Monday, 14 January 2008

Induced Coupling & Chirality

As a sequel to my last entry, I’m glad to post here a letter from my colleague and friend Stan Sykora about induced chilarity and coupling.


Dear Carlos,
I wonder whether you could put this as an entry on your blog. Originally, it ought to be a comment on your last entry, but then it grew …

Concerning the chains of CH2 groups with convergent chemical shifts, I would like to point out another mechanism which has similar NMR consequences and leads inevitably to strong couplings even at the highest fields (present and future):

Consider a chain of the -(CH2)n-R attached to a chiral carbon:



with the substituents S1,S2 and S3 all distinct. The chirality of the first carbon is explicit, so I will say that it is first-order. Now, when R is not a proton, Ha and Hb have different chemical shifts. But this means that the two protons are not equivalent and therefore the second carbon is also chiral (let us call this the induced chirality of second-order). Which means that the protons Hc and Hd in the second formula are also chemically distinct, etc. This logic carries on iteratively, leading to induced chiralities of ever higher orders.

Naturally, the chemical shifts Ha-Hb, Hc-Hd, He-Hf, etc. decrease rapidly with induced chirality order, while their geminal couplings stay more or less fixed at about -20 Hz (geminal J's are always negative and quite large). Consequently, even in a relatively short chain of methylenes attached to a chiral carbon, the protons in the first one will be non-equivalent and well resolved (maybe even first order), but the shifts of the protons in the subsequent methylenes will be much closer (hence leading to strong coupling patterns), and finally each CH2 will approximate an equivalent A2 spin group. Again, moving to a higher field may help a bit, but not too much - at best it might push up by one the limit of induced chirality treatable by first-order approach.

This also brings to my mind another problem: the predictions of the chemical shifts due to the above cases of chirality. I have noticed that the Modgraph and ACD softwares both do some physically funny things there, so I do not trust them. ACD apparently just drops-in a fixed value (like 0.01 ppm) every time there is any chance of induced chirality, while Modgraph, more appropriately, makes the shifts 0 whenever they don't know better. Since I have a cute idea about what could/should be done, could you tell me (or ask an expert like Erno Pretsch) what is the current state-of-the art on this point? As often happens with my cute ideas, it might be well known since 50 years (in which case, I will keep my mouth shut). But if it is not, it might be a thesis assignment for a student.

BTW, I wonder whether the adjective "induced" might not be more proper also in the context of virtual coupling you were writing about in your last entry. Maybe "induced coupling" would be a bit less controversial. Though I am not sure. You could propose it for an e-poll.
Ciao, Stan


Robert said...

Describing the '2nd carbon being chiral' in a system such as (S1S2S3)C-CHaHbX [where X can be an alkyl chain of any length, or a heteroatom] is not quite correct. The molecule depicted above is chiral and for this case it is also asymmetric since its point group is C1. Therefore, one should not expect to find any symmetry relationship between subunits in such a case. This means that the geminal pair of protons are 'diastereotopic' and hence 'anisochronous'. Diastereotopic nuclei will always couple, although as we go out along a long alkyl chain, the geminal coupling constant will probably disappear into the line-width. It has nothing to do with 'hindered rotations', it is simply due to symmetry considerations. Rather than describing the 2nd carbon as 'chiral', one should describe its local point symmetry as being 'chirotopic'- chiral site symmetry. All points [and all nuclei] in any chiral molecule molecule are always chirotopic. This means that their local sites have symmetry elements belonging ONLY to the First Kind [identity and rotation]. Thus, all the methyene carbons in the straight chain are non-stereogenic chirotopic nuclei NO matter how far away they are from the (S1S2S3)C-carbon atom AND how fast the chain is rotating. Only the (S1S2S3)C-carbon atom is a stereogenic chirotopic nucleus where an exchange of any two substituents will give rise to a new stereoisomer. As the chain becomes more mobile, the difference in magnetic environments between the Ha and Hb nuclei should become less and less and we will go from two less-intensity distorted doublets in an AX-system to two more-intensity distorted doublets in an AB-system, i.e. from first order weak coupling to second order strong coupling. The weighted time-average environment of Ha will never be the same as that of Hb. Perhaps as we approach the terminus of a very long chain, we might arrive at a 'deceptively simple' pattern which gives the 'false appearance' of being A2, but even here magnetic environment of one of the geminal pair can can never be IDEALLY equal to that of its partner.

stan said...

Hi Robert, you are right when saying that once a molecule is chiral, all its methylene group are chirotopic, regardless of internal rotations. I was simplifying (maybe too much), trying to say essentially the same thing as you: once there is a primary source of chirality, namely the stereogenic (S1S2S3)C-carbon atom, it gives rise to non-stereogenic chirotopic situations (which I have called 'induced' chirality) all around.
My point was not to discuss the subtleties of stereochemistry and its terminology but just point out that situations where chemical shifts within CH2 groups due to chirality decrease rapidly as one moves away from the stereogenic center. Consequently, one is likely to pass in just a few steps through three types of geminal couplings: weak (1st order), strong, and 'too strong to observe'. I take it that we agree on this.
We probably also agree on the fact that internal rotations HELP to 'equalize' the two protons in a chirotopic CH2 group, though you are right that they, in principle, can never make them fully equivalent.