In an oral exam for physical organic chemistry, one student was asked to explain the differences in the ionization potential ($IP$) of syn and anti tricyclo[$^{2,5}$]octa-3,7-diene (unsaturated 3-ladderane, see Figure).

Apparently, the $IP$s were as follows:



$\Delta IP_{syn}=0.34$




$\Delta IP_{anti}=0.96$

Koopman's Theorem $IP=-\epsilon_{HOMO}$ might help us here. I think it leads us to an orbital diagram. How do I have to combine the $\sigma$-bonds with the $\pi$-system to be able to explain the difference in the $IP$? Maybe you know a paper for it? I checked out Joseph J. Gajewski's Hydrocarbon Thermal Isomerizations (2004), but he only discusses mechanistic features, no theoretical aspects.


  • 1
    $\begingroup$ Any through-bond interaction that occurs between the cyclobutane sigma bonds and the two pi bonds will be the same for the syn and anti isomers. Only the through-space interaction between the two pi bonds will differ in the syn and anti isomers. In the through-space case the syn isomer will show a greater splitting between the two pi bonds. Are you sure the IPs you've assigned to the anti isomer aren't really the IPs for the syn isomer? $\endgroup$
    – ron
    Aug 28, 2016 at 19:16
  • $\begingroup$ I copied it from an exam protocol. It might very well be that the examinee mixed it up. Do I follow you correctly? You are saying, that the splitting in the syn-config should be larger because of the antisymmetric through-space overlap (through bond being equal). Consequently, the HOMO of the syn would lie higher, i.e. less stabilized, i.e. smaller IP (which would be less negative according to Koopman). This would indeed be indicative of mixing the two IPs. $\endgroup$
    – ste
    Aug 28, 2016 at 19:38
  • $\begingroup$ Yes,the HOMO would be slightly destabilized and the orbital below the HOMO (HOMO-1) would be slightly stabilized in the syn isomer due to any through space orbital interaction. In the anti isomer the HOMO and HOMO-1 would occur at roughly the same energy. The difference in IP1 and IP2 in the anti isomer is attributable to the electrostatic repulsion from placing a second charge in the molecule. In the syn isomer you also have an even greater electrostatic effect plus the through-space effect. It would be interesting to separate these two effects. I wouldn't be surprised if the $\endgroup$
    – ron
    Aug 28, 2016 at 20:30
  • $\begingroup$ electrostatic effect alone accounted for most of the difference. $\endgroup$
    – ron
    Aug 28, 2016 at 20:31
  • $\begingroup$ I found a paper, that investigates this very question: 10.1002/cber.19731060105 $\endgroup$
    – ste
    Sep 1, 2016 at 11:00


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.