[Kyriee]

Hi, I'm a chemist sophomore who recently took OChem1.

Given these four molecules ( https://imgur.com/ez7yDTD ), my professor asked me to guess how much each one of them would run on a TLC. I said that, because of the -COOH group they are pretty polar, even tho they have a big non-polar group. I also said that (b) and (d) would run less because of their relatively higher polarity due to the fluorine atoms, but he only said "no".


I passed OChem, but apparently I haven't completely understood the concept of polarity.

Thank you all!

[rolnor]

I agree with you, they are pretty polar. Fluorine bound to carbon is not polar. I would say that the stronger the acid, the more polar molecule. Is aliphatic acid stronger than aromatic, or?

[Kyriee]

I would say that the aromatic ring helps stabilizing the conjugated base of the acid, while the aliphatic is less acidic.
But in a TLC we don't have a water solution, so we can't predict if that effect will contribute to the acidity of the molecule.
Also, while the aromatic ring can stabilize the conjugated base, it also make it less polar (overall, looking at the resonance forms) and so less likely to bond with the stationary phase of the TLC.
Am I somehow right? Because I would like to know the answer to his question.

[rolnor]

I am not sure actually wich one is more polar in TLC, it would be interesting to see experimental. The fluorine atoms will probably not affect the RF too much but again, I am not sure. The difference in pKa between benzoic (4,2) and acetic (4,6) acid is not so big.
The fluorine will make the acid stronger but maybe not so much. I think the fluoro acids in your example can be difficult to find commersially so perhaps we will not get eperimental data wich is a pity. It could be that the acids will be tailing on TLC and the effect of fluorines and the aromatic or aliphatic part of the molecules will have little effect on RF value. This depends in wich mobile phase is choosen. Maybe its neccesary to use HPLC to se the difference.

[hollytara]

the acidity or pKa can make a difference in TLC / HPLC - but it depends a lot on mobile phase also (and if you are doing your HPLC with reversed phase columns....) 

The set of compounds mixes three factors: acidity (pKa), aromatic vs aliphatic, and F vs H.  The problem is that they are not entirely independent.

If you asked me benzene vs cyclohexane (using standard hexane/ethyl acetate solvent system), I would think benzene is more polar (and polarizable) and runs slower.  So that is aromatic vs. aliphatic. 

If you asked me m-difluorobenzene vs benzene or 1,3-difluorocyclohexane vs cyclohexane, the fluorination can actually make the compounds less polar - I would predict the fluoro compounds run faster.  (This is seen the most with perfluorinated compounds - which often have unexpectedly rapid passage through chromatographic systems.)

If it was just the acids, the pKa of cyclohexanecarboxylic is 4.9, benzoic is 4.2, 2,6 difluorobenzoic is 2.85 so the isomer drawn here is somehwere around 2.85-3.0  The 1,3-difluorocyclohexanecarboxylic acid maybe 3.8-4.0? The electronegativity of the F's is transmitted differently through the aliphatic and aromatic skeletons.  Based on acidity alone I would predict cyclohexanecarboxylic is fastest, then benzoic, then difluorocyclohexane, then difluorobenzoic. 

So the difficulty is this - fluorination by itself can mae Rf larger, but the interaction with the acidity can make Rf smaller.  Which factor is more important?  Hard to say.  If you move to different mobile phases it can make prediction even more difficult.

The technique where it is pretty simple is actually capillary electrophoresis - these are all about the same size so pKa would be about the only factor!