Spectroscopy Tutorial: Examples

Example 7


MW 120

Calculate the degree of unsaturation: the answer is 5. If the degree of unsaturation is 4 or greater, look for an aromatic ring, which has a degree of unsaturation of 4 (3 double bonds plus 1 ring). In addition to an aromatic ring, the molecule can have a carbon-carbon double bond, a carbonyl, or another ring.

IR Spectrum

Look for a carbonyl, since the degree of unsaturation indicates that the compound could have a double bond and we know that the molecule has an oxygen. There is a band at 1703, suggesting an alpha, beta unsaturated aldehyde or ketone(1710-1665). To see if the compound might be an aldehyde, look for bands in the region 2830-2695. In the spectrum below, note the two bands in this region, suggesting that the compound is indeed an aldehyde.

The IR can also help determine whether or not the compound is an aromatic (although the NMR is a better diagnostic method for this). Look for the C–H stretch in aromatics from 3100-3000. Note that compounds that are not aromatic show C-H stretch from 3000-2850.

Consult the section on Aromatics for more information on IR spectroscopy of aromatics.

From the IR alone, we have determined that the molecule probably has aromatic, alkyl, and aldehydic functional groups. Now let's look at the NMR.

Proton NMR Spectrum

Since the IR spectrum indicates an aldehyde, look for this functionality in the NMR spectrum. The aldehydic proton appears in the NMR from 9-10, usually as a small singlet. The singlet of one proton in the spectrum below verifies the presence of an aldehyde:

Aromatic protons show up from 6.5-8.5 ppm. The four protons in the region 7.3-7.8 ppm indicate four aromatic protons; thus the aromatic ring has two substituents. Benzylic protons - hydrogens on a carbon adjacent to an aromatic ring - show up from 2-3 ppm; the three protons in the singlet peak at 2.4 ppm are likely benzylic protons. Since there are three protons, it is a methyl group.

Therefore we know that the molecule is a di-substituted aromatic ring with a -CHO group and a -CH3 group. Double check that we have the proper number of carbons: 6 (in ring) plus 1 (aldehyde) plus 1 (methyl) equals 8, the number given in the molecular formula.

These hydrogens correspond as follows with the peaks in the NMR:

The molecule is 4-methylbenzaldehyde: a para substituted aromatic. In the organic chemistry teaching laboratory courses, you are not held responsible for proper assignment of ortho-, meta-, or para-substituted rings. However, para substituted rings almost always show a symmetrical pattern in the aromatic region, as in the spectrum above. (See the section on Aromatics.)


Example 7 is 4-methylbenzaldehyde: