# Why the dissolution of hydrophobic compounds decreases the entropy of water molecules in the vicinity of the solute?

The following is a quote from Lehninger's Principles of Biochemistry, 4th edition, pg.52:

(...) dissolving hydrophobic compounds in water produces a measurable decrease in entropy. Water molecules in the immediate vicinity of a nonpolar solute are constrained in their possible orientations as they form a highly ordered cagelike shell around each solute molecule.

Why is this so? My intuition tells me that dissolving polar molecules (hydrophilic) would cause a higher ordering of water molecules in the vicinity of the solute than the ordering caused by dissolving nonpolar compounds. This is because I think that the polar solutes should align with the dipoles of the water molecules, thus constraining the alignment of water molecules in the vicinity of the hydrophilic solute.

Is there a simple explanation for the opposite behavior?

Edit: I want to understand, at least intuitively, the physical mechanism behind the entropy decrease of the water molecules around an hydrophobic molecule.

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• Related (if not a duplicate) chemistry.stackexchange.com/questions/7034/… – Satwik Pasani Dec 2 '13 at 16:49
• @SatwikPasani I read it. Thanks for pointing it out, but it is not a duplicate. Actually, if you read the answer given there, it seems to support my intuition that hydrophilic compounds attach to water molecules, reducing their freedom and thus decreasing entropy. However, my question is why the decrease in entropy associated with the water molecules in the vicinity of an hydrophobic compound is greater. – becko Dec 2 '13 at 16:59
• Agreed. Not a duplicate. Just related:) – Satwik Pasani Dec 2 '13 at 17:21
• The concepts of kosmotropy and chaotropy might be useful for an explanation, but I have no answer myself. Here's another link with more in-depth content. – Nicolau Saker Neto Dec 2 '13 at 22:33
• I want to understand the physical mechanism behind the entropy increase of dissolving a nonpolar compound in water. I think that's more suited to Physics.SE. – becko Dec 3 '13 at 14:44

• @becko Unfortunately I cannot provide a quantitative analysis, but I imagine that at some point it must happen, even if it takes something like $\ce{Hf\ ^{4+}}$. For ions with sufficiently high charge/radius ratio, the water molecules in the first solvation shell have such a constrained geometry that they become ligands in a coordination compound ($\ce{Hf(H2O)^{4+}_6}$), which can be thought of as the ultimate spatial constraint (and hence maximum entropy decrease) relative to a molecule of water in the bulk liquid. – Nicolau Saker Neto Dec 3 '13 at 23:36