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When using gradient method, the usual K doesn't apply. Gradient has it's own reention factor - the K'.
K' is calculated by
For a good separation, K' should be around 5.
However, keep in mind that the ultimate goal is resolution > 1.5
Once the method passes all validation tests, the method can be used for the investigation
After that, change the method (eg; solvent type, %B, etc) and see if the new method passes the validation test.
HPLC - flow chart
Result
-Chromatogram
-Peak table
-etc.
What is happening in the column?
Ayame Saito 25133381
Why do we do this?
Additional notes
In this flow chart, we will use "Lorenzo's problem" as an example.
Lorenzo's problem is that he might have contaminated a batch of ketoprofen tablets with other compounds
Might contain Sulindac, Piroxicam, Naproxen, Diclofenac, and Phenylbutazone
What is....
Total run time?
the time it takes for all of the sample molecules to leave the HPLC column
Resolution?
measure of separation of peaks
calculated by
Back pressure?
pressure in the column
Result - chromatogram
HPLC specifications given by Lorenzo;
Scouting gradient chromatogram
Try changing the solvent. If this still doesn't work, try changing the column type and/or to isocratic method.
CHROMacademy
Have you got a chromatogram that fulfilles the criteria? in this case;
Not at all
Gradient
Isocratic run is possible if;
Δtg < 0.25tg
What is happening in the column?
Q. What else changes K'?
CHROMacademy
A. Flow rate, tg (gradient time), and column dimensions (diameter and length)
CHROMacademy
What is happening in the column?
Nearly!
Total run time < 15minutes
Selectivity > 1
Resolution > 1.5
Back pressure < 4000psi (for HPLC)
Number of plate > 4,000 preferably aorund 10,000
Why do we do this?
To see whether the gradient method can give good chromatogram with good resolution. Run time should not be a problem since in gradient, the total run time is determined by yourself.
%B is different throughout the column.
When the molecules start to move out in to the mobile phase, they are "pushed" by the high %B behind them. However, in front of them, the %B is lower than their preference, which prevents them from moving forward.
This focuses the molecules and this is seen as sharp peaks on the chromatogram
Why do we do this?
Using gradient run sounds great, but it takes time (as you must wait for re-equilibration time) and isocratic run is much more simple.
Try changing flow rate, tg, △ϕ or even the column dimensions to get K' nearly 5.
But you should also keep in mind that we are aiming for resolution >1.5.
For explanation on how compounds separate, read additional notes on Step 2(b)
Yes
C 18 column
Scouting gradient
What is K?
Before method development....
Congratulations, start method validation
Isocratic
Pick your column
We need method development so that when we run the contaminated tablets in HPLC, we can identify which peak corresponds to what drug molecule.
Method
Development
What other columns are available? And what are their characteristics?
K is the retention factor of any non-uracil peak.
K = (tR - t0)/t0
Generally, K should be between 2- 20 (or 1-10 depending on literature).
Small K means the compound did not interact much with the stationary phase and eluted out of the column fast
Big K means the compound had lots of interactions with the stationary phase and was retained.
K only involves the retention time (tR) and dead time (t0) and many factors affect tR. Therefore, many factors can change K and even the slightest change in something can cause a big change in K.
Biphenyl - retains molecules with aromatic residues
PFP (pentafluoro phenyl) - retains basic compounds
Polar embedded - retains ionised molecules
Aqueous C18 - able to withstand high % of water
100%B
Other column
Result - chromatogram
Result - chromatogram
Multiplying factor rule
Uracil is often used to measure the dead time - the time that any compound will stay in the column.
Uracil is very polar and in reverse phase chromatography, it never interacts with the stationary phase. Uracil doesn't like to be in the mobile phase (as mobile phase is non-polar), but the stationary phase is even more non-polar which makes uracil hate the stationary phase.
Have you got a chromatogram that fulfilles the criteria? in this case;
CHROMacademy
What is happening in the column?
Not at all
By applying the multplying factor rule, it looks like 70~50% would give a good K.
Let's try a run with 70% rather than 50%, just in case multiplying factor is at or close to 3.
This was obtained under 100%B. Since only 1peak is shown, let's assume that the start of the peak (1.05 mins) is the uracil and the end of the peak (around 1.3 mins) is our last compound
In reality the result was quite different - 70%B gave total run time of 2.3mins!
Use the last peak retention time from this chromatogram to calclutate the "real multiplying factor".
From the calculations, it was estimated that 30%B would give K around 11 and run time of about 13 mins. 30%B is too low; let's try 50%B.
Q. So what changes t0?
Q. What else changes K?
Worked example:
Conditions;
5 - 100%B in 15 mins
tR (av) = 8.5 mins
Gradient slope = rise (%)/run (min) = 95/15
= 6.3% increase every minute
%B at 8.5mins = 6.3 x 8.5 = 53.55%
A. Any chemical parameter.
Changing chemcal parameter will change the type of interaction that occurs in the column, so tR will be changed. However, t0 remains constant (as it does not have any chemical interactions in the column), so K is altered.
A. All mechanicam parameters; column length, column diameter, flow rate, and linear velocity, and maybe viscosity.
Column length will change t0 because in a longer column, uracil must travel a longer distance to reach the detector.
Column diameter, flow rate and linear velocity are all related: column diameter and flow rate determines linear velocity.
Faster the linear velocity, the faster uracil will reach the detector. To achieve this, a smaller diameter and a higher flow rate is required.
Viscosity may change t0, but not very significantly.
50%B already gave a run time over 15 minutes. This was not expected from the calculations however, keep in mind that multiplying factor is 2 ~ 3, it is not constant every time 10%B decreases.
Nearly!
Rs is about 1.4
Why do we do this?
Due to the nature of the analytes and mobile/stationary phase, no molecules should be retained in the column so we only see one peak. Also, if the any compound takes long (eg; >4 minutes), there is no way that compound can be eluted out when we decrease %B.
What is happening in the column?
Total run time < 15minutes
K between 2 - 20
Selectivity > 1
Resolution > 1.5
Back pressure < 4000psi (for HPLC)
Number of plate > 4,000 preferably aorund 10,000
Anayte 1 is rather polar and has more preference to the mobile phase - short retention time*
Analyte 2 has equal preference between mobile phase and stationary phase - medium retention time*
Analyte 3 is rather non-polar has more preference to the stationary phase - long retention time*
Yes
Why do we do this?
Instead of changing %B bit by bit and seeing what happens, this is a quick way of estimating the run time. The isocratic run estimation must be validated though, as it is only an estimate and the actual chromatogram may show something different to what was expected.
*In relative terms
Congratulations, start method validation
What is N?
What is ɑ?
Result - Peak table
Q. What else changes N?
ɑ is the selectivity between two peaks. It is the relative distances of two peaks.
ɑ = K2/K1
Small ɑ means the two peaks are close together, with poor resolution.
Big ɑ means the two peaks are quite far apart, with good resolution.
Because ɑ is dependent on K, and K relies on chemical parameters, ɑ can only be changed by changing chemical parameters.
N is the efficiency of the column.
N = 16・(tR/width)^2
or
N = length of column/HETP
N refers to the number of theoretical plates in a column. The more plates, the more efficient separation of compounds.
Small N means the peaks are wide, due to insufficient separation. Wide peaks increase the chance of peak overlap.
Big N means the peaks are sharp. Sharp peaks decrease the chance of peak overlap.
N can only be changed by mechanical parameters. The aim is to make the peaks sharper, but not change the interaction that occurs between the compounds and the stationary and mobile phase.
These two were taken under the same condition, except the column diameter.
The narrower column shows a much better N than a wider one.
You might need to change the column and start method development all over again.
Have you got a chromatogram that fulfilles the criteria? in this case;
Not at all
Peak table from 100x4.6x5 column
Peak table from 100x3.2x5 column
What is happening in the column?
N = 16・(tR/width)^2
If tR increases more than width does, efficiency goes up. Therefore, it is not always wider peak = low efficiency.
A. Mechanical parameters such as length, flow rate, particle size, pore size, and injection volume. Although not a mechanical parameter, sample solvent can also affect N.
Increased column length increases N as more plates are able to fit into a longer column.
Flow rate will change N via linear velocity.
Increased LV increases mass transfer but increased LV decreases longitudinal diffusion.
Increasing particle size increases Eddy diffusion.
Increased pore size means more chance of mass transfer.
Increased injection volume gives increased longitudinal diffusion.
Sample solvent type can affect N as the use of immiscible solvents can lead to cavitations in the column, which gives poor efficiency.
Total run time < 15minutes
K between 2 - 20
Selectivity > 1
Resolution > 1.5
Back pressure < 4000psi (for HPLC)
Number of plate > 4,000 preferably aorund 10,000
Yes
Congratulations, start method validation
Why do we do this?
We don't want to change the type of interaction that occurs in the column - we kno that what we have now already works and gives somewhat good separation. We just want to make the peaks sharper to improve resolution.
Not at all
Might need to change solvent or even the column
This test involves all general criteria for a "good chromatogram" is met under the method developed.
Have you got a chromatogram that fulfilles the criteria? in this case;
Nearly!
Rs is about 1.4
Total run time < 15minutes
K between 2 - 20
Selectivity > 1
Resolution > 1.5
Back pressure < 4000psi (for HPLC)
Number of plate > 4,000 preferably aorund 10,000
In this step, do an injection of a compound(s) which already has been done before. Compare the results of what you had injected previously with the one you have just done now. Are they different? If they are, there is something wrong and you should not carry on with your investigation.
This step also requires the result to satisfy the following;
K > 2.0 with all peaks separated
Repeatability : RSD of less than or equal to 1% for n>5 (desirable)
Resolution >2 between the peak of interest and the closest eluting peak
Tailing factor < 2
N > 20,000
Yes
Congratulations, start method validation
LOD is the limit of detection - the lowest concentration that can be detected ie signal:noise ratio of 3:1
Robustness measures how much deliberate change in conditions the method can withstand
Keep diluting the sample until you get signal:noise of 3:1
The deliberate changes reflect the likely changes in condition that can occur when operating a procedure.
Such changes are concentration of B (+/- 2–3%), buffer concentration (+/- 1-2%), buffer pH (+/- 0.1-0.2), temperature (+/- 3C), etc.
The results obtained under these deliberate changes should not be different to the original method.
LOQ is the limit of quantification - the lowest concentration that can be quantified with enough accuracy and precision (signal:noise = 10:1)
Q. What else can change selectivity?
Result - chromatogram
Keep diluting the sample until you get signal:noise of 10:1
This is what we had obtained in step 2(b) - 50% ACN. What if we changed the organic phase (solvent B) to a mix of ACN and MeOH?
This was done with 30% ACN and 30% MeOH (=60%B altogether). The critical pair has not resolved completely, but is looking better.
But wait! a different pair of peaks have come closer now. We must now work out the composition of %B (ie how much of ACN and MeOH).
When changing type of B, you must consider - isoeluotropicity.
Isoeluotropicity refers to the ability of 2 different solvents having the same solvent strength in other words, they give "overall separation in the same timeframe" - CHROMacademy.
Isoeluotropic solvent percentages can be found on the solvent nomogram
A. Chemical parameters such as pH, %B, and ligand type, and also temperature
pH changes selectivity via ionisation state of analytes.
%B will change selectivity via K according to the multiplying factor rule.
Note: change in selectivity by %B is a very minor effect - it may not be enough to change the order of the peaks
Temperature does not have a direct effect on selectivity, but it will change the pKa of acidic/basic compounds.
Ligand type will also change selectivity because you are changing the whole stationary phase. Different stationary phase ligands have different properties which means that they interact differently with the sample compounds
Note: changing ligand type will require you to start the method development all over again.
We have worked out (by trial and error) that 20% ACN, 40% MeOH and 40% water gives a nicely separated chromatogram.
What is happening in the column?
CHROMacademy
This nomogram states that 53% of ACN has the same solvent strength as 64% MeOH
Why use mix of solvents?
Accuracy is achieved when the method developed gives a response that corresponds to the actual numerical value
Specificity is how well the method can detect the compound of interest in the presence of other molecules
Why do we do this?
Selectivity will change the chemical interactions that occur between the sample, mobile phase and stationary phase. This will change the relative position of all the peaks which might change the order of the peaks eluted out. This might solve any critical pairs or give you a better resolution overall.
It is not shown here, but results (chromatograms) were obtained using MeOH only and was compared to the ACN chromatogram. It showed that ACN had critical pair which was resolved by using MeoH. However, MeOH had created a different critical pair which was not seen on ACN. It was estimated that using a mix would resolve all the critical pairs.
CHROMacademy
A response given by the method (eg; AUC) should calculate back to the actual concentration of analyte injected.
For example, if 1mg of sample was injected and the AUC of the peak is given. Using the AUC, the mass of sample injected is calculated to be 0.9mg. This means that 0.1mg, or 10%, was "lost", therefore the recovery is 90%.
The recovery should be 98-102%. The example above fails accuracy test as only 90% was recovered.
This test must be done over at least 3 concentrations (50, 100, and 150%) and each must be done in triplicate (9 tests). The recovery is reported as an average of all 9.
Other compounds (eg; degradants, impurities, etc) may be present in the sample which might interfere with the identification of our compound in interest.
The 2 most closest peaks are often a compound and its' degradants/impurities. This test requires that the method/assay is unaffected by the degradants/impurities. This means that all peaks should be separated with adequate resolution.
A chromatogram with the all the compound to be assessed (eg; active ingredient, degradant, excipients, impurities, etc) should show all peaks resolved.
Method validation makes sure that the method produced during method development is working as good as it's supposed to be.
Method
Validation
References
Range is the minimum and maximum concentration determined to produce response with appropriate accuracy, precision and linearity.
Repeatability is being able to obtain same results using the same machine, same batch, same person, but different times (eg; throughout the day)
It is the range of concentration in which the graph (produced in the linearity test) shows r^2 greater than or equal to 0.99
At least 6 injections should be done and the standard deviation in the results obtained should be less than or equal to 2%
What if the method fails validation test?
Intermediate precision is being able to obtain same results using different machine, different person on a different day.
Linearity is the ability of the method to produce linear relationship between response and concentration.
All chromatograms taken from the Chromatogram database available on Moodle
Method validation tests:
http://www.pharmtech.com/hplc-method-development-and-validation-pharmaceutical-analysis?id=&pageID=1&sk=&date=
Pictures from CHROMacademy
http://www.chromacademy.com/hplc-training.html
Other drawings made by me
The criteria to pass this test is standard deviation of less than or equal to 2% in the results obtained.
The response (eg; AUC) is plotted against different concentration of the sample. At least 5 different concentrations (150, 100, 50, 25, and 12.5%) are required to plot of the graph.
The graph should have r^2 of equal to or greater than 0.99