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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    $\begingroup$ Related (if not a duplicate) chemistry.stackexchange.com/questions/7034/… $\endgroup$ – stochastic13 Dec 2 '13 at 16:49
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    $\begingroup$ @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. $\endgroup$ – becko Dec 2 '13 at 16:59
  • $\begingroup$ Agreed. Not a duplicate. Just related:) $\endgroup$ – stochastic13 Dec 2 '13 at 17:21
  • $\begingroup$ 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. $\endgroup$ – Nicolau Saker Neto Dec 2 '13 at 22:33
  • $\begingroup$ 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. $\endgroup$ – becko Dec 3 '13 at 14:44

I can't say I give this answer with great confidence, and I'll have to resort to some hand-waving. Think of the hydration shell around a hydrophobic substance. To minimize the local free energy, the water molecules will avoid any interaction with the hydrophobe and will seek to maximize attachment to the neighbouring water molecules, creating a sort of hydrogen-bonded encapsulating "net" where all molecules avoid pointing inwards to the hydrophobe too much. This limits the number of low energy positions available for the water molecules, so they have constrained geometry relative to the bulk of the liquid, and therefore lower entropy.

Edit: Now I realize I practically just reworded the paragraph that was already quoted, so let me add a little more. Though it seems that hydrophobic entities in general will decrease the local entropy, the effect of adding hydrophilic ones can produce an effect either way. Very polar molecules with strong dipoles and small ions with high charge cause the water molecules to align in a very specific way to maximize attractive forces, so indeed they also lower the local entropy by forcing the hydration shell to assume a specific configuration. However, less polar molecules and larger ions with low net charge don't require the hydration shell to assume a very specific geometry, and indeed the requirements may even be more relaxed compared to the hydrogen bonding network in liquid water. In that case, the hydrophilic substances could conceivably increase entropy.

Maybe the source of your confusion is that you expected the entropic effect of solvating hydrophobes and hydrophiles would be opposite, but in fact hydrophiles can have either the same or the opposite effect.

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  • $\begingroup$ So you're saying that some very strongly polar hydrophiles can decrease the entropy of water molecules in their vicinity even more than hydrophobic compounds? That's more in accordance with my intuition... $\endgroup$ – becko Dec 3 '13 at 23:10
  • $\begingroup$ @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. $\endgroup$ – Nicolau Saker Neto Dec 3 '13 at 23:36

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