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Problem 3

Formula: C4H10O

Spectroscopy Reference

IR spectrum NMR spectrum

Rule 2, omit O, gives C4H10
4 - 10/2 + 1 = 0 degrees of unsaturation.
No pi bonds or rings.

The broad band at 3339 indicates an O-H stretch, probably an alcohol. The bands at 3000-2850 indicate C-H alkane stretches. The band at 1041 is 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

A and B do not couple because of A's ability to H-bond.

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