Q18 / 180 · COMEDK 2025 (10 May Shift 3) · Chemistry
Which of the following is the most stable free radical?
- A$\mathrm{CH}_2=\dot{\mathrm{C}} \mathrm{H}$
- B$\dot{\mathrm{C}} \mathrm{H}_3$
- C$\left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}}$
- Correct answerD$\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H}$
Solution
The stability of free radicals is determined by factors such as resonance, hyperconjugation, and inductive effects.
Option (1) is a vinyl radical, which is highly unstable due to the $sp^2$ hybridized carbon atom having the unpaired electron in an $sp^2$ orbital.
Option (2) is a methyl radical, which has no alkyl groups to provide stability through hyperconjugation.
Option (3) is a tertiary butyl radical, which is stabilized by hyperconjugation from nine $\alpha$-hydrogen atoms.
Option (4) is a diphenylmethyl radical, which is highly stabilized by extensive resonance with two phenyl rings. The unpaired electron is delocalized over the entire conjugated system of both benzene rings, making it significantly more stable than the tertiary butyl radical.
Answer: $\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H}$
Option (1) is a vinyl radical, which is highly unstable due to the $sp^2$ hybridized carbon atom having the unpaired electron in an $sp^2$ orbital.
Option (2) is a methyl radical, which has no alkyl groups to provide stability through hyperconjugation.
Option (3) is a tertiary butyl radical, which is stabilized by hyperconjugation from nine $\alpha$-hydrogen atoms.
Option (4) is a diphenylmethyl radical, which is highly stabilized by extensive resonance with two phenyl rings. The unpaired electron is delocalized over the entire conjugated system of both benzene rings, making it significantly more stable than the tertiary butyl radical.
Answer: $\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H}$
Hence, the correct answer is option: 'D'.