1H spectrum of natural abundance NH4Cl (1.5 M) in 1M HCl/H2O showing coupling to 14N as a triplet and coupling to 15N as a weak doublet. Coupling constants for triplet and doublet are 52.4 Hz and 73.5 Hz, respectively, as shown in figure.
Please explain me how to index to these peaks to protons (or origin of these peaks).
Thanks!
All the protons in a NH4+ are equivalent so, in the absence of coupling, should give a singlet.
15N is spin 1/2 so shows spin-spin coupling( just like a proton would) so splits the resonance into a doublet.
14N is spin 1 so can exist in three spin states +1,0,-1 so spits the peak into a triplet. Coupling is rarely seen to nuclei with spin >1/2 it is seen here because of the symmetric electron distribution in the ammonium group (giving no electric field gradient around the nucleus).
You may also find that the centre of the triplet (14NH4+) is not quite in the same place as the centre of the doublet (15NH4+). This is called an isotope shift (and is much more difficult to explain away)
All the protons in a NH4+ are equivalent so, in the absence of coupling, should give a singlet.
15N is spin 1/2 so shows spin-spin coupling( just like a proton would) so splits the resonance into a doublet.
14N is spin 1 so can exist in three spin states +1,0,-1 so spits the peak into a triplet. Coupling is rarely seen to nuclei with spin >1/2 it is seen here because of the symmetric electron distribution in the ammonium group (giving no electric field gradient around the nucleus).
You may also find that the centre of the triplet (14NH4+) is not quite in the same place as the centre of the doublet (15NH4+). This is called an isotope shift (and is much more difficult to explain away)
I should have also mentioned that the size of the coupling is in the ratio of the magnetogyric ratios (15N -27.13 and 14N 19.34 Mrad s-1 T-1)and the intensities depend on the natural abundance 15N 0.37% and 14N 99.63%.
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