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Problem 7

Formula: C8H14O3

Spectroscopy Reference

IR spectrum NMR spectrum

Rule 2, omit O, gives C8H14
8 - 14/2 + 1 = 2 degrees of unsaturation.
Look for 2 pi bonds or aliphatic rings, or 1 of each.

The bands at 1745 and 1716 indicate that there are two carbonyls, probably an aliphatic ester and an aliphatic ketone. 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

Even though A, B, C and D are all 2H peaks, they can be distinguished by chemical shift and splitting. B is outside the normal range for protons next to carbonyls, because it's adjacent to both carbonyls and the combined deshielding is higher than normal.

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