# Polar compounds that doesn't dissolve in water? [closed]

I'm studying for my exam introduction to chemistry. In the course, there's an exercise where we have to explain why something dissolves in water and why other compounds doesn't.

For me, if the substance is polar, than it dissolves in water because H2O is polar to. (and non-polar dissolves in non-polar).

The first example is AluminumOxide. In the course, it's mentioned that this doesn't dissolve in water. They say that the forces are to high? But it's a polar compound? How do you know if a polar compound will dissolve?

## closed as too broad by Mithoron, Todd Minehardt, A.K., Jon Custer, Jannis AndreskaAug 22 '18 at 16:28

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You may accidentally have chosen an awkward example because I'm not sure that $\ce{Al2O3}$ is ever properly molecular. It is usually crystalline and may even be in the form of nano-crystals, in which case there will be considerable lattice binding energy to overcome. In its doped form, with the odd Cr$^{3+}$ ion, it is the mineral Ruby which is clearly not soluble in water.

In the general case polarity is not the only consideration for solubility. There is entropy change of mixing which is always favourable when two species are mixed and also enthalpy change when placing molecules of one type into the other. The enthalpy is where the polarity part comes in.

In water in particular there are other entropy changes to consider because water is so hydrogen bonded. The H-bonding structure is broken up when a solute is added and water has then to 'bridge the gap' as it were round the solute. This forms a more structured water arrangement than before adding the solute and so entropy is reduced. This is primarily responsible for insolubility in water.

We measure the process by considering the free energy change as $\Delta G = \Delta H - T\Delta S$ where $\Delta H$ is the change in enthalpy, $\Delta S$ that of the entropy. When the process if favourable $\Delta G$ will be negative.

Surprisingly adding, for example, non-polar cyclohexane to water has a change in enthalpy ($\Delta H$) that is close to zero, whereas for a polar molecule $\Delta H$ will be negative. The H-bonding bridging with cyclohexane will make the change in entropy negative so making the free energy positive, which is unfavourable.

With a polar molecule in water H bonding bridging is not necessary to such an extent as for a non-polar molecule and the change in entropy is smaller and so with its negative $\Delta H$ the polar molecule has a negative $\Delta G$ and can dissolve.

It depends on how polar is polar; phosphine is technically polar, but its actual dipole moment is ... C'mon man!

Such cases aside, partially chlorinated hydrocarbons such as dichloromethane are good at not dissolving in water despite a respectably large polarity. Having hydrogens attached only to carbon and chlorine as the electronegative element means dichloromethane and similar compounds can't replace the strong hydrogen bonds that they would have to break upon dissolving in water.