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Rank, from the most stabilized to the least stabilized, the following free radicals according to their stabilization energies:
(I) CH3CH2 (II) CH2CH3 (III) (CH3)2CH (IV) (CH2=CH—CH2)
(I) CH3CH2 (II) CH2CH3 (III) (CH3)2CH (IV) (CH2=CH—CH2)
- a)IV > III > II > I
- b)I > IV > III > II
- c)III > IV > I > II
- d)III > IV > II > I
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Rank, from the most stabilized to the least stabilized, the following ...
To rank the free radicals according to their stabilization energies, we need to consider the factors that contribute to stabilization.
The factors that contribute to stabilization of free radicals include:
1. Hyperconjugation: The presence of adjacent sigma bonds can provide stabilization through the overlap of electron density from the sigma bond into the empty p orbital of the free radical.
2. Inductive effect: The presence of electron-donating alkyl groups can stabilize the free radical.
3. Resonance: If the free radical can form resonance structures, it will be more stabilized.
Now, let's analyze each free radical:
(I) CH3CH2: This free radical has one adjacent sigma bond and one alkyl group, providing some stabilization through hyperconjugation and the inductive effect.
(II) CH2CH3: This free radical also has one adjacent sigma bond and one alkyl group, providing the same level of stabilization as (I).
(III) (CH3)2CH: This free radical has two adjacent sigma bonds and one alkyl group, providing more stabilization through hyperconjugation and the inductive effect compared to (I) and (II).
(IV) (CH2=CH: This free radical has one adjacent sigma bond and one pi bond. The presence of the pi bond allows for resonance stabilization, which is a stronger stabilizing factor compared to hyperconjugation and the inductive effect.
Therefore, the ranking of the free radicals from most stabilized to least stabilized is as follows:
1. (CH2=CH
2. (CH3)2CH
3. CH3CH2
4. CH2CH3
The factors that contribute to stabilization of free radicals include:
1. Hyperconjugation: The presence of adjacent sigma bonds can provide stabilization through the overlap of electron density from the sigma bond into the empty p orbital of the free radical.
2. Inductive effect: The presence of electron-donating alkyl groups can stabilize the free radical.
3. Resonance: If the free radical can form resonance structures, it will be more stabilized.
Now, let's analyze each free radical:
(I) CH3CH2: This free radical has one adjacent sigma bond and one alkyl group, providing some stabilization through hyperconjugation and the inductive effect.
(II) CH2CH3: This free radical also has one adjacent sigma bond and one alkyl group, providing the same level of stabilization as (I).
(III) (CH3)2CH: This free radical has two adjacent sigma bonds and one alkyl group, providing more stabilization through hyperconjugation and the inductive effect compared to (I) and (II).
(IV) (CH2=CH: This free radical has one adjacent sigma bond and one pi bond. The presence of the pi bond allows for resonance stabilization, which is a stronger stabilizing factor compared to hyperconjugation and the inductive effect.
Therefore, the ranking of the free radicals from most stabilized to least stabilized is as follows:
1. (CH2=CH
2. (CH3)2CH
3. CH3CH2
4. CH2CH3
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Rank, from the most stabilized to the least stabilized, the following free radicals according to their stabilization energies:(I) CH3CH2 (II) CH2CH3 (III) (CH3)2CH (IV) (CH2=CH—CH2)a)IV > III > II > Ib)I > IV > III > IIc)III > IV > I > IId)III > IV > II > ICorrect answer is option 'A'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Rank, from the most stabilized to the least stabilized, the following free radicals according to their stabilization energies:(I) CH3CH2 (II) CH2CH3 (III) (CH3)2CH (IV) (CH2=CH—CH2)a)IV > III > II > Ib)I > IV > III > IIc)III > IV > I > IId)III > IV > II > ICorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Rank, from the most stabilized to the least stabilized, the following free radicals according to their stabilization energies:(I) CH3CH2 (II) CH2CH3 (III) (CH3)2CH (IV) (CH2=CH—CH2)a)IV > III > II > Ib)I > IV > III > IIc)III > IV > I > IId)III > IV > II > ICorrect answer is option 'A'. Can you explain this answer?.
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