Talk:Module1:Dima1407

Q1. Your energy values are all correct. The dimerisation is indeed under kinetic control, the reason is that the transition state leading to the endo product is stabilised by a secondary orbital interaction. In general, there is no need to give the energies to so many significant figures. Consider the accuracy of the calculation, at any given time, the values you get back may differ by as much as 0.001 units so there is some error in the numbers obtained. As your answer suggests, bending strain does come from deviation from ideal bond angle; in the case of the monohydrogenated compounds there is more bending strain for molecule 3 because its angles are further from the ideal 120 degrees for sp2 carbon bonds.

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Q2. Since this problem mainly involved finding different conformational isomers and then reporting the lowest energy conformation, it would have been useful for you to discuss the way in which you modified the starting points for calculations to find different conformers. Your explanations for the stereoselectivity in the reactions are right; the aniline clashes with the carbonyl group specifically because there is repulsion between the electron rich atoms.

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Q3. Your energy for cpd 9 is good and you found the right twist-boat 6-ring conformation. The structure of compound 10 is incorrect – the double bond geometry should have the gem-dimethyl-bridge and hydrogen in a cis relationship. Yes, this is a hyperstable alkene and your explanation is right – strain in the alkane derived by hydrogenation is unusually high so the process is unfavourable.

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Q4. The MOs look good and the syn double bond is indeed the most nucleophilic with the largest part of the HOMO located on it. The IR stretches you got are as expected. The reason that the C-Cl bond is weaker in the diene is that the pi orbital of the anti double bond (which you remove to get the monoalkene) can interact with its antibonding orbital.

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MP. This is a good choice for a project. Your NMR calculation is good and the analysis shows that computational chemistry can be used to distinguish between the two isomers. There was also the question of how these compounds can be distinguished experimentally. The ultimate answer would come from X-ray diffraction analysis of a product crystal; in this case there are NMR techniques that could solve this problem because every carbon atom and proton can be assigned and the relative distances of some key components can be determined by correlation spectroscopy. This technique shows the distance in terms of number of bonds certain atoms are away from each other. The reason that there are fewer peaks reported in the experimental NMR spectrum than number of carbon atoms is that some of the atoms are equivalent (due to rotation they can interchange and the NMR is an average of the shifts in the two positions). A good way to analyse the fit of calculated NMR shifts of possible isomers to experimental data is to find the error in each shift for all calculated isomers and see if any particular structure fits the data best.