Talk:Organic:HughLaurie

Q1: The energy values are correct and the dimerisation reaction is indeed under kinetic control; the reason for this kinetic control is stabilisation of the transition state leading to the endo product due to a secondary orbital interaction that is not possible in the transition state leading to the exo product. The hydrogenated products are indeed differentiated by their bending strain and this is well explained. As for the hydrogenation itself, it is fair to say that a metal catalyst would be used. The reaction could be under thermodynamic control if the intermediates in the catalytic cycle can interconvert. You are right to point out that the activation barrier would be lowered by a catalyst, but it is probably the case that the uncatalysed reaction (e.g. direct addition of hydrogen to the double bond) has a massively high activation barrier.

Q2: The calculations are good and the reasons for the stereoselectivity in the reactions are correct – perhaps a discussion of the carbonyl angle would have been beneficial. It is important to note that these reactions are not stereospecific, but are stereoselective (as you correctly termed them in the end). A stereospecific reaction can ONLY have one product (e.g. an SN2 reaction always proceeds with inversion of the reacting centre), whereas a stereoselective reaction favours one product over the other.

Q3: You have correctly found the most stable isomer, but your energy values are higher than expected. One of the isomers in fact has a chair conformation for its 6-ring in its most stable form; it would have been good to see some of the different energies you found along the way to the minimum and this would also be reflective of the work that goes into this optimisation, with so much tweaking necessary. The explanation for the slow olefin reactivity is good; it should be noted that not all bridgehead alkenes display this feature, it is just that some of them are hyperstable (of course for small polycycles, bridgehead alkenes are forbidden – e.g. Bredt’s rule.

Q4: Your MOs look good and are presented well. Indeed, it is the double bond on the same side as the chlorine that is the most nucleophilic (you would normally describe this disposition as syn rather than endo). The changing bond stretches are nicely discussed and you were one of the few people to mention bond length, but perhaps you could have demonstrated this by reporting the C-Cl lengths for the compounds in question. I think in the case of the double bond, the stretch doesn’t really change significantly; it is maybe just a fairly unpredictable consequence of the structural modification.

MINI-PROJECT: The reaction seems like a great choice to me; it is good to see a reaction scheme, but perhaps some discussion of the mechanism would have padded out the introduction a little, and help you explain the product distribution observed. You have shown some molecular orbitals, but perhaps it would be better to discuss the HOMOs of the two isomers in terms of the relative energies. In fact I think the energy calculations on the molecules themselves were probably enough for the discussion about which is the thermodynamic product. What happened to your calculated spectroscopic data? If you have more to submit, then please contact me at the email address from which your mark was sent.