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I wonder, why does benzoic acid have a higher melting point than stearic acid?

Here is the data: Benzoic acid melting point = 122.3 °C
Stearic acid melting point = 69.3 °C

I do understand that organic acids can bond with both London dispersion forces (van der Waals) and hydrogen bonds (due to the -COOH group). I have also learned that the longer carbon chains, the higher melting point due to the increase of the London dispersion forces.

However, in the case with benzoic acid vs stearic acid, I can't figure out an explanation as to why benzoic acid has a higher melting point. I suspect it is because of the resonance structure in the benzene ring in benzoic acid?

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    $\begingroup$ Consider the dimer $$\ce{Ph-C(OHO)2C-Ph}.$$ The respective dimer for stearic acid is less likely to be formed, because of entropy. The BA dimer would be nicely planar, with great intermolecular cohesive forces. $\endgroup$ – Poutnik Jan 30 at 10:50
  • $\begingroup$ @Poutnik - Thank you! Please excuse me if I am very wrong in my following guess: So I can say that benzoic acid has a higher melting point because larger contact area gives stronger intermolecular forces? $\endgroup$ – Csharpyikes Jan 30 at 10:59
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    $\begingroup$ Well the resonance within the aromatic ring can be seen as a reason in the sense that results in a regular flat structure that can orderly pack; perhaps a source of electromagnetic interaction between molecules, ecc. At the end you are comparing very different things. The "longer the molecule..." rule of thumb is OK for comparing terms of the same class of compounds that differ, eg, by a few carbon atoms. Roughly, think of a tile versus a flexible spaghetti. The first il likely to pack and interact with the neighbours tightly. Cyclohexanecarboxylic acid is a liquid, to let you finish.... $\endgroup$ – Alchimista Jan 30 at 11:03
  • $\begingroup$ @Alchimista - Thanks, but I still find it difficult to understand. Is it the shape of the molecule that makes benzoic acid more prone to arrange itself more tightly? $\endgroup$ – Csharpyikes Jan 30 at 12:32
  • $\begingroup$ @Csharpyikes more tightly than what? All this comes from you comparing benzoic acid to stearic acid. They are different enough to render the more than reasonable rule of thumb not applicable here. I should write an answer? It would be tedious for me and at the end convey the message that everything should be taken with a grain of salt. Even aromatics are not lining down on each other because flat, normally a kind of tilted arrangement called herring-bone is found. This point to interactions - more specific than generic dispersive/London/VdW forces - associate with the aromatic rings. Still $\endgroup$ – Alchimista Jan 30 at 13:05
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Puzzles are more puzzling when the pieces almost fit together, but for some unexplained reason, just don't fit. In other words, a comparison should be made between two almost (but not quite) similar things.

In this case, stearic and benzoic acids are quite different. They have some carbons (quite different, and not even the same number!), and one similar group of four atoms ($-COOH$) - not very much similarity.

How about we compare the melting points of octadecane and stearic acid - and their structures? There's a lot of similarity there, and a little dissimilarity. Octadecane melts at 28-30 C (Ref. 1) vs. stearic acid at 69.3 C (Ref. 2). Obviously the difference is caused by the added stickiness of the carboxyl groups for one another, because they form dimers in the crystal, a structural feature (Ref. 3) lacking in the paraffin.

enter image description here

That establishes that a carboxyl group raises the melting point of a paraffin chain. An x-ray structure for $C_{24}H_{50}$ (octadecane not available) looks just like the one for stearic acid, but without the carboxyls (Ref. 4).

enter image description here

Now let's compare benzoic acid with toluene. The numbers of carbons are the same, and we just convert one $CH_3$ group to a $COOH$, as in the previous comparison. Toluene stacks in the solid (Ref. 5) with energies about 3-4 kcal/mol, which isn't much, considering that toluene melts at -95 C. But benzoic acid stacks nicely (Ref. 6) and melts at 122 C.

enter image description here

Now there's a puzzle: why does a carboxyl group raise the mp of a paraffin by 40 degrees, but raises the mp of an aromatic ring by 217 degrees? Looking at the formulas, looking at the drawings of a molecule (or two) doesn't help. Looking at actual data (because those little molecules have a mind of their own, and we have a hard time thinking like they do) like x-ray crystal data, suggests that putting a little planar group ($COOH)$ onto a long chain might strengthen intermolecular bonding a little, but extending a ring's planarity and adding in some hydrogen bonding, which amplifies the extent of planarity could do some powerful intermolecular stickiness.

I guess the wiggly straight chains don't have such good contact with their neighbors as the benzoic acid dimers do. Who would have guessed, without the crystal structure data?

Will puzzles never cease? If octadecane melts at 28-30 C, and its mp is raised by turning a methyl into a carboxyl, you might wonder what a phenyl group would do. Apparently, not much. The mp of 1-phenyloctadecane is 29 C (Ref. 7).

Ref.1 https://en.wikipedia.org/wiki/Octadecane

Ref.2 https://en.wikipedia.org/wiki/Stearic_acid

Ref. 3 https://www.researchgate.net/figure/Crystal-packing-of-left-the-A-super-14-and-right-the-A-1-13-two_fig1_233992144

Ref. 4 https://www.researchgate.net/figure/Crystal-structure-of-C-24-H-50_fig1_270344720

Ref. 5 https://www.x-mol.com/paper/6024417

Ref. 6 https://www.researchgate.net/figure/Selected-intermolecular-contacts-in-the-benzoic-acid-crystal-viewing-the-dimers-from-the_fig2_51049984

Ref. 7 https://pubchem.ncbi.nlm.nih.gov/compound/1-Phenyloctadecane#section=Experimental-Properties

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  • $\begingroup$ +1 I particularly liked your first paragraph. Many budding chemistry students don't get that they are asking "Why is an apple different than an orange?. They are both fruit." $\endgroup$ – MaxW Jan 30 at 19:17
  • $\begingroup$ Thanks James. It made it all clear $\endgroup$ – Csharpyikes Jan 31 at 10:42

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