Chem. 230 – 9/25 Lecture. Announcements I. Quiz 1 Results Solutions have been posted See class distribution Grade cut-offs have not been set, will depend somewhat on class scores Based on Quiz 1, I might expect cut-offs a couple of % below 90/80/70%. Announcements II.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
where H = Plate Height
u = linear velocity
and A, B, and C are “constants”
Most efficient velocity
Inside of column
(one quarter shown)
Shaded area = cross-sectional area = area*porosity
Low flow conditions
Higher flow conditions
5% by mass ea.
20% by mass ea.
Tailing peak (up fast, down slow)
Fronting peak (up slow, down fast)
Other Reasons for Odd Peak Shapes
Injection plug time = 0.1 min = 6 s (so no peaks narrower than 6 s unless on-column trapping is used)
Will not partition to stationary phase until mobile phase mixes in
In strong solvent
Analytes stick on column until stronger mobile phase arives
In weak solvent
Interactions by decreasing strength
Ion – Ion Interactions
Strong attractive force between oppositely charged ions
Of importance for ion exchange chromatography (ionic solute and stationary phase)
Also important in ion-pairing used in reversed-phase HPLC
Very strong forces (cause extremely large K values in absence of competitors)
From a practical standpoint, can not remove solute ions from stationary phase except by ion replacement (ion-exchange)
Ion – Dipole Interactions
Attractive force between ion and partial charge of dipole
Interactions by decreasing strength – cont.
Ion – Dipole Interactions – cont.
Determines strength of ionic solute – solvent interactions, ionic solute – polar stationary phase interactions, and polar solute – ionic stationary phase interactions
Important for some specific columns (e.g. ligand exchange for sugars or Ag+ for alkenes)
Metal – Ligand Interactions
ion – ion or ion – dipole interaction, but also involve d orbitals
Interactions by decreasing strength – continued (non-ionic interactions = van der Waal interactions)
Van der Waals Forces
dipole – dipole interactions (requires two molecules with dipole moments)
important for solute – solvent (especially reversed phase HPLC) and solute – stationary phase (especially normal phase HPLC)
Hydrogen bonding is a particularly strong dipole-dipole type of bonding
dipole – induced dipole interactions
induced dipoles occur in molecules with no net dipole moment
larger, more electron rich molecules can get induced dipoles more readily
induced dipole – induced dipole interactions (London Forces)
occur in the complete absence of dipole moments
also occur in all molecules, but of less importance for polar molecules
Somewhat of a one-dimensional model can be made by assigning a single value related to polarity for analytes, stationary phases, and mobile phases (See section 4.3)
These models neglect some interactions however (e.g. effects of whether an analyte can hydrogen bond with a solvent)
Describe the dominant forces involving the molecules to the right in interacting with non-polar molecules? in interacting with polar molecules
How does going from DB-1 (100% methyl stationary phase) to DB-17 (50% methyl – 50% phenyl) in GC affect elution of fatty acid methyl esters? (e.g. C16 vs. C18 vs. C18:1)
Silica has many SiOH groups on the surface (pKa ~2). What interactions will occur with the analyte phenol, C6H5OH, if the eluent is a mixture of hexane and 2-propanol?
Sugars are often separated on amino columns. A sugar that has a carboxylic acid group in place of an OH group will have extremely large retention times (at least at neutral pH values). What does this say about the state of the amino groups?
In reversed phase HPLC with a C18 column, benzene and methoxybenzene (anisole) have very similar retention times. What are the differences in the interactions between the two solutes and mobile phases and stationary phases?
A heavily used non-polar GC column is used to separate non-polar to polar columns. Polar compounds are observed to tail. A new column replaces the old column, tailing stops, and the polar compounds elute sooner. Explain the observations.
How does “method development” work?
Goal of method development is to select and improve a chromatographic method to meet the purposes of the application
Specific samples and analytes will dictate many of the requirements (e.g. how many analytes are being analyzed for and in what concentration?, what other compounds will be present?)
Coarse method selection (e.g. GC vs HPLC and selection of column type and detectors) is often based on past work or can be based on initial assessment showing problems (e.g. 20 compounds all with k between 0.2 and 2.0 with no easy way to increase k)
Optimization then involves making equipment work as well as possible (or limiting equipment changes)
Flow rate at minimum H vs. higher flow rates (covered with van Deemter Equation) – low flow rate not always desired because of time required and sometimes smaller S/N
Maximum flow rate often based on column/instrument damage – this can set flow rate
Trade-offs in reducing H
In packed columns, going to small particle sizes results in greater back-pressure (harder to keep high flow)
In GC, small column and film diameters means less capacity and can require longer analysis times
Trade-offs in lengthening column (N = L/H)
Longer times due to more column (often not proportional since backpressure at same flow rate will be higher)
Compounds X and Y are separated on a 100 mm column. tM = 2 min, tX = 8 min, tY = 9 min, wX = 1 min, wY = 1.13 min, so RS = 0.94. Also, N = 1024 and H = 100 mm/1024 = 0.097 mm
Let’s increase L to 200 mm. Now, all times are doubled:
tM = 4 min, tX = 16 min, tY = 18 min. So DtR (or d) now = 2 min. Before considering widths, we must realize that N = L/H (where H is a constant for given packing material).
N200 mm = 2*N100 mm. Now, N = 16(tR/w)2 so w = (16tR2/N)0.5
w200 mm/w100 mm = (tR200 mm/tR100 mm)*(N100 mm/N200 mm)0.5
w200 mm/w100 mm = (2)*(0.5)0.5 = 21-0.5 = (2)0.5
w200 mm = 1.41w100 mm
RS = 2/1.5 = 1.33
Or RS 200/RS 100 = d/wave = (d200/d100)*(w100/w200)= (L200/L100)*(L100/L200)0.5
So RS is proportional to (L)0.5
Increasing column length is usually the least desired way to improve resolution (because required time increases and signal to noise ratio decreases)
Alternatively, k values can be increased (use lower T in GC or weaker solvents in HPLC); or αvalues can be increased (use different solvents in HPLC or column with better selectivity) but effect on RS is more complicated
Note: above equation is best used when deciding how to improve RS, not for calculating RS from chromatograms