Why does hydroquinone possess a non-zero dipole moment? The $\ce{OH}$ groups present at para positions on the benzene ring should cancel the effect of each other... if there is a plane change then exactly when do atoms in a molecule change planes?


According to this paper:

Planar configurations for resorcinol and hydroquinone are more acceptable on comparison of experiment and calculation than are the structures assuming either free or partially restricted rotation of the hydroxyl groups.

It states that the dipole is present because there is restricted rotation with the hydroxyl groups. The preferred conformation for hydroquinone is a flat molecule with the hydrogens in the plane.

  • 1
    $\begingroup$ there are two possible planar configurations for hidroquinone, one having zero dipole moment. Why one with non-zero dipole moment dominates? $\endgroup$ – permeakra Dec 1 '14 at 18:04
  • 2
    $\begingroup$ @permeakra It doesn't. The measured dipole is 1.4 D and the paper reports 2.97 D for the second conformation. So, I get a concentration of about 47.1% for the second conformation. $\endgroup$ – LDC3 Dec 2 '14 at 1:29
  • $\begingroup$ I don't understand why in table II, of the linked paper, the calculated value of dipole moments for all three molecules is equal. What kind of calculation would result into that? $\endgroup$ – Aumkaar Pranav Mar 2 '20 at 11:08
  • $\begingroup$ Why can't I access the link to the paper. $\endgroup$ – Desai Jan 28 at 7:54

If you look at the two possible planar structures of hydroquinone, one will have both hydroxyl hydrogens on the same side of the molecule (with a nonzero net dipole moment since the two OH dipoles are additive as vectors), whereas the other will have them on opposite sides (and a zero dipole moment since the OH dipoles will cancel each other out).


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