At first I thought that both the OH groups would cancel out their dipole moments in hydroquinone. On reading further, I realized that because of the two different planar structures possible, it has a non-zero dipole moment. My textbook says the dipole moment of para-dichlorobenzene is zero. But by the same logic we used for hydroquinone, shouldn't this have a non-zero dipole moment as well?

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    $\begingroup$ Cl isn't the same as an OH group. For one, there's an extra hydrogen atom that can rotate out of the plane. $\endgroup$ Commented Aug 29, 2017 at 8:00
  • $\begingroup$ Possible duplicate of Non-zero dipole moment of hydroquinone $\endgroup$ Commented Aug 29, 2017 at 8:00
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    $\begingroup$ Probably not a dupe although very related. $\endgroup$
    – Jan
    Commented Aug 29, 2017 at 8:20
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    $\begingroup$ OP, please take a quick look at this meta question. There is no need to apply boldfaced type that much in such a short post. $\endgroup$
    – Jan
    Commented Aug 29, 2017 at 8:32
  • $\begingroup$ There is exactly one possible conformation of p-dichlorobenzene. It is stiff as a board. The protons at the OH groups can move, largely independently. $\endgroup$
    – Karl
    Commented Aug 29, 2017 at 18:23

1 Answer 1


To facilitate this discussion, here are the two structures.

para-hydroquinone and para-dichlorobenzene

You may remember from the beginning of organic chemistry classes that free rotation can occur around single bonds. This is mostly taught for $\ce{C-C}$ bonds but true for any σ symmetric bond. On the other hand, rotation is restricted around (partial) double bonds which rely on exactly one π system such as the phenyl ring.

This means that all $\ce{C-O, O-H}$ and $\ce{C-Cl}$ bonds in the above image are freely rotatable. Rotating either $\ce{C-O}$ bond of hydroquinone will move the corresponding hydrogen out of the phenyl ring plane. (In more scientific terms: the dihedric angle will no longer be $0^\circ$ or $180^\circ$.) Therefore, planarity may be lost. Also, if both hydrogens point to the same side, inversion symmetry is lost. In both cases, a nonzero dipole moment will be observed, as highlighted in this related question.

In para-dichlorobenzene’s case, you can also rotate the $\ce{C-Cl}$ bond. However, the result is entirely philosophic in nature, as only the positions of atomic nuclei are relevant for symmetry considerations and such. However you rotate that ball at the end of a stick, the end result will be the same. There is no way in which you could reduce symmetry. Hence the molecule has zero dipole moment.


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