Talk:Mod:jcl08M1
Q1. Your energy values are good for the dimerisation isomers and the monohydrogenated endo-dimers and you correctly say that the reaction is under kinetic control. You should be careful in reviewing what you write up because you actually said, “This indicates that Endo dimer has less strained conformation(mainly due to less torsion in the structure), hence thermodynamically more stable.” Whereas you meant to say the exo-product is more stable! Your diagram to show the secondary orbital interactions that make the endo-transition state more stable is very good and clearly shows the important frontier molecular orbitals. For the monohydrogenation of the endo-isomer, you are right to say that bending strain is the major factor that differentiates the two products and this is indeed due to different deviations from ideal sp2 bond angles. Since compound 4 is the lowest in energy it would, as you say, be the preferred product if the reaction is under thermodynamic 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. Your energy values are fine and the way the energies of different conformations is presented is a nice way of doing it. You say that vdW interactions and bending are the key differences between some of your conformers; a further point for discussion would be why there are these differences (i.e. what specific features are different).
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Q3. This question was very well answered. Your energies are spot on and you correctly identified the lowest energy isomer. The discussion of the differences in the isomers and the description of different (higher energy) conformations you found were excellent. Also the definition of a hyperstable alkene was exactly right.
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Q4. The MOs look a little odd, you would expect a pi centred HOMO to look like a pi bonding orbital (e.g. some density above the alkene some below it with an orbital node in between. The question was which of the double bonds is more nucleophilic (dichlorocarbene is an electrophilic carbene), this therefore means which has the largest share of the HOMO. You should be careful to use the right phrases when describing stereochemical features. The two double bonds are described in the text as being syn (same side) or anti (opposite) to the C-Cl bond. Endo and exo are terms that relate to a groups relative position on the inside or outside respectively. Both of the double bonds in this tricyclic system are “endo” because the are inside the 6-ring, if there is a double bond between one of the atoms in the 6-ring and another that isn’t in the ring it could be described as exo. The IR stretches you found are good and the reason for the weakening o the C-Cl bond in the hydrogenated system is correct.
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MP. This system is a good choice because the molecule is conformationally restricted due to its rigid structure and the reaction gives a mixture of isomers whose properties you can compare. Analysis of the isomer energies is a good extension of the tools you used in the earlier problems - it seems reasonable to suggest thermodynamic control in this reaction, because the conjugate addition that sets the stereochemistry may well be reversible (retro-Michael reactions are commonly observed reactions). You mention at the end that no mechanism is proposed but transition states can actually be calculated by similar methods to those you have used as you will find out in module 3. The bar chart is a good graphical representation of the error in the NMR calculations, it seems as though your calculated results are a good match to the experimental and also it is apparent that you can tell the isomers apart because the 13C sift at the key difference (the configuration at the 4-position) is accurately simulated.