General Chemistry

Vitamin Solubility

Molecular Basis for Water Solubility and Fat Solubility

The solubility of organic molecules is often summarized by the phrase, "like dissolves like." This means that molecules with many polar groups are more soluble in polar solvents, and molecules with few or no polar groups (i.e., nonpolar molecules) are more soluble in nonpolar solvents. (You encountered these concepts in the "Membranes and Proteins" experiment and the related tutorial, "Maintaining the Body's Chemistry: Dialysis in the Kidneys".) Hence, vitamins are either water-soluble or fat-soluble (soluble in lipids and nonpolar compounds), depending on their molecular structures. Water-soluble vitamins have many polar groups and are hence soluble in polar solvents such as water. Fat-soluble vitamins are predominantly nonpolar and hence are soluble in nonpolar solvents such as the fatty (nonpolar) tissue of the body.

What makes polar vitamins soluble in polar solvents and nonpolar vitamins soluble in nonpolar solvents? The answer to this question lies in the types of interactions that occur between the molecules in a solution. Solubility is a complex phenomenon that depends on the change in free energy (ΔG) of the process. For a process (in this case, a vitamin dissolving in a solvent) to be spontaneous, the change in free energy must be negative (i.e., ΔG<0). The green box below describes the thermodynamic processes that govern solubility.

Thermodynamics of Dissolution (Solubilization)

The dissolution of a substance (solute) can be separated into three steps:

  1. The solute particles must separate from one another.
  2. The solvent particles must separate enough to make space for the solute molecules to come between them.
  3. The solute and solvent particles must interact to form the solution.

The free energy (G) describes both the energetics (i.e., the enthalpy H) and the randomness or probability (i.e., the entropy S) of a process ( ΔG=ΔH-TΔS, where T is the absolute temperature). The enthalpy and entropy changes that occur in the dissolution process are shown in Figure 2, below. In the dissolution process, steps 1 and 2 (listed above) require energy because interactions between the particles (solute or solvent) are being broken. Step 3 usually releases energy because solute-solvent interactions are being formed. Therefore, the change in enthalpy (ΔH) for the dissolution process (steps 1 through 3) can be either positive or negative, depending on the amount of energy released in step 3 relative to the amount of energy required in steps 1 and 2. In terms of the change in entropy (ΔS) of the dissolution process, most dissolution processes lead to a greater randomness (and therefore an increase in entropy). In fact, for a large number of dissolution reactions, the entropic effect (the change in randomness) is more important than the enthalpic effect (the change in energy) in determining the spontaneity of the process.

Enthalpy Entropy

Figure 2

The figure on the left schematically shows the enthalpy changes accompanying the three processes that must occur in order for a solution to form: (1) separation of solute molecules, (2) separation of solvent molecules, and (3) interaction of solute and solvent molecules. The overall enthalpy change, ΔHsoln, is the sum of the enthalpy changes for each step. In the example shown, ΔHsoln is slightly positive, although it can be positive or negative in other cases.

The figure on the right schematically shows the large, positive entropy change, ΔSsoln, that occurs when a solution is formed. (Although ΔSsoln is generally positive, this value could be negative in certain situations involving the dissolution of strong ions.)

In general, if the solute and solvent interactions are of similar strength (i.e., both polar or both nonpolar), then the energetics of steps 1 and 2 are similar to the energetics of step 3. Therefore, the increase in entropy determines spontaneity in the process. However, if the solute and solvent interactions are of differing strength (i.e., polar with nonpolar), then the energetics of steps 1 and 2 are much greater than the energetics of step 3. Hence, the increase in entropy that can occur is not enough to overcome the large increase in enthalpy; thus, the dissolution process is nonspontaneous.

To illustrate the importance of ΔH and ΔS in determining the spontaneity of dissolution, let us consider three possible cases:

  1. The dissolution of a polar solute in a polar solvent.

The polar solute molecules are held together by strong dipole-dipole interactions and hydrogen bonds between the polar groups. Hence, the enthalpy change to break these interactions (step 1) is large and positive (ΔH1>0). The polar solvent molecules are also held together by strong dipole-dipole interactions and hydrogen bonds, so the enthalpy change for step 2 is also large and positive (ΔH2>0). The polar groups of the solute molecules can interact favorably with the polar solvent molecules, resulting in a large, negative enthalpy change for step 3 (ΔH3<0). This negative enthalpy change is approximately as large as the sum of the positive enthalpy changes for steps 1 and 2; therefore, the overall enthalpy change (ΔH1+ΔH2+ΔH3) is small. The small enthalpy change (ΔH),together with the positive entropy change for the process (ΔS), result in a negative free energy change (ΔG=ΔH-TΔS) for the process; hence, the dissolution occurs spontaneously.

  1. The dissolution of a nonpolar solute in a polar solvent.

The nonpolar solute molecules are held together only by weak van der Waals interactions. Hence, the enthalpy change to break these interactions (step 1) is small. The polar solvent molecules are held together by strong dipole-dipole interactions and hydrogen bonds as in example (a), so the enthalpy change for step 2 is large and positive (ΔH2>0). The nonpolar solute molecules do not form strong interactions with the polar solvent molecules; therefore, the negative enthalpy change for step 3 is small and cannot compensate for the large, positive enthalpy change of step 2. Hence, the overall enthalpy change (ΔH1+ΔH2+ΔH3) is large and positive. The entropy change for the process (ΔS) is not large enough to overcome the enthalpic effect, and so the overall free energy change (ΔG=ΔH-TΔS) is positive. Therefore, the dissolution does not occur spontaneously.

  1. The dissolution of a nonpolar solute in a nonpolar solvent

The nonpolar solute molecules are held together only by weak van der Waals interactions. Hence, the enthalpy change to break these interactions (step 1) is small. The nonpolar solvent molecules are also held together only by weak van der Waals interactions, so the enthalpy change for step 2 is also small. Even though the solute and solvent particles will also not form strong interactions with each other (only van der Waals interactions, so ΔH3 is also small), there is very little energy required for steps 1 and 2 that must be overcome in step 3. Hence, the overall enthalpy change (ΔH1+ΔH2+ΔH3) is small. The small enthalpy change (ΔH), together with the positive entropy change for the process (ΔS), result in a negative free energy change (ΔG=ΔH-TΔS) for the process; hence, the dissolution occurs spontaneously.

The principles outlined in the green box above explain why the interactions between molecules favor solutions of polar vitamins in water and nonpolar vitamins in lipids. The polar vitamins, as well as the polar water molecules, have strong intermolecular forces that must be overcome in order for a solution to be formed, requiring energy. When these polar molecules interact with each other (i.e., when the polar vitamins are dissolved in water), strong interactions are formed, releasing energy. Hence, the overall enthalpy change (energetics) is small. The small enthalpy change, coupled with a significant increase in randomness (entropy change) when the solution is formed, allow this solution to form spontaneously. Nonpolar vitamins and nonpolar solvents both have weak intermolecular interactions, so the overall enthalpy change (energetics) is again small. Hence, in the case of nonpolar vitamins dissolving in nonpolar (lipid) solvents, the small enthalpy change, coupled with a significant increase in randomness (entropy change) when the solution is formed, allow this solution to form spontaneously as well. For a nonpolar vitamin to dissolve in water, or for a polar vitamin to dissolve in fat, the energy required to overcome the initial intermolecular forces (i.e., between the polar vitamin molecules or between the water molecules) is large and is not offset by the energy released when the molecules interact in solution (because there is no strong interaction between polar and nonpolar molecules). Hence, in these cases, the enthalpy change (energetics) is unfavorable to dissolution, and the magnitude of this unfavorable enthalpy change is too large to be offset by the increase in randomness of the solution. Therefore, these solutions will not form spontaneously. (There are exceptions to the principle "like dissolves like," e.g., when the entropy decreases when a solution is formed; however, these exceptions will not be discussed in this tutorial.)

In general, it is possible to predict whether a vitamin is fat-soluble or water-soluble by examining its structure to determine whether polar groups or nonpolar groups predominate. In the structure of calciferol (Vitamin D2), shown in Figure 3 below, we find an –OH group attached to a bulky arrangement of hydrocarbon rings and chains. This one polar group is not enough to compensate for the much larger nonpolar region. Therefore, calciferol is classified as a fat-soluble vitamin.

Calciferol

Figure 3

This is a 2D ChemDraw representation of the structure of calciferol, Vitamin D2. Although the molecule has one polar hydroxyl group, it is considered a nonpolar (fat-soluble) vitamin because of the predominance of the nonpolar hydrocarbon region.

 


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This page created by Matt Traverso, Washington University in St Louis.
© 2004, Washington University.
Materials and Information present may be reproduced for educational purposes only.

Revised: 2004-08-08