In my self-study, I recently came across the following question:

"Choose the solute of each pair that would be more soluble in hexane ($\ce{C6H14}$). Explain your answer.

(a) $\ce{CH3(CH2)10OH}$ or $\ce{CH3(CH2)2OH}$ ..."

Undecanol is more soluble in hexane because it is apparently less polar than propanol. Further Internet searches revealed that alcohols decrease in polarity as the chain length increases (assuming a basis in an alkane; don't know if this rule generalizes which I guess is a sub-question) but it is still not clear why. None of the sources I found explained this. The way I imagine things, the carbon-hydrogen bonds should add a bunch of zero vectors to the the O–H vector that both molecules share, giving the same polarity in both cases. But this is apparently wrong.

So why is undecanol less polar than propanol then? Does it have something to do with a more advanced bonding theory?


2 Answers 2


If you're looking at the overall polarity of the solvent, you're not only consider the value of the individual dipole moment on a single molecule, but also their “density”, i.e. the number of these dipoles per unit of volume. You are right that the dipole moment of an individual linear-chain alcohol will, in first approximation, be independent of the chain length. However, in a given volume, you'll fit fewer of those long molecules, and the polarity of the overall solvent is thus lower.


In the context of a single molecule, the distance between the poles is inversely proportional to the polarity. The farther apart they are, the less polar the molecule is if they have the same electronegative potentials. This Link explains it well.


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