# Dipole moment in aromatic compounds

Compare the dipole moment of the following:

• o-nitrophenol
• o-dichlorobenzene
• o-xylene

What I thought was:
o-dichlorobenzene should have the maximum, due to the acute angle and being more electronegative than the I- effect of $$\ce{CH3}$$ groups, and that o-nitrophenol would be the least since OH shows M+ and $$\ce{NO2}$$ is M- effect.

But the data from the internet says:
o-nitrophenol (3.01 D) > o-dichlorobenzene (2.67 D) > o-xylene (0.64 D)
How?

• Think of it this way: o-xylene is a virtually undisturbed aromatic ring, o-dichlorobenzene is an aromatic ring with electrons slightly pulled towards chlorine, and o-nitrophenol is an aromatic ring with electrons pulled towards NO2 with great force, and also pushed to it from another direction (should the direction be the same, maybe the effects would mostly negate each other, but this is not the case). – Ivan Neretin Jan 6 at 14:52
• @IvanNeretin any reason why those effects dont cancel or atleast reduce the net moment? – Yogeshwara Krishna Jan 6 at 15:08
• Why, of course they do reduce it. In p-nitrophenol the dipole moment would be even greater. As to why they don't cancel out: why would they? Different directions, like I said. – Ivan Neretin Jan 6 at 15:22
• @IvanNeretin Shouldn't the dipole moments in o-nitrophenol cancel out? I mean, -OH is strong donor, -NO2 is strong acceptor. Now if I try to draw vectors, they seem to slightly opposite to each other (not exactly! But at around 60 degrees from -OH dipole moment's head to -NO2 dipole moments tail.). A reasoning I found from an unreliable source suggest that, "o-nitrophenol has high dipole moment because of its ionic resonating structure". Could you please enlighten me? – McSuperbX1 Aug 19 at 16:54
• Like I said, of course they do partially cancel out. – Ivan Neretin Aug 19 at 16:55