Free Radical Stability
The correct order of stability of free radicals is:
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Solution
(CH3)3C•>(CH3)2CH•>CH3CH2•>CH3•
Free radical stability increases with increasing substitution at the radical center, primarily due to hyperconjugation. Adjacent C–H sigma bonds can overlap with the half-filled p orbital on the radical carbon, delocalizing the unpaired electron and lowering the energy of the system. The number of C–H bonds available for hyperconjugation determines the degree of stabilization: (CH₃)₃C• (tertiary, 9 adjacent C–H bonds) > (CH₃)₂CH• (secondary, 6 adjacent C–H bonds) > CH₃CH₂• (primary, 2 adjacent C–H bonds) > CH₃• (methyl, 0 adjacent C–H bonds for hyperconjugation). The +I inductive effect of alkyl groups also contributes modestly. Therefore: (CH₃)₃C• > (CH₃)₂CH• > CH₃CH₂• > CH₃•.
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About This Question
- Subject
- chemistry
- Chapter
- some basic principles of organic chemistry
- Topic
- free radical stability
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
(CH3)3C•>(CH3)2CH•>CH3CH2•>CH3•
Free radical stability increases with increasing substitution at the radical center, primarily due to hyperconjugation. Adjacent C–H sigma bonds can overlap with the half-filled p orbital on the radical carbon, delocalizing the unpaired electron and lowering the energy of the system. The number of C–H bonds available for hyperconjugation determines the degree of stabilization: (CH₃)₃C• (tertiary, 9 adjacent C–H bonds) > (CH₃)₂CH• (secondary, 6 adjacent C–H bonds) > CH₃CH₂• (primary, 2 adjacent C–H bonds) > CH₃• (methyl, 0 adjacent C–H bonds for hyperconjugation). The +I inductive effect of alkyl groups also contributes modestly. Therefore: (CH₃)₃C• > (CH₃)₂CH• > CH₃CH₂• > CH₃•.
This hard difficulty chemistry question is from the chapter some basic principles of organic chemistry, covering the topic of free radical stability. It appeared in the 2025 exam.
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