Talk:Mod:AM8709mod1

As your submission is so late, the marking is very complicated so I actually include the mark of each round and your overall mark to explain why you get the mark at the end. We don't normally release it here, but I hope this help you focus on the next two modules.

First marking round at the deadline 17.00 25/11/2011

Q1

"However, Once endo isomers"
"That is, in thermodynamically controlled conditions exo products are more likely to be formed as a major product, not exo products."
I'm under the impression that you mix the endo-exo geometry of the products with endo-exo transition states.
"Because the molecular wavefunctions of the species at the transition states have to be taken into account". No, because MM cannot form and break bonds!
Your energies quote are beyond the accuracy limit of the technique.
Wrong understanding of torsion energy.

Mark: 3/15 Total mark: 3/100

Second marking round at 17.00 26/11/2011, 15% penalty

Q1

Same Jmol structure for 3 and 4.
Still too many digits after the dot for energies. Correct energies though.
Correct analysis of bend energy contribution "(Guozhu Liu 2005 Ind Eng Chem Res)" you should have learnt how to quote a reference in the write up page. What's the page number of the paper in question? and the volume?
"Unlike the dimerisation, the hydrogenation of the endo dimer is a thermodynamically controlled reaction." Hydrogenation is almost never under thermodynamic control.

Mark: 5/15

Q2

"(Bringmann G, Mortimer AJP 2005 Angew Chemie Int)" same problem as in question 1.
"two chair conformations A and B" I think they're one chair, one twisted-boat.
All your Jmols in Table 3 belong to question 1. Both MM2 and MMFF94 data are wrong, most like also taken from question 1.

Mark: 2/15 Total mark: 7/100. Applying 15% penalty gives 6/100. Adding the previous mark gives 9/100.

Third marking round at 17.00 27/11/2011, 30% penalty

Q2

Jmols are now updated. But 9B and 10B are mixed up.
The optimised energies seem correct, but still too many digits.
" Therefore, we can conclude that the atropisomer with the carbonyl pointing below the ring 9 is thermodynamically more stable than 10. " Wrong, 10A is more stable than 9A.
" The conformers 9A and 10A have larger dihedral angles compared to the the conformers 9B and 10B." where is the proof?
" the dihedral angle for 10A is 167° while for 9A is 56°" A picture is required here. Your steric hindrance argument is not correct. The stabilisation comes from the near antiperiplanar arrangement of the alpha C-H and the C=O group, and is explained by MOs, which may not have been factored in MM2. The answer for the energy difference is elsewhere.
"Hyperstable alkenes": you only provided bits of information from literature, but don't seem to understand the real reason here. Calculation of the energy of the hydrogenated product is required.

Mark: 6/15 Total mark: 6/100. Applying 30% penalty gives 4/100. Adding previous mark gives 13/100.

Fourth and final marking round, 45% penalty. We don't usually go beyond 72h, as the penalty by then is too high.

Q3

MOs are ok, but I can only see the chlorine in some structures.
Correct prediction of reactivity.
IR frequencies correct.
However, you completely failed to spot that the two double bonds are different in bond strength, bond length and vibrational frequencies. The reason for this was included in one of the references in the question.

Mark: 8/15. No Dspace link for SCAN calculation so that part receive zero mark, which leaves 4/15.

Q4

You didn't explain your choice of the R group.
your A and B' Jmols are wrong. The OAc groups are nowhere near the oxonium.
PM6 energies are all wrong. How can it be the better method with those numbers?
No calculation for C, C', D and D'.

Mark: 2/15

Mini project

No Dspace link, no mark.
No defined question. What structure are you calculating your C13 NMR for? Why?

Total mark: 6/15. Applying 45% penalty gives 3/15. Adding to previous mark gives 16/100.