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Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: Telracs on March 09, 2012, 07:04:00 PM

Title: assigning chemical shift in H NMR to 5 or 6 membered mono-sub aromatic rings
Post by: Telracs on March 09, 2012, 07:04:00 PM

I did a search on here but couldn't find this particular topic.  In my Organic Chemistry 2 lab we did an H NMR (among other analysis) for acetylferrocene.  The spectrum shows a peaks at 2.4, 4.2, 4.45, and 4.75ppm and I've assigned them with 3H, 5H, 2H and 2H respectively. I know the singlet at 2.4ppm is from the methyl group on the first aromatic ring, and the singlet at 4.2ppm is from the 5 equivalent H on the second aromatic ring.  So my question is how do I tell which Hydrogens (the 2 closest to the substitution (labeled "a") or the 2 furthest from the substitution(labeled "b")) represent each signal?  My lab requires me to choose which specific H goes with each signal but from searching online for the past week I got the feeling its not within Organic chemistry 2 lab to be able to separate these.  I was thinking maybe it had something to do with resonance but I'm pretty confused at this point.  I ran across the very same problem when assigning  peaks to symmetric 6-membered rings because "the closer to the electronegative group = more downfield" doesn't seem to work for aromatics (according to my TA but that's all the help he offers).

Title: Re: assigning chemical shift in H NMR to 5 or 6 membered mono-sub aromatic rings
Post by: argulor on March 09, 2012, 10:58:28 PM
Your TA is correct, it is more difficult to assign aromatic protons without further analysis. In the case of acetylferrocene you should have two distinct triplets for your aromatic protons. We know that they should be furthest downfield because of the fact that they are aromatic. Typically aromatic protons appear between 7-9ppm, in the case of acetlyferrocene they are not, any ideas why?

You should still be able to assign the peaks based on the conventional (proximity to EN atom). It just gets more hairy in the aromatic region when you have more aromatic protons.

I recommend doing some searching here; they have an extensive collection of spectra.