stability of radicals

[inaccensa]

---

Members don't see this ad.

The radicals follow tertiary >secondary>primary>methyl. Is this because the R group attached to the tertiary carbon donate their electron density and makes the radical more stable.?

Also, carbocations cannot form on a benzene ring, but carbanions can, right?

Why is the activation energy required to form tertiary radicals less than the secondary? I 'd think that since the tertiary are more stable (they are at a lower energy level), more energy will be required to reach the transition state.

 

[inaccensa]

I think I know the answer to my last question, the energy required for a compound lets say -t-butane to become t-butyl radical will be less since tertiary radicals are at a low energy on the energy profile diagram. Can you please look at the other questions.