Why the hydrogen bonds are more intense than Van der Waals forces in the case where the molecules with the Van der Waals forces have a stronger dipolar moment than the molecules with the hydrogen bond?

For example, ethanol (hydrogen bond) has a boiling point of 78 ºC and ethanal 20 ºC (Van der Waals forces) but their dipole moments are respectively 1,69 D and 2,69 D.

I thought the intensity of the hydrogen bond was due to the high polarization of the bond between a small and very electronegative atom like N, O, F, but this is not the case in the previous example, as I understand.

  • $\begingroup$ So, you're asking how dipole moment works, or about boiling point? I think title doesn't reflect the main point. $\endgroup$
    – Mithoron
    Dec 4, 2020 at 2:31
  • $\begingroup$ I have reformulated the question. Hope in this way is clearer. $\endgroup$
    – fich
    Dec 4, 2020 at 3:31
  • 2
    $\begingroup$ "Dipole moment" not "momentum". It's a fair question. $\endgroup$
    – Buck Thorn
    Dec 4, 2020 at 9:09
  • $\begingroup$ You are mixing different interactions, or the other way around things are mixed and coexisting. 1) VdW forces, or whatever goes under the umbrella of dispersive forces, do not require any permanent electrical dipole over the whole molecule nor over part of it; 2) H bond involves a dipole along the X-H and another electronegative atom, dipoles along the bonds. In principle you can imagine a molecule with two such occurrences and no el dipole mom. 3) shape and size of a molecule are involved in the overall strength of local dispersive forces. This point influences also the B.P.,togheter with Mw $\endgroup$
    – Alchimista
    Dec 4, 2020 at 9:32
  • $\begingroup$ See London dispersion, Debye and Keesom forces, that together are covered by the broader term van der Waal's force, sometimes with Casimir effect. OTOH, the hydrogen bond is the true chemical bond. $\endgroup$
    – Poutnik
    Dec 4, 2020 at 10:39

3 Answers 3



In vacuum, the calculated (classical) interaction between to aligned dipoles decreases with the square of the distance. If the dipoles can't come very close to each other, the interaction will be weak. If you treat hydrogen bonds as a special case of dipole-dipole interaction, you will find that a N-H ... O=C interaction is much stronger than a C-H ... O=C interaction. For one, the N-H group has a larger dipole moment than the C-H group. For another, the distance between the donor and acceptor are larger for the weak C-H hydrogen bond.

When the dipoles of two ethanol molecules approach, they can come fairly close because the hydrogen on the hydroxyl group lacks inner electrons. For ethanal, in contrast, the carbonyl carbon of one molecule has to come close to the oxygen atom of the other. Both atoms have filled inner shells (and more electrons in the outer shell), so that approach is limited by electronic repulsion.

Covalent character

The hydrogen bond has covalent character, explaining why it is a directional interaction. This covalent character (like all covalent bonds) can not be fully explained using classical (Coulomb interaction) arguments. So in a hydrogen bond, we have an additional attractive term that is lacking in pure dipole-dipole interactions. In extreme cases, you might think of the hydrogen bond as a 4 electron-3 center bond (the 4 electrons are the lone pair of the acceptor and the bonding electrons of the covalent bond of hydrogen with the rest of the molecule; the three atoms are the hydrogen, the electronegative atom it is covalently bound to, and the acceptor atom).


Some use Van der Waals forces as an umbrella term for (permanent) dipole-dipole, dipole-induced dipole, and temporary dipole-induced dipole interactions. Others use Van der Waals forces (or dispersion forces or London interactions) as a separate and weaker kind of interaction and contrast it with the stronger dipole-dipole interaction.


When a $\ce{H}$ atom is bound to an electronegative atom like oxygen, its only electron is mostly placed between the two nuclei. Outside of this $\ce{H}$ atom, the $\ce{H}$ nucleus is nearly "naked". Speaking naively, there is no other electron to occupy the space around the nucleus towards outside. It looks as if the nucleus $\ce{H}$ is not protected any more against approaching negative charges. On the contrary it can attract external negative charges, like the outer doublets of the oxygen atom from another molecule. This attraction is purely electrostatic, and it is strong, much stronger than Van der Wals forces, which are the result of attracting and repulsing forces.

This happens in water and ethanol.

In ethanal $\ce{CH3CHO}$, the $\ce{H}$ atom is bound to an atom (carbon) which has nearly the same electronegativity. The electron of this $\ce{H}$ atom is not moved so strongly between the two atoms $\ce{C}$ and $\ce{H}$. So this electron protects the nucleus from approaching negative charges. The H-bond in ethanal is quite weak.

  • $\begingroup$ But @Maurice ethanal has a high dipole moment due to the C-O bond that is highly polariced (C and O atoms have very different electronegativity). In both molecules there are C-H weak bonds that do not do not contribute to the total moment of the molecule (almost). $\endgroup$
    – fich
    Dec 4, 2020 at 10:57
  • $\begingroup$ @fich Yes the dipole moment C=O is important. It has no effect on the possibility of H from CHO to form a H-Bond. The H atom is protected by its electron which is not "only" in the region between C and H. $\endgroup$
    – Maurice
    Dec 4, 2020 at 11:03
  • $\begingroup$ Yes, I understand that in ethanal there is no highly polarice -H bond. But how do you explain then that the intermolecular forces in ethanal are weaker than in ethanol being ethanal a more polarized molecule than ethanol (as we can see by the dipolar moments)? $\endgroup$
    – fich
    Dec 4, 2020 at 11:10
  • $\begingroup$ @fich Dipolar forces are not important with respect to pure Coulomb-type electrostatic forces. In dipole interactions, there is always an attraction AND a repulsion between charges. In electrostatic forces like in H-bonds there is only a plus-minus attraction between H nucleus and electron from next Oxygen atoms. This is much stronger than dipole - dipole interactions. $\endgroup$
    – Maurice
    Dec 4, 2020 at 13:02
  • $\begingroup$ Thanks for the responses and effort. I have selected Karsten response because I did not understand why do you treated H-bonds as if they were not polarized (they are strong dipoles in fact). The key point is distance between dipoles, as Kasrten stated. Maybe you wanted to explain this. $\endgroup$
    – fich
    Dec 4, 2020 at 22:24

Dipole moments are considered only in the absence of intermolecular hydrogen bonding.

Hydrogen bonding is much stronger than other dipole-dipole interactions (Van der Waals forces of attraction). Then energy liberated by dipole-dipole interactions pales in comparison to that liberated by hydrogen bonding. The stronger dipole moment argument is used where neither of the molecules that we are comparing has any form of hydrogen bonding (inter or intra).

Hope it helps!

  • $\begingroup$ Thanks for the response. But I want to know why happens what you explain. The energy of the molecules with hydrogen bond is not due to the electromagnetic interaction between their highly polar molecules? $\endgroup$
    – fich
    Dec 4, 2020 at 3:49
  • $\begingroup$ Dipole-dipole interactions occur between any two atoms having a partially positive and a partially negative charge. On the other hydrogen bonding is a specific subset of this where the atom having a partial positive charge is a hydrogen atom bound to a highly electronegative atom and the atom having a partial negative charge is another highly electronegative atom, which greatly increases the polarity of the bond. Greater the polarity of the bond, stronger is the bond. $\endgroup$
    – Adiboy
    Dec 4, 2020 at 4:20
  • $\begingroup$ This is the main reason why hydrogen bonding is mostly seen in molecules that contain hydrogens bonded to N, O or F and N, O or F. The high polarity means that hydrogen bonds are very strong (at more than 10% strength of a typical covalent bond length) and hence we use the terms 'bonds' and not 'interactions' for them. $\endgroup$
    – Adiboy
    Dec 4, 2020 at 4:24
  • $\begingroup$ @Aldiboy But as you say, the high polarity implies a very strong bond in the hydrogen bond, but in the example I wrote, the polarity is greater in ethanal than ethanol and only the last has hydrogen bond and therfore greater boiling point. This is the point I do not get. $\endgroup$
    – fich
    Dec 4, 2020 at 4:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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