High-Performance Liquid Chromatography (HPLC) purification technology is a crucial purification technique in the biopharmaceutical industry, used to separate various components in a mixture based on their different chemical properties. In the field of pharmaceutical research and development, comprehensive impurity studies and effective impurity control are directly related to the safety and efficacy of a drug, making them one of the key factors for successful drug approval.
However, the challenge in impurity research lies in identifying and controlling those unknown and persistent impurities that arise during the production process or stability studies. The formation mechanisms of these impurities are complex, and their structures are unknown, making them difficult to trace and analyze using conventional methods. In the face of this challenge, preparative liquid chromatography technology becomes a powerful tool for overcoming these difficulties. As an efficient separation technique, preparative liquid chromatography (PLC) can isolate trace impurities from complex samples, enabling further structural identification and toxicity evaluation. This not only addresses critical challenges in impurity research but also provides a stronger foundation for ensuring the safety and efficacy of pharmaceuticals.
Medicilon Cloud Lecture Hall invites Mo Li from the Process Analysis Department to provide an in-depth analysis of the key considerations in HPLC operation, along with strategies and methods to solve common operational challenges.
Q1. You just mentioned that lowering the column temperature can increase compound retention, but I've encountered situations where increasing the column temperature also resulted in increased compound retention. What could be the reason for this?
Mo Li: In chromatography analysis, lowering the column temperature typically increases the retention time of the solute. In some cases, the retention time of a compound may increase with temperature. This could be due to several factors: the solute molecules may become ionized, the structure of the solute molecules may be affected, or changes in the stationary phase due to temperature variations could also contribute to the increased retention time.
Analytical Testing Center
Q2. How should the solubility of a sample for preparation be tested? Do you have any good suggestions?
Mo Li: First, after receiving the sample, it is advisable to consult with colleagues from the synthesis department to find out what solvent the compound was evaporated from, so you can understand the compound’s original solvent. Alternatively, you can conduct a solubility test by weighing out several samples of approximately 10 milligrams each and placing them in injection vials. Add 100 microliters of a diluent to each sample and observe their solubility. If the sample does not dissolve, you can gradually increase the amount of diluent. If the sample remains insoluble under neutral conditions, you might consider adding a small amount of acid or base to aid in dissolution. Commonly used solvents can generally be methanol and acetonitrile. If these solvents fail to dissolve the sample, you can try more versatile solvents such as dimethyl sulfoxide (DMSO), dimethylformamide (DMF), tetrahydrofuran (THF), or N-methylpyrrolidone (NMP). Samples used for solubility testing can subsequently be utilized for method development, ensuring that the sample is not wasted.
Q3. What is presaturation? What is its purpose?
Mo Li: Presaturation techniques play an important role in chromatography analysis. Their primary purpose is to protect the chromatography column and maintain its performance stability. Presaturation can prevent the loss of stationary phase within the chromatography column. Before the mobile phase enters the chromatography system, the presaturated column treats the mobile phase to ensure it has sufficient contact with the stationary phase and reaches a saturated state. After saturation, when the mobile phase enters the analytical column, it helps prevent the loss of stationary phase from the column, thereby maintaining the column’s performance.
Q4. How should you clean the protection column that cannot be ultrasonically cleaned?
Mo Li: It was mentioned during the live broadcast that a protection column should not be subjected to intense ultrasonic cleaning because it might cause damage to the packing material. If the protection column becomes contaminated, you can unscrew and disassemble it, remove the column core, and clean it with a 50% isopropanol solution to remove any adsorbed samples from the surface. In cases of severe contamination, you can perform a separate washing procedure similar to what you would do with a regular chromatographic column to flush out the contaminants. It’s important to regularly check the theoretical plate number of the protection column to assess whether its performance has declined. Additionally, monitor for any unusual increases in pressure, as these could be signs that the protection column needs to be replaced. Since protection columns are consumables, you only need to replace the column core when performance declines, rather than replacing the entire chromatographic column. This approach helps save costs and maintain the continuity of analysis.
Q5. In practice, what types of columns are used for separating structural analogs?
Mo Li: The choice of chromatographic column depends on the chemical properties, size, polarity of the molecules to be separated, and the required resolution and selectivity. For aromatic compounds, particularly halogenated benzenes (such as chlorobenzene, bromobenzene, etc.), using a phenyl column can improve separation. For different substituent positions on the benzene ring, you might also consider using a pentafluorophenyl column, among others.
Q6. How can you address issues with abnormal peak shapes, such as split peaks?
Mo Li: In chromatographic analysis, we might encounter common issues such as sample overloading, improper solvent use, column contamination, or column failure. These issues can lead to abnormal peak shapes and affect the accuracy of the analytical results. To address these problems, you might consider reducing the sample load, such as lowering its concentration, decreasing the injection volume, or changing the sample solvent. If the chromatographic column is contaminated or its performance has declined, replacing it with a new column can restore accuracy and reproducibility in the analysis.
Mo Li’s special share: The derivation process of the retention factor 𝐾 formula.
The retention factor K is obtained by dividing the total amount of solute molecules in the stationary phase by the total amount of solute molecules in the mobile phase. The amount of solute molecules in both the mobile phase and the stationary phase is equal to the sample concentration (denoted as Cs and Cm, respectively) multiplied by the volume of the phase (denoted as Vs and Vm, respectively). Therefore, the equation can be derived as follows:
k=(CsVs)/(CmVm)=( Cs/Cm)/( Vs/Vm)
=KΨ（k is the equilibrium constant, and Ψ is the ratio of the mobile phase to the stationary phase.）
Solute molecules will indeed be present in either the stationary phase or the mobile phase. Assuming the mole fraction of the solute in the mobile phase is R, the mole fraction of the solute in the stationary phase would be 1−R. Therefore, the retention factor k can be derived from these mole fractions as follows:
k=(1-R)/R
Or
R=1/(1+k)
The retention time tR of a solute molecule X can be defined as the distance traveled divided by the velocity. Here, the distance is the length of the chromatographic column L, and the velocity of the chromatographic band is ux. Thus, the retention time can be expressed as:
tR=L/ux
Similarly, the retention time of the solvent peak is given by:
t0=L/u
If u is the average velocity of the mobile phase, and you remove the column length L from the equation, you derive the following relationship:
tR=t0u/ux
The moving rate or velocity ux of solute molecule X as it travels through the chromatographic column can be determined by calculating the mole fraction R of the solute present in the mobile phase at any given time. Generally, ux is equal to R multiplied by the flow rate or velocity U of the solvent molecules:
ux=Ru0
So using ux=Ru and R=1/(1+k) to replace tR=t0u/ux, we get the following equation
k=(tR/t0)-1
In the equation, tR is the compound retention time, t0 is the dead volume time
