User talk:Xx108 (module 1)

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Q1: Your energy values are good – although you should have shown at least images of your optimised structures! Too many jmols can cause the wiki page to crash, but jpgs are no problem. Also, it is not necessary to quote values to so many decimal places – There is error to account for in the calculation which is an estimation to start with. You are right to say that the lowest energy exo-dimer is the thermodynamic product and the endo-dimer is the kinetic product. It should be noted however that it is NOT the kinetic product because it is the highest in energy (all that means is that it isn’t the thermodynamic product). The kinetic product of a reaction is the one which has the lowest energy transition state that leads to it. In this case it so happens that the endo-dimer is higher in energy but has a lower energy transition state due to the secondary orbital interaction you mentioned. Typically, the kinetic product is the same as the thermodynamic product (i.e. lowest energy product comes from the lowest energy transition state). For the mono hydrogenated dimer the analysis of strain contributions is good – the major difference is indeed due to that double bond in the bicycle deviating from the ideal sp2 bond angles more than the one in the 5-ring. Again, higher energy product is not synonymous with kinetic product: In this case it isn’t really possible to comment on which product is formed by hydrogenation without considering the conditions employed. For the reduction of a double bond there are a variety of different conditions, mostly involving a metal catalyst and hydrogen gas – usually these reactions are irreversible and kinetically controlled. Without calculating transition states (not possible with MM2), it is not possible to make computational assessments of selectivity.

Q2. Again, some pictures or jmols would be useful. Are you sure you mean boat and not twist-boat? The boat conformation of a 6-ring is usually much higher in energy than chair or twist-boat. Your optimised energies are spot on and the lowest energy atropisomer is correctly assigned to be the down form. Trying different forms of the 6-ring is a good main focus for this question, but you could have commented a bit more on what else you tried to do in order to get the lowest energy. A comparison of the strain contributions would also have been good. The definition of a hyperstable alkene is good.

Q3. Did you attempt the symmetrisation in GaussView to avoid asymmetrical orbitals? As you correctly pointed out they should be symmetrical, but this calculation can be unreliable using lower level computational methods. You should give your energy values for all calculation methods so that the energies can be assessed. Your description of the pi-sigma* interaction is a little confused; the stretching frequency of the C-Cl bond increases when the double bond is removed, indicating a stronger bond because there is no longer donation into the sigma*. You can also see the impact of this interaction in the C=C stretching frequencies – lower for the anti double bond which donates its bonding electrons and is hence weakened.

Q4. The methyl group is a good choice for this calculation and in general for computational chemistry on molecules with generic alkyl groups. Also I agree that semi-empirical methods are better in this case. You should have found that for MOPAC calculations A=C and B=D and the cation is treated more as a non-classical carbocation. It is correct that all of A’/B’/C’/D’ are higher in energy than A/B/C/D. Your actual energy values are a little high and it is difficult to work out why without some jmols to look at. The reason the reactions are diastereoselective is in part due to the dominance of A and B over A’ and B’ and also due to unfavourable orientation of A’ and B’ for nucleophilic attack on the oxonium carbon.

MP. You are right to that you can monitor the reaction by the disappearance of the C=O stretch in the IR spectrum, but optical rotation would not be the first choice for distinguishing alcohols 6 and 7. In this case 2D NMR techniques could be employed such as NOESY which could be used to show the proximity of the hydrogen next to the alchol group to other hydrogens in the molecule. In your analysis of the NMR data, it is better to compare the differences in ppm rather than % differences. The use of ppm makes this analysis arbitrarily smaller for high field (low ppm) resonances. If you considered the actually frequencies involved in MHz, the percentage differences would be much different. Generally, 2-3 ppm is seen as a large error in an NMR calculation, so your values are not too bad. The key question is whether this error is small enough that you could distinguish between the two possible isomers. You could have calculated the spectrum for the other alcohol and compared those values to the reported data – hopefully it would be significantly different and you could therefore use calculated NMR to tell the two compounds apart. In order to discuss the selectivity of the reaction, it would have been useful to see a scheme illustrating the reaction mechanism; you suggest that trapping of an intermediate with tBuOH is stereodetermining – I presume this means protonation of an initially formed radical anion, but this is not clear.