For example, diethyl ether ($\ce{C2H5OC2H5}$) has a limited solubility of $6.05\rm~\frac{g}{100~mL}$ water at $25\rm~^\circ C$. However, diethylamine ($\ce{(C2H5)2NH}$) is a lot more soluble in water. Why are ethers less soluble than amines? Both can form 2 hydrogen bonds with water (ethers accept two H-bonds, and amines donate an H-bond and accept an H-bond). Oxygen is more electronegative than nitrogen, so shouldn't ethers be more soluble?
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3$\begingroup$ Consider chemical interactions between water and each of these, not just physical interactions. $\endgroup$– SendersReagentMar 19, 2016 at 19:26
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Diethyl Ether can accept hydrogen bonds readily, and you are correct in pointing out it would be better than diethylamine at doing such. However, diethylamine can also donate a hydrogen bond, giving it two modes of solvation, possibly both simultaneously.
Also, there is some minor acid-base reactions that may occur forming diethylammonium hydroxide, which being an ion, would certainly be more soluble than diethyl ether.
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1$\begingroup$ I'm guessing having an H-bond donor is probably more than the acid-base stuff. Actually, I'm not sure. Constant equilibrium between all the amine molecules with water is probably fairly important. $\endgroup$ Mar 19, 2016 at 19:36
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$\begingroup$ Because they both can form a total of 2 H-bonds with water, would having both donating and accepting capabilities be better than just having accepting capabilities? Or is the reason for more solubility mainly that amines are bases? $\endgroup$ Mar 19, 2016 at 20:21
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$\begingroup$ I would expect the hydrogen bonding is more important, but that is a guess. $\endgroup$ Mar 20, 2016 at 3:53
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$\begingroup$ So why is having one donating and one accepting better than having two accepting? $\endgroup$ Mar 20, 2016 at 14:20
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1$\begingroup$ A molecule won't accept two hydrogen bonds. Once the electron density is donated, the molecule is somewhat electron deficient and which won't donate more. OTOH if a molecule is donating a hydrogen bond it is more able to also accept one. $\endgroup$ Mar 20, 2016 at 22:48