Is it possible to run a 13C spectra in which one carbon is coupled (with 1H) and remaining carbons are decoupled, using Auto-XDB probe with VNMRJ 3.2 software? Thanks , Praveen.T asked Oct 04 '13 at 23:47 praveen |
The experiment you ask for perhaps can be done in vnmrj, but would even so it would be tricky. Note that it is which proton(s) is(are) coupled to the carbon that counts. One would have to decouple all the other protons except that(those) coupling to that carbon. The reverse experiment would be more straightforward, decoupling selectively. Here you would put a low power cw pulse on the proton you believe is coupling to the carbon of interest and it will remove that coupling selectively. Compare that to a fully coupled carbon and you should have what you want. Another alternative would be to run a coupled HSQC, which works fine for getting 1-bond C-H couplings. For long-range couplings, sequences such as HETLOC are useful. As usual with NMR, many variation exist depending on the specific compound and information needed. answered Oct 09 '13 at 15:57 Charlie Fry |
If One of the C-H coupling is significantly of different value than others, may be one order of magnitude or close to, then you may select by appropriate delay times for retrieving selectively a particular coupling. "COLOC" type of sequence answered Oct 11 '13 at 01:20 |
You may modify a 'standard' 2D HSQC pulse sequence, such that the 180 proton pulse in the middle of the 13C evolution period is preceded by a soft/shaped 180 pulse that is selective for the proton's that are coupling to the carbon of interest. This would be equivalent to a 360 pulse for the proton of interest, leaving the remaining unchanged. Alternately, a shaped pulse specifically designed for inversion of all proton peaks other than the peak of interest may be used to replace the 'hard' 180 proton pulse in the middle of the 13C evolution period of HSQC. These methods will work only if the chemical shifts of the proton/s coupling to the carbon of interest are well resolved from those of the remaining protons. answered Oct 27 '13 at 04:11 sekhar Talluri |