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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?

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    $\begingroup$ 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, 2017 at 2:04
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    $\begingroup$ 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, 2017 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, 2017 at 2:21
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    $\begingroup$ 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, 2017 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, 2017 at 3:37

2 Answers 2

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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.)

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If you want to ponder this look up the Boiling Points and Melting Points of the isomeric pentanes and butyl alcohols; their shapes are similar; an oxygen atom with two unshared pairs substituting for a carbon with two C-H bonds. [Should you want to complicate things a bit throw in the 4-carbon ethers or ketones too.] Their Van der Waals forces are reasonably similar and the alcohols add hydrogen bonding. A careful perusal will give a reasonable idea of what is going on.

At the phase equilibrium Delta G = 0 = Delta H - [T Delta S]; T = Delta H/Delta S. This means an analysis of intermolecular forces, really heats of fusion and vaporization, and the probable entropies of melting and vaporization will give the trends in the transition temperatures.

For simple molecules linear molecules have more surface area and less conformational interactions so Van der Waals forces are greater and H-bonding is less restricted and MPs and Bps should be raised. This is moderated by a greater entropy change from solid to liquid since many rotational modes are freed up. Secondary and iso molecules should have similar entropy changes and lower Heats so should have lower MP and BP. The very symmetrical tertiary isomers have very different heats of melting because of lattice arrangements, although a lower entropy change should also contribute, and lower vaporization energies because of less surface area and possible inhibition of H-bonding for steric reasons. The entropy changes seem to be lower. Symmetrical molecules have higher MPs and lower BPs.

These are general trends that I have surmised. As molecules become more complicated the analysis must become more detailed. This is shown in liquid crystals, additional liquid phases; polymers, glass transition phase changes; and of course, proteins, enzymes and other biomolecules that have exquisite foldings and conformations.

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  • $\begingroup$ Will someone please give me the reason for the down vote. $\endgroup$
    – jimchmst
    Dec 25, 2022 at 21:14

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