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The 2 Hydrogens near the carbonyl (the lower 2 $\ce{H}$s are not bonded to the nitrogen) should be the most deshielded so I think they appear as the rightmost peak in the NMR. Are these 2 $\ce{H}$s a doublet of doublets? I think this because they are split by 2 different $\ce{H}$s.

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    $\begingroup$ That's glutamine and not tryphophane in your molecular formula $\endgroup$ – Mad Scientist Apr 8 '13 at 10:42
  • $\begingroup$ sorry, I meant glutamine $\endgroup$ – user176105 Apr 8 '13 at 13:30
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You're thinking about deshielding the wrong way around, the most deshielded signal appears at the highers ppm value, so to the left. In this case it would be the signal at around 3.8 ppm, as the signal at 4.7 ppm is residual water.

The two signals at the right are multiplets, if you take a look they're obviously caused by rather complicated coupling patterns that you can't easily resolve. The reason for that is that the protons you see in this spectrum are not magnetically equivalent, even the CH2 protons are not magnetically equivalent. This leads to the complicated couplings you can observe here.

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  • $\begingroup$ thank you. Also, is it true that the H on the dash corresponds to the triplet at about 3.7ppm? $\endgroup$ – user176105 Apr 8 '13 at 19:48
  • $\begingroup$ @user176105 That one is the H-alpha and it is indeed at 3.7 ppm. And it is a doublet of doublets with a very similar coupling constant, which looks like a triplet. $\endgroup$ – Mad Scientist Apr 8 '13 at 20:01
  • $\begingroup$ thanks again. just to clarify, if i write doublet of doublets on the beta hydrogens that would be wrong? $\endgroup$ – user176105 Apr 8 '13 at 20:16
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The signal at ~3.8ppm is looks like a triplet[t], C-alpha The signal at ~2.5ppm looks like a triplet of doublets [td], C-gamma. The signal at ~2.2ppm looks like a doublet of doublet of doublets [ddd], c-beta.

The 2.2 protons are assigned 2.2 because they are the farthest from carbonyls and thus the least deshielded by these functions.

The ddd signal indicates clearly that both protons on C-beta are non-equivalent, because you need 2 3-bond couplings to make a ddd when combined with the 2-bond geminal coupling.

Given the C-beta protons are non-equivalent, the triplet is actually a superimposed doublet of doublets, and the td is actually a similarly superimposed double of doublets [the superimposed-apparent triplet pair and then...] of doublets.

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  • $\begingroup$ thanks. so even though the 3.8ppm peak appears as a triplet should I still write doublet of doublets? $\endgroup$ – user176105 Apr 8 '13 at 20:59
  • $\begingroup$ In my research days I wrote that as (dd [apparent t]) but you should confer with your faculty member about her/his preference. In my teaching days I tended to accept either asnwer. $\endgroup$ – Lighthart Apr 8 '13 at 21:46

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