Timeline for Why do cyclic ethers have higher boiling points than their acyclic isomers?
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| Jun 5, 2015 at 13:29 | comment | added | Jan Jensen | 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 | |
| Jun 5, 2015 at 13:21 | comment | added | Nicolau Saker Neto | 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... | |
| Jun 5, 2015 at 11:55 | comment | added | Jan Jensen | 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. | |
| Jun 5, 2015 at 11:39 | history | edited | Jan Jensen | CC BY-SA 3.0 |
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| Jun 5, 2015 at 11:39 | comment | added | Nicolau Saker Neto | 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 $0.88\ \mathrm{g\ ml^{-1}}$. | |
| Jun 5, 2015 at 11:23 | history | answered | Jan Jensen | CC BY-SA 3.0 |