*Dipole moment of benzene and its derivatives*

*Dipole moment of benzene and its derivatives*

Zero dipole moment of the benzene suggested that it has regular hexagonal planer structure and it confirms Kekule’s form.

Regular Planer Hexagonal Structure of Benzene having Centre of Symmetry |

However, if hydrogen atom is substituted by another atom or group, it acquires polar character.

Examples of such derivative of benzene are

**C₆H₅Cl**,**C₆H₅NO₂**,**C₆H₅OH,**etc.When di-substituted benzene are considered it can be shown that o-isomer will have the highest value of dipole moment then other two isomers. p-derivative has the lowest value while m-derivative has the value between the two.

The values can be calculated using the vector addition principle

**μ**= m

_{1}

^{2}+ m

_{2}

^{2}+ 2 m

_{1}m

_{2}cosθ

provided the group must also lie on the same plane of the benzene ring.

__Di-substituted derivative of benzene:__

__Di-substituted derivative of benzene:__

Di-substituted derivative of benzene |

μ² = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθHere, m _{1} = m_{2} = m and θ = 60°Thus, μ² = 2 m²(1 + cos60°) = 2 m² (1 + 1/2) ∴ μ= √3 m |

μ² = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθHere, m _{1} = m₂ = m and θ = 120°Thus, μ² = 2 m²(1 + cos120°) = 2 m² (1 – 1/2) ∴ μ = m |

μ² = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m₂ cosθHere, m _{1} = m_{2}= m and θ = 180°Thus, μ² = 2 m²(1 + cos180°) = 2 m² (1 – 1) ∴ μ = 0 |

Thus

**μ of o-isomer〉****μ of m-isomer〉**

**μ of p-isomer**

**(this has been confirmed for C₆H₄Cl₂ and C₆H₄(NO₂)₂.**

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

- Problem 1:

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

- Answer:

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

__(v) Cis and Trans forms of Geometrical Isomerism:__

__(v) Cis and Trans forms of Geometrical Isomerism:__

Presence of double bond in C – C link restricts the free rotation and geometrical isomerism develops.

Dipole Moment of Cis and Trans Isomers |

The Measurement of dipole moment helps to distinguish the two forms of geometrical isomers.

__(vi) Change of Dipole Moment with Temperature:__

__(vi) Change of Dipole Moment with Temperature:__

For molecule CH₂Cl – CH₂Cl,

__several confirmations are also changed and so dipole moment (μ) of the molecule is also changed.__- Problem 2:

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

- Answer:

For o – xylene the θ = 60° and μ = √3 m. Thus, m = 0.693/√3 = 0.4 D

Again θ = 120° for toluene and dipole moment(μ) of toluene = m.

Thus the dipole moment of toluene is 0.4 D.