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Why is fat nonpolar while water is polar?

In water there is the $\ce{H-O}$ bond and the $\ce{O}$ is more negative, we have $\text{dipole = charge}* \text{distance}$

A fat molecule is like $\ce{H-O-C-C-C-C-C-C...}$ but the carbon atoms don't matter, the polarity is still ${q*d}$, as in the water molecule, so they should have the same amount of polarity.

So why do people say fat is a nonpolar molecule?

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    – M.A.R.
    Commented Mar 30, 2015 at 11:59

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The polarity of a single bond might indeed be the same for some different molecules. When speaking of polarity of molecules however, the whole molecule should be taken into account, since the macro properties are dictated by all the parts of a molecule.

Regarding the fatty acids: The 'head' of such a molecule is indeed polar, and in specific cases this means that the interactions and reactivity will be like polar molecules. This effect however is only contained to the head of the molecule. Fatty acids as a whole are still considered non-polar because this type of molecule will not be dissolved in water. Note that the main reason for this is not due to the interactions of water with the fatty acid. The fatty acid is just as 'happy' interacting with water as it is with other fatty acid chains. Water, on the other end, does not 'want' to interact with fatty acids, since it will have to rearrange around the fatty acid molecule to ensure the same number of hydrogen bonds are made. This is an entropic cost, that makes the total Gibbs Free Energy change unfavorable.

The result is that water molecules do not rearrange the hydrogen bonds with other water molecules and the fatty acids interact with each other. Due to this effect, water not being willing to rearrange hydrogen bonds, fatty acids are considered non-polar.

Hope this has helped, and if I used to much terminology or jargon, please let me know!

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    $\begingroup$ "Water, on the other end, does not 'want' to interact with fatty acids, since it cannot form nearly as many hydrogen bonds as it can with other water molecules." - this is probably not the case; most hydrophobic compounds have favourable enthalpy of solution, but unfavourable entropy of solution due presumably to highly ordered water structure around the solvated hydrocarbon chain. $\endgroup$
    – J. LS
    Commented Mar 30, 2015 at 16:45
  • $\begingroup$ You are right in the sense that for many small non-polar entities the enthalpy is still favorable. Sorry for taking that shortcut. In this case the Entropy is the reason for the total Gibbs Free Energy to be unfavorable. Would you like to see an edit on that, stating the reason for not 'wanting' to interact is an entropic effect? Further more I would like to add that this holds for small molecules. In the literature I consulted, an example case of neopentane was discussed. I could imagine, and I do not have numbers to back this up, that for long fatty acids, this does not still hold. $\endgroup$
    – Eljee
    Commented Mar 30, 2015 at 18:16
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    $\begingroup$ The hydrophobic effect is really complex and most explanations are heavy with conjecture. For a long-chain fatty acid I would expect positive enthalpy of solution; for example from this paper: "In other words, a large positive enthalpy of mixing of a non-polar solute with water was compensated by a large negative enthalpy of iceberg formation that resulted in a higher solubility than that for a regular solution. This explanation should be applicable to the dissolution of a fatty acid in water at a temperature below 150 degrees C." $\endgroup$
    – J. LS
    Commented Mar 30, 2015 at 18:33
  • $\begingroup$ "Iceberg formation" just refers to the reorganization of the hydrogen bonding network to a more ordered state. $\endgroup$
    – J. LS
    Commented Mar 30, 2015 at 18:35
  • $\begingroup$ Okay, so the paper states the numbers I did not have (or looked for). So under 150 Celsius. All clear I would say. Thanks for looking that up. Regarding the question, I can imagine that this is a bit to deep into the subject, though it is of course best to have a proper, full explanation is stead of an incomplete one. I'll make an edit to my answer. $\endgroup$
    – Eljee
    Commented Mar 30, 2015 at 18:38
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While there is nothing wrong with @eljee's answer there is a more obvious reason why fats are not as polar as you assume.

Fats are not long chain alcohols: they are the triglyceride esters of long chain carboxylic acids. So, rather than being terminated by a relatively polar carboxylic acid or alcohol group, they are terminated by a relatively non polar glycerol ester group. Only when saponified (which breaks the free acids from the ester) do they become the more schizophrenic long chain acids which act as surfactants because one end of the molecule is polar enough to like water while the other is fatty enough to prefer oily, non-polar environments. Hence soap!

The wikipedia page summarises this well.

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