Relating normal vibrational modes to local vibrational modes: benzene and naphthalene

Wenli Zou1, Robert Kalescky, Elfi Kraka

  • 1Department of Chemistry, Southern Methodist University, 3215 Daniel Avenue, Dallas, TX 75275-0314, USA.

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Simple unsubstituted benzene has six aromatic protons, all chemically equivalent. Therefore, benzene exhibits only a singlet peak at δ 7.3 ppm in the 1H NMR spectrum. The observed shift is far downfield because the aromatic ring current strongly deshields the protons. Any substitution on the benzene ring makes the aromatic protons nonequivalent, and the protons split each other. The peak is, therefore, no longer a singlet and the splitting pattern and their associated coupling constants depend...
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According to the molecular orbital (MO) model, benzene has a planar structure with a regular hexagon of six sp2 hybridized carbons. As shown in Figure 1, each carbon is bonded to three other atoms with C–C–C and H–C–C bond angles of 120°. The C–H bond length is 109 pm, and the C–C bond length is 139 pm which is midway between the single bond length of sp3 hybridized carbons (154 pm) and sp2 hybridized carbons (133 pm).
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When Infrared (IR) radiation passes through a covalently bonded molecule, the bonds transition from lower to higher vibrational levels. The fundamental vibrational motions that result in infrared absorption can be classified as stretching or bending vibrations.
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Aromatic compounds can be identified or analyzed using proton NMR and carbon‐13 NMR. Typically, aromatic hydrogens or hydrogens directly bonded to the aromatic rings are strongly deshielded by the aromatic ring current. Therefore, they absorb in the range of 6.5–8.0 ppm in proton NMR spectra. For instance, aromatic hydrogens directly bonded to the benzene ring absorb at 7.3 ppm. However, aromatic hydrogens of larger rings absorb farther upfield or downfield than the ideal range. Consider...
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