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What factors determine the melting point of a fatty acid?

  1. Chain length
  2. The number of methylene group
  3. The ionized state of the fatty acid
  4. Its degree of saponification
  5. Its ability to alter the entropy of water
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Are you looking for one answer of the 5, or are multiple answers possible? If it's the latter, then it's 1, 4, 5 from a quick recollection. –  MCM Feb 18 '13 at 20:13
    
i was looking for multiple - thanks! –  Abigailb55 Feb 18 '13 at 21:58
    
You'll want to wait for a full answer. All of the answers will affect the melting point to some degree, but whether it's a significant amount is another thing. –  MCM Feb 19 '13 at 0:08
    
You might look at this question and answer which was posted to SE.Biology but moved to SE.Chemistry. Logically I would expect that your question will also be moved. –  Alan Boyd Feb 19 '13 at 7:32
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1 Answer 1

(1) Chain Length

Will definitely affect melting point, as this website explains pretty well:

"Melting point principle: as the molecular weight increases, the melting point increases."

(2) Number of Methylene groups.

This is another way of describing unsaturated from saturated fats. The more saturated a fat is, the straighter it is. Methylene groups cause kinks, which disrupts the Van der Waals forces along the rest of the carbon chain.

As such, from the link above again:

"On the other hand, the introduction of one or more double bonds in the hydrocarbon chain in unsaturated fatty acids results in one or more "bends" in the molecule. The geometry of the double bond is almost always a cis configuration in natural fatty acids. These molecules do not "stack" very well. The intermolecular interactions are much weaker than saturated molecules. As a result, the melting points are much lower for unsaturated fatty acids."

(3) Ionized state of the fatty acid.

This will have a very minor affect. The fatty acid generally has an unpaired ester (-ate at the end) which can have a negative charge. However, from the link above again:

"However, in fatty acids, the non-polar hydrocarbon chain gives the molecule a non- polar character."

So even if the ester had a charge, the negative character is miniscule compared to the intermolecular forces exerted by the non-polar tail. Particularly since the charge can distribute amongst the two oxygen molecules which are conjugated, reducing the reactivity further.

(4) Degree of saponification.

I'm not super-familiar with the degree of saponification, but from a quick overview of the process I'd say this wouldn't affect melting point - counter to my comment. The process of making soap involves only the acidic portion of the fatty-acid triglycerides. That portion of the macromolecule is going to be pretty much the same regardless of the fatty acid, so it will have nearly the same reactivity regardless of its chain length and conjugation.

What will not saponify are usually waxes - which are pretty much fully-saturated hydrocarbon chains with very few (if any) acidic sites. Parafin Wax, for instance, does not saponify, and has the formula ${C_{31}H_{64}}$.

(5) Ability to alter entropy of water.

Like the degree of saponification option above, the ability of a fatty acid to alter the entropy of water correlates to the number of reactive sites throughout the molecule. As-such, unsaturated fats (those with methylene groups) are going to be slightly more reactive as pi-bonds are more reactive than sigma bonds.

So, given that this answer relies on a previous option, it's probably better to go with the previous option.

Ultimately, if I were answering the question I'd choose 1, 2 since the rest either depend on those two or are miniscule. However, beware that it's your question to answer and not mine.

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