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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7Cl isn't the same as an OH group. For one, there's an extra hydrogen atom that can rotate out of the plane. – orthocresol Aug 29 '17 at 08:00
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Possible duplicate of Non-zero dipole moment of hydroquinone – orthocresol Aug 29 '17 at 08:00
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2Probably not a dupe although very related. – Jan Aug 29 '17 at 08:20
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2OP, please take a quick look at this meta question. There is no need to apply boldfaced type that much in such a short post. – Jan Aug 29 '17 at 08:32
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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. – Karl Aug 29 '17 at 18:23
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To facilitate this discussion, here are the two structures.
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.
Jan
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