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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

  • Allow a few minutes isocratic run at the lowest %B at the start to make sure all analyte compounds are stuck on the stationary phase and not travelling through the tubings
  • Allow a few minutes isocratic run at the highest %B at the end to make sure all analyte compounds have eluted out

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.

How to read the flow chart

HPLC - flow chart

Response to be changed

Result

-Chromatogram

-Peak table

-etc.

Main points of this step

What is happening in the column?

  • Methods used to obtain the chromatogram in the results section

Ayame Saito 25133381

  • Principles behind the chromatogram
  • Orange = important notes

Why do we do this?

Additional notes

Problem to solve

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

HPLC can be used qualitatively or quantitatively.

Whether quantitaive or qualitative, you must establish a few things first;

1. Know the physicochemical properties of the sample to be injected

2. Determine the concentration of a sample associated with a particular AUC under certain wavelength

Result - chromatogram

STEP 2(a) - Find optimum K' (gradient)

HPLC specifications given by Lorenzo;

  • 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

!Decision making step!

Scouting gradient chromatogram

Try changing the solvent. If this still doesn't work, try changing the column type and/or to isocratic method.

STEP 1(a) - Scouting gradient

CHROMacademy

the first step in optimising K' is changing △ϕ, the organic range.

As stated in Step 1(a), use the %B of initial peak as the minimum %B, and use the %B of last peak as the maximum %B.

Or you may wish to go all the way up to 100%B then bring it down by looking at the resultant chromatograms.

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?

Scouting gradient can tell whether the run should be isocratic or gradient. If gradient is to be used, the %B of which the first peak came out should be used as the initial %B

Q. What else changes K'?

CHROMacademy

A. Flow rate, tg (gradient time), and column dimensions (diameter and length)

CHROMacademy

  • %B is changed throughout the run
  • Compounds elute out when the %B is high enough for them
  • Every compound has different optimum %B
  • Peaks are nice and sharp
  • Analyte molecules accelerate through the column as %B increases

What is happening in the column?

  • 5-10%B to 100%B over a period of time (usually 20 minutes)

Nearly!

Total run time < 15minutes

Selectivity > 1

Resolution > 1.5

Back pressure < 4000psi (for HPLC)

Number of plate > 4,000 preferably aorund 10,000

  • %B changes throughout the run
  • When the %B is enough for a compound to move out of the stationary phase, it will elute out
  • Shallow graidnet slope will allow fine separation

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.

Chapter 1

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

Optimise K

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

STEP 1(b) - 100% B

!Decision making step!

Result - chromatogram

STEP 2(b) - Find optimum K (isocratic)

Multiplying factor rule

In reverse phase chromatography, everything (except water and uracil) is non-polar. In this step, we expect to see only 1 peak eluting out fairly quickly.

From Step 1(a) - Scouting gradient

Find the %B of the average retention time. Average retention time is calcuated by (tf - ti)/2

This can be done by looking at the chromagram like this

Alternatively, calculate the slope of the gradient then calculate the %B

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.

  • Use 100% B for mobile phase
  • B = acetonitrile (ACN)
  • Analyte = composite containing; Uracil, Ketoprofen, Sulindac, Piroxicam, Naproxen, Diclofenac, and Phenylbutazone
  • Flow rate = 1.5mL/min
  • Tem = ambient
  • Column = Restek Pinnale 150x4.6x5

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%

From Step 1(b) - 100%B

The multiplying factor rule - every 10%B drop, K is mltiplied by 2~3 times. Therefore the total run time is also multiplied by 2~3 times.

Use this rule to figure out the optimum %B to give total run time less than 15minutes.

  • Analytes, stationary phase and mobile phase are all non-polar
  • Analytes travel to the column with the mobile phase
  • Non-polar compounds prefer to be around non-polar compounds - "Like likes Like"
  • The analyte molecules don't "care" whether they are in mobile or stationary phase because both are non-polar
  • Analyte molecules stay in mobile phase and don't interact with the stationary phase -> one peak

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

1

  • %B decreases -> it gets more polar
  • Analyte molecules prefer to be in stationary phase more than in mobile phase
  • Different molecules have different preferences to stationary phase over mobile phase -> polar molecules prefer to be in mobile phase more than stationary phase and vice versa

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*

2

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.

3

*In relative terms

Congratulations, start method validation

Optimise N

Optimise ɑ (alpha)

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.

STEP 3(b) - Optimise efficiency

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.

!Decision making step!

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

Efficiency (N) is referring to the wideness of the peaks. Wide peaks have increased chance of peak overlap, and hence are said to have low efficiency and vice versa.

To change N, any mechanical parameter can be changed. For example, column diameter.

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.

  • Change column diameter to 3.2mm

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

  • tR and t0 remains proportional to each other as no chemical parameters are changed
  • Decreaed diameter = increased LV (at the same flow rate)
  • Higher LV means that longitudinal diffusion is minimised - however effect mass transfer is increased
  • Net effect = narrower peak
  • Narrower peak with increased LV is only up until a certain point - at this certain point, increased LV gives increased peak width

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.

!Decision making step!

System suitability

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, LOQ

Robustness

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

STEP 3(a) - Change selectivity

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

If the peaks haven't been separated, the selectivity may need to be changed - selectivity is a chemical respone and it depends on K (retention factor) being changed disproportionately between 2 peaks.

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.

  • ACN and MeOH mix
  • When doing mix, work out the composition of the two solvents to give you the best resolution (obtained practically)

What is happening in the column?

CHROMacademy

This nomogram states that 53% of ACN has the same solvent strength as 64% MeOH

Accuracy

Specificity

  • When the type of mobile phase is changed, the chemical (namely acidic, basic and dipolar) interactions are changed
  • This change in interactions will change the retention times of each individual compound
  • Uracil (t0) is not affected because it doesn't have any chemical interactions with the mobile and stationary phase
  • This results in change in K

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

Repeatability

Range

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

Linearity

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

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