Dipole Moment of Benzene Molecule
Dipole moment of benzene molecule is zero suggested that the benzene ring has a regular hexagonal planner aromatic structure with sp2 hybridized atomic orbitals in the carbon atom. The dipole moment of benzene and substituted benzene molecules are calculated by different theoretical or experimental formulas in chemistry. In learning chemistry, substituted benzene molecule shows aromatic properties with (4n + 2)π electrons. The aromatic hydrocarbon benzene was first isolated by Faraday in 1825 in cylinders of the compressed illuminating gas molecule from the pyrolysis of whale oil. The chemical bond in the benzene molecule has double bond properties with five canonical structures having different bond energy.
If a hydrogen atom in the benzene ring is substituted by another atom or group, this molecule acquires bond polarity. Examples of such benzene derivatives which show dipole moment are chlorobenzene, nitrobenzene, phenol, benzyl alcohol, etc. The dipole moment of o, p, m-isomers of the benzene ring can be calculated from the formula, μ2 = m12 + m22 + 2 m1m2 cosθ.
Ortho para meta position in benzene ring
When one group or element introduces into the ring, only one product is produced. The ortho, para, meta or o, p, m isomers formed by the introduction second group in the benzene ring given above the picture. The dipole moment of ortho isomer of benzene molecule will have the highest value and p isomer has the lowest value equal to zero like methane.
Dipole moment and structure of molecules
The resultant dipole moment of molecules is equal to the vectorial sum of individual bond or group moments. It depends on the structure of the molecules. The dipole moment of the di-substituted molecules like xylene, dichlorobenzene, dinitrobenzene can be calculated by vectorial addition formula, μ2 = m12 + m22 + 2 m1m2 cosθ. Where m1 and m2 are individual group moments and θ = angle at which m1 and m2 are inclined to each other.
Dipole moment of xylene
Ortho, para, and meta-xylene are the substituted derivatives of benzene having the molecular formula C6H4(CH3)2.
Dipole moment of o-xylene
For o-xylene molecule, the groups moment of CH3 group = m and θ = 60°. From dipole moment formula, μ2o-xylene = 2m2(1 + cos60°) = 3m2. Therefore, dipole moment of o-xylene = √3m.
Dipole moment of p-xylene
For o-xylene molecule, θ = 180°. From dipole moment formula, μ2p-xylene = 2m2(1 + cos180°) = 0. Therefore, dipole moment of o-xylene = 0.
Dipole moment of m-xylene
For o-xylene molecule, θ = 120°. From dipole moment formula, μ2p-xylene = 2m2(1 + cos120°) = m2. Therefore, the dipole moment of o-xylene = m. From the above three calculations, the dipole moment di-substituted molecules of a benzene ring or xylene, μortho > μmeta >μpara. Similar way, we can calculate the dipole moment of o, p, and m-dinitrobenzene or dichlorobenzene.
Dipole moment of 1, 3, 5 trinitrobenzene
1, 3, 5 trinitrobenzene being symmetrical non-polar molecule. When it is dissolved in benzene, a charge transfer bonding involves some kind of donner acceptor interaction. Due to the formation of charge-transfer complexes in the C6H6 solution, the compound shows a significant dipole moment.
Problem: If dipole moment of o-xylene molecule = 0.693D, calculate the μ for toluene molecule.
Answer: For o-xylene, θ = 60° and μ = √3 m. Therefore, mxylene = 0.693/√3 = 0.4D. Again for toluene, θ = 120° and the μtoluene = 0.4D.
Problem: Why p-dinitrobenzene has μ = 0 but para – dihydroxybenzene μ ≠ 0?
Answer: In p-dinitrobenzene, the substituted nitro group is the same plane of benzene ring but in p-dihydroxybenzene, the substituted hydroxy groups are not on the same plane. Therefore among these two substituted molecules, the dipole moment of p-dinitrobenzene is zero but the p-dihydroxybenzene molecule is polar.