## The dipole moment of the benzene ring

Zero dipole moment of the hydrocarbon like benzene ring suggested that it has regular hexagonal planer molecules and it confirms Kekule’s form. Regular planer hexagonal structure of benzene ring having the center of symmetry.

However, if hydrogen atom substituted by another atom or group, it acquires polarity.

Examples of such products of benzene are C_{6}H_{5}Cl, C_{6}H_{5}NO_{2}, C_{6}H_{5}OH, etc. When di-substituted benzene considered it can be shown

- ortho-isomer will have the highest value of dipole moment than the other two isomers.
- para-derivative has the lowest value while m-derivative has the value between the two.

### A di-substituted derivative of benzene

The resultant dipole moment of a molecule is equal to the vectorial sum of individual bond or group moments. Thus for a di-substituted product of benzene C_{6}H_{6}X_{1}X_{2}, the dipole moments can be calculated by the following formula of vectorial addition

μ = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθ

#### Ortho-isomer of benzene

μ_{2} = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθ

When m_{1} = m_{2} = m and θ = 60^{0}

Thus, μ2 = 2 m^{2}(1 + cos60^{0})

= 2 m^{2}(1 + 1/2)

∴ μ= √3 m

#### Meta-isomer of benzene

μ^{2} = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθ

When, m_{1} = m_{2} = m and θ = 120^{0}

Therefore, μ^{2} = 2 m^{2}(1 + cos120°)

= 2 m^{2} (1 – 1/2)

∴ μ = m

#### Para-isomer of benzene

μ^{2} = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθ

Here, m_{1} = m_{2}= m and θ = 180^{0}

Therefore, μ^{2} = 2 m^{2}(1 + cos180°)

= 2 m^{2} (1 – 1)

∴ μ = 0

Thus dipole moment of o-isomer〉 μ of m-isomer〉μ of p-isomer (this has been confirmed for C_{6}H_{4}Cl_{2} and C_{6}H_{4}(NO_{2})_{2})

This determination of the dipole moment of molecules helps to determine the orientation of the groups in the benzene ring.

Problem

Para – dinitro benzene has μ = 0 but para – dihydroxy benzene μ ≠ 0. Explain.

Answer

Through the para-dinitro benzene has μ = 0. p-dihydroxy benzene has dipole moment(μ ≠ 0). This is due to the fact that the two substituted hydroxy groups are not on the same plane of the benzene ring but are inclined to the ring.

Problem

The dipole moment of o-xylene = 0.693 D. Finds the dipole moment of toluene.

Answer

For ortho – xylene the θ = 60^{0} and μ = √3 m.

Thus, m = 0.693/√3 = 0.4 D

Again θ = 120^{0} for toluene and dipole moment(μ) of toluene = m. Thus the dipole moment of toluene is 0.4 D.