NMR Spectroscopy of Aromatic Compounds (#1e)

1H NMR Spectroscopy of Aromatic Compounds Erich Hückel’s study of aromaticity in the 1930s produced a set of rules for determining whether a compound is aromatic. According to the Hückel criteria, a molecule must be cyclic, nearly planar, possess a p orbital on each carbon, and have 4n + 2  electrons, where n is an integer. These factors allow one to label a molecule as “aromatic,” giving the molecule some special reactive characteristics that will be discussed in detail in your Chemistry 302 lecture. For now, the classic example of an aromatic compound is benzene, seen below with its 1H NMR spectrum.

7.33

benzene

6.00 9 8 7 6 5 4 3 2 1 0 In aromatic compounds like benzene, the protons on the aromatic ring are shifted downfield. For example, the six protons in benzene are magnetically and chemically equivalent and appear at 7.33 ppm. This is farther downfield than alkene protons, which appear between 4.5-6.5 ppm. NMR serves as a useful tool to determine whether a compound is aromatic. For example, the protons in cyclooctatetraene (C8H8), which is shown below, appear at 5.78 ppm indicating it is in the typical alkene region, not the aromatic region near 7 ppm.

5.78

cyclooctatetraene

0.00

8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 -0.5 37

If a hydrogen atom on the benzene ring is replaced with a different substituent, such as a methyl group, some of the hydrogens become magnetically inequivalent. For example, in the compound toluene, there are three different kinds of hydrogens on the benzene ring. The hydrogens neighboring the methyl (Ha) are magnetically inequivalent to the hydrogens two carbons away from the methyl (Hb) and the hydrogen directly across the ring from the methyl group (Hc). Ha, Hb, and Hc all couple to each other and have J values that correlate with the number of bonds between coupling protons. Ha and Hb are ortho to one another (adjacent); Jortho = 6-10 Hz. Ha and Hc are meta to one another (two carbons apart); Jmeta = 1-3 Hz. Although there is not an example in this molecule, if two protons are three carbons apart, they are para; Jpara = 0-1 Hz. As you can see, the J value decreases as the number of bonds between hydrogens increases. The peak for the aromatic protons is a complex multiplet as a result of the many J values and overlapping peaks. With greater resolution, it is possible to see these separate peaks. Ha

Hb CH 3

Ha Hc

Hb

toluene

7.25 7.20 7.15 7.10

5.00 3.00

7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0

If the monosubstituted benzene contains an electron-withdrawing group such as Br or Cl, the protons adjacent to the electron-withdrawing group will be shifted slightly downfield of the other aromatics, producing separate peaks. The NMR of bromobenzene is shown below. Notice the peaks that are shifted downfield (7.4-7.5 ppm). These Ha couple to Hb (Jortho = 6-10 Hz) which results in a doublet. The doublet is further split by coupling to Hc with a very small J value (Jpara ~ 0-1 Hz).

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Br

Ha Ha

Hb Hb

Hc

7.55 7.50 7.45 7.40 7.35 7.30 7.25 7.20 7.15

2.00 3.14

8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 -0.5

For disubstituted benzenes, the aromatic peaks are often separated as with bromobenzene. There are a few exceptions. For instance, the compound para-dimethylbenzene will have only a singlet in the aromatic region because all of the hydrogens are magnetically equivalent due to the symmetry in the molecule. This will be true of any para-disubstituted benzene having two identical substituents. However, ortho-dimethylbenzene will have separated aromatic peaks.

An important splitting pattern that identifies a specific type of substituted benzene is that of a para-disubstituted benzene having two substituents with different electron-withdrawing effects. An example is shown below with para-bromoethylbenzene. The distinctive pattern is the pair of doublets, which can be widely spaced or close together depending on the groups on the ring. This pair of doublets is only seen with a para-disubstituted benzene, not ortho- or meta- disubstituted benzenes. The latter display much more complex patterns.

Ha Hb CH 3

Br Ha

Hb 0.00

8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5

1.99 1.94 2.00 2.95 8 7 6 5 4 3 2 1 0 -1 39

13C NMR Spectroscopy of Aromatic Compounds As with other 13C NMR spectra, aromatic compounds display single lines for each unique carbon environment in a benzene ring. Aromatic carbons appear between 120-170 ppm. The 13C NMR spectra of bromobenzene and p-bromoethylbenzene are shown below for comparison. There are four different carbon environments in bromobenzene, and four different peaks. Similarly, for p- bromoethylbenzene, four environments appear but the pattern of the peaks is quite different and more separated. Note that one of the carbons appears much farther downfield at 143 ppm.

131.49

Cb Br 129.97 Cc Ca 126.81

Cd Cb Cc

122.49

200 150 100 50 0

131.28129.58

Cb Cc Ca CH 3 Cd Cb Cc Br 15.45

28.32

0.00 119.24 143.07

200 150 100 50 0

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DEPT spectra can also be obtained for aromatics; as with earlier examples, CH’s appear as upright singlets in the DEPT-90 and DEPT-135. Quaternary carbons appear in the proton decoupled spectra but not in DEPT spectra.

Postlab You should complete the following assignment in your lab notebook along with the other spectroscopy assignments. A table of 1H chemical shifts is located on page 44. Solvent peaks typically appear at 0 ppm or 7.24 ppm. A printable copy of this assignment can be found at www.unm.edu/~orgchem.

1. How many separate peaks would be observed in the 13C spectra of the following compounds? a. 1,2-dinitrobenzene b. 1,3-dinitrobenzene c. 1,4-dinitrobenzene

2. Propose structures for the compounds that would be expected to generate the following 1H NMR spectra. Calculate the degree of unsaturation of each compound and assign the protons to their respective peaks.

A. 1.29

C H Br 10 13

7.38 7.25 7.40 7.35 7.30 7.25

1.78 1.89 9.00 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0

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B.

C9H12

1.30 1.29 1.28 1.27 1.26 1.25 1.24 1.23 1.22 1.21 1.20

7.35 7.30 7.25 7.20 7.15

2.98 2.97 2.96 2.95 2.94 2.93 2.92 2.91 2.90 2.89 2.88 2.87 2.86 2.85 2.84 2.83 2.82 2.81 2.80

5.10 1.03 6.00

9 8 7 6 5 4 3 2 1 0 -1

C.

C6H4BrCl

7.60 7.55 7.50 7.45 7.40 7.35 7.30 7.25 7.20 7.15 7.10 7.05 7.00

2.00 2.03

8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0

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