Henry Rzepa's Blog

It is always rewarding when one comes across a problem in chemistry that can be solved using a continuous stream of rules and logical inferences from them. The example below[cite]10.1039/P19930000299[/cite] is one I have been using as a tutor in organic chemistry for a few years now, and I share it here. It takes around 50 minutes to unravel with students.

The narrative is that attempted preparation of 1 resulted instead in a mysterious compound [A], which when heated extruded S=C=O to give 2, and upon further heating gave 3. The challenge is to identify [A] with the help of the spectroscopic information provided, to infer the mechanism of its formation and further to suggest what the stereochemistry of the methyl group in 3 might be.

The ¹H NMR of [A] is set out below for future reference: δ 1.70 (3H,d,6Hz), 2.23 (1H, t, 3Hz), 3.73 (2H, d, 3Hz), 4.84 (1H, dd, 7,8Hz), 5.15 (1H, d, 10Hz), 5.27 (1H, d, 17Hz), 5.51 (1H, dd, 8,16.5 Hz), 5.77 (1H, dq, 6,16.5 Hz), 5.88 (1H, ddd, 7,10,17Hz). 

As usual, one has to start somewhere, and here the task is to number the atoms, and then try to “reaction map” them to the products.

1. The first real decision is how to map S9 or S10. Occam’s razor suggests that the sulfur in the SCO comes from S9 (this would allow C10-C11 to be left alone), but if that hypothesis is wrong, we can always return and try the alternative. Let us go with the simpler option first.
2. Another relatively simple decision is to map C12-C13 as shown in 3, since this only changes its bond order by one (few mechanisms require a change in bond order of > 1 in any single mechanistic step). 

Analysis of the ¹H NMR starts with the most obvious (marker) group, the methyl:

1. The methyl is J-coupled to C2-H (6Hz), and hence this is assigned to 5.77 ppm.
2. C2-H is J-coupled to C3-H (16.5 Hz) and hence this is assigned to 5.51 ppm
3. C3-H is J-coupled to C4-H (8 Hz) and hence this is assigned to 4.84 ppm. 
4. We now encounter a problem. C4-H has a chemical shift which suggests it is not attached to an sp²-C, but has become sp³-hybridized. But the relatively high chemical shift suggests that this carbon may be attached to electronegative substituents. C4 is flagged for attention below.
5. C4-H is J-coupled via J 7 Hz to the peak at 5.88 ppm. The chemical shift is typical of sp²-C, and is assigned as C5-H.
6. C5-H is J-coupled via two couplings of 10 and 17 Hz to peaks at 5.15 and 5.27 ppm. Both these are also sp²-C, which may be assigned as C6-H. As such it can only carry three attached atoms (two Hs and a C-C) and so the C6-O7 bond cannot be retained. C6 is flagged for attention below.
7. The remaining peaks can be assigned as C11-H and C13-H from their mutual ⁴J coupling of 3Hz.

Armed with these inferences, a list of to-dos can now be assembled.

1.  For the transform 1 → [A], break C6-O7
2. Form a bond to C4 using if possible an electronegative atom.

This pattern of break one σ-bond/form one σ-bond, reminds of a sigmatropic pericyclic reaction. A typical example is the Cope rearrangement, in which a bond forms between the termini of two double bonds separated by three σ-bonds. The penny drops when one re-draws the original compound by rotating about a single bond (a perfectly allowed operation):

A [3,3] Cope is now exposed. The (re)-numbering in red shows the pattern described above, and completes the assignment of the bond forming above to C4 as C4-S9. The next step is to find out how to extrude S=C=O. 

1. To get to 2, one needs to create the C6-S10 bond (it is  sp²-C in [A]). 
2. The O8-S10 bond needs to break.
3. The recently formed C4-S9 bond needs to break again, with the result of extruding the required S=C=O.

This pattern of forming and breaking bonds, but in unequal number reminds of the so-called ene class of pericyclic reaction. Both the Cope and now the ene are six-electron thermal pericyclic processes.

We can now turn our attention to the last reaction shown above. Since we have both structures now, we can do a retrosynthetic analysis, which reveals that in the final step, C2-C13 and C5-C12 have both got to form. Such a pattern is another six-electron pericyclic reaction, the Diels-Alder _π2_s + _π4_s cycloaddition. Again, we have to rotate about the C3-C4 single bond (green arrow) to get the diene of the reactant into a conformation capable of undertaking this reaction. We are helped in this by ensuring that the trans hydrogens at both C2-C3 and C4-C5 (which we inferred from the values of the J-couplings above) are not transformed during our redrawing of this conformation.

The conclusion to this tutorial comes in assigning the stereochemistry of that methyl group. The _π4_s component of the cycloaddition mandates that the two bonds forming to C5 and to C2 must both form suprafacially across this four-carbon unit. We know that the bond to C5 must form on the bottom face, so as to rotate the C5-H up. Therefore it must form on the bottom face also of C2, likewise rotating the attached hydrogen up. Therefore the methyl must point  down in the final product.

QED.

But not quite, since nowadays, one can take the NMR analysis one step further. In another post, I will perform a full quantum mechanical prediction of the above NMR spectrum to see how well it matches what is reported above.

Tags: final product, pericyclic, tutor, Tutorial material

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