atom with more hydrogen atoms attached (the more substituted carbon atom) in the double bond.
- This leads to the formation of the more stable carbonium ion intermediate.
Reason for the Major Product:
- In the case of propene, when it reacts with HI, the less substituted carbon atom in the double bond undergoes addition with the hydrogen atom of HI.
- This forms a more stable secondary carbonium ion intermediate.
- The more substituted carbon atom, which would have formed the n-propyl iodide product, forms a less stable primary carbonium ion intermediate.
- The more stable secondary carbonium ion intermediate is formed due to the greater stability of the secondary carbocation compared to the primary carbocation.
- Therefore, the major product formed is isopropyl iodide (2-iodopropane) rather than n-propyl iodide (1-iodopropane).
Summary:
- The addition of HI to propene proceeds through a more stable carbonium ion intermediate.
- The less substituted carbon atom in the double bond undergoes addition with the hydrogen atom of HI, forming a more stable secondary carbonium ion intermediate.
- This leads to the formation of isopropyl iodide (2-iodopropane) as the major product.
When propene reacts with HI, the addition of HI to the double bond can occur in two ways: Markovnikov addition and anti-Markovnikov addition. In the case of propene, the major product formed is isopropyl iodide (2-iodopropane) rather than n-propyl iodide (1-iodopropane). This is because the addition proceeds through a more stable carbonium ion.
Markovnikov Addition:
- In Markovnikov addition, the hydrogen atom of HI adds to the carbon atom with fewer hydrogen atoms attached (the less substituted carbon atom) in the double bond.
- As a result, the iodine atom adds to the carbon atom with more hydrogen atoms attached (the more substituted carbon atom) in the double bond.
- This leads to the formation of the more stable carbonium ion intermediate.
Reason for the Major Product:
- In the case of propene, when it reacts with HI, the less substituted carbon atom in the double bond undergoes addition with the hydrogen atom of HI.
- This forms a more stable secondary carbonium ion intermediate.
- The more substituted carbon atom, which would have formed the n-propyl iodide product, forms a less stable primary carbonium ion intermediate.
- The more stable secondary carbonium ion intermediate is formed due to the greater stability of the secondary carbocation compared to the primary carbocation.
- Therefore, the major product formed is isopropyl iodide (2-iodopropane) rather than n-propyl iodide (1-iodopropane).
Summary:
- The addition of HI to propene proceeds through a more stable carbonium ion intermediate.
- The less substituted carbon atom in the double bond undergoes addition with the hydrogen atom of HI, forming a more stable secondary carbonium ion intermediate.
- This leads to the formation of isopropyl iodide (2-iodopropane) as the major product.
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Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. Can you explain this answer?
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Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. 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 Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. Can you explain this answer?.
Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. 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 Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. Can you explain this answer?.
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Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Addition of HI to double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product, because addition proceeds througha)a more stable carbonium ion.b)a more stable carbanion.c)a more stable free radical.d)homolysis.Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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