Dipole Moment of Benzene Molecules
The dipole moment of molecules like benzene zero suggested that the benzene ring has a regular hexagonal planner aromatic structure with sp2 hybridized atomic orbitals in the carbon atom. Therefore, the dipole moment of benzene and substituted molecules are calculated by different theoretical or experimental calculation formulas in chemistry. In learning chemistry, substituted benzene molecule shows acid properties in phenols or aromatic alcohol but it shows base properties in aromatic amines. Hydrocarbon like benzene was the first isolated by Faraday in 1825 in cylinders of the compressed illuminating gas molecule from the pyrolysis of whale oil. All the chemical bonds in the C6H6 molecule have 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 molecules of which shows dipole moment is chlorobenzene, nitrobenzene, phenol, etc, and o, p, m-isomers of the benzene ring.
When one group or chemical element introduces into the ring, only one product is produced but for the second group, ortho-, para- and meta- or o, p, m isomers forms. The ortho-isomer benzene molecule will have the highest dipole moment value than the p and m-isomers molecules but p-derivative has the lowest value of bond polarity equal to zero like methane molecule.
Substituted Derivative of Benzene ring
The resultant dipole moment of molecules equal to the vectorial sum of energy of individual bond or group moments. Therefore, the dipole moment of the di-substituted molecules of the benzene ring ( C6H6x1x2) can be calculated by vectorial energy addition formula, μ2 = m12 + m22 + 2 m1m2 cosθ.
Dipole Moment of o, m, p-isomers Molecules
Experimental work show that the o, p-substitution activated benzene nucleus for chemical reaction. But m-substitution deactivated the nucleus for chemical reaction. The electric polarization value of o-derivative is highest but p-derivatives are lowest.
μ2 = m12 + m22 + 2 m1 m2 cosθ
For o-isomer, m1 = m2 = m and θ = 60°
Therefore, μ2 = 2 m2 (1 + cos60°)
= 2 m2(1 + 1/2) = 3m2
∴ μortho = √3 m
For m-isomer, θ = 120°
Therefore, μ2 = 2 m2 (1 + cos120°)
= 2 m2 (1 – 1/2) = m2
∴ μmeta = m
For para-isomer, θ = 180°
Therefore, μ2 = 2 m2 (1 + cos180°)
= 2 m2 (1 – 1) = 0
∴ μpara = 0
The dipole moment di-substituted molecules of benzene ring are, μortho > μmeta >μpara. These determination of the dipole moment of molecules helps to determine the orientation of the groups or periodic table chemical elements in ring.
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 organic benzene molecules, the dipole moment of p-dinitrobenzene is zero but the p-dihydroxybenzene molecule is polar in chemistry.