Talk:Mod:kc8
Nice introduction: I imagine the fact that you took a bit of time to become familiar with the types of calculation helped you a lot.
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Q1: The energy values are spot on for the dimers and hydrogenated compounds. Everything you said about thermodynamic vs. kinetic control was correct. The reason the transition state that gives the endo product is more stable is that a stabilising “secondary orbital interaction” is possible. For the discussion of contributions to strain: Bending strain relates specifically to deviation from ideal bond angles. This is the main factor differentiating the energies of the hydrogenated compounds is bending strain because the double bond in the bicyclic unit is further from its ideal 120 deg angles.
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Q2. Your energy values are fine and the application of the different calculation methods is a good way of reducing the chance of coming to erroneous conclusions with the error in the molecular mechanics modelling. Your explanation for the stereoselectivity is all correct. You mention a 6-membered intermediate that leads to the product of the Grignard addition (I think you mean transition state here – you should be careful to make sure your definitions are accurate when talking about intermediates or transition states because they are significantly different species). Some discussion of how you arrived at the lowest energy conformations might have been useful here – maybe there is a relatively high energy conformer with the CO group pointing in the other direction, but it only exists in small amounts compared to the lowest energy conformation. The reason you can’t include the Grignard reagent into your calculations is that magnesium is not a recognised atom type for MM2 (the parameters aren’t known to the MM2 data). It is possible to get around this by using MOPAC or DFT calculations or by manually defining the magnesium cation parameters and saving them to the MM2 force field.
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Q3. The lowest energy conformations are correctly assigned and your structures look good (this calculation can be problematic in terms of the molecule rearranging itself into a different diastereoisomer but you managed to avoid this problem). This is indeed an example of a hyperstable alkene (usually medium ring/bridgehead alkenes). Your definition is almost correct: the point is that the alkane generated by hydrogenation would contain very high levels of strain so its formation is disfavoured even though the alkene itself also appears strained (being at a bridgehead).
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Q4. Your calculated energy is as expected and your MO diagrams look good. The double bond that has the largest proportion of the HOMO located on it is indeed the most nucleophilic. The IR stretch values look good but your explanation of the difference with other substituents is only half correct. Only some of those groups are electron-donating, for example the –CN group is electron withdrawing and will in fact draw electron density out of the double bond. If you look at the C-Cl bond stretch – it is significantly larger (stronger bond) when the exo-double bond is removed. This is due to an interaction between the exo-double bond and the C-Cl antibonding orbital which weakens the C-Cl bond.
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MP. This is an interesting choice of project with the thermodynamic control being a good reason for study by computational chemistry – it is a shame that in the end the NMR calculation wasn’t able to distinguish the isomers. Energy values in Hartrees are usually converted to kcal and also you would have noticed that they are pretty huge numbers: 1 Hartree = 627.5 kcalmol^-1. What you also have to realise is that the energy you have calculated is the total energy (according to the Schroedinger equation) and is thus much larger than an enthalpy of formation etc. Normally these energies are reported with reference to one molecule (set at 0 kcalmol^-1) so that you can see the difference in energy (and the numbers are then more manageable). You mentioned that the lit. data has fewer peaks than your calculation if you look at the ppm values that are not reported they are in the range for aromatic carbons. In the molecule there is the phenyl side-chain which due to free rotation has 4 sets of equivalent atoms (ipso, ortho, meta, para); for the calculation the molecule is “frozen” into a set conformation making 6 inequivalent atoms. The best way to account for this is to take the average value for atoms which are in reality equivalent. It is good that you applied the correction in an attempt to make the results more accurate. One general problem is that the differences in NMR shifts are very small, and the best error you can expect from calculation is 1-5 ppm.