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Problem 9

Formula: C11H14O2

Spectroscopy Reference

IR spectrum NMR spectrum

Rule 2, omit O, gives C11H14
11 - 14/2 + 1 = 5 degrees of unsaturation.
Look for an aromatic ring, plus 1 other pi bond or aliphatic ring.

The band at 1603 or 1894 might indicate a carbonyl outside the normal range; the small peak around 1800 indicates it may be an aldehyde. The band at 1510 is an aromatic ring. The bands at 3000-2850 indicate C-H alkane stretches.

Structure answer

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

NMR answer

Based on what is covered, you shouldn't need to distinguish between B and C, but you should be able to tell that the ring is para-substituted.

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