Problem 16

C₁₀H₁₄O
Rule 2, omit O, gives C₁₀H₁₄
10 - 14/2 + 1 = 4 degrees of unsaturation.
Look for an aromatic ring.

The broad dip at 3250 indicates an OH group. The peak at 2963 indicates an alipahtic group. The peaks at 1599 and 1514 indicate an aromatic ring.

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

You are not expected to tell the aromatic Hs apart from each other, but you should be able to tell that the ring is para-substituted.