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In 3F.19 b), it says that H bonding in water causes the molecules to be held together more tightly than in diethyl ether. But if there are two lone pairs on the central O atom, then doesn't diethyl ether have H bonding too? What makes the H bonding in water stronger than the H bonding in diethyl ether?
Hello Alexa Mugol 3I! To answer your question, diethyl ether is not capable of hydrogen bonding. When you draw the lewis structure of diethyl ether, the Oxygen atom is in the center of the diagram and is not connected to any hydrogens. If the oxygen were covalently bonded to a hydrogen, then diethyl ether would exhibit hydrogen bonding. However, this is not the case. The strongest intermolecular force that diethyl ether can have is dipole-dipole interactions, which are weaker than the hydrogen bonding exhibited by water. Because diethyl ether has weaker intermolecular forces, it requires less energy to break the attraction between two diethyl ether molecules. Therefore, it has a higher vapor pressure than water because more of the diethyl ether molecules can be broken off and enter the gas state. I hope this helps!
sarahsalama1G wrote:doesn't the strength of london disperson forces take priority over dipole-dipole forces when determining which compound has a higher boiling point?
Not really. You have to first compare IMF strength before comparing different London dispersion force factors like size or number of electrons. In this case, diethyl ether has no actual hydrogen bonds while H2O has hydrogen bonds. We can assume that we aren't considering potential hydrogen bonding sites in this problem. Diethyl ether is nonpolar and participates in London dispersion interaction while H2O is polar and participates in hydrogen bonding. Therefore H2O has a higher boiling point.
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