Ubbelohde effect within weak C-H···π hydrogen bonds: the rotational spectrum of benzene-DCF3

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...

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).

The vibrational frequency of a bond is directly proportional to its bond strength. As a result, stronger bonds vibrate at higher frequencies, while weaker bonds vibrate at lower frequencies. The stretching vibration of the strong O–H bond in alcohols and phenols (very dilute solution or gas phase) appears as a sharp peak at 3600–3650 cm−1.
However, the extent of hydrogen bonding influences the observed stretching frequency and band broadening. Intermolecular or intramolecular hydrogen bonding...

In aromatic compounds, such as benzene, the circulation of (4n + 2) π-electrons sets up a diamagnetic or diatropic ring current around the perimeter of the molecule. This current induces a magnetic field that opposes the external field inside the ring and reinforces it on the outside. The protons in benzene are deshielded and exhibit high chemical shifts in the range 6.5–8.5 ppm. The shielding effect at the center of the ring is evident in complex aromatic molecules, such as annulenes. In...

The inscribed polygon method is consistent with Hückel’s 4n + 2 rule and helps to learn whether the given cyclic compound is aromatic or not. The compound is stable and aromatic if every bonding molecular orbital (MO) is completely filled with a pair of electrons. However, if the non-bonding or antibonding orbitals are filled with electrons, the compound is unstable and not aromatic. Consider the Frost circle diagrams for cycloalkenes containing 4 to 8 carbons.

In bromoethane, the three methyl protons are coupled to the two methylene protons that are three bonds away. In accordance with the n+1 rule, the signal from the methyl protons is split into three peaks with 1:2:1 relative intensities. The methylene protons appear as a quartet, with the relative intensities of 1:3:3:1.
Qualitatively, any spin plus-half nucleus polarizes the spins of its electrons to the minus-half state. Consequently, the paired electron in the hydrogen–carbon bond must have a...