### Example 13

C7H5NO3

MW 151

Calculate the degree of unsaturation: the answer is 6. The molecule probably has an aromatic ring and two double bonds.

### IR Spectrum

Of course, the molecule could be an amine. However, it can also have a -NO2 group. An NO2 group has a degree of unsaturation of 1, so in the IR we should look for:

• aromatic ring: 1600-1585, 1500-1400, 900-675
• -NO2 group: 1550-1475, 1365-1290
• carbonyl or double bond: 1760-1665 or 1680-1640

As well as the C-H stretches:

• aromatic, alkene: 3100-3000
• alkane: 3000-2850
• aldehyde: 2830-2695

From the IR, we know that there definitely is a carbonyl group, either an aldehyde or a ketone, but probably an aldehyde because of the C-H stretch band in the region 2830-2695. From the position of the carbonyl, it is probably an alpha, beta-unsaturated aldehyde or ketone. Probably, the carbonyl is on an aromatic ring. The compound is likely to be aromatic because of the many bands in the aromatic region and th C-H stretch above 3000. A nitro group is also supported because of the bands in the nitro region.

### NMR Spectrum

The four protons from 7.2-8.7 ppm indicate 4 aromatic protons, which in turn indicates a di-substituted aromatic ring. The singlet of one proton at 10.1 ppm indicates an aldehyde, as hypothesized in the IR discussion. This aldehyde group must be on the ring (as supported by the IR position of the carbonyl), since there are only 7 carbons in the molecule and 6 are in the ring. The remaining ring-substituent must be a nitro group, -NO2.

The compound is meta-substituted. Study the compound drawn below. Given the integral values shown below, you should be able to correlate the singlet, two doublets, and the triplet with the four hydrogens on the ring with what you have been taught so far.

### Summary

Example 13 is 3-nitrobenzaldehyde: