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HPLC Method development and validation flow chart

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a sai

on 10 June 2015

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Transcript of HPLC Method development and validation flow chart

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
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
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
STEP 1(b) - 100% 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
HPLC - flow chart
Method
Development
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
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.
Might contain Sulindac, Piroxicam, Naproxen, Diclofenac, and Phenylbutazone
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
Result - chromatogram
We need method development so that when we run the contaminated tablets in HPLC, we can identify which peak corresponds to what drug molecule.
Problem to solve
What is happening in the column?
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
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.
1
2
3
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*
*In relative terms
Before method development....
Pick your column
C 18 column
Other column
STEP 1(a) - Scouting gradient
5-10%B to 100%B over a period of time (usually 20 minutes)
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
Scouting gradient chromatogram
What is happening in the column?
%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?
Using gradient run sounds great, but it takes time (as you must wait for re-equilibration time) and isocratic run is much more simple.
Isocratic run is possible if;

Δtg < 0.25tg
STEP 2(b) - Find optimum K (isocratic)
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.
Result - chromatogram
What is happening in the column?
%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
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.
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.
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.
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
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%

Isocratic
Gradient
STEP 2(a) - Find optimum K' (gradient)
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
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.
Result - chromatogram
%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
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.
Multiplying factor rule
%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
What is happening in the column?
!Decision making step!
Have you got a chromatogram that fulfilles the criteria? in this case;
Total run time < 15minutes
Selectivity > 1
Resolution > 1.5
Back pressure < 4000psi (for HPLC)
Number of plate > 4,000 preferably aorund 10,000
STEP 3(b) - Optimise efficiency
Change column diameter to 3.2mm
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.
Result - Peak table
What is happening in the column?
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
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.
STEP 3(a) - Change selectivity
ACN and MeOH mix
When doing mix, work out the composition of the two solvents to give you the best resolution (obtained practically)
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.
Result - chromatogram
What is happening in the column?
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 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.
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).
We have worked out (by trial and error) that 20% ACN, 40% MeOH and 40% water gives a nicely separated chromatogram.
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.
This nomogram states that 53% of ACN has the same solvent strength as 64% MeOH
Q. What else can change selectivity?
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
.
Q. So what changes t0?
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.
Yes
Not at all
Nearly!
Optimise K
Chapter 1
What is K?
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.
What other columns are available? And what are their characteristics?
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
Optimise ɑ (alpha)
What is ɑ?
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
.
!Decision making step!
Have you got a chromatogram that fulfilles the criteria? in this case;
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
Not at all
Nearly!
Rs is about 1.4
Optimise N
What is N?
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!
Have you got a chromatogram that fulfilles the criteria? in this case;
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
Not at all
Nearly!
Rs is about 1.4
Peak table from 100x
4.6
x5 column
Peak table from 100x
3.2
x5 column
These two were taken under the same condition, except the column diameter.
The narrower column shows a much better N than a wider one.
Q. What else changes N?
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.
Q. What else changes K?
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.
N = 16・(tR/width)^2
If tR increases more than width does, efficiency goes up. Therefore, it is not always wider peak = low efficiency.
Congratulations, start method validation
Congratulations, start method validation
Congratulations, start method validation
!Decision making step!
Have you got a chromatogram that fulfilles the criteria? in this case;
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
Not at all
Congratulations, start method validation
You might need to change the column and start method development all over again.
Try changing the solvent. If this still doesn't work, try changing the column type and/or to isocratic method.
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.

Method
Validation

Method validation makes sure that the method produced during method development is working as good as it's supposed to be.
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
Q. What else changes K'?
A.
Flow rate
,
tg
(gradient time), and
column dimensions
(diameter and length)
Specificity
Repeatability
Accuracy
Linearity
Range
LOD, LOQ
Intermediate precision
Robustness
System suitability
Specificity is how well the method can detect the compound of interest in the presence of other molecules
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.
Linearity is the ability of the method to produce linear relationship between response and concentration.
Range is the minimum and maximum concentration determined to produce response with appropriate accuracy, precision and linearity.
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
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
LOD is the limit of detection - the lowest concentration that can be detected ie signal:noise ratio of 3:1
Accuracy is achieved when the method developed gives a response that corresponds to the actual numerical value
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.
Repeatability is being able to obtain same results using the same machine, same batch, same person, but different times (eg; throughout the day)
At least 6 injections should be done and the standard deviation in the results obtained should be less than or equal to 2%
Intermediate precision is being able to obtain same results using different machine, different person on a different day.
The criteria to pass this test is standard deviation of less than or equal to 2% in the results obtained.
Keep diluting the sample until you get signal:noise of 3:1
LOQ is the limit of quantification - the lowest concentration that can be quantified with enough accuracy and precision (signal:noise = 10:1)
Keep diluting the sample until you get signal:noise of 10:1
Robustness measures how much deliberate change in conditions the method can withstand
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.
This test involves all general criteria for a "good chromatogram" is met under the method developed.
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
Once the method passes all validation tests, the method can be used for the investigation
What if the method fails validation test?
After that, change the method (eg; solvent type, %B, etc) and see if the new method passes the validation test.
References
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

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Ayame Saito 25133381
100%B
Scouting gradient
Might need to change solvent or even the column
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.
How to read the flow chart
Response to be changed
Methods used to obtain the chromatogram in the results section
Main points of this step
Result
-Chromatogram
-Peak table
-etc.
What is happening in the column?
Principles behind the chromatogram
Orange = important notes
Why do we do this?

Additional notes
For explanation on how compounds separate, read additional notes on Step 2(b)
Why use mix of solvents?
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.
Full transcript