1H-NMR and 13C-NMR spectra of 1,2,2-trichloropropane

The molecule has 3 different carbons, so the proton-spin decoupled 13C-NMR spectra will have 3 peaks. The molecule has only two different types of protons that are not coupled so the 1H-NMR will have two unsplit singlet signals.





Off-resonance-decoupled 13C-NMR Spectra of 1,2,2-trichloropropane. The CCl2 group appears as a singlet, the CH2Cl group appears as a triplet, and the CH3 group as a quartet.

The off-resonance-decoupled carbon spectrum will show the spin-spin splitting between the carbon and the hydrogen it has attached to it. The N+1 rule applies so a carbon with 3 hydrogens on it will give rise to a quartet, while a carbon with only 1 hydrogen will have a doublet. Carbons with no hydrogens attached to them will have singlet signals.





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1H NMR VERSUS 13C NMR

 

 

 

 One of the greatest advantages of ¹³C-NMR compared to ¹H-NMR is the breadth of the spectrum - recall that carbons resonate from 0-220ppm relative to the TMS standard, as opposed to only 0-12 ppm for protons. 

 

Because of this, ¹³C signals rarely overlap, and we can almost always distinguish separate peaks for each carbon, even in a relatively large compound containing carbons in very similar environments. 

 

In the proton spectrum of 1-heptanol, for example, only the signals for the alcohol proton (H_a) and the two protons on the adjacent carbon (H_b) are easily analyzed.  The other proton signals overlap, making analysis difficult. 

In the ¹³C spectrum of the same molecule, however, we can easily distinguish each carbon signal, and we know from this data that our sample has seven non-equivalent carbons. (Notice also that, as we would expect, the chemical shifts of the carbons get progressively smaller as they get farther away from the deshielding oxygen.)

 

This property of ¹³C-NMR makes it very helpful in the elucidation of larger, more complex structures.

NMR PROBLEM 3.....1H-NMR and 13C-NMR for a heterocyclic aldehyde

1H-NMR and 13C-NMR for a heterocyclic aldehyde


1H-NMR and 13C-NMR for a heterocyclic aldehyde. Notice the correlation between the chemical shifts in the two spectra. The proton spectrum has a sweep width of 10 ppm, and the carbon spectrum has a width of 200 ppm


The peak areas of the 13C-NMR are not proportional to the number of carbons giving rise to the peaks. Carbons with no protons attached to them usually give rise to small peaks. In the proton spin decoupled 13C-NMR each carbon signal appears as a unsplit singlet. The heterocyclic aldehyde has 5 carbons all in different environments so the spectrum will show 5 peaks.