enter image description here Which of the above compounds has a lower bond rotation energy?

Here is my progress: We basically need to find which compound has more single bond character. Now both compounds will break the double bond, and give the electron pair so as to make both cycles aromatic. But what do I do after that?

  • $\begingroup$ related chemistry.stackexchange.com/questions/95249/… $\endgroup$
    – Mithoron
    Commented May 23, 2020 at 12:41
  • $\begingroup$ Huh, there was this exact case recently, but it might be deleted now. I wager that double bond to cyclopropenyl has more unfavorable angles (60 when 120 deg. is usual). $\endgroup$
    – Mithoron
    Commented May 23, 2020 at 12:46
  • $\begingroup$ Tropylium cation(the top ring on the right image after bond rotation) is EXTREMELY stable. Probably the most stable carbocation, among the "usual" examples. It's even more stable than cyclopropyl methyl carbocation. $\endgroup$
    – user600016
    Commented May 23, 2020 at 17:05
  • $\begingroup$ @user600016 Is there a reason to why it is so stable? $\endgroup$ Commented May 24, 2020 at 3:34
  • $\begingroup$ Highly conjugated aromatic system. PS: take a look at. chemistry.stackexchange.com/questions/48264/… $\endgroup$
    – user600016
    Commented May 24, 2020 at 4:50

1 Answer 1


Cleavage of the double bond will lead to both compounds having two rings that are aromatic, however the amount of resonance energy(hence stability) will not be the same.

Both systems will have a Cyclopentadienyl anion, but one will have a Cyclopropenium cation while the other has a Tropylium cation . As the Tropylium cation has more resonance structures, it will have a higher resonance energy and hence greater stability than the Cyclopropenium cation. So (b) has a lower bond rotation energy.


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