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I know like dissolves like, but I am still unclear. 9-fluorenone has a polar $\ce{C—O}$ bond.

Is it because 9-fluorenone can only form London forces with other 9-fluorenone molecules and hexane can break these forces?

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Firstly, as a general rule, organic compounds with more then 6 carbon have higher London forces then polar forces. This is a very general rule and doesn't work in all circumstances. Thus, this molecule is not as polar.

Secondly, cyclic compounds have much weaker London forces then molecules that have straight chains. Thus, this makes the London forces not so high(relative).

Thirdly, to have something dissolve in hexane it must have weak intermolecular forces, but not weaker then hexane itself, as hexane only has to suspend the solute molecules in solution and prevent the solute from clumping up and falling to the bottom.

So, quite simply, what happens is that the hexane molecules surround the 9-Fluorenone molecules, which diffuse through the solution.

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    $\begingroup$ I am curious about your claim that “cyclic compounds have much weaker London forces then molecules that have straight chains”. I don't believe it is true for aromatic rings, but in any case, do you have a reference or explanation of that statement? $\endgroup$ – F'x Oct 5 '12 at 18:05
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    $\begingroup$ Aromatic rings have an additional attraction due to their permanent quadrupoles. See my answer to another question. I am not sure if I would classify the pi-stacking interaction as a dispersion force since it is neither induced nor transient. $\endgroup$ – Ben Norris Oct 6 '12 at 0:34
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I would argue differently from Mike. 9-fluorenone has one polar $\ce{C=O}$ bond, but it also has two aromatic rings. Dispersion interactions are strong for such aromatic rings, and the point can be made that this factor will drive the solvation of 9-fluorenone in hexane.

This is confirmed by the fact that 9-fluorenone is insoluble in water.

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