2
$\begingroup$

Radical cation derived from cyclopropene

Suppose I have a cyclopropene molecule with a + charge on the singly bonded carbon.I think the singly bonded carbon is $\mathrm{sp^2}$ hybridised, isn't it? To obtain this species from cyclopropene, one can consider removing two hydrogen atoms, one homolytically (the one bonded to pure p orbital of carbon) and the other heterolytically (the one bonded with $\mathrm{sp^2}$ orbital of carbon).

So, finally, there is one electron in the pure p orbital and one unit positive charge on one $\mathrm{sp^2}$ hybridised orbital, shouldn't the electron in the pure p shift to the $\mathrm{sp^2}$ orbital as $\mathrm{sp^2}$ has more s character (so it is near the nucleus), does this shift happens?

I am unclear about this scenario, can someone please explain where the electron and the positive charge reside in the C atom.

$\endgroup$
  • $\begingroup$ Correction: "I removed one H atom from the singly bonded Carbon heterolytically" Yes, it's heterolytically, but then it's the "hydride ion" that you have removed and not the hydrogen atom. $\endgroup$ – Gaurang Tandon Jan 14 '18 at 13:39
  • $\begingroup$ chemistry.stackexchange.com/questions/69925/… $\endgroup$ – Mithoron Jan 14 '18 at 22:12
2
$\begingroup$

You do not really have a singly bonded carbon atom. When you remove the hydrogens from the originally saturated carbon the pi electrons delocalize to make an aromatic ring. In effect you have a cyclopropenyl cation with one of the hydrogen atoms homolytically removed. The positive charge is thus delocalized in the ring but the odd electron is localized to the carbon with no hydrogen, in the outward pointing $sp^2$ hybridized orbital.

$\endgroup$
  • $\begingroup$ Thanks, what is the hybridisation of carbons here, is it sp2? $\endgroup$ – drake01 Jan 15 '18 at 13:19
  • $\begingroup$ All carbons are sp2. $\endgroup$ – Oscar Lanzi Jan 15 '18 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.