Vollhardt  Capítulo 15 (Benzenos e Aromaticidade)

Vollhardt Capítulo 15 (Benzenos e Aromaticidade)

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686 C h a p t e r 1 5 B e n z e n e a n d A r o m a t i c i t y

d 5 7.23 ppm as a doublet with a 9-Hz ortho coupling. The hydrogen fl anked by the two
nitro groups (at d 5 8.76 ppm) also appears as a doublet, with a small (3-Hz) meta coupling.
Finally, we fi nd the remaining ring hydrogen absorbing at d 5 8.45 ppm as a doublet of
doublets because of simultaneous coupling to the two other ring protons. Para coupling
between the hydrogens at C3 and C6 is too small (,1 Hz) to be resolved; it is evident as
a slight broadening of the resonances of these protons.

In contrast with 1H NMR, 13C NMR chemical shifts in benzene derivatives are domi-
nated by hybridization and substituent effects. Because the induced ring current fl ows
directly above and below the aromatic carbons (Figure 15-9), they are less affected by it.
Moreover, the relatively large 13C chemical shift range, about 200 ppm, makes ring current
contributions (only a few ppm) less noticeable. Therefore, benzene carbons exhibit chemi-
cal shifts similar to those in alkenes, between 120 and 135 ppm when unsubstituted (see
margin). Benzene itself exhibits a single line at d 5 128.7 ppm.

13C NMR Data
of Two Substituted

Benzenes (ppm)











Aromatic Coupling

 Jortho 5 6 – 9.5 Hz
 Jmeta 5 1.2 – 3.1 Hz
 Jpara 5 0.2 – 1.5 Hz

Exercise 15-7

Working with the Concepts: Using Spectral Data to Assign the
Structure of a Substituted Benzene

A hydrocarbon shows a molecular-ion peak in the mass spectrum at myz 134. The associated
[M 1 1] ion at myz 135 has about 10% of the intensity of the molecular ion. The largest peak in
the spectrum is a fragment ion at myz 119. The other spectral data for this compound are as
 follows: 1H NMR (90 MHz) d 5 7.02 (broad s, 4 H), 2.82 (septet, J 5 7.0 Hz, 1 H), 2.28 (s, 3 H),
and 1.22 ppm (d, J 5 7.0 Hz, 6 H); 13C NMR d 5 21.3, 24.2, 38.9, 126.6, 128.6, 134.8, and
145.7 ppm; IR n˜ 5 3030, 2970, 2880, 1515, 1465, and 813 cm – 1; UV lmax(e) 5 265(450) nm. What
is its structure?
The spectra supply a lot of information, so your fi rst problem is Where do I start? Typically,
researchers look at the most important data fi rst — the mass and the 1H NMR spectrum — and then
use the other spectra as corroborating evidence. The following solution employs only one of several
strategies that you could apply. It is often not necessary to dwell in such depth on the information
embedded in the individual spectra, and you may fi nd it more useful fi rst to identify the most
important pieces, such as the molecular ion in the mass spectrum, the spectral regions and/or split-
ting patterns in the 1H NMR spectrum, the number of 13C signals, and a characteristic IR frequency
or UV absorption. Whenever you are ready to formulate a (sub)structure, you should do so, and
keep validating (or discarding) it as you analyze additional data. Trial and error is the key!
• The mass spectrum is already quite informative. There are not many hydrocarbons with a mole-
cular ion of myz 134 for which a sensible molecular formula can be derived. For example, the
maximum number of carbons has to be less than C12 (mass 144, too large). If it were C11, the
molecular formula would have to be C11H2, ruled out by a quick look at the

1H NMR spectrum,
which shows more than two hydrogens in the molecule. (Besides, how many structures can you
formulate with the composition C11H2?) The next choice, C10H14, looks good, especially since
going down once again in the number of carbons to C9 would require the presence of 26 hydrogens,
an impossible proposition, because the maximum number of hydrogens is dictated by the generic
formula CnH2n12 (Section 2-4).
• C10H14 means four degrees of unsaturation (Section 11-11), suggesting the presence of a combi-
nation of rings, double bonds, and triple bonds. The last option is less likely because of the absence
of a band at ,2200 cm – 1 in the IR spectrum (Section 13-3). This leaves us with double bonds
and/or rings.
• A check of the 1H NMR spectrum does not show signals for normal double bonds, but instead
an aromatic peak. A benzene ring has the postulated four degrees of unsaturation. Hence we are
dealing with an aromatic compound.

Exercise 15-6

Can the three isomeric trimethylbenzenes be distinguished solely on the basis of the number of
peaks in their proton-decoupled 13C NMR spectra? Explain.

 C h a p t e r 1 5 687

In Summary Benzene and its derivatives can be recognized and structurally characterized
by their spectral data. Electronic absorptions take place between 250 and 290 nm. The
infrared vibrational bands are found at 3030 cm21 (Caromatic – H), from 1500 to 2000 cm

(C – C), and from 650 to 1000 cm21 (C – H out-of-plane bending). Most informative is NMR,
with low-fi eld resonances for the aromatic hydrogens and carbons. Coupling is largest
between the ortho hydrogens, smaller in their meta and para counterparts.

15-5 Polycyclic Aromatic Hydrocarbons
When several benzene rings are fused together to give more extended p systems, the mole-
cules are called polycyclic benzenoid or polycyclic aromatic hydrocarbons (PAHs). In
these structures, two or more benzene rings share two or more carbon atoms. Do these
compounds also enjoy the special stability of benzene? The next two sections will show
that they largely do.

There is no simple system for naming these structures, so we shall use their common
names. The fusion of one benzene ring to another results in a compound called naphthalene.
Further fusion can occur in a linear manner to give anthracene, tetracene, pentacene, and so on,

Exercise 15-8

Try It Yourself

Deduce the structure of a compound C16H16Br2 that exhibits the following spectral data: UV lmax
(log e) 5 226sh(4.33), 235sh(4.55), 248(4.78), 261(4.43), 268(4.44), 277(4.36) nm (sh 5 shoulder);
IR n˜ 5 3030, 2964, 2933, 2233, 1456, 1362, 1060, 892, 614 cm21; 1H NMR (300 MHz) d 5 7.75
(s, 1 H), 7.45 (s, 1 H), 2.41 (t, J 5 7.0 Hz, 4 H), 1.64 (sextet, J 5 7.0, 4 H), 1.06 (t, J 5
7.4 Hz, 6 H); 13C NMR d 5 13.51, 21.54, 21.89, 78.26, 96.68, 124.4, 125.2, 135.2, 136.8 ppm.
(Caution: Look closely at the NMR spectra with reference to the molecular formula. Moreover,
the IR spectrum contains important information. Hint: There is symmetry in the molecule.)

• Looking at the fragmentation pattern in the mass spectrum, the mass ion at m/z 119 points to
the loss of a methyl group, which must therefore be present in our molecule.
 Turning to the 1H NMR spectrum in more detail and remembering that we have 14 hydrogens
in our structure, we can now probe how these hydrogens are distributed. The 1H NMR spectrum
reveals 4 H in the aromatic region, all very close in chemical shift, giving rise to a broad singlet.
The remainder of the hydrogen signals appear to be due to alkyl groups: One of them is clearly
a methyl substituent at d 5 2.28 ppm; the others are part of a more complex array, containing
two sets of mutually coupled nuclei. Closer inspection of this array shows it to be a septet (indi-
cating six equivalent neighbors) and a doublet (indicating a single hydrogen as a neighbor), point-
ing to the presence of a 1-methylethyl substituent (Table 10-5). The relatively large d values for
the single methyl and the tertiary hydrogen of the 1-methylethyl groups suggest that they are close
to the aromatic ring (Table 10-2).
• Moving on to the 13C NMR spectrum, the preceding analysis requires the presence of (at least)
three sp3-hybridized carbon peaks (Table 10-6). Indeed, there are three signals at d 5 21.3, 24.2,
and 38.9 ppm, and no additional peaks in this region. Instead, we see four resonances in the aromatic
range. Since a benzene ring has six carbons, there must