Problem 17

C₁₃H₁₀O₃
Rule 2, omit O, gives C₁₃H₁₀
13 - 10/2 + 1 = 9 degrees of unsaturation.
Look for 2 aromatic rings plus another pi bond or aliphatic ring.

The band at 1682 indicates a carbonyl, probably an ester. The band at 3192 indicates C-H aromatic stretch; aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-of-plane). The wide band just to the left of the 3192 band indicates O-H stretch (alcohols and phenols).

This is the structure. See if you can assign the peaks on your own.

You're not expected to be able to tell all the aromatic Hs apart. In fact, there are several structures that could match this pattern, as far as you can tell from the information on this site: