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So I've come across this lovely spectrum of a compund where only its chemical formula is given with C5H9NO. I have made the pictures of the spectrum available below. obviously there is some pretty hefty coupling going on but I cant really figure out what it could be. So far I think we definitely have a CH3 group that could be attached to either heteratom. On the other side, the single proton at 6.66 ppm could also be an H attached directly to the N-atom (maybe?). Anyways, I hope you guys can help me here and maybe also explain me the couplings that are going on.

enter image description here enter image description here enter image description here EDIT: to not get marked as homework, because it isnt, I suggesting that it should be some sort of pyrrole or pyrrolidine structure type. Further I think that the CH3 group must be shielded from other protons and is either near or is sitting at a heteroatom. Next I have narrowed down two structural possibilities: either we have a proton sitting at the nitrogen and the other sitting on a double bound, or we have diastereotropic protons. My best candidate so far is the picture below (please let me know if it would be possible):enter image description here

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    $\begingroup$ Sorry, but if you dont want this closed as "homework", at least try to identify the couplings ("ddt" etc.), and calculate the coupling constants. Btw. marking the individual peaks of a splittings in ppm is just wrong. And one more hint, your normalisation of the integrals is rather weird. The methyl group consists of three protons, so it should be 3, not "100". $\endgroup$ – Karl Oct 19 at 21:07
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    $\begingroup$ I admit the one at 5.2pmm is kind of weird. There is a clear roof with the one at 5.8, however. Forget figuring out the exact splittings there. Thats a real higer order spectrum. $\endgroup$ – Karl Oct 19 at 21:37
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    $\begingroup$ @Karl I´ve now thought about it for quite a while, please look at my edited version. I hope it makes it clear, that its not homework $\endgroup$ – TheChemist Oct 19 at 22:26
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    $\begingroup$ Methoxy resonates at ~3.1 ppm so its not consistent. $\endgroup$ – Buck Thorn Oct 19 at 23:00
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    $\begingroup$ The signal at 5.2 ppm is not a multiplet, but two signals. Thus, the suggested structure with two CH2 groups is not consistent with the spectrum. chem.wisc.edu/areas/reich/nmr/Notes-05-HMR-v26-part2.pdf $\endgroup$ – Karsten Theis Oct 20 at 2:45
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Finding a structure with the correct formula that matches the chemical shifts reasonably well is not so difficult.

The multiplet integrals, normalized such that the area corresponding to one $\ce{H}$ is $\approx 33$ suggests the following number of $\ce{H}$ (moving upfield): 1,1,2,2,3.

The upfield singlet suggests an uncoupled terminal methyl group, so we begin with the simplest assumption, that we have a molecule with a linear alkane.

For a linear alkane, the 4 remaining $\ce{C}$ require 7 bonds in addition to $\ce{3 C-C}$, $\ce{1 C-N}$, and $\ce{1 C-O}$ single bonds. The $\ce{N}$ requires 2 additional bonds, the O 1. Since there are $\ce{6 H}$ remaining, either unsaturation or multiple bonds such as $\ce{C=N}$ or $\ce{C=O}$ bonds are required (we dismiss ring formation for the time being).

The high chemical shift and simple coupling pattern suggests the methyl group sits next to an electron withdrawing group without protons, carbonyl would fit.

If we include a carbonyl, we now have remaining $\ce{3 C}$ requiring 6 bonds in addition to $\ce{2 C-C}$ and $\ce{1 C-N}$ single bonds. The $\ce{N}$ requires 2 additional bonds. Since there are $\ce{6 H}$ remaining, either unsaturation or a multiple bond such as either $\ce{C=N}$ or $\ce{C=O}$ bond is still necessary. The broad singlet at $\pu{6.7 ppm}$ suggests a single $\ce{N-H}$ proton, so we assume the $\ce{N}$ is part of the chain. Let's try an amide.

If we include the amide, we now still have 3 remaining $\ce{C}$ requiring 6 bonds in addition to $\ce{2 C-C}$ and $\ce{1 C-N}$ single bonds. Since there are $\ce{5 H}$ remaining, unsaturation is necessary. We assume an ethylene group is present with an intervening methylene, we arrive at the following proposed structure with the right formula, which I checked with an online simulator:

enter image description here

The match of the chemical shifts is pretty good, but the detailed splittings are not explained. Hans Reich provides tutorials on his site showing you how to decompose complex multiplets, you may want to look at that. Other simulation software are also available.

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    $\begingroup$ Thank you so much. I knew that the structure cant be too difficult... very nice explanation of how one should think when analyzing these structures $\endgroup$ – TheChemist Oct 20 at 7:48
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This is an interesting question. However, disadvantage is having only $\ce{^1H}$-$\mathrm{NMR}$ to deal with. Yet, it has pretty good resolution (probably using $\pu{400 MHz}$ machine) so we can resolve the splitting pattern very easily and predict the structure.

First, OP has correctly assigned two degree of unsaturation. For molecular formula, $\ce{C5H9NO}$: $$u=\frac{5\cdot2+2+1-9}{2}=2$$

However, OP's suggested structure was incorrect since chemical shift of -$\ce{OCH3}$ is usually greater than $\pu{3 ppm}$. Because of the chemical shift of -$\ce{CH3}$ group in the given $\ce{^1H}$-$\mathrm{NMR}$ is about $\pu{2.0 ppm}$, one can conclude that it is either olefinic or one $\alpha$ to a cabonyl group. Nonetheless, OP's suggestion of proton resonance around $\pu{6.66 ppm}$ is due to $-\ce{NH}-$ group is correct, but due to its deshielding nature ($\pu{6.66 ppm}$), one can suggest that it is an amido $-\ce{NH}-$ group. However, OP correctly assign it next to a $-\ce{CH2}-$ group.

Since I suspect non cyclic nature of the unknown, I have assigned splitting patterns and calculate relevant coupling constants assuming the spectrum is taken in $\pu{400 MHz}$ machine. All splitting patterns of remaining resonances are $ddt$ as depicted in fallowing diagrams:

allyl acetamide_1

allyl acetamide_2

Accordingly, these patterns and chemical shifts well match with N-allyl acetamide (see insert in first diagram).

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    $\begingroup$ thank you as well for your answer! Its very much appreciated and now I understand the coupling as well. good to see, that I wasnt completely wrong. the -OCH3- I should have caught to be at 3ppm, that was a rookie mistake $\endgroup$ – TheChemist Oct 20 at 7:51
  • $\begingroup$ One question though: what is this formula for the degree of unsaturation? $\endgroup$ – TheChemist Oct 20 at 7:57
  • $\begingroup$ It is a little complected to write in details in a comment. But, you can find it in any textbook used in spectroscopic analysis of organic compounds course or just google it. For a simple hydrocarbon, $\ce{C_nH_a}$: $u = \frac{n \times 2+2-a}{2}$. $\endgroup$ – Mathew Mahindaratne Oct 20 at 8:05
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    $\begingroup$ The triplet in the methylene group must be a $dd$ of 1 Hz each (fitting to the two 1 Hz $t$ around 5.2 $\endgroup$ – Karl Oct 20 at 10:53
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    $\begingroup$ Thanks for showing the details of the doublet of triplets at 5.21 ppm that looks like a quartet because the coupling constants are related by a factor of 2. Is there any way to tell them apart? $\endgroup$ – Karsten Theis Oct 21 at 11:30

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