I am currently struggling with an assignment question regarding structural determination using both $\ce{^1H}$ and $\ce{^{13}C}$ NMR data.

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The first thing I have done is calculate my double bond equivalences which I believe is 3. I believe the IR data corresponds to a $\ce{C=O}$ bond which is supported by the number of DBEs. As it states this is a cyclic compound I believe that the there is a ring containing a double bond and a carbon double bonded to an oxygen which accounts for the 3 DBEs.

Putting together this information which I believe to be correct with the spectroscopic data above is where I am really struggling.

I am quite confident in saying that the 6.8 triplet is due to an aromatic carbon with two hydrogens on the adjacent carbons so I have a fragment there that I believe is correct but I cannot get anything more than that.


1 Answer 1


Your calculations of the double bond equivalents seems correct, and there is also another hint towards a $\ce{C=O}$ double bond: the corresponding $\ce{^{13}C}$ chemical shift.

Also, while it says you have $\ce{^1H}$ decoupled $\ce{^{13}C}$ data, this is clearly wrong, because they’re showing you multiplets. The $\ce{^{13}C}$ data is $\ce{^1H}$ coupled data, so you can use the information there to deduce the number of carbons with which number of hydrogens.

A third hint given to you is the amoun of signals: You see signals that add up to six carbons and eight hydrogens, meaning there is no symmetry in your molecule.

The 6.8 hydrogen could, by its chemical shift, be aromatic, but you do not have enough DBE’s for an aromatic ring (you would need at least 4). Do you know other types of protons that give chemical shifts in similar ranges, and are consistent with both your DBE count and the structure element(s) you have already found out?

Further ideas to think of:

  • You have certain integrals on your hydrogen signals, especially one that stands out (not the $6.8\,\mathrm{ppm}$). What could this integral mean, and how does it affect the rest of the structure?
  • Make a list of the carbons you have and the amount of hydrogens attached to them.
  • How can you turn all that into a cyclic compound?
  • Once you have, you can check if the multiplets of the hydrogen signals can be fitted to that structure. Hint: You seem to have no H–H couplings other than across three bonds. (So no long-range couplings.) And no H–C couplings other than one bond length.

Right, in this first spoiler tag I’m just gonna state the different signals. Check the second one, if you really want to know the solution.
The carbon at 209.6 (singlet) is most likely not an aldehyde but a ketone. If it had a proton connected, it would need to be a doublet.
The singlet at 1.8 is a methyl group, belonging to the carbon at 12.1.
The triplet at 6.8 needs to be next to a double bond (strong low-field shift). The corresponding carbon needs to be 157.9 (doublet).
Since it’s a triplet, it couples to two hydrogens. There is no other hydrogen on a double bond, but there is a triplet and a quartet left. One of these two must couple to the proton integral one at 6.8. Luckily they are both integral 2. A little further reasoning tells us, it must be the quartet, because the other triplet (2.4) has to couple to two protons, which have to couple back, therefore no other possibility.
There is another carbon at a double bond (142.1) and it has no hydrogens attached to it (singlet). It needs to have the methyl group on it.


The structure I deduced is 2-methylcyclopent-2-en-1-one.

  • $\begingroup$ So far I think I have it that the (1.8 ppm singlet) could be in a position where it is at the end of a carbon chain because I believe it looks like a methyl group? Because it is a single does that mean it is likely part of a fragment such as CH3-(C=O)-C-? But the C13 data definitely has a shift that indicates an aldehyde so I don't know how the two different NMR data fits in. Also so as you have suggested I have made a list of my carbons and their attached hydrogen but I think I have something wrong because I cannot see how this can be cyclic. I will keep trying though thank you! $\endgroup$
    – Michael
    May 20, 2015 at 16:04
  • $\begingroup$ @Michael Are you sure that the $\ce{^{13}C}$-data implies an aldehyde? And yes, 1.8 ppm is definitely a methyl group, but remember that you have a cyclic molecule … $\endgroup$
    – Jan
    May 20, 2015 at 16:36
  • 1
    $\begingroup$ Sorry I had to have a quick look but I'm a bit unsure so now I have had a look at the structure and can better understand the steps working backwards thank you. I think I need to practice lots more of these and probably start off on basic ones. Thank you very much, I will try to solve more problems like this using the method you have posted! :) $\endgroup$
    – Michael
    May 20, 2015 at 23:31

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