Short background

As it is known there are interactions between:

  1. Charges
  2. Charge and dipole
  3. Hydrogen bonding
  4. van der Waals (VDW) forces

From stronger (1) to weaker (4).

VDW forces are divided into dipole-dipole, dipole-induced dipole, induced-induced dipole; also from stronger to weaker.

The problem

Reading about solutions I found interesting the intermolecular forces involved on aniline and dichloromethane mixing. Below both structrures are shown:

![anilineAniline ![dichloromethane Dichloromethane


What is the most important intermolecular force between those molecules and why? It might be interesting to see an AIM analysis of aniline, but I have no idea how to do that. Also, I'm not sure about what experiment could give valuable information on this case.

My attempt

Those previous interactions aren't mutually exclusive. $\ce{-NH2}$ might be a negative density region of the molecule (inductive attraction), but it is also donor by resonance, so the ring has partially a negative charge density. In consecuence, it is not easy to see the dipole. Dichloromethane, on the contrary, is an easy-to-see dipole.

It might result that London dispersion is the most relevant interaction here, but I can't very well justify it.


1 Answer 1


According to this NIST publication (caution: link leads directly to a biggish PDF file) the dipole moment in the gas phase of dichloromethane and aniline are quite similar (1.6 and 1.5 D respectively). These are respectable dipole moments, and neither molecule is very big, so offhand I would say an argument can be made that the dipole-dipole interaction is most important, particularly because it's orienting.

Note by the way that rating dispersion forces as the weakest should be done cautiously. It works for small molecules, but since dispersion forces scale with the number of electrons in a molecule and no other interaction does, once you get above a fairly modest size of molecule dispersion forces (along with excluded volume forces) dominate. That's why eicosane is a solid at room temperature while water is a liquid.

  • $\begingroup$ Very interesting, do you find it easy to see the dipole direction on aniline? I'm in trouble with that... $\endgroup$
    – user43021
    Commented Jun 14, 2018 at 3:01
  • 1
    $\begingroup$ Nope. That's why I looked it up. For what it's worth, I could rationalize it by arguing that the entire C6-N system has some pi delocalization going on because it lowers the energy of the pi electrons, to be able to spread out over all 7 atoms. But that would necessarily mean the N has a partial + and the ring a partial - because the lp is not always on the N, roughly speaking. Hence, a dipole. But that's all ex post facto hand waving. One would have to do a quantum chemistry calculation to know whether it checks out. $\endgroup$ Commented Jun 14, 2018 at 5:17
  • 1
    $\begingroup$ From a VB point of view it would suggest a contribution from the resonance structure with NH2+ on one end (with a double bond to C1) and a carbanion at the para position of the ring. My o-chem is rusty, but isn't that res str the reason NH2 is para (as well as ortho) directing for electrophilic substitution? $\endgroup$ Commented Jun 14, 2018 at 5:25
  • $\begingroup$ Cool, thanks :) What is VB? Yes you are right with substtutions, that could be seen in some calculations also, I suppose. So the dipole is possibly positive on N and negative on para... interesting. $\endgroup$
    – user43021
    Commented Jun 14, 2018 at 14:14
  • $\begingroup$ Sorry, VB = "valence bond theory, " I mean if you are trying to explain the electronic structure by some combination of Lewis structures. $\endgroup$ Commented Jun 15, 2018 at 4:09

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