Let be the dipole moment of a standard chlorobenzene molecule. In an idealized regular hexagonal pla — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Treat the $C-Cl$ bonds as vectors originating from the center of a regular hexagon. In $1,2,3$ -trichlorobenzene, the chlorine atoms are at adjacent positions. Let the central $C-Cl$ bond (position 2) align with the positive y-axis, providing a vector $\vec{v}_2 = \mu \hat{j}$. The $C-Cl$ bonds at positions 1 and 3 are separated by $60^\circ$ from position 2. Vector $\vec{v}_1$ is at an angle of $30^\circ$ above the negative x-axis (or $60^\circ$ from the y-axis), and $\vec{v}_3$ is at $60^\circ$ from the y-axis on the positive x-side. However, it is mathematically simpler to align position 2 with the x-axis ($0^\circ$). Then $\vec{v}_2 = \mu \hat{i}$. Position 1 is at $60^\circ$, so $\vec{v}_1 = \mu(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j})$. Position 3 is at $-60^\circ$, so $\vec{v}_3 = \mu(\cos(-60^\circ) \hat{i} + \sin(-60^\circ) \hat{j})$. The sum of vectors 1 and 3 is $2\mu\cos(60^\circ) \hat{i} = 2\mu(0.5) \hat{i} = \mu \hat{i}$. Adding vector 2 gives a total resultant of $\mu \hat{i} + \mu \hat{i} = 2\mu \hat{i}$. The magnitude is $2\mu$, thus $k = 2$.