Given two gas chromatography stationary phases the first one being pure poly(dimethylsiloxane) and the second one being 5%phenyl/95% poly(dimethylsiloxane) and two analytes: n-octane and n-octanol. Assuming equivalent column dimensions and chromatographic conditions, which analyte would be expected to elute last on both phases?Why?

The boiling point of n-octanol is 194.8 degree C and the boiling point of n-octane is 125 degrees C. In the case of gum or liquid stationary phase like polysiloxanes, three prime kinds of interactions play an essential role, that is, dipole, dispersion, and hydrogen bonding. Of all these, the prime kind of interaction for all the kinds of polysiloxanes stationary phase is dispersion.  

It is based on the concept that the solute or the analyte, which is more volatile will elute quickly from the column, that is, it is reliant on the concept of volatility, and the volatility is determined based on its boiling point. Of the analytes used, n-octane exhibit less boiling point in comparison to n-octanol. Thus, when the pure polydimethylsiloxane stationary phase is used, one can expect n-octanol to elute at the end, as it is less volatile in comparison to n-octane. Between the n-octanol hydroxyl group and the stationary phase, no intermolecular force takes place.  

On the other hand, when 5 percent phenyl 95 percent polydimethylsiloxane stationary phase is used n-octanol will elute at the end. Due to the presence of 5 percent phenyl, the stationary phase will become more polar in characteristic and will possess the tendency to combine with more polar groups, and as n-octanol is a more polar analyte than n-octane, therefore, it will retain n-octanol for a prolonged time, and thus, will get elute at the end.  

answersilicon (si); oxygen (o);