tag:blogger.com,1999:blog-6953549091784501422.post3538359384798945145..comments2024-03-18T00:31:58.501-07:00Comments on NMR Analysis, Processing and Prediction: Induced Coupling & ChiralityCarlos Cobashttp://www.blogger.com/profile/13500275318435740775noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6953549091784501422.post-81428940723574592192008-04-17T00:59:00.000-07:002008-04-17T00:59:00.000-07:00Hi Robert, you are right when saying that once a m...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.<BR/>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.<BR/>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.stanhttps://www.blogger.com/profile/10235838368996981815noreply@blogger.comtag:blogger.com,1999:blog-6953549091784501422.post-16131672512124839942008-04-14T17:20:00.000-07:002008-04-14T17:20:00.000-07:00Describing the '2nd carbon being chiral' in a syst...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.Unknownhttps://www.blogger.com/profile/01738976109401580542noreply@blogger.com