Bruice - Chimica Organica

Spectroscopy Problems

Now that all the fragments of the compound have been identified, we can put them together. The compound

ethyl 4-hydroxybenzoate

OCH

The signals at 1.1 and 1.8 ppm have been magnified and are shown as insets on the spectrum (the

and

represent the ppm scale) so that you can better see the splitting. The triplet (1.1 ppm) that integrates to

3 protons and the quartet (1.8 ppm) that integrates to 2 protons are characteristic of an ethyl group. (The

peak to the right of the quartet is actually the beginning of the adjacent signal that integrates to 6 protons.)

The singlet (1.7 ppm) that integrates to 6 protons indicates that there are two methyl groups in the same

environment. Because the signal is a singlet, the carbon to which they are attached cannot be bonded to any

hydrogens. The only atom not accounted for in the molecular formula is Br.

Therefore, the ethyl group and the bromine must be the two substituents that are attached to the carbon.

Thus, the compound is

2-bromo-2-methylbutane

C CH

A major clue comes from the IR spectrum. The strong absorption at

1710 cm

indicates the presence

of a carbonyl

“

group. Because the compound has only one oxygen, we know that it must be an

aldehyde or a ketone. The absence of absorptions at 2820 and

2720 cm

tells us that the compound is not

an aldehyde.

The absorptions at 2880 and

2970 cm

are due to

stretches of hydrogens attached to

carbons.

The

H NMR

spectrum has two unsplit signals. One integrates to 9 protons and the other to 3 protons.

A signal that integrates to 9 protons suggests a

tert

-butyl group, and a signal that integrates to 3 protons

suggests a methyl group. The fact that they are both singlets indicates that they are on either side of the

carbonyl group. Therefore, the compound is

3,3-dimethyl-2-butanone

C C

That the methyl group shows a signal at

2.1 ppm

reinforces this conclusion because that is where a

methyl group attached to a carbonyl group is expected to occur.