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Both are chains of the same length, except diethyl ether has an oxygen atom in the middle instead of a carbon. Diethyl ether has a slight dipole from the oxygen atom, so shouldn't the intermolecular forces be stronger than for pentane which give it a higher boiling point?

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    $\begingroup$ Curious: b.p. of propane: −42.04 °C; b.p. of dimethyl ether: −24 °C $\endgroup$ – Kenny Lau May 14 '16 at 6:09
  • $\begingroup$ Heptane: 98.7 °C; Dipropyl ether: 90 °C $\endgroup$ – Kenny Lau May 14 '16 at 6:10
  • $\begingroup$ Nonane: 151.0 °C; Dibutyl ether: 142.4 °C $\endgroup$ – Kenny Lau May 14 '16 at 6:11
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    $\begingroup$ Maybe this reduces the van der waals/ hyfrophobic interactions? Maybe the oxygen atom is interfering? $\endgroup$ – Polisetty May 27 '16 at 18:09
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    $\begingroup$ The disruption of hydrophobic interactions has probably a greater effect than the establishment of polar interactions. $\endgroup$ – Marko Aug 17 '16 at 22:17
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This is perhaps not a very scientific answer, but consider how branched alkanes have a lower boiling point than straight chain alkanes. This is due to the inability of the individual molecules to "stack" (if you want to get technical it's to do with surface area). Now imagine stacking diethyl ether molecules. The problem arises that two oxygen molecules (2 lone pairs each) on two separate diethyl ether molecules would not want to stack directly on top of eachother as the oxygen atoms have decently high electron densities. The repulsive force would likely be minimal, as demonstrated by the insignificant difference in boiling points between diethyl ether and n-pentane. However, this is only an educated guess.

To strengthen my argument, check out Diethyl sulfide, which has a much higher b.p. than both n-pentane and diethyl ether. This could be due to the larger atomic radius (and therefore larger surface area) of Sulfur in comparison to Oxygen. The charge has a larger area to be distributed over, resulting in less repulsion between "stacked" sulfur atoms.

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  • $\begingroup$ None of the answers explain why alkanes with more than four carbon atoms show the irregularity, while the smaller ones don't. $\endgroup$ – Krishnanand J Oct 22 '17 at 16:18
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Even though diethyl ether is a polar molecule but because of steric hindrance,only weak dipole-dipole interactions exist between its molecules.So it have lower boiling point than pentane

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    $\begingroup$ Perhaps, I'm missing something here, but how being more polar correlates with a lower b.p.? $\endgroup$ – ChemistryHelpCenter Aug 17 '16 at 18:21
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First of all this effect is very small. It is on the same order as the b.p. difference of H2O and D2O.

The effect is due to slightly smaller surface area of ether relative to pentane.

Oxygen atom in ether has 2 lone pairs. Lone pairs occupy more volume than hydrogen atoms. As a result the C-O-C angle in ether (110$^o$) is smaller than C-C-C angle in pentane (112$^o$-113$^o$). As a result pentane molecules are more stretched out, have more surface area to interact with neighbors.

For a similar reason neo-pentane is a gas while other pentane isomers are liquids. More "ball-like" structure, less surface area, lower boiling point. For the same reason SF6 and WF6 (bp = 17$^o$C) are two other gases worth mentioning.

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