Re: What is the effect on enthalpy during hydrophobic 'bond' formation?

Thanks for another great question Jithesh.

So-called hydrophobic "bonds" are the result of the "hydrophobic effect", which is driven by changes in entropy rather than enthalpy.

In solution, water molecules are free to rapidly exchange hydrogen bonds with their neighbors, but because they cannot make hydrogen bonds with hydrophobic molecules, those water molecules close to the surface of a hydrophobic molecule are constrained in the number of neighbors with whom they can exchange hydrogen bonds. As a result, the water molecules near the surface of hydrophobic molecules remain hydrogen bonded with their neighbors much longer than they would have in solution (because they don't have any other neighbors to exchange hydrogen bonds with), and hydrophobic molecules are said to be surrounded in an ice-like "shell" of water molecules.

The number of water molecules in these "shells" is a function of the total surface area of the hydrophobic molecules in the solution. So, when hydrophobic molecules aggregate, the total hydrophobic surface area is lower than when those molecules are separate, and the number of water molecules not involved in these "shells" increases. This is demonstrated in the figure to the right, where two large hydrophobic molecules are shown in blue, and the water molecules are shown in red. You can see the "frozen" water molecules "trapped" between the two hydrophobic molecules in figure A. In figure B, where the two hydrophobic molecules are together in an aggregate, those water molecules that were "trapped" are free to join the rest of the water molecules in solution. So, when hydrophobic molecules aggregate, the number of water molecules in ice-like "shells" is minimized and entropy in that solution is maximized (assuming that the temperature and the molecules themselves stay the same).

With this in mind, we can calculate the change in the Gibbs free energy of such a system, which is expressed as G = H-(T*S) , where G is the Gibbs free energy, H is the enthalpy, T is the temperature, and S is the entropy. A negative G indicates that a reaction will proceed, and you can see that as G decreases as S increases, with no change necessary to H. Now, the second law of thermodynamics tells us that entropy tends to increase, which explains why the hydrophobic effect results in the spontaneous formation of so-called hydrophobic bonds.

So now, I hope you can see why changes in enthalpy are not necessary for the formation of so- called hydrophobic bonds.

You can find more information about the hydrophobic effect by reading The hydrophobic effect: formation of micelles and biological membranes, second edition by Charles Tanford. (Krieger publishing company, Malabar, Florida, August 1991)