In my organic chemistry lab, we ran a TLC plate on silica to determine the identity of an unknown compound. The two possibilities were benzoic acid and acetanilide. I know that the general rule is, the more polar the compound is, the more it absorbs onto the silica and slower and lower it would travel (corresponding to a lower $\mathrm{R_f}$ value). However, one question on my lab report asks:

In practice, benzoic acid tends to have larger $\mathrm{R_f}$ values than acetanilide. Attempt to find an explanation why this is. And note that spots from more concentrated benzoic acid leas to larger $\mathrm{R_f}$ values.

My professor said the answer has something to do with the $\ce{-OH}$ group on benzoic acid.


1 Answer 1


The answer is the dimer formation of the benzoic acid. This is why higher concentrations lead to more extended dimer formation, thus higher $\mathrm{R_f}$ values as the carboxyl group "gets shielded."

One thing of note is that $\mathrm{R_f}$ values are not directly dependent on the polarity. They also depend for example on the size of the molecule (Oswald radius). And the relation between properties of the molecules (as stated for example in the form of molecular descriptors like $\log P$) and the resulting $\mathrm{R_f}$ is very complicated indeed. This is also illustrated by the fact that there are predictors for IR spectra, NMR spectra, solubility and lots of molecular descriptors but I have never heard of a good $\mathrm{R_f}$ value predictor.

This is why I would always be careful to talk about it like "the polarity" of the compound, especially as "the polarity" is not a clearly defined quantity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.