As I understand, the intermolecular forces of ethyl cyanide are of van der Waals nature (dipole-dipole) but in the propanol molecule there are hydrogen bonds which are in principle much stronger. Why then their boiling points are equal (97 ºC)? Their molar mass is very similar too.

Obviously, the triple bond of the cyanide group must be the cause, but I do not know how it can affect the boiling point, as the C-N bond is less polar than the O-H bond.

  • $\begingroup$ Van der Waals bonds usually describe a different class of (weaker) interactions than dipole-dipole interactions (ultimately induced dipole interactions between molecules that don't have strong dipolar features). Whether this is a factor here where both molecules have strong dipoles is less clear. $\endgroup$ – matt_black Dec 11 '20 at 12:30
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    $\begingroup$ @matt_black The term van der Waals force ...... is sometimes used loosely for all intermolecular forces. The term always includes the London dispersion force between instantaneously induced dipoles. It is sometimes applied to the Debye force between a permanent dipole and a corresponding induced dipole or to the Keesom force between permanent molecular dipoles. $\endgroup$ – Poutnik Dec 11 '20 at 16:29
  • $\begingroup$ It goes in the direction of being a coincidence. You can ask the same question for a huge number of pairs of substances. All you need is picking up values from tables or literature. That is because you have a number of factors (even interconnected) which contributes to the phase diagram. From molecular mass to molecular shape, from packing to polarity, etc. In the specific case, cyanide molecules are quite polar. CN is an electron withdrawing group, you should look at its effect on the whole molecule. $\endgroup$ – Alchimista Dec 12 '20 at 10:58
  • $\begingroup$ Use the term propiononitrile rather than ethyl cyanide lest someone gets the idea it is toxic in the sense of cyanide. $\endgroup$ – user55119 Dec 12 '20 at 15:49
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    $\begingroup$ And, as you are a teacher, in the case you proposed and for those purpose, it is not important how much che CN bond are polar, but how much the molecule is. $\endgroup$ – Alchimista Dec 13 '20 at 17:03

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