Rule 2, omit O, gives C5H10
5 – 10/2 + 1 = 1 degree of unsaturation.
Look for 1 pi bond or aliphatic ring.
The band at 1727 indicates a carbonyl, probably an aldehyde; an aldehyde is also suggested by the band at 2719 which is likely the C-H stretch of the H-C=O group. The bands at 3000-2850 indicate C-H alkane stretches.
Proton NMR Spectrum
Since the IR spectrum indicates an aldehyde, look for this functionality in the NMR spectrum. The aldehydic proton appears in the NMR from 9-10, usually as a small singlet.
The spectrum above shows a small singlet corresponding to one proton at 9.2 ppm, confirming that the compound is an aldehyde. Protons on the carbon adjacent to the aldehyde carbonyl will show up at 2-2.7 ppm; this is the triplet peak of 2 protons at 2.4 ppm on the above spectrum. Thus, so far we know that there is an aldehyde group next to a methylene group which is next to a carbon that has two hydrogens:
This accounts for 3 of the 5 carbons in the molecule. The un-colored hydrogens in the above structure could correspond to the peak of 2 hydrogens centered at 1.6 ppm; this peak is a pentet indicating that these protons are adjacent to carbons with a total of 4 hydrogens. The peak centered at 1.35 ppm has two hydrogens and is a sextet, indicating it is next to carbons that have a total of 5 hydrogens. Finally, the peak at 0.9 ppm has 3 hydrogens and is a triplet, indicating it is a methyl group adjacent to a carbon that has 2 hydrogens. Therefore, it looks like the molecule is a straight-chain of 5 carbons with the aldehyde group at one end:
Note that the closer a group is to the carbonyl function, the further downfield it is shifted. Here is how the NMR correlates to the structure: