Talk:Mod1: Lukas Miseikis
Q1: Your energy values are all correct and your analysis of strain contributions is good. For the hydrogenation products the major difference is the bending strain due to different alkene bond angles which deviate from ideality to a different extent. You state “in this case Endo is the thermodynamic product” when I think you mean the opposite on the basis of the rest of the paragraph and your results – be careful that the key statements are worded correctly! What you say about kinetic vs thermodynamic control is good. The requirement for kinetic control is not really that certain products are inaccessible, it is more that the reaction is irreversible (under the time period of the reaction equilibrium between the products and starting materials is not reached); i.e. reactions can be poorly selective but still under kinetic control.
Q2. Your global minimum energy values and structures are correct and the general approach to answering this question is good (detailing the way you came to the lowest energy conformation); the 6-ring is an obvious starting point because of the known conformations it can adopt. These compounds are indeed examples of hyperstable olefins and your calculation of OS is a nice extra (although it would have been better to use DFT methods to ensure an accurate comparison). Olefin strain is mostly an effect of unusual strain exhibited in the hydrogenated compounds (which would be reflected in both the product and transition state energies). The alkenes in this type of ring system are actually surprisingly flat and close to ideal dimensions in general.
Q3. The MOs look good and the IR stretches are correct. All of the explanations are reasonable: The C-Cl is indeed weakened when there is an exo double bond because of the pi-sigma* interaction and the endo alkene is the most reactive towards electrophiles. One little point – it is worth including all energy values calculated (whether for MM or MOPAC methods these are sometimes used as a gauge of the calculation accuracy during marking).
Q4. R=methyl is the correct choice here and your energy values are good. Some of the MM2 structures have lower energy conformations available, but they are harder to find (cf Q2). Your discussion about the differences between MM2 and MOPAC is correct – MOPAC can find bonding interactions even if you don’t explicitly draw them (hence A=C and B=D). Having calculated energies of C, C’, D and D’ you could have commented further on the reaction selectivity. The relative abundance of these key intermediates is an important factor in determining the reaction outcome as they each favour one product after nucleophilic attack. As well as this, the C and D configurations are more reactive because the trajectory for nucleophilic attack is better.
MP. This looks like a good choice for a mini-project and the question you are tackling is a good one – whether it is possible to distinguish between different isomers using calculated NMR. Presenting the data in tables is fine, but there are some better, graphical ways to do it (typically bar charts are used). You have stated that the lit data more closely matches the calculated data for the correct isomer and this type of analysis can help to present the case because it is easier to assess the deviations visually than by looking at a set of numbers. For the coupling constant, the ability to calculate it is only really useful if you can show that one isomer is substantially different from the other. One other aspect that would have been interesting to discuss is the way in which the authors of the report differentiated the isomers experimentally (usually 2D NMR techniques). A summary at the end of the mini project would have been helpful – also it may have been useful to you to set out your aims before answering the question.