Calculate the degree of unsaturation: the answer is 1. The molecule has one double bonds or one ring.
Look for C=C (1680-1640) or C=O (1760-1665) stretching vibrations - there are none. Since the molecule has a nitrogen, look for a band in the region 3400-3250 - there is a band at 3335, which probably indicates the N-H stretch of a primary or secondary amine. Since the molecule has an oxygen, look for O-H stretch: the band in this region more resembles N-H stretch than O-H stretch. (See examples 3, 11, and 12). This compound is a good candidate for containing a non-aromatic ring.
The singlet at 1.6 is in the region 0.5-3.0 ppm, indicating a single amine proton. The NMR shows 2 sets of equivalent methylene (-CH2-) groups, each of which is next to a carbon with 2 hydrogens.
You can try drawing different structures for this compound. When you are done trying, the correct structure is drawn below. Hydrogens on a carbon which is adjacent to a nitrogen show up from 1.5-2.0 (see Amines); while hydrogens on a carbon adjacent to an oxygen show up from 3.3-4.0 ppm (see Ethers).