Talk:Mod:SJN2511

Correct energies for all compounds.
" 1.12 kcal.mol-1"
The second half could use some visual aid. I know what you're talking about, but a new reader would have struggled to follow.

"the "up" conformer (9) is less energetically favourable on account of inducing a higher-energy twist-boat conformation on the adjacent cyclohexane ring" the structure you showed actually is chair for the cyclohexane ring, as you pointed out later.
"Which may be due to strain on the carbonyl sp2 carbon: even assuming ideal C-Ccarbonyl-C bond angle of 120° (it's usually a bit less), compound (9) exhibits 126°, compared to 120° for compound (10). " It's hard to ignore the opposite trend for the C=C double bond here.
Comparing 10 and hydrogenated 10: you must balance the equation!
Good explanation of the hyperstability of the alkene.

Your MOs don't have the correct symmetry. There is an error in the software during optimisation process, which you're expected to spotted and rectify or discuss with us.
"the LUMO is located around the anti-alkene bond, meaning that this bond will more readily undergo nucleophilic attack instead of acting as an nucleophile. The LUMO+1 and LUMO+2 calculations serve to show that even the chlorine-carbon bond would undergo nucleophilic attack before the endo-alkene bond would" you need to read the reference included in the question.
Vibrational frequencies for 12 are slightly off. Again, this has to do with the optimisation process.
Bond length discussion missing. The discussion on the interaction between the anti double bond and the C-Cl anti-bonding orbital is too brief.

"diastereoisomery"
"the oxocarbenium and dioxolonium conjugated systems"
Correct choice of the Me group.
B' Jmols are wrong diastereomers. So are D and D'.
Much more analysis can be done here For example the difference between MM2 and PM6 results. The energies of intermediates and the diastereoselectivity of the reaction.

Distinguishing between 5 and 6 is so trivial, it doesn't require computational chemistry. The diastereoselectivity of the reduction (to give either 5 or 6) is worth studying and you were meant to do that.
Optical rotation: what you have for 6a is the epimer of 6, not enantiomer. Hence the same sign of rotation for both compounds.
I'm not following your HOMO/LUMO argument here. In any case, modelling radical reactions, particularly the transition states, is extremely tricky and require much higher level of theory. Brave attempt though.