In terms of structural basis, alkanes are insoluble or immiscible in water because of such presence of C–H...O interactions that are in regime of van der Waals interaction, which tends to be regarded as dispersion contact rather than as hydrogen bond. Benzenes and other planar molecules can make stronger O–H...π interactions that render small solubility of the planar molecules.

Then, I found these statements from a lecture note (sadly I can not find the proper sources in the lecture notes and I can not find any good result with varying keywords in Google).

There are several steps of solubilization of a molecule:

  1. Solutes must overcome solute:solute interactions in order to separate and hydrate solute molecules from one another.
  2. Solvents must overcome solvent:solvent interactions in order to create a cavity within which a solute molecule can fit.
  3. Both solutes and solvents must overcome solute:solvent interactions in order to reorganize its ice formation in the solvent shell.

From those above statements, it can be clearly seen that solubility of molecules is not suffice to be explained solely by looking only at single thermodynamic state of solute:solvent interaction.

Here are my questions:

  1. To be honest, I have not found any paper mentioning crystal structure showing water orientation among larger planar molecules. Based on my intuition from above, the water molecules around planar molecules should orient to the planar surfaces because of presence of O–H...π interactions in hydrogen bonding regime. On similar reasonings, the water molecules around cyclic 3D (cyclopropanes, bicyclopropanes, etc.) should orient away from these cyclic molecules because of presence of C–H...O interactions in van der Waals regime. Are my reasonings correct?

  2. If such solute:solvent interactions could make an ice network (or a cluster of water), does it mean thermodynamically solute:solvent interactions are trapped by themselves and can not proceed to make further solute:solute and solvent:solvent interactions? Can this particular solute:solvent interaction alone rationalize physical observation of very small solubility of larger planar molecules?

  3. I find that this is a bit paradox for this case. Some drug discovery campaigns begin to design new chemotypes that have high sp3 characters because planar sp2 molecules have poor solubility and metabolization, although they have to put more efforts on making molecules with quaternary stereocenters. In this view, alkanes are seen to be more soluble than planar molecules although both of them should be insoluble in water. I feel that it is too confusing to say alkanes are both soluble and insoluble at same time. How should I make such words so that I do not make this confusion while keeping both chemical bonding and physical phenomenons are agree to each other.

  • $\begingroup$ Regarding the steps you list, I am confused because at least 1 and 2 seem to be the first steps for the process of hydration, not dehydration. Step 3 on the other hand could be argued to be a step involved in dehydration. Also keep in mind that the "iceberg" hypothesis (formation of an ice-like solvation layer) is not a universally accepted description of the structure of a hydration layer. $\endgroup$
    – Buck Thorn
    Jun 19 '21 at 13:29
  • $\begingroup$ @BuckThorn The first step is the tendency for a planar molecule to make a stacking to other planar molecule (like π-π stacking), creating an aggregation that makes it is difficult to be solubilized although I am not too sure if colloidal aggregation fits into the context for the driving force of such solubilization of aggregates. I think the first step is a bit similar to sublimation of crystal lattice. $\endgroup$
    – làntèrn
    Jun 19 '21 at 13:47
  • $\begingroup$ @BuckThorn I think so for the second step, the second step is like seeing two sides of a coin – a transition state to hydrate molecules generated from the first step and at same time, water network restructuring for the third step. $\endgroup$
    – làntèrn
    Jun 19 '21 at 13:55
  • $\begingroup$ The enthalpy of dissolving in water is often approx zero or slightly negative, it is the entropy changes that cause insolubility. Water has to bridge over the non-polar molecule which greatly restricts its H bonds. $\endgroup$
    – porphyrin
    Jun 19 '21 at 14:14
  • $\begingroup$ Ok, I understand now that the third point in the top list is the last step, involving solvent rearrangement about a solute, but those three points then describe hydration (or solvation), not dehydration. It's a minor point but note that you may want to edit your post to make it clearer. $\endgroup$
    – Buck Thorn
    Jun 20 '21 at 2:38

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