Chromatography Forum

I'm looking for some feedback on a way that we recently determined a rough relative response factor for an impurity. We did not have a pure standard for the impurity, but we could generate it. It was known that by mixing the API with anoher chemical we could produce the impurities we were seeing. So, in a volumetric flask the two chemicals were mixed and the flask was allowed to incubate. After the incubation, the flask was qs'd and a dilution was made to bring the mixture into the linear range of the method. The mixture was injected along with a standard. The mixture showed a 70% assay for the API and the sum of the remaining peaks totaled to 90% of the API standard. We know that by MS we should expect the same number of peaks that we saw, therfore we determined that the RRF for the impurities (all the peaks are positional isomers) is 3. Does this sound like a reasonable approach?

I think it works. Let me rephrase it to make sure I got it right:

You started with the API and a "reagent".
Only one other significant peak was generated, so you can safely assume that the API was converted only to this "impurity".
The API peak decreased by 30%
The area of that "impurity" peak was 90% of the original API peak.
You are ignoring the area of the "reagent" peak.

Assigning a value of 1.00 to the original standard API, you now have:
API = 0.7
impurity = 0.9

That 0.9 represents 30% of the original API, so your response factor is 0.9/0.3 = 3

That would assume that no gunk formed?

Tom & Uwe,
Because of the reactants, an aldehyde and a secondary amine, we are fairly sure that nothing other than our our reactants and degradants were in the flask. From a spectroscopic point of view, the response factor makes sense too (more highly conjugated product).

It will remain an assumption until you do a mass balance.

Hans