[Telamond]

I'm stuck at a spectra that I'm supposed to solve. I only received the 1H-NMR and 13C-NMR since it was supposed to be an easy structure.
I was trying to solve it for 30 minutes, I couldn't get the number of carbons to add up, so I suggested that the spectrum was of a heteroaromatic (2,4,6-trimethylpyridine). The TA said that it was correct, and I told them how I was having difficulties figuring out that it was a heteroaromatic.

Then he told me that heteroaromatics's aromatic carbon has chemical shifts far from eachother, so that was an easy way to tell...

13C-NMR for 2,4,5-trimethylpyridine looked like this:

Chemical shift    Multiplicity
156.3                    s
146.3                    s
120.2                    d
23.3                     q
19.9                     q

The top three peaks are the supposedly aromatic region...

Can anyone explain why they're far apart from eachother for heteroaromatics?
I have no idea... it isn't written in any of my text books and I can't find anything about it googling it. 

Thanks in advance.

[nj_bartel]

I don't know for sure as I haven't been taught, but if I had to say I'd say it would be due to the fact that the nitrogen in the ring is a strong inductive and resonance EWG, and thus strong deshields proximal carbons, less strongly so more distant carbons.

[frenchy]

As far as I know, you shouldn't have multiplicity in a ¹³C NMR

Heteroarmatics tend to deshield the closest atoms beside the heteroatoms making the distance between the different environments larger compare to "just" an aromatic ring.

Can you give all the data for your ¹H and ¹³C NMR?

[Squirmy]

As far as I know, you shouldn't have multiplicity in a ¹³C NMR

Not in the common ¹H-decoupled spectrum, but you can run a ¹³C without decoupling. This gives rise to splitting by attached hydrogens and since I=+1/2 for ¹H, the N+1 rule applies.

More commonly, if you want to know how many hydrogens are attached to a carbon, you would run an APT or DEPT.