[Compaq]

Hey!

To put my question shortly: how would one decide whether aniline was orto/meta/para substituted, based no the proton nmr, C13 nmr and the IR spectra provided on this site?

http://www.chem.ucla.edu/cgi-bin/webspectra.cgi?Problem=bp14&struct=on

It would probably help to have a piece of paper in front of you.

Here is how I went about it:

1) molecular formula is given: C6H6NBr
2) Calculate DBE=4
I think aromatic ring when I see numbers of 4 or higher. It may be something else, but that is my first thought, and I note it down to check for something that might confirm in proton nmr
3) I take a quick look at the IR spactrum:
I see a medium/strong absorption at around 3400cm-1. There are no oxygens here, so I check the table and see that -NH2 fits nicely. I note it.
There are also something around 1600cm-1. This might be C=C bonds within the aromatic system. I note it.

4) I then take a look at the proton nmr
Here, I see

3,65ppm (s) ==> NH2?
6,8-6,9ppm (m) ==> aromatic protons?
6,58ppm (some sort of dd?) ==> aromatic protons?
7,1ppm (t) ==> aromatic protons?

Already now, I'm confident that I have an aromatix 6 carbon ring, which is disubstituted. Br and NH2.

I check the C13-nmr. Six different peaks tells me no symmetry. That at least tells me that the aniline isn't para substituted.

I draw up the meta substituted aniline. Then there would be, from my logic, a singlet in the proton nmr spectrum, relatively low field. But there isn't. Also, the degree of symmetry would be bigger in a meta, rather than a orto substituted system (I would think).

I have no COSY or HMQC or anything to further let me look for coupling patterns. So, how exactly can I tell that this is meta substituted, rather than orto substituted?

Thank you in advance for taking the time to take a quick look at this rather simple spectroscopy task

[fledarmus]

Usually it is the splitting pattern that is the most distinctive indicator that you have a meta or an ortho subsituted benzene ring, if there is enough of a chemical shift difference to show all four protons distinctly (as there is here). You have what looks like a very narrow triplet at ~6.8, which is characteristic of an aromatic proton with a substituent on each side, then another proton on each side. Meta splitting. There is little if any para splitting. The peaks at ~6.6 and ~6.9 are the two protons which are meta to the one at 6.8 - each of them shows a doublet from splitting with the one ortho to them, and a much smaller doublet from the meta one. Finally, the proton between the two at 6.6 and 6.9 appears to be sort of a triplet at ~7.0, because it is being split by two ortho protons. For a meta substituted benzene ring, if you are lucky, this is what you see - a very narrow triplet, two doublet of doublets (one of which is very narrow), and a wider triplet or doublet of doublets with both being fairly wide.

An well-separated ortho substituted system will show doublet of doublets (one of which is very narrow) for the outside two protons, and either triplets or doublet of doublets (both wide) for the inner two protons.

If you compare your spectrum to the attached spectrum for o-bromoaniline, it might make more sense: the outer protons are at ~6.7 and ~7.4 , the inner at ~6.6 and ~7.1

[Compaq]

Thanks a lot. Very helpful.

So the narrow triplet at ~6.8 is due to coupling. I didn't think about that, I just assumed it would come out as a singlet.

That is a problem I've been having. Geminal coupling and vicinal coupling both happens, giving more complicated coupling patterns. So, should I assume that all protons will couple with unequal protons within three bonds? And allylic coupling whenever that's a possibility?

And the little nerve wrecking thing is that the problem I mentioned here is much, much easier than the exam problems. At least I'm starting solving problems 1.5 months before the exam

Thanks again for your *delete me*

[fledarmus]

That is a problem I've been having. Geminal coupling and vicinal coupling both happens, giving more complicated coupling patterns. So, should I assume that all protons will couple with unequal protons within three bonds? And allylic coupling whenever that's a possibility?

Yes. And for aromatic protons, strong ortho, weak meta, and very weak but sometimes present para coupling.

There are a few rare cases where you will not see the coupling. Rapid proton exchange with solvents or other molecules (if your molecule contains for example -OH, -NH-, etc.) is the most obvious group that you won't see coupled (most of the time anyway - there are exceptions where due to sterics there is little exchange and you can actually see an -HN-CH- coupling or others). If two protons are at right angles to each other on adjacent carbons, they also won't couple, but that happens very rarely.

[Compaq]

If two protons are at right angles to each other on adjacent carbons, they also won't couple, but that happens very rarely.

And now we're on the Karplus rule, right? I remember our Professor mentioning that being aware of this, and being able to utilize such graphs to determine coupling patterns, was A-student stuff.

I feel I lack a good overview over chemical shifts. My book ("Spectroscopic methods in Organic Chemistry", Sixth edition, Williams/Fleming) has a large overview, but it's hard to orient oneself. The same goes for the IR section. And I'm not able to find any overview for C13 at all.