Chromatography Forum

The link below (which I hope works) shows a method which I have been working on. The chromatography is very consistent. No carry over, great linearity/reproducibility/ruggedness/accuracy.

The resolution for the photo below is 2.1 using eq'n:

R = (t1-t2) / (0.5 * (tw1 + tw2)

Question 1: Although the degradant peak (peak of interest) is well resolved, it is still sitting on the shoulder of the large earlier eluting peak (although t=0 baseline is almost achieved). Are there any regulatory (ICH/FDA/etc) guidelines that discuss this? Can anyone comment on why I should/n't proceed to qualification and then validation?

Question 2: Because the size of the peaks are so drastically different, does the aforementioned resolution equation still work, or is there a more appropriate equation. I didn't really find anything in Snyder, Kirkland ed. 2 that would lead me to believe that there is a more appropriate calculation.

Finally (sorry about the length): My boss has previously asked me to demonstrate peak purity using a spectral evaluation. When the resolution is good and the peak of interest's baseline has stabilized, sepctral purity is a gimme. With this particular component, and manipulating the software (setting baseline reference spectra to a time that is milliseconds before or after peak elution) I can force the software to generate a spectrally pure peak. Is this fair?

http://pg.photos.yahoo.com/ph/mtnshawn2 ... pg&.src=ph

There are other factors that may be more appropriate and would indicate the problem. One would be the discrimination factor where you divide the peak height - the height of the valley by the total peak height. In you case, since the baseline never returns to 0, this value would be far from 1. In the ideal case of baseline resolution it would be 1 (El Fallah and Martin, Chromatographia 24 (1987), 115). You can also find a discussion of this in my book on HPLC Columns in the method development section.

From a practical point of view, the problem might come from the way the sample is made up. Is it dissolved in mobile phase?

Uwe-
Thanks for the response.

Baseline - although the sample has not returned to baseline (@ t=0, abs = 0 mAU), the baseline abs is ~ = to 1 mAU in the valley b/t the 2 peaks. The height of the degradant (small) peak is 2.9. If I understand the eq'n you are providing, the resolution would be ~ 2.9. Is this correct?

The sample is not made up in MP. The reason for this is that the API (active component) would readily degrade in the MP (99:2:0.0125, H2O:ACN:H2SO4). The sample diluent is 0.05% sodium metabisulfite in order to minimize oxidation of the API.

shawn

I did a lousy job explaining the formula.

Assume that the small peak named degradant with the area 45 has the height H(45). Draw a line for its baseline towards the larger peak with the area 1300, and measure the height of this valley H(V). Then calculate the following:

d = (H(45) - H(V))/H(45)

If there were no H(V), the value would be 1. This value is not at 1 in your chromatogram, which means you got significant overlap, and a peak purity determination will not succeed.

In addition, you have also a significant overlap between all your peaks and the API, which increases the difficulty for the peak purity determination (and this may actually be the primary reason why the peak purity determination does not work.

If you would use the same calculation as above, but using the baseline before the main peak as the reference for calculating the height, the factor d would be as little as 0.5, making a purity determination useless. I do not know how to resolve the problem if your active is not at all stable in the mobile phase, as you appear to indicate. I assume that the H2SO4 is the problem. The long tail of the API could very well be caused by degradation of the API in the mobile phase during the chromatographic run.

I am sure that you did not want to hear this...

Uwe-
Thanks for the detail on the calculation. In the end it still may not help and better resolution (more development) will need to be achieved.

I need to correct my degradation/MP statement. I don't think (but haven't completely proven) that the API is degrading while in the MP. The sodium metabisulfite is in the diluent so that as samples sit on the LC, degradation is minimized.

Thanks muchly for your thoughts.
shawn