This is one of the most drastic differences in a physical property I've ever seen for two such similar molecules, and in a simplistic sense anyway the difference lies in the opposite direction from what one might expect. tert-butyl alcohol is completely sterically hindered from participating in H-O hydrogen bonding so it would seem to be limited to van-der-Waals interactions. sec-butyl alcohol interactions should be dominated by van-der-Waals forces, but should also have some ability to participate in hydrogen bonding. Furthermore, sec-buty alcohol has a slightly larger van-der-Waals surface.

These are such common reagents and the unusually high melting point of tert-butyl alcohol is so well known that there is a surprising dearth of information on theoretical or experimental evidence available (in my searching anyway) explaining the reason for this $\pu{140^oC}$ disparity in melting points.

Does anyone know of a concise, coherent explanation for this observation?

  • 1
    $\begingroup$ Melting point is not just about intermolecular forces. The ability of a compound to form a regular solid can play a major role as well. Symmetry is often a deciding factor in otherwise similar molecules. $\endgroup$
    – Ben Norris
    Jan 9 '17 at 2:04
  • 1
    $\begingroup$ The most important point is that all C-C and C-O bond angles are fixed with respect to each other. Only the X-H bonds can move (OH and methly groups can rotate), and that is also possible in the crystalline state. $\endgroup$
    – Karl
    Jan 9 '17 at 2:13
  • $\begingroup$ What gives you the idea that there can be no hydrogen bonding? Check the crystal structure of tBuOH and you'll find that you're wrong. $\endgroup$
    – Karl
    Jan 9 '17 at 2:21
  • 1
    $\begingroup$ Two things to consider are that: tert-butyl alcohol has higher symmetry that sec-butanol and also sec-butanol is chiral. $\endgroup$
    – A.K.
    Jan 9 '17 at 2:37
  • $\begingroup$ Big kudos to all the symmetry and bond angle responses, I think that's right on track. I have an idea along those lines that I've always thought was a bit nebulous which is why I've asked for a better explanation. And Karl, yes I should have left out the word "completely" and just said "sterically hindered from participating in H-O hydrogen bonding". It is there. Still, that alone doesn't begin to explain the nearly 2-fold difference in absolute melting point temperature relative to sec-butyl alcohol. $\endgroup$
    – airhuff
    Jan 9 '17 at 3:37

I am not aware of a solid scientific publication on this matter, but the following points should elucidate what is happening:

  • The high melting point sort of proves that there is strong hydrogen bonding in tBuOH. I also don't see how the tBu-group could be so bulky that each OH group cannot form two hydrogen bonds.
  • n-, sec- and iso-butanol all have pairs of methyl- and/or OH-groups that can be in gauche or anti position with respect to each other. The energy difference is in the range of 4 kJ/mol, which is not much higher than the thermal energy $k_BT$ at 300K. Those alcohols exists more or less as a mixture of conformational isomers, which precludes crystallisation. In tBuOH, there is only one conformation.
  • Rotation of the methyl groups has a low excitation barrier, but it can also rotate in the solid phase (there is some tunneling involved), so this does not preclude crystallisation.

(The energy of ~4kJ/mol is from n-butane, but I expect no large difference. The gauche conformation of the OH group is probably even lower, energetically.)


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.