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Problem 4

Formula: C6H14O

Spectroscopy Reference

IR spectrum NMR spectrum

Rule 2, omit O, gives C6H14
6 - 14/2 + 1 = 0 degrees of unsaturation.
No pi bonds or rings.

The broad band at 3350 indicates O-H stretch, probably an alcohol. The bands at 3000-2850 indicate C-H alkane stretches. The bands from 1320-1000 indicate C-O stretch, consistent with an alcohol.

Structure answer

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

NMR answer

As noted, C and D are not equivalent even though they're on the same carbon.

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