# Why is the melting point of tert-butyl alcohol 140 °C higher than that of sec-butyl alcohol?

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?

• 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. – Ben Norris Jan 9 '17 at 2:04
• 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. – Karl Jan 9 '17 at 2:13
• 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. – Karl Jan 9 '17 at 2:21
• Two things to consider are that: tert-butyl alcohol has higher symmetry that sec-butanol and also sec-butanol is chiral. – A.K. Jan 9 '17 at 2:37
• 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. – 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.)