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Problem 14

Formula: C8H14O

Spectroscopy Reference

IR spectrum NMR spectrum

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

The band at 1718 indicates a carbonyl, probably a ketone. The bands at 3000-2850 indicate C-H alkane stretches. Since the compound is an alkene, one would expect to see C=C stretch at 1680-1640; these weak bands are not seen in this IR (according to Silverstein, "the C=C stretching mode of unconjugated alkenes usually shows moderate to weak absorption at 1667-1640"). Since the compound is an alkene, C-H stretch should appear above 3000 (not seen: the absorption for this single hydrogen must be too weak).

Structure answer

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

NMR answer

Here, the two E methyl groups have different chemical shifts because they are permanently locked into different positions with respect to the alkene.

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