In the picture, you can see the bridge whose end points are in para relation (for making the image clear).

Now, what my friend said is that there is no possibility of tautomerisation as there is no H to the right of =O and to the left of =O, on removing H, the bridge will be loaded(?!?).

But according to me, positive charge on the bridge end-points may be unstable as the bridge is out of the plane of the planar molecule and making it sp2 hybridized isn't a kind thing to do. But on removing H from the bridge end point (for tautomerisation), there will be negative charge on the bridge end point, which is perfectly normal.

So will it tautomerise or not?

  • 1
    $\begingroup$ Have you taken non-classical ions into account? Your molecule is well suited for it: en.wikipedia.org/wiki/2-Norbornyl_cation $\endgroup$ – ssavec Mar 18 '16 at 8:33
  • $\begingroup$ Is there any kind of anomaly related to Norbonyl derivatives? Please answer if you know. It will be a great help. $\endgroup$ – Reeshabh Ranjan Mar 18 '16 at 9:49

If you're asking specifically about the tautomerisation, then I made this comparative computational study using B3LYP theory at the 6-31g* level, and the results are:

Effects of the bridge on tautomerisation

So as is evident, the bridge causes the alkene to be dramatically strained. You can compute the Boltzmann ratio and find that the tautomerisation in the first structure is completely negligible at room temperature.

If you're talking about the formation of an enolate ion, the results shouldn't differ much, because the alpha carbon still needs to become sp2 hybridised, in order to delocalise the carbanion into the carbonyl group. And this, as we know, is the reason for the increase in acidity of protons adjacent to carbonyl groups.

All this is of course to the best of my knowledge.

  • $\begingroup$ Excellent answer, completely got what I was looking for! So finally, the intermediate step of formation of carbanion shall not be much affected due to bridging, as the carbanion(-ed) carbon will be still sp3 hybridised, right? $\endgroup$ – Reeshabh Ranjan Mar 18 '16 at 15:15
  • $\begingroup$ @ReeshabhRanjan Just to be clear, the difference in these two compounds has everything to do with bridging. $\endgroup$ – jerepierre Mar 18 '16 at 15:46
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    $\begingroup$ Note that the increased strain of an alkene at a bridgehead is well-known enough to have been named: Bredt's rule $\endgroup$ – Ben Norris Mar 18 '16 at 21:13
  • $\begingroup$ Well as per your answer, giving enough heat will make the bridge carbon sp2 hybridised. I think the bridge will break easily later? $\endgroup$ – Reeshabh Ranjan Mar 19 '16 at 10:09

I can confirm Brian's findings, who beat me to it. On the DF-BP86/def2-SVP level of theory, the energy difference is $\Delta G = 238.8~\mathrm{kJ\, mol^{-1}}$. Which means, that the enol tautomer does virtually not exist.

The straining can be quite easily explained by two pictures.

The newly created $\pi$ bond will certainly prefer a planar arrangement. Due to the rigidity of the whole structure, this is not possible. In one case the angle deviates by $35.5^\circ$ from the ideal conformation. In the other case it deviates by $58.3^\circ$. And this is just looking at the most distorted angles.
Since the geometrical distortion is so big, there cannot be a good enough overlap between the p-orbitals of the new $\ce{C=C}$ bond and the oxygen, hence delocalisation and therefore stabilisation is not possible.

enter image description here enter image description here


Under sufficiently vigorous conditions (strong base, high temperatures}, 3,3-dimethyl[2.2.2]bicycloctan-2-one will probably undergo homoenolisation and a ring opening. The product most likely is a bicyclo[3.2.1]octanone.


  • $\begingroup$ But under normal conditions, the strain should not be enough for the ring to open. Please consider normal condition only for tautomerisation. $\endgroup$ – Reeshabh Ranjan Mar 18 '16 at 9:51

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