Linear-shaped molecules of a solute with a large non-polar section and a small polar head are not soluble in a polar solvent. Why is this? Why don't the attractive forces between a molecule of polar solvent and the small polar head of the solute molecule cause the polar head of the solute molecule to dissociate?

The typical explanation I find on google is that the non-polar section "outweighs" or "is greater than" the polar head, but I'm a little confused as to what that means.


During the disolution of a compound C in a solvent S, intermolecular forces between molecules of C and between molecules of S are broken (which requires energy), and intermolecular forces between each molecule of C and several molecules of S around it are formed (which releases energy). As such, dissolution of C into S is only favorable if the released energy is greater than the required energy.

Let's consider octanol $\ce{CH3(CH2)7OH}$ in water. We need to break the hydrogen bonds between octanol molecules and the London dispersion forces in them to separate the molecules, and we need to break the hydrogen bond network in water to accommodate each octanol molecule. Then, we surround each octanol molecule by water molecules. Only the polar head ($\ce{OH}$) will be able to form new hydrogen bonds (and account for the broken ones), but the interaction between water and the hydrophobic tail is merely permanent dipole-induced dipole, which are weak compared to all the hydrogen bonds we had to break between the water molecules to be able to accomodate the large hydrophobic portion, so we're losing energy rather than gaining it. Thus, this will be poorly soluble in water.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.