Question

5. Liquid Chromatography. b. For a reverse-phase HPLC separation with a mobile phase pH = 3.00, circle which of the following compounds would elute first and briefly describe why: Propanoic acid (CH3CH2COOH, PK. = 4.87) or Propylamine (CH2CH2CH2NH2, pks = 3.43). b. If one uses gradient elution with normal phase chromatography, would it be better to go from a less polar to a more polar solvent or a more polar to a less polar solvent? Why?

Answer 1

Solution :

A) The propanoic acid having pka = 4.87 elutes first. In reverse phase chromatography, mobile phase is polar and stationary phase is non-polar. Propylamine being the less polar molecule interacts with non-polar stationary phase, hence its elution becomes slow. Thus, a polar molecule eluted faster than a non-polar molecule. Propanoic acid is more polar because of its strongest acidity (lower pka) than propylamine (pka = 14 - pkb = 14 - 3.43 = 10.67).

B) Propylamine elutes faster than propanoic acid.

In the normal phase chromatography, the mobile phase is non-polar and stationary phase is polar. Thus, a less polar molecule elutes faster than a more polar one. Propanoic acid being the more polar interacts with polar stationary phase, hence its elution is slower than propylamine.