Question

25-17. (a) Nonpolar aromatic compounds were separated by HPLC on an octadecyl (CD bonded phase. The eluent was 65 vol% meth- anol in water. How would the retention times be affected if 90% methanol were used instead? (b) Octanoic acid and 1-aminooctane were passed through the same column described in (a), using an eluent of 20% metha- nol/80% buffer (pH 3.0). State which compound is expected to be eluted first and why Octanoic acid CH CH CH CH,CH CH,CH CH2NH2 c)Polar solutes were separated by hydrophilic interaction chroma- tography (HILIC) with a strongly polar bonded phase. How would retention times be affected if eluent were changed from 80 vol% to 90 vol% acetonitrile in water? (d) Polar solutes were separated by normal-phase chromatography on bare silica using methyl 1-butyl ether and 2-propanol solvent. How would retention times he affected if eluent were changed from 40 vol% to 60 vol% 2-propanol? (Hin: See Table 25-4.)

Answer 1

(a) Methanol is less polar thn water.

If we were to replace the eluting solvent from 65% MeOH in water to 90% MeOH in water, we are reducing the polarity of the eluting solvent, thus the non-polar components woudl stay n the non-polar stationary phase for longer periods. The retention times gets longer.

(b) octanoic acid is more polar than aminooctane.

With 20% MeOH in water, the solvent is more polar and would elute octanoic acid first from the column at pH 3 as more polar solvent dissolves more polar component preferentially.

(c) acetonitrile is less polar than water so increasing the solvent from 80% to 90% acetonitrile in water as eluent, would make the solvent less plar and thus non-polar compounds would slute first from the column.

Retention time woud increase.

(d) Propanol is more polar than ether.

Changing the solvent from 40% to 60% propanol would increase polarity of the eluting solvent. So less polar compounds would elute faster in such a solvent while decreasing the reatention time for polar components.