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Problem 16

Formula: C10H14O

Spectroscopy Reference

IR spectrum NMR spectrum

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.

Structure answer

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

NMR answer

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.

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