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TL;DR version is the question title. Some context and data follow.

I was creating an assignment for my organic chemistry students in which they would need to draw as many isomers as they could from a simple formula, for example $\ce{C3H6O}$ (which fits 7 structures - 9 if you count minor enol tautomers and 11 if your count stereo). In this set (ignoring stereo and enols) there are:

Two alcohols

Two carbonyl compounds

  • Propanal - boiling point $=46\ ^\circ\text{C}$
  • Acetone - boiling point $=56\ ^\circ\text{C}$

Three ethers

The boiling point of methyl vinyl ether immediately stood out as anomalous. The fully saturated ethyl methyl ether has a boiling point of $7.4\ ^\circ\text{C}$.

Here are some more data comparing cyclic ethers to their acyclic isomers (and some analogs). The gap appears to be closing as the number of carbons increases.

Four carbons

Five carbons

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  • $\begingroup$ I've wondered myself about this. It seems to me that it 's connected with bigger accessibility of oxygen atom (sterical effects). $\endgroup$
    – Mithoron
    Commented Jun 2, 2015 at 21:19
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    $\begingroup$ Apparently this behaviour is not limited to ethers; n-hexane ($\mathrm{68.5\ °C}$), 1-hexene ($\mathrm{63\ °C}$) and cyclohexane ($\mathrm{80.7\ °C}$). Did you check any other functional groups? $\endgroup$
    – Nicolau Saker Neto
    Commented Jun 2, 2015 at 21:23
  • $\begingroup$ No, I only thought about ethers - there this effect seems to be biggest. $\endgroup$
    – Mithoron
    Commented Jun 2, 2015 at 23:40
  • $\begingroup$ Ah, so penicillin isn't the only accidental finding now... :) $\endgroup$
    – M.A.R.
    Commented Jun 3, 2015 at 12:05
  • $\begingroup$ For anyone reading this question, I highly recommended also taking your time to go through its spinoff and the answers therein. $\endgroup$
    – Nicolau Saker Neto
    Commented Jun 3, 2015 at 21:41

2 Answers 2

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The data I could find suggests that cyclic ethers have higher dipole moments than their acyclic counterparts.

$$\mathbf{Four~carbons}$$ \begin{array}{c @{} c} \hline \text{THF} & \mathrm{1.63~ D ^1, 1.75~ D ^2} \\ \text{1-butene oxide} & \mathrm{1.89~ D^2} \\ \text{ethyl vinyl ether} & \mathrm{1.26~ D ^2} \\ \text{diethyl ether} & \mathrm{1.15 ~D ^3, 1.15 D ^2} \\ \text{methyl propyl ether} & \mathrm{1.11 ~D ^2} \\\hline \end{array}

$$\mathbf{Five~carbons}$$ \begin{array}{c @{} c} \hline \text{Tetrahydropyran} & \mathrm{1.58~D ^2, 1.87~D ^4} \\ \text{2-Methyltetrahydrofuran} & \mathrm{1.38~ D ^5} \\ \text{dipropyl ether} & \mathrm{1.21~D^2, 1.00~D^6} \\\hline \end{array}

This seems consistent with intuitive expectations based on conformational models where, in the cyclic ethers where the molecular geometry is constrained by the ring, the lone pairs are pointing in the opposite direction from the carbon skeleton. Whereas the lone pairs and carbon chains in the conformationally more mobile acyclic isomers are less geometrically fixed and therefore less "directed" in space.

It seems likely that the higher dipole moments in the cyclic compounds would lead to greater alignment / ordering in the liquid phase, which in turn would lead to higher boiling points.

References:

1: http://en.wikipedia.org/wiki/Tetrahydrofuran
2: https://physicalchemistryrosamonte.wordpress.com/material-balances/material-balances-on-a-crystallizer/physical-properties-of-pure-methanol/dipole-moment/
3: http://en.wikipedia.org/wiki/Diethyl_ether
4: http://www.drugfuture.com/chemdata/tetrahydropyran.html
5: http://www.stenutz.eu/chem/solv28.php?s=1&p=20
6: https://books.google.com/books?id=G6jaBwAAQBAJ&pg=PA50&lpg=PA50&dq=propyl%20ethyl%20ether%20dipole%20moment&source=bl&ots=VA52gqJ0kn&sig=GM8tua_QpXccFW3VXpy9qiNf7rk&hl=en&sa=X&ei=rzRvVZ__Cc32yQT9gIPACQ&ved=0CDsQ6AEwBQ#v=onepage&q=propyl%20ethyl%20ether%20dipole%20moment&f=false

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    $\begingroup$ The conformationally-influenced dipole explanation seems pretty consistent with observations like the similarity in boiling point between pentane $(35.9\ ^\circ\text{C})$ and diethyl ether $(34.6\ ^\circ\text{C})$, while cyclopentane $(49.2\ ^\circ\text{C})$ has a lower boiling point than THF $(66\ ^\circ\text{C})$. $\endgroup$
    – Ben Norris
    Commented Jun 3, 2015 at 17:47
  • $\begingroup$ While this is an interesting suggestion, I think the finding may be more general than for ethers. In particular, alkanes/alkenes also seem to show a similar boiling point increase upon cyclization (for example, this link says "Cycloalkanes have boiling points which are about 10 - 20 K higher than the corresponding straight chain alkane."). What would be the reason in this case? Dipoles may not be the whole explanation. I think it may also have to do with vaporization entropy, but have not been able to find data to back it up. $\endgroup$
    – Nicolau Saker Neto
    Commented Jun 3, 2015 at 17:53
  • $\begingroup$ Another complication I see is 1,4-dioxane ($0.45\ \text{D}$, $101 \text{°C}$) and dimethoxyethane ($1.71\ \text{D}$, $85 \text{°C}$, which anti-correlate somewhat strongly. $\endgroup$
    – Nicolau Saker Neto
    Commented Jun 3, 2015 at 18:09
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    $\begingroup$ In 1,4-dioxane the dipole moment may be "artificially" lowered since the local dipoles in the two halves will somewhat cancel. Still, the local dipole moments should still facilitate ordering even though "overall" the molecule has a low dipole moment. In the case of dimethoxyethane, perhaps the conformational mobility is somewhat restricted due to the electrostatic repulsive effect the adjacent lone pairs exert on "preferred" conformations. $\endgroup$
    – ron
    Commented Jun 3, 2015 at 18:20
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    $\begingroup$ In my mind, molecules in general are more conformationally restricted in liquids due to mutual proximity. Cyclic molecules naturally have fewer conformational degrees of freedom in both phases, and so would not gain quite as much entropy upon vaporization. In an acyclic molecule, the chain would be much more free in the gas phase. This could make it entropically more favourable for acyclic molecules to evaporate, thus decreasing their boiling point relative to cyclic ones. This effect has been suggested elsewhere, with no confirmation. $\endgroup$
    – Nicolau Saker Neto
    Commented Jun 3, 2015 at 19:53
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There are many, sometime competing, some-time reinforcing, factors that affect the boiling point of these molecules:

  1. the polarity (e.g. approximated by the dipole moment)
  2. dispersive interactions
  3. increase in entropy upon evaporation

Some illustrative examples (data taken from the NIST Webbook unless otherwise noted). The boiling point of tetrahydropyran is 361 K while that of ethyl propyl ether is 337 K. The corresponding standard enthalpies of vaporization ($\Delta H^\circ_{\text{vap}}$) are 35 and 31 kJ/mol. This is consistent with the higher dipole moment of cyclic structures mentioned in @ron's answer.

Unfortunately I could not find the dipole moment of ethyl propyl ether, but it is probably similar to dipropyl ether and this is backed up by MolCalc calculations.

However, while the dipole moment of dipropyl ether (1.0-1.2 D) is lower than tetrahydropyran (1.6-1.8 D; taken from @ron's answer) the boiling point is 363 K - slightly higher than that of tetrahydropyran, as is the $\Delta H^\circ_{\text{vap}}$ (36 kJ/mol). So polarity cannot be the whole story in general. Dipropyl ether is slightly larger so larger dispersion interaction is the most likely explanation for the high boiling point here.

Finally, the boiling point of allyl ethyl ether is lower than tetrahydropyran (340 K) while the $\Delta H^\circ_{\text{vap}}$ is the same (35 kJ/mol). (The dipole is smaller than for tetrahydropyran but bigger than for ethyl propyl ether - according to MolCalc - but the double bond is more polarizable and will lead to an increase in dispersion interactions compared to tetrahydropyran). Since the enthalpy change is the same, allyl ethyl ether must have a higher $\Delta S^\circ_{\text{vap}}$. It's a more "floppy" molecule than tetrahydropyran so the gas phase vibrational and conformational entropy will be higher. (According to MolCalc the vibrational entropy of ethyl propyl ether is 54 J/molK higher than for tetrahydropyran).

So, the higher boiling point of tetrahydropyran compared to ethyl propyl ether is likely due to a combination of factors:

  1. stronger interactions between tetrahydropyran molecules due to it being more polar
  2. a lower increase in entropy upon vaporization, due to tetrahydropyran being less flexible.

The fact that allyl ethyl ether has roughly the same boiling point as ethyl propyl ether but the roughly the same $\Delta H^\circ_{\text{vap}}$ as tetrahydropyran suggest that point 2 contributed the most to the difference in boiling point between tetrahydropyran and ethyl propyl ether.

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  • $\begingroup$ Great work helping with a numerical analysis and bringing everything together! I just think it might be useful to point out that yet another factor may be simple molecular packing in the liquid, as discussed in the spinoff thread. For example, ethyl propyl ether has a density of up to $0.75\ \mathrm{g\ ml^{-1}}$, while tetrahydropyran is significantly denser at $0.88\ \mathrm{g\ ml^{-1}}$. $\endgroup$
    – Nicolau Saker Neto
    Commented Jun 5, 2015 at 11:39
  • $\begingroup$ Thanks! I am not sure how predictive the density is. For example, n-hexane has a density of 0.65 g/ml but a higher boiling point than ethyl propyl ether, while the enthalpies of vaporization is the same. $\endgroup$
    – Jan Jensen
    Commented Jun 5, 2015 at 11:55
  • $\begingroup$ Interesting pair for comparison. I wonder what the explanation in that case would be. Polarity would seem to favour a higher boiling point for the ether, while molecular mass, dispersion interactions, vaporization entropy and liquid packing seem similar. This is complicated stuff to predict... $\endgroup$
    – Nicolau Saker Neto
    Commented Jun 5, 2015 at 13:21
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    $\begingroup$ If the enthalpy of vaporization is the same, then n-hexane must have a higher entropy of vaporization. F.eks. according to MolCalc the vibrational entropy is 4 J/molK higher than ethyl propyl ether. Also, the conformations of n-hexane will be closer in energy so the conformational entropy will be higher $\endgroup$
    – Jan Jensen
    Commented Jun 5, 2015 at 13:29

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