A question on a past exam for a course I'm studying for asks:

What's the relation between Van der Waals forces and hydrophobic interactions?

From what I understand, Van der Waals forces are just a name to summarize a bunch of little weak forces that only come into play when molecules are placed close to one other, based off this previous answer to my question.

I understand that hydrophobic interactions arise because non-polar components of a molecule are unable to form bonds with hydrogen in water, but disrupt the hydrogen, forcing them to make a cage-like structure (although I'm not clear on why), based off this entry in the Chemistry wiki.

Is the relation between these two forces that Van der Waals forces can be encouraged by hydrophic forces? Is there some other relation I'm missing here? Does it have something to do with the interactions denaturing at high temperatures?


Van der Waals forces are the attractions/repulsions - the forces - between molecules or atoms, other than attractions like ionic attractions, and covalent attractions. These forces are:

  1. Keesom Effect - This is an effect caused by two polar atoms interacting with each other. Two permanent dipoles are involved, meaning the molecules/atoms involved are polar. This may be attractive or repulsive, depending on the dipoles involved.
  2. Deybe force - This is an effect caused between a molecule with is polar, and one that is not. Because the one that is polar affects the electrons on the second (non-polar) molecule, it creates an attractive force between the two molecules.
  3. London dispersal effect - This is a force that acts between two non-polar molecules/atoms. Because the electrons around each molecule/atom repel each other, it creates a redistribution of charge, inducing an instantaneous dipole moment. This is the force that acts in liquid noble gases, to make them liquid, or between methane molecules in liquid form.

Hydrophobic forces are caused because molecules like ethane and other hydrocarbons - petrol etc. and oils like sunflower oil etc. are non-polar, and therefore don't "like" being dissolved in a polar solvent like water. In water, there are hydrogen bonds between the individual molecules, which is why water is a liquid - the van der Waals forces are far too small in water to bind the water as a liquid at room temperature. If we introduce non-polar substances into the water - polar solvent - it upsets these hydrogen bonds, and creates an increase in enthalpy because of this. Therefore the lowest energy state is for the hydrophobic hydrocarbons to separate themselves from the water.

The relationship between van der waals forces and hydrophobic interactions is that the van der waals act to bind the hydrophobe - non-polar substance - together, to separate from the polar solvent/water, and these contribute to the energy needed to separate the two substances. The seperation causes a decrease in the entropy of the system. To counter this decrease in entropy, there must be some decrease in enthalpy. Because the hydrophobe disrupts the hydrogen bonding in water, when they separate, the hydrogen bonding then causes an decrease in enthalpy, because of the favorable interactions. This is also where the van der Waals forces come in. They are favourable interactions, so cause a decrease in enthalpy, and this helps the separation of the hydrophobe and the water into two separate phases, because it makes the separation more energetically favourable. They are not necessary, but make separation more favourable, because the enthalpy change is greater for the separation process. This makes it more favourable.

  • 2
    $\begingroup$ This is a good answer, but the last part about van-der waals forces being required to pull the hydrophobic solvent together is incorrect. If you had particles with no attractive forces between them in a solvent that has attractive forces, as long as the enthalpy drop for phase separation is larger than the entropic penalty, they will separate. Van der waals forces help to make the enthalpy drop larger, but they aren't necessary. $\endgroup$ – thomij Jun 23 '14 at 19:02
  • $\begingroup$ There are also some more vdW forces, like pi-pi-stacking, etc, pp. $\endgroup$ – Martin - マーチン Jun 25 '14 at 8:36
  • $\begingroup$ @Martin. That's interesting. I didn't know about them! $\endgroup$ – Swedish Architect Jun 25 '14 at 14:03
  • 2
    $\begingroup$ The definition is a little bit fluffy and you are right, that the named three are the prominent ones. In a more common and/ or general sense it refers to all kinds of weak interactions. This answer has listed some more. $\endgroup$ – Martin - マーチン Jun 26 '14 at 1:34
  • $\begingroup$ Atoms cannot be described as "polar". Polar molecules, polar bonds, not polar atoms... $\endgroup$ – Tan Yong Boon Jun 14 '18 at 14:38

All sorts of dispersion force and polarization interactions are around the 5 kcal/mole of a hydrogen bond. Recent hype is the "halogen bond"


Long before that, the curious co-crystallization of $\ce{CBr4}$ and $\ce{CPh4}$ was noted. Tetra-(4-bromophenyl)methane gives the same structure.


Clathrates with water (methane, $\ce{SF6}$, $\ce{CO2}$, xenon). Benzil is interesting. Pi-stacking and lattice forces make the solid state conformation very different from gas phase, solution, or molten, Benzil xtal structure

If you are in water, hydrophobic exclusion from the hydrogen-bonded polar continuum affords a lot of energy to play with, then diddle with the medium,

kosmotropes and chaotropes

  • $\begingroup$ Let me see if I'm understanding correctly here, because my experience in chemistry is quite limited. You're saying, that like the Van der Waals force, hydrophobic interactions aren't really as a result of a single force, so it's wrong to talk about them that way. Consequently, saying that hydrophobic interactions and Van der Waals reactions are related is correct, since there's probably something in the two collections that overlaps? $\endgroup$ – Seanny123 Apr 25 '14 at 0:59
  • $\begingroup$ Perhaps it is worth noting that I am not coming from a chemistry background, but an electrical engineering background and that the context of my question is studying for a Biophotonics course? $\endgroup$ – Seanny123 Apr 25 '14 at 3:22
  • $\begingroup$ van der Waals is one class of interactions. People must publish. Low energy interactions become more minutely differentiated. imm.buct.edu.cn/mlei/wp-content/uploads/2012/08/… intenal page 36; epj.eg.net/articles/2013/12/2/images/… For biology, view ncbi.nlm.nih.gov/books/NBK21726 $\endgroup$ – Uncle Al Apr 25 '14 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.