Chromatography Forum

Hello,

I have a gradient method:

20% MeOH, pH 2.5 => 80% MeOH, pH 2.5 in 6 minutes
Flow: 1.2 mL/min
Column dimensions: 150 x 4.6 mm (column volume: 1.7 mL)
Retention times: 3.5 minutes for compund A and 5.3 minutes for compound B

(after this program: in 2 minutes back to 20% MeOH pH 2.5 and reequilibrate for 2 minutes; total time = 10 min)


According to JW Dolan's equation (http://www.chromatographyonline.com/lcg ... =&pageID=2), I can calculate k* assuming a value of 5 for S:

k* = (6 * 1.2)/(1.15*1.7*0.6*5) = 1.23

In isocratic separations, k' should ideally be between 2 and 10. Is the same also true for k* in gradient separations and thus, should I modify a part of my variables (e.g. using a 2.1 mm column or start at 40%)?

Additional questions:
- why do retention times are not part of the k* equation?
- if I apply the approach of S = 5: does k* characterizes my gradient system in general?
- if I want to calculate S for both of my compounds out of a log k versus % MeOH diagram (linear): how to do this?
for example: 37.5% MeOH results in k = 20, log k = 1.3 / 48.5% MeOH results in k = 2, log k = 0.3
=> delta of log k = 1, delta of % MeOH = 11% = 0.11
=> S = 1/0.11 = 9 ???

Thank you very much for your help.

Regards

Florian

According to JW Dolan's equation (http://www.chromatographyonline.com/lcg ... =&pageID=2), I can calculate k* assuming a value of 5 for S:

This may sound a bit nit-picky, but it would be better to say "estimate" rather than "calculate".

Is the same also true for k* in gradient separations

Pretty much. In our Advanced Method Development course we usually tell people to shoot for about 2 - 8. You have to remember that this is only a rule of thumb, not an iron-clad requirement (hence my earlier comment about "estimate" instead of "calculate"). The only reason for not wanting k' or k* to be too high is that it's a poor tradeoff of slightly better resolution for much longer run time.

and thus, should I modify a part of my variables (e.g. using a 2.1 mm column or start at 40%)?

Easier to just change the time. Increasing the initial %B can cause problems if you have any peaks of interest that elute early in the gradient.

Additional questions:
- why do retention times are not part of the k* equation?

Because you cannot, in fact, calculate k* from a single run (you have to know S, which means that you need information from two runs).

- if I apply the approach of S = 5: does k* characterizes my gradient system in general?

Yes, all of the peaks in your separation will have approximately the same k* if they have similar S values. Note my first comment above though: all of this is only a rule of thumb.

- if I want to calculate S for both of my compounds out of a log k versus % MeOH diagram (linear): how to do this?
for example: 37.5% MeOH results in k = 20, log k = 1.3 / 48.5% MeOH results in k = 2, log k = 0.3
=> delta of log k = 1, delta of % MeOH = 11% = 0.11
=> S = 1/0.11 = 9 ???

That's it. 9 would be a bit on the high side (albeit not outrageously high) for a small molecule, and would be on the low end of typical for a peptide.

Dear Tom,

thanks a lot for your prompt feedback.

I have two compounds with S = 2 and 9

My current situation results in a k* value below 1 for compound 2. As I cannot change column dimensions (haven't another column here) and I don't intend to increase time or flow, I see a potential in decreasing the delta % B as my current separation is "much too good".

Are my calculations/considerations correct?



Regards

Florian

Okaaaaay. The reason for the rule of thumb that k* should be >2 is that resolution is proportional to k*/(1+k*). In this case, your resolution is good enough, so that rule of thumb isn't relevant.

Dear Tom,

thank you for your reply.

I have only two components and I apply gradient only due to the fact that isocratic runs lead to a too low k for compound 1 with S = 2 when k for compound 2 (S = 9) is in a suitable range and the other way round (suitable k for compound 1, but much too late elution of compound 2). The resolution itself is therefore not a problem and I am glad that in these circumstances I can "forget" the k* rule.

Regards

Florian