If you look at an IR spectrum of 1-butanol, you will see:

• there are sp³ C-H stretching and CH₂ bending modes at 2900 and 1500 cm^-1.
• there is a strong C-O stretching mode near 1000 cm^-1.
• there is a very large peak around 3400 cm^-1. O-H peaks are usually very broad like this one.

Peak shapes are sometimes very useful in recognizing what kind of bond is present. The rounded shape of most O-H stretching modes occurs because of hydrogen bonding between different hydroxy groups. Because protons are shared to varying extent with neighboring oxygens, the covalent O-H bonds in a sample of alcohol all vibrate at slightly different frequencies and show up at slightly different positions in the IR spectrum. Instead of seeing one sharp peak, you see a whole lot of them all smeared out into one broad blob. Since C-H bonds don't hydrogen bond very well, you don't see that phenomenon in an ether, and an O-H peak is very easy to distinguish in the IR spectrum.

Problem IR5.1.

Even though there are only two C-O bonds in dibutyl ether, the C-O stretching mode is even stronger than the peak at 2900 cm^-1 arising from 10 different C-H bonds. Explain why.

Problem IR5.2.

The IR spectrum of methyl phenyl ether (aka anisole) has strong peaks at 1050 and 1250 cm^-1.

a) Identify the type of bond corresponding to these two peaks.

b) Why are there two peaks for this type of bond in this molecule, and not just one?

c) Draw a second, zwitterionic resonance structure for methyl phenyl ether.

d) Use the zwitterionic resonance structure to explain why one of these bonds shows up at a higher frequency than the other one.