Keto-enol tautomerism in a bicyclo compound

Question

Question

Which one is more stable- the keto or enol form. The compound given is this:

Solution

The enol form would violate Bredt's Rule and would therefore be unstable. Hence the keto form is more stable

My doubt

But I am not entirely convinced. Can't we have alpha hydrogens migrating from either of the carbon atoms I've indicated by arrows? Probably they're expecting us to assume that the alpha hydrogen would migrate from the bridge-head carbon, but I just wanted to know if there's any loophole in my concepts. Any help would be appreciated.

Answer 1

Most of the times, the keto-tautomer is lower in energy than the enol-tautomer of a given compound. In all cases this is also very dependent of the actual conditions.
There are few cases, where the enol-tautomer can be dominant. One of the most common examples is malondialdehyde, and, of course, its derivatives.
I am fairly certain that the original question is trying to trick you into thinking this way; and testing your 3D awareness. Or something like this; I'm not sure.

With these structures it's always a good idea to visualise then. I recommend a molecule kit for hands on approaches, but clay and matches are great, too. Even chestnuts and toothpicks are great tools. If you are technically inclined, there are great programs, such as Avogadro (avogadro.cc), which can do basic modeling.
For the following, I have used gfn2-xtb as an optimizer; that is a semi-empirical level of theory, see xtb-docs.readthedocs.io.

You can clearly see that if you migrate the bridgehead-hydrogen, the resulting lone pair would not be able to form resonance with either of the carbonyl groups. That is known as the empirical observation named Bredt's rule.

On the other hand you are quite correct that you could form a tautomer from the other α-carbon. This carbon can easily distort to have planar coordination and the lone pair can be in resonance with the carbonyl groups. (Provided of course that the framework is localised bond orbitals; and that the model is a valid approach.)

On this level of theory, the enol form is about 64.5 kJ/mol higher in energy than the keto-form. So while there certainly is a theoretical possibility, the likelihood is very small. (Just for giggles: the bridge tautomer is 246.6 kJ/mol higher in energy.)

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Appendix

1. Keto-Tautomer

   17
   symmetry c1
   C       -0.232194517     -0.455059284     -0.108971891
   C        0.435695625      0.521858137      0.853516654
   H        0.456616659     -1.075279085     -0.674274662
   C       -0.652035075      0.924669592      1.832141805
   O        1.572314637      0.893979605      0.826125495
   C       -1.888025507      0.221717603      1.265101729
   H       -0.393829692      0.553384752      2.825493466
   H       -0.740240162      2.008868613      1.896308913
   C       -2.240914528      0.890781562     -0.065758217
   H       -2.719883618      0.149774125      1.962761267
   C       -1.080374859      0.487832943     -0.956781929
   H       -3.167688540      0.496107389     -0.485905106
   H       -2.336524789      1.975006913     -0.012902862
   O       -0.862082222      0.837168772     -2.079995835
   C       -1.289181703     -1.103984849      0.787745832
   H       -1.992066775     -1.719717994      0.228564710
   H       -0.851574934     -1.695108793      1.590840632
   

2. Enol-Tautomer

   17
   symmetry c1
   C       -1.252989400     -1.048025267      0.775079043
   C       -0.235131476     -0.354975263     -0.135182749
   C        0.316010839      0.694441373      0.817231157
   C       -0.666608713      1.048338273      1.645393318
   C       -1.895763330      0.268945971      1.228776635
   C       -2.331880748      0.853259922     -0.128837161
   C       -1.153613858      0.494907935     -1.014278443
   O        1.568858497      1.186824174      0.741278311
   O       -0.952399200      0.822318914     -2.150244908
   H        0.493388572     -0.958265049     -0.672260873
   H        2.056130434      0.748941273      0.034835696
   H       -0.640184178      1.790739752      2.416980529
   H       -2.686913718      0.203266237      1.970705846
   H       -3.231277420      0.373221210     -0.518677687
   H       -2.502755602      1.928070454     -0.108020068
   H       -1.934290934     -1.708837720      0.239732785
   H       -0.777205764     -1.591075191      1.590309569
   

3. Bridgehead-enol-tautomer

   17
   symmetry c1
   C       -1.375174317     -1.103606241      0.843045628
   C       -0.248454950     -0.489532946     -0.004463589
   C        0.359288830      0.442230533      0.837984097
   C       -0.605216725      0.791584850      1.950720322
   C       -1.905351826      0.244045543      1.301925006
   C       -2.166063721      0.867188742     -0.087057960
   C       -0.935024925      0.450040319     -0.933532366
   O        1.456979798      1.143062588      0.565484195
   O       -0.594744149      0.929934770     -1.975783635
   H        1.590334610      1.860425911      1.199389402
   H       -0.422954467      0.294599984      2.906280842
   H       -0.604773151      1.872841199      2.115837534
   H       -2.773451234      0.269475098      1.961354663
   H       -3.086355462      0.469282331     -0.513923137
   H       -2.229917288      1.956721084     -0.092669863
   H       -2.077588583     -1.677749691      0.243281460
   H       -0.985680441     -1.728640075      1.644336401
   

   

Answer 2

I think you are correct, the α-H at the positions you marked would be able to migrate and form an enol tautomer. However, the keto form would still be more stable because in the enol form we do not gain aromaticity or conjugation, but lose out on the strong intermolecular H-bonding that the keto form was undergoing.