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.
