A 13C 1D spectrum of sample KGM5-75-1c2 was obtained using "standard" acquisition parameters (Figure 1). In particular, the pulse width (pw) was 8.0, and the relaxation delay (d1) was 4.5 sec. This spectrum shows 6 large signals that can be assigned to 5 carbon nuclei in the rings and the 1 trimethyl group. But the signals for 2 carbon nuclei are missing, or are as small as the small peaks arising from impurities in the sample.

How can these 2 missing peaks be identified???

How can we assign the existing and missing peaks???

Carbon nuclei that are not directly bound to hydrogens often require LONG relaxation delays to allow them to relax back to equilibrium before the next pulse-acquire transient is conducted. If the relaxation delay is too short and the relaxation is incomplete, carbon signals may be small or absent. Therefore, it would be reasonable to repeat this experiment with a longer d1 delay, since the chlorinated carbon and carbonyl nuclei have no hydrogens.

However, chlorine nuclei often cause carbon nuclei to relax FASTER than carbon nuclei directly bound to hydrogens. This causes their signals to be broader, and therefore shorter in height. A second 13C 1D spectrum was obtained with a SHORTER pulse width (pw=2) and relaxation delay (d1=0.7) so that NMR signals that relax quickly would be enhanced relative to "normal" signals (Figure 2). A comparison of the two spectra shows that the signal at 167 ppm and 80 ppm are relatively larger in this second spectrum, and therefore must be the two missing signals. Comparing these two spectra also aids in the assignment of these signals to the chlorinated carbon and carbonyl carbon.

Note also that the chloroform carbon signal is relatively larger in the second spectrum than in the first spectrum. This is further evidence that chlorinated carbons relax faster. In fact, we rely on this phenomenon to reduce the height of the chloroform solvent signal relative to solute signals in "standard" 13C spectra.