Chemistry Forums for Students > Physical Chemistry Forum
coupling constant (spin-spin)
(1/1)
Morphic flip:
If J 1(H, 13C) = 208 MHz
Where do I get the figures/info needed to work out a value for 1 J (31P - 53Cr) ?
Can`t can`t find reference for what to useĀ ???
Another problem I have ::) is using pascal`s triangle,
how do I work out "a spin 1/2 nucleus coupling to a one spin 3/2 nucleus, how many lines the resonance is of the spin 1/2 nucleus split into, and the relative intensities"
Multiplet Intensity Ratio
Singlet 1
Doublet 1:1
Triplet 1:2:1
Quartet 1:3:3:1
Quintet 1:4:6:4:1
Sextet 1:5:10:10:5:1
Septet 1:6:15:20:15:6:1
Dan:
--- Quote from: Morphic flip on May 24, 2006, 07:44:44 AM ---Another problem I haveĀ ::) is using pascal`s triangle,
how do I work out "a spin 1/2 nucleus coupling to a one spin 3/2 nucleus, how many lines the resonance is of the spin 1/2 nucleus split into, and the relative intensities"
--- End quote ---
Pascal's triangle only works for coupling to spin-1/2 nuclei.
For coupling to nuclei with I>1/2 (where I is nuclear spin), you must consider the number of space quantisations of of mI. The number of these space quantisations is 2I+1 - the signal is split into this number of peaks of equal intensity (just as coupling to one I=1/2 nucleus, where 2I+1 = 2, splits the signal into a 1:1 doublet).
Just to clarify, if you couple to several I>1/2 nuclei, you will have to do a tree/stick diagram or something similar to work out the intensity ratio.
So to answer your second question, coupling to one I=3/2 nucleus you will give a 1:1:1:1 quartet.
As for your first question, I have no idea
Morphic flip:
Thank`s for the well explained reply.
So if the quartet is 1:1:1:1, I take it the intensities are all the same?
Cheers.
Dan:
yeah, 1:1:1:1 is the intensity ratio of the peaks in the quartet
Navigation
[0] Message Index
Go to full version