Talk:Mod1:shl108
Q1. The values you report for the energies of the dimerisation products and the monohydrogenated endo-isomers are all good and the dimerisation is under kinetic control as you said. Secondary orbital interactions are indeed the reason for this kinetic control, stabilising the endo-transition state. Your diagram shows these interactions and is good except for the fact that orbitals meant to show bonding for the endo-dimer are actually of opposite phase – for a bonding interaction, the orbitals must be in the same phase. The difference in energies for the monohydrogenated endo-products is largely due to bending strain as you say. One of the alkenes has more bending strain than the other because its bond angles deviate from the ideal bond angles for sp2 centres to a greater extent. For this reaction, the lowest energy product is formed, but this does not necessarily tell you that the reaction is under thermodynamic control. The lowest energy product could be formed under thermodynamic OR kinetic control. You know that the dimerisation reaction is under kinetic control as the higher energy product is formed and this could only be true under kinetic control. Alkene hydrogenation is usually carried out with a metal catalyst (e.g. palladium on charcoal) and hydrogen gas; under these conditions, reactions are usually under kinetic control (i.e. the products are not in equilibrium with the starting materials) and selective hydrogenation of alkenes is due to the steric effects that allow coordination of one alkene more easily than another.
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Q2. The energy values you got and the carbonyl dihedrals are all close. The origin of stereoselectivity is well explained (assistance vs. repulsion) and as you have said, magnesium cannot be included in an MM2 calculation because the force field doesn’t have parameters for this atom. I like how you presented the different conformations you found, but perhaps you could have described the differences between them that makes one more favourable than another as a further discussion point.
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Q3. The energies are right and you have correctly identified the lowest energy atropisomer. The importance of the 6-ring conformation mentioned and a comparison of energy terms is given which is good. Also, putting the energy values in a table is a nice way to present it. The definition of a hyperstable alkene is spot on.
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Q4. Your MO and IR stretch calculations are good and the HOMO shows that the syn-alkene is the more nucleophilic (although you said anti – syn means on the same side so the double bond on the same side as the C-Cl bond and anti means opposite). The C-Cl bond is indeed stronger in the hydrogenated example because that interaction between the pi and sigma* orbitals is no longer possible. The presentation of the data in the table is a nice way to display it all together.
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MP. The reaction is a good choice for this type of study because there are two isomeric products that are conformationally constrained and different enough in structure to expect differences in physical properties. You have listed all of your calculated data but it might be better here to compare calculated to experimental data directly. It seems as though you have a reasonable fit to the real data, a good way to represent this is graphically (with a bar chart for example which shows the general magnitude and spread of error. It is worth considering how to differentiate the two isomers experimentally, you point out that the alkene is the most different part of the two isomers with significant differences in 13C NMR shifts. In practise the best method to unambiguously determine which structure you have is X-ray crystallography, otherwise 2D NMR techniques can be used that tell you which hydrogen atoms and carbon atoms are connected (HSQC) or 3 bonds apart (HMBC).