# How does a mostly non-polar molecule prevent a polar head from being pulled off by other polar molecules?

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.

• The same way the Sun is powerless to rip our Moon away. The Moon stays close by because the strength of Earth's gravity is great enough at this close distance to ward off the Sun. Similarly, covalent bonds at close range are typically strong enough to prevent other molecules ripping them apart unless these also come into close contact. – Oscar Lanzi Aug 2 at 17:57

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.