Chemistry 322

Spring, 1999

Final Examination Answer Key

May 13, 1999

You may use a calculator, ruler, ONE sheet of paper with equations or anything YOU wrote or typed, and the Miller textbook. You may ask questions of Prof. Scheeline or the TAs, but should not communicate with anyone else during the exam.

Grades WILL NOT be posted. Please email Scheeline no sooner than May 14 if you want to learn your grade prior to receiving the official grade from the University.

According to University rules, after grades are issued you may review your exam by coming to Scheeline's office, where they will be kept for one year, then destroyed. You may not remove your exam booklet from Scheeline's custody.

1. Interpretation of Chromatograms. You've now spent almost four months interpreting chromatograms. Here is a chromatogram from the separation of the enantiomers of baclophenlactam. Tell us everything you can from interpreting this chromatogram. Your answer should be a table in the form: variable, meaning, value. "Meaning" should be a short phrase describing the variable, not some long explanation. Grading: 3 points per correct answer (there are at least 11 possible answers -- maybe more). Example:

tv

void time

1.2 minutes

Here are 11 possible answers (maybe you'll find more!)

	Number of components	>= 2
tr,1	Retention time 1	7.4 min
tr,2	Retention time 2	10.6 min
k1	Capacity factor 1	6.2
k2	Capacity factor 2	8.8
w1	Peak width 1 (base)	1.1 min
w2	Peak width 2 (base)	1.1 min
R	Resolution	16
	Relative retention	1.42
N	Number of theoretical plates	~1000
Vvoid	Column void volume	2.4 mL

How I got the numbers: 12 minutes = 8.9 cm, so 1 minute = 0.74 cm. Retention times were straight units conversions: d/0.74 cm/min = t. k = t_r/t_v. Peak width is found by extrapolating the peak position and half-maximum points to baseline and measuring the width (again using t = d/0.74 cm/min as a conversion factor). R = (t₁+t₂)/(2w). . N = 16(t_r/w)² = 16*(8.8/1.1)² = 1024 or round off to 1000 (if we use the second peak; as long as your number is consistent with the rest of the table, that's fine). Finally, void volume = flow rate times void time = 1.2 min * 2 mL/min.

2. Interpretation of TLC and Gel Electrophoresis. Shown below are a TLC plate (a) and an electrophoresis gel (b).

a) Compute the partition coefficient for compound A, and suggest what you'd do to modify the experiment to get rid of the streak and instead get a well-resolved spot for compound B. The stationary phase is alumina and the solvent a mixture of 50/50 mixture acetonitrile and pH 7 phosphate buffer in water. (10)

The solvent front moved 6.9 cm. A is at 2.1 cm. R thus = 2.1/6.9 = 0.30. R = 1/(1+k) so k = 1/R-1 = 3.3-1 = 2.3. B is an equilibrium mixture of two different pH species (so it spends half its time eluting with k = k_acid, while the rest of the time it elutes with k = k_base). To focus the spot, either increase or decrease the pH by at least 2 units so that that B spends all of its time in a single pH form. Chose which direction to change so that A stays focused.

b) This gel was developed by first using a pH gradient along the vertical direction and then using the size-exclusion properties of the gel in the horizontal direction. b1) Given that the MW range of the proteins is 1,000 - 20,000, mark on the diagram where an protein with isoelectric pH of 5.4 and molecular weight of 18,000 would appear. (5) b2) In spotting the protein mixture on the plate, does it matter where along the left edge the spot is placed? Explain. (3) Does it matter whether isoelectric focusing is done before or after size-exclusion elution? Again, explain. (2)

The arrow points at the circle indicating where the 18 kD, pI 5.4 protein be. Vertically, it's at the point where pH 5.4 would appear in the pH gradient. Horizontally, the gel will trap larger proteins more effectively than smaller proteins, so the heavy protein doesn't migrate very far (this is in contrast to size exclusion chromatography, where the ability of light molecules to penetrate the gel slows them down, while the heavier molecules stay in the mobile phase). The spotting location is irrelevant -- molecules migrate to their isoelectric pH, so they can start at any pH point along the gradient and still end up in the same place. Isoelectric focusing should be done first, since one can use a highly open gel to allow focusing to proceed quickly. If size-exclusion is done first, the development in the pH gradient could be quite slow for the largest proteins, allowing excessive diffusion and reduced resolution for the lighter constituents.

3. GC Detectors. Explain why each of the following choices of detectors is appropriate for the application listed. (5 points each)

a) Thermal conductivity detector/Chem 224 (Quantitative Analysis) teaching laboratory

b) Flame ionization detector/Arco Petroleum Co. research laboratory

c) Electron capture detector/U. S. Environmental Protection Agency Pesticide Monitoring

d) Mass spectrometer/Chem 337 (Advanced Organic Synthesis) teaching laboratory

a) Universal detector, inexpensive; "good enough for what it's for."

b) FIDs are good at detecting reduced carbon, and petroleum is largely made of hydrocarbons

c) Pesticides typically contain either halogens or phosphorus or both, which are highly electronegative and thus can be sensed with ECDs

d) One wants to know the structure of newly synthesized compounds, and mass spec. is a good way to get strong confirmation of proposed structures (if one can match both molecular weight and fragmentation patterns to hypothesized structures). GC/MS is among the most sensitive methods for determining structures of small quantities of materials.

4. LC Separations. LC probably has the greatest variety of parameters of any of the methods we discussed. Suppose you needed to separate sugars in a fermentation bath for ADM (just down the road in Decatur). You must separate glucose, fructose, galactose, arabinose, sucrose, lactose, and maltose. One separation that appears to work well is ion-exchange in unbuffered solution.

a) Why is this a surprising choice of both mobile and stationary phases? Considering that sugars are neutral hydrocarbons, what does this tell you about aqueous sugar chemistry? (10)

Ion-exchange only separates charged analytes. Thus, the sugars must be ionized to a significant degree in water, even though we usually draw them as undissociated molecules. They're probably anions. It's unusual that buffering isn't needed. Evidently, the pK_a is low enough that any reasonable pH won't suppress ionization.

b) Suggest what detector to use and why you think it will work. If your detection method requires derivatization, specify whether you'd do pre-column or post-column derivatization. (5)

Not all of these sugars are reducing sugars, so electrochemistry won't work. Because of the range of sugars desired, enzyme-coupled methods probably won't work. I'd actually probably use near-infrared absorption, but we didn't talk about this in class so that's not cricket. Post-column derivatization with fluorescence detection OR chemiluminescence using luminol, a heme catalyst and a mixture of saccharide oxidases would work. If no gradient elution is required, refractive index might be adequate as well.

c) The ion-exchange resin is available in 3 micrometer, 5 micrometer, and 10 micrometer diameters. Columns are available in 50, 100, and 250 mm lengths. Diameters of either 2, 5, or 10 mm are available. How would you go about choosing what combination of packing and column to use? (10)

How big a sample is needed? If we don't need to isolate the separated species, go for the smallest column that will give adequate resolution; if preparative work is being done, choose the largest diameter column. Smaller particles give both shorter HETP and higher backpressure. The pressure limit for most columns is 6000 psi, so the smallest packing could probably only be used with the shortest columns. Thus, for preparative work, we're likely to choose 10 micron particles in a 250 mm by 10 mm column, while for analysis we'd choose 3 micron particles in a 50 mm by 2 mm column.

5. Capillary Electrophoresis. Listed below are two possible ways to set up a capillary electrophoresis system to separate polypeptides produced by a combinatorial synthesis experiment. In one setup, every parameter value is undesirable for some reason. In the other, the parameter is more reasonable. Explain what's wrong in each case. 5 points per explanation.

Parameter	Setup 1	Setup 2
Elution Voltage	100 V	10,000 V
Capillary diameter	200 microns	25 microns
Buffer	1 M citrate	0.1 mM citrate
Detector	Fluorescence	Amperometry

Elution voltage: resolution is proportional to electric field which in turn is proportional to applied voltage. If one takes the rule of thumb that one can get 10⁶ at 30 KV, then setup 1 would only give 3300 plates -- not many at all, while setup 2 would give 330,000, a lot better.

A wide-bore capillary will have such low resistance that solvent heating and convective instabilities will lower resolution.

Buffer must have low enough concentration that there's a high zeta potential and thus an electro-osmotic flow. Using the lowest concentration of buffer consistent with maintaining accurate pH is desirable.

Fluorescence can only detect some amino acids (such as tryptophan), but not the simplest amino acids (such as glycine, alanine, and isoleucine). Amperometry at a sufficiently reducing potential can detect the amide moities found in all peptides and proteins.

6. SEC and Field-Flow Fractionation.

a. We saw in the second hour exam that for SEC, one must use molecules with similar functionality and density to calibrate the mass vs. retention time scale. What molecular properties must match between samples and standards for sedimentation FFF? (10)

Retention time is based only on molecular diameter and the difference in density between sample and mobile phase. Thus, matching density is the ONLY parameter required.

b. In principle, which method (SEC or S-FFF) should give better resolution and dynamic range? Is this method actually more commonly used? If not, why not? (10)

S-FFF can give much better resolution since one can arbitrarily decrease HETP by increasing the gravitational forces and by adjusting the density of the mobile phase to differ just slightly from that of the analyte. S-FFF isn't as commonly used as SEC due to cost and the specialized nature of the equipment required.

7. Programmed Elution. Explain why one is unlikely ever to see methods with the following programming sequences: (5 points each)

a. GC column temperature ramped from 300 C to 100 C

Starting at high temperature, lots of compounds would elute together, including all the compounds which could elute at the lower, but not the higher, temperature. As the temperature decreased, any higher-boiling materials would condense on the column.

b. LC solvent mixture starts as pure hexane and gradually switches to pure water

Hexane and water can't mix. Thus one would not be able to create a homogeneous intermediate mobile phase. Imagine trying to run ANY detector with a two phase mobile phase passing through it -- what a noisy mess!

c. Capillary electrophoresis voltage starts at 25 KV and is gradually decreased to 10 KV

Why elute rapidly when nothing has yet migrated down the capillary, then drop the voltage just when high resolution is desirable? Actually, if you wanted to move a group of similar compounds down a capillary, then slowly let them drift out the end and be detected, one might construct some pathological situation where you actually would use this programming. But I've never seen it.

8. Most important word in the course: The most important physical phenomenon occurring in most chromatographic separations, influencing the speed of separation, the scale of the columns and stationary phases, the resolution, and practically every other important characteristic is Diffusion (5) As often as we mentioned this, EVERYONE should have gotten it!

9. Ion Exchange Chromatography.

a. The order of elution of ions in IEC is the same, regardless of which stationary phase you choose. Explain why this is so. (5)

Separation is based on pure electrostatics. The size of an ion and its charge are inherent in its nature, so while one can tune resolution and elution time, the fundamental source of the type and magnitude of interaction can not be changed (unless one complexes the ions, but then they're no longer the original moity).

b. Amino acids can be separated by ion-exchange chromatography, reverse phase liquid chromatography, and capillary zone electrophoresis. Suppose you're trying to separate lysine from tryptophan. Useful data:

If your specimen comes as a bodily fluid (pH close to 7), explain how you'd select a method and which one you think you'd actually use. Your explanation should be based on the fundamental characteristics of the methods you are considering, and should specify the expected elution order if you are in a position to predict this. (5 points per method)

The species are both charged at neutral pH, so ion-exchange or CZE would both be feasible means of separation. LC would require that one choose a stationary phase that had differential interactions with the side chains on the amino acids. This could probably be done, but why take the trouble to fool around with columns and conditions when the other two methods are available?

If one is looking for small amounts of material, choose CZE. If one is working on a preparative scale, choose ion-exchange. In CZE, lysine comes out first (highest charge => faster migration), while in IEC, tryptophan comes out first (lower charge gives weaker interaction with cation exchange resin).

10. Fluid Mechanics and Mass Transfer Kinetics

a. You'd like to run 1 mL/min through an LC column. You connect the outlet of your pump to the inlet of the column through a capillary tube with a length of 1 m and an inside diameter you have no way to measure. You find that the pressure drop across the capillary is 1000 psi. You desire the pressure drop be less than 50 psi. What dimensions would an appropriate connecting tube between the pump and the column have? (10)

Considering the Poissieulle flow relationship, one can either cut the length to 50/1000* 1 m = 50 mm or, if that's impractical, increase the diameter by (1000/50)^1/4. The fourth root of 20 is 2.11, so just a little over doubling the tube's diameter would solve the problem. Of course, combinations of increasing tube i.d. and shortening it are also plausible.

b. You have a capillary that is coated on the inside with a long-chain hydrocarbon on the end of which is a -COOH moity. At pH 7, the capillary works well, giving good electrophoretic separations. At pH 2, elution times increase by an order of magnitude, and the peaks broaden. Explain why there was such a dramatic change. (5)

At pH 2, the COOH becomes protonated, the zeta potential thus drops, the electro-osmotic flow ceases, and diffusion then broadens the electrophoretic peaks.

c. Two GCs are used to separate hydrocarbon samples on C18 columns. Both have flow rates of helium carrier gas of 10 mL/min. Yet one has poorly resolved peaks that appear between 3 s and 10 s after sample injection, while the other has well-resolved peaks that appear between 1 min and 3.3 min after injection. Explain how this can be, describing how the columns differ and any other characteristics of the elution behavior that would help someone "fix" the poorly operating GC. (10)

The first column is a capillary column. The flow rate is so high that one is far to the right on the van Deemter plot, giving a large HETP and poor resolution. The second column is an ordinary column, and one is operating near the optimum flow rate. The void volume is larger in the second case. Here's a sketch:

Flow is unitless, as it's in flow rate (ml s^-1)/void volume (ml). Similarly, HETP is normalized to column diameter. Slowing the flow rate on the first column will improve resolution.