[mobfesk2]

Hi chem forumers,

I have a question about coupling constants (J) in proton NMR.

Am I right in calculating J as follows:

1. For a doublet = the difference between the frequency of the 2 peaks x frequency of NMR machine e.g. 400MHz

2. For a triplet = the difference between the peak frequencies of the outer 2 peaks (not middle one) x 400MHz

3. For a double doublet = don't know how to explain this one! Maybe best to look at the pic

4. Singlet = No J value can be calculated as it is only 1 peak

Any other types of peak I've missed out???

Also, what would be the best way to show these constants and chemical shifts in a thesis???

Thank you very much for any suggestions

[DavidW]

1. Yes.

2. Not quite.  For triplet, you take the difference in ppm between the *middle* peak and an outer peak, and multiply by Mhz.

3.  The way you drew the red lines is the correct way to calculate the two J values.

4. Yes

5. Well there are lots of other variations (double of triplets, etc.), but they are more or less the same idea.

[movies]

For number one, yes, but make sure you are looking at the frequency in ppm when you multiply by the spectrometer frequency.  For most spectrometers you can also have the chemical shift displayed in Hz, in which case you can just calculate the difference between the two signals.  This is more accurate because the ppm value is usually rounded to a tenth or hundreth of a ppm, which becomes significant when you multiply by 400!

For number three, do you really mean a doublet of doublets, or do you mean a quartet?  For a quartet (a resonance split by three other equivalent nuclei) there would be only on J value.  For a doublet of doublets, you would have two, as you indicated in the image.

[DavidW]

Good point Movies, I didn't realize he could be referring to a quartet and not a doublet of doublets.

[bluepanda]

how does one know when to calculate J values?

[Babcock_Hall]

I calculate them to try to decide which atoms are coupled to which other atoms.  Sometimes having the value of the coupling constant can sort this out.  I am not sure that I am answering your question.

[PhDoc]

Babcock_Hall, by any chance did you go to the University of Wisconsin?