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Well this might seem like a dumb question but I was just wondering if any other factor would apply here as well?


By simply looking at the number of hyperconjugative structures, I could easily tell that the isopropyl would be the most stable followed by ethyl and the vinyl radical at the last.
But for the vinyl, I didn't find applying hyperconjugation satisfactory.
Would we apply some other concept while comparing the vinyl radical with the other two?

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  • $\begingroup$ How would we describe the hybrization of the orbital containing the electron in the ethyl radical, the vinyl radical. Which of those two orbitals would an electron be more stable in? $\endgroup$ – ron Mar 19 at 15:42
  • $\begingroup$ @ron I believe the hybridization of the orbital containing the electron in ethyl free radical would be $sp^2$ while in vinyl free radical it would be $sp$ and since sp has more s character so it would keep the electron more closely held. But how would this help us determine whose more stable? $\endgroup$ – Prajwal Tiwari Mar 19 at 15:59
  • $\begingroup$ But according to that, the vinyl free radical would be more stable than ethyl free radical?! But that's not true right? Vinyl should be least stable $\endgroup$ – Prajwal Tiwari Mar 19 at 16:47
  • $\begingroup$ @ron can we say that since the electron is in sp hybridised orbital in vinyl free radical, it would be pulled more towards the already existing C-C sigma and pi bonds which might lead to its destability. Hence making the vinyl free radical the least stable? $\endgroup$ – Prajwal Tiwari Mar 19 at 16:49
  • $\begingroup$ Your thinking is correct. The orbital containing the lone electron is electron deficient and does not want to be closer to the nucleus. $\endgroup$ – ron Mar 19 at 17:05

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