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I believe that it can form 2 hydrogen bonds, one with oxygen and the other with hydrogen but a classmate said that oxygen can form 2 hydrogen bonds? Is this true? I can't find anything on google to support or refute that claim.

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    $\begingroup$ I don't know whether this can be proven or disproven. Hydrogen bonds are not as stable as covalences. They can interchange constantly. But well ! In principle, the Oxygen atom of CH3OH has two unused doublets. Each doublet can attract one H atom from foreign molecules. $\endgroup$ – Maurice Apr 10 at 10:59
  • $\begingroup$ It depends of the state of the sample, the pressure applied, and the criteria of hydrogen bonds applied (e.g., Phys. Chem. Chem. Phys., 2016, 18, 2736, doi 10.1039/c5cp06583f; researchgate.net/publication/…). Using both a distance criterion (CH-O < 2.6 A) and an angular one (> 130 degree), their model predicts an an increase of H bonds with growing higher pressure, though. $\endgroup$ – Buttonwood Apr 10 at 18:08
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    $\begingroup$ Water makes four hydrogen bonds in ice, so methanol should be able to form three when bonding partners are available. $\endgroup$ – Karsten Theis Apr 11 at 2:44
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Although this was an MD study (and might be said to be somewhat dated), quoting from Matsumoto and Gubbins (Ref. 1):

Most of the molecules are in the two-bonding state, but a small fraction of molecules are in the one- or three-bonding state, as already pointed out in previous studies. 7 ,8

According to this study most of the methanol molecules form 2 H-bonds but ~4-7% form 3.

References

  1. Matsumoto, M., Gubbins, KE. The Journal of Chemical Physics 93(3):1981-1994 (doi 10.1063/1.459076).
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    $\begingroup$ You would do good by pointing out first that yes, technically a molecule can make three H-bonds, but no, in the bulk compound you can't expect much more than 2 per molecule, for purely arithmetic reasons. $\endgroup$ – Ivan Neretin Apr 10 at 16:49
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    $\begingroup$ @IvanNeretin There is an publication by Etter on my table (Acc. Chem. Res. 1990, 23, 120-126, doi 10.1021/ar00172a005). With its illustrations about hydrogen bonding patterns, e.g. of p-terephthalic acid, trimesic acid, and in figure 1 (p. 125) about 2-aminopyrimidine and succinic acid co-crystal [ccdc.cam.ac.uk/structures/search?pid=ccdc:1257588&sid=IUCr], I speculate your comment could be improved. For clarity, I suggest replacing "you can't expect more than 2 per molecule" by the more explicit / specific "you can't expect more than 2 per molecule of methanol". $\endgroup$ – Buttonwood Apr 10 at 17:22
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    $\begingroup$ @Buttonwood Yeah, sure, but a comment can't be edited. It is the answer that is intended to be improved by the joint power of our comments. $\endgroup$ – Ivan Neretin Apr 10 at 19:50
  • $\begingroup$ @IvanNeretin By arithmetic I suppose you mean there aren't enough donors and acceptors for all molecules to have 3 H-bonds.Luckily the MD simulation was self-consistent. The average number of H-bonds per molecule in that study was below 2 (see Table II). $\endgroup$ – Buck Thorn Apr 10 at 19:54
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    $\begingroup$ @BuckThorn Exactly. $\endgroup$ – Ivan Neretin Apr 10 at 19:55
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If we consider n methanol molecules such that intermolecular H bonding exists between all( value of distance has value that allows intermolecular H bonding). One molecule has 1 O atom that bonds with n-1 H atoms. This case is for n atoms. So, number of bonds should be n(n-1). The H atom on 1molecule is counted as this bond is formed by other molecules having O.

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