[chmjmhi]

Could someone please point me towards a fool-proof (read: viva voce-proof) theoretical explanation for the simple fact that unfunctionalised phenyl groups sometimes appear as three 13C signals rather than the four I would expect (ipso, ortho, meta and para).

[Ψ×Ψ]

Could it just be an intensity problem?  The ipso signal probably appears much less intense than the others...

[kiwi]

Could it just be an intensity problem?  The ipso signal probably appears much less intense than the others...

they often do

[Mitch]

Sounds like a signal problem. Can you see it if you really pack your nmr-tube?

[kiwi]

i assume that you are familiar with the nuclear overhauser effect, which is why the ipso signal is lower. From a practical viewpoint, jamming a whole bunch of compound in the tube is a good way. but if you are limited in compound, you can encourage the peaks to appear by increasing d1, the relaxation delay. your experiment time will also increase however.

[Ψ×Ψ]

now THAT i didn't know.  i guess it would make sense to run a quick array to optimise d1 if you really, really needed those signals to show up.

[kiwi]

now THAT i didn't know.  i guess it would make sense to run a quick array to optimise d1 if you really, really needed those signals to show up.

yeah optimising it is a good idea, but can be a real pain too. sometimes a metric guesstimate is the best approach. 

[chmjmhi]

Ooops, seems like I posted this question twice due to the server downtime. Sorry about that. Thanks for the suggestions, if I cannot make the peaks resolve by changing solvent I shall try fiddling with the relaxation times. My original question however concerned the theoretical explanation, not the solution to the problem. But I guess I have to specify the problem better before you can advise me how to explain it, right?

[hmx9123]

Quaternary carbons come up much weaker than their brethren, mostly due to the fact that having a proton attached helps the nucleus to relax faster.  Better than increasing d1, though, is to reduce the flip angle.  I deal with a lot of quaternary carbons, and you can actually pulse faster with a lower flip angle if you want to shorten your time and increase your signal to noise.  There's a good article on this trade-off between d1 and flip angle in an old JMagRes somewhere.  I'll see if I can dig up the article for you.

[kiwi]

Quaternary carbons come up much weaker than their brethren, mostly due to the fact that having a proton attached helps the nucleus to relax faster.  Better than increasing d1, though, is to reduce the flip angle.  I deal with a lot of quaternary carbons, and you can actually pulse faster with a lower flip angle if you want to shorten your time and increase your signal to noise.  There's a good article on this trade-off between d1 and flip angle in an old JMagRes somewhere.  I'll see if I can dig up the article for you.

i'd be interested in that article too, thanks

[movies]

i assume that you are familiar with the nuclear overhauser effect, which is why the ipso signal is lower. From a practical viewpoint, jamming a whole bunch of compound in the tube is a good way. but if you are limited in compound, you can encourage the peaks to appear by increasing d1, the relaxation delay. your experiment time will also increase however.

Isn't it the case that the NOE gives an enhancement to the signal of protonated carbons, not a decrease in signal to the quaternary carbons?  I am almost positive that this is the case.

[Yggdrasil]

The sign of an NOE depends on the molecular correlation time and your field strength.  When your correlation time << 1/(larmor frequency), the NOE will enhance your signal.  This is the case for most small molecules.  However, when correlation time >> 1/(larmor frequency), you don't have any more double quantum transitions, so zero quantum transitions become the primary means of cross-relaxation.  This leads to a negative NOE, decreasing the signal from coupled nuclei.  This is the case for protein NMR (beause proteins are huge).

I've never done small molecule NMR, though so I don't really know how large a molecule has to be to give negative NOEs.

[kiwi]

Isn't it the case that the NOE gives an enhancement to the signal of protonated carbons, not a decrease in signal to the quaternary carbons?  I am almost positive that this is the case.

i'm coming from the small molecule angle, the NOE gives an enhancement to those nuclei that can relax by it (in practice usually everything but quaternary carbons). as mentioned above, for (much) larger molecules, the situation breaks down. and from doing a whole bunch of molecules, up to probably 800 daltons i have never noticed a breakdown in this relation. And afaik, the effect comes about because of the decreased relaxation time for those nuclei, resulting in increased signal only because a higher proportion of nuclei will relax in a given relaxation delay d1. if you wait for longer (up d1), the quat signals will relax too.

[chmjmhi]

@hmx9123: Thank you, I'll give it a read out of general interest if you turn it up.

@kiwi and the discussion in general: Yep, I'm familiar with NOE. In fact I can see all my quaternary carbons. In the current example I am looking at 1-methyl-3-phenyloxindole and its derivatives and because of really nasty overlap in the aromatic region of the 1H spectrum it is hard to get much information out of the HMQC, which means I don't know which of the tertiary carbons I am missing. I lean towards the 4'-aromatic (overlapping the 2',6' or 3',5' signal), since I have a full account of carbons in 1,3-dimethyloxindole.