# Question d4b59

Apr 29, 2014

The dipole moment of 1,4-dihydroxybenzene is not zero because resonance locks the $\text{C-O-H}$ groups in the plane of the ring.

#### Explanation:

If 1,4-dihydroxybenzene (IUPAC name benzene-1,4-diol) had the linear structure below, we would predict a dipole moment µ = 0.

If the $\text{O}$ atom were ${\text{sp}}^{3}$ hybridized, the $\text{C-O-H}$ bond angle would be about 109°.

With free rotation about the $\text{C-O}$ bond, the opposing $\text{O-H}$ bond dipoles would still cancel each other.

The dipole moment would be zero.

It appears that the $\text{O}$ atoms are ${\text{sp}}^{2}$ hybridized for resonance with the benzene ring.

This prevents free rotation about the $\text{C-O}$ bonds and gives rise to cis and trans isomers.

The cis isomer has the $\text{O-H}$ bonds pointing to the same side of the ring.

The $\text{O-H}$ dipole moments (red) reinforce each other to give a resultant molecular dipole (blue).

The calculated dipole moment of the cis isomer is 2.97 D.

The trans isomer has the $\text{O-H}$ bonds pointing to opposite sides of the ring.

It may look as if the $\text{O-H}$ dipole moments (red) do not cancel each other.

However, if we slide them to put their origins in the centre of the ring (blue), we see that they are pointing in exactly opposite directions.

The bond dipoles cancel, so the predicted molecular dipole moment is µ = 0#.

The observed dipole moment of benzene-1,4-diol is 1.4 D.

It appears that the observed dipole moment is a weighted average of the dipole moments of the two conformations.