The increasing order of stability of the following free radicals is
- $\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}}$
- $\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H}$
- $\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H}$
- $\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H}$
Solution
Asked in: JEE Main 2006
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