Chemical Bonding Quick Quiz 3

The dipole moment of the dichlorobenzene is proportional to the $\cos \frac{\theta}{2}$ , where $\theta$ is the angle between the two $\ce{Cl}$ bonds. Therefore, the smaller the $\theta$ , the larger the dipole moments. The order is p- < m- < o-dichlorobenzene. Since the angle for p-dichlorobenzene $\theta_p = 180^\circ$ and $\cos 90^\circ =0$ , the dipole moment of p-dichlorobenzene is $0$ . That of o- is $\frac{\cos 30^\circ}{\cos 60^\circ} = \sqrt{3} = 1.732$ that of m-dichlorobenzene.

Due to the partial charge of $\ce{H}$ atoms, the electronegativity of $\ce{CH3}$ substituent is less than that of $\ce{Cl}$ . Therefore, the dipole moment of toluene is less than that of chlorobenzene and m-dichlorobenzene.

Therefore, the order is $\boxed{4<1<2<3}$ .

Searches on the internet confirm the answer. The dipole moments of the molecules are as follows:

• p-dichlorobenzene $\mu_p = 0 \space D$
• toluene $\mu_t = 0.4 \space D$ ch 24.1
• m-dichlorobenzene $\mu_m \approx 1.55 \space D$ pg 92
• o-dichlorobenzene $\mu_o \approx 2.68 \space D$ (estimation)