[Anthasci]

Hey,

I wasn't quite sure where to put this thread, feel free to move it to a more appropriate section if I got it wrong. So, I have 4 compounds and have to predict what would happen with different mobile phases on SiO₂ stationary phase (TLC).



1. Mobile phase consists of CH₂Cl₂:CH₃OH:H₂O (17:2:1). I think this is more or less a no brainer, compound 1 travels the shortest distance and Rfs increase in the order of 1 4, since MP is relatively non-polar.

2. But now instead of water, we have HCOOH (1 part). I'd reckon this means all of the basic centres in the compounds above get protonated and travel a shorter distance, proportional to the quantity of protonated N-atoms in the amide bond. Is it really this simple? I'd say that compound 1 is, relatively speaking, the least polar of the four after protonation. But, since the resonance structure of an amide bond clearly shows a partial positive charge on the nitrogen, does this perhaps mean it is hardly basic at all? But then, the TLC would look exactly the same or am I missing something here?
The task basically requires me to draw TLCs 1) and 2) parallel to one another for comparison of Rf.

Thanks for any and all clarification!

[Babcock_Hall]

Do you know the typical pK_a values for an amide and for formic acid?

[Anthasci]

A quick internet search reveals that pKa of formic acid is about 4 units bigger than that of amides. So that means the -NH moiety in the amide bond gets ... deprotonated?

[Babcock_Hall]

No, it means that formic acid is too weak an acid to protonate an amide bond to any significant degree.

[Anthasci]

So, the chromatogram is more or less the same?

[Babcock_Hall]

The presence of acetic acid makes a given mobile phase more polar, and I suspect that the same is true of formic acid.