Problem Set 5 Solutions CH332 (SP ’06)
1. Determine the structure that gave rise to the following mass spectrum, and identify the fragments responsible for all peaks above 20% relative abundance (these appear at m/z 77, 105, 111, 139, 141). The parent ion appears at mass 216. The empirical formula has been determined by high-resolution mass spectrometry to be C13H9ClO. Discuss how isotope peaks aid in fragment identification.
First we need to generate possible structures based on the empirical formula. Then we can compare the masses of fragments in the mass spectrum to the potential structures and attempt identification. The empirical formula indicates that this molecule is highly unsaturated (9 degrees of unsaturation). We should therefore expect plenty of pi bonds and rings. A fragment at mass 77 is immediately apparent, so we know that the molecule contains at least one benzene ring. Given that as a starting point, I believe that in this case there is only one rational structure (more correctly 3 structures, since the position of the chlorine cannot be determined absolutely) that is possible given the high degree of unsaturation:
Now we need to see if the peaks in the mass spectrum are consistent with fragmentation of this structure. Some fragments we might expect are:
m/z = 77 m/z = 105
m/z = 139 m/z = 111
Note that each ion containing Cl should show a relatively intense isotope peak at 2 mass units higher due to the 24% natural abundance of 37Cl (these show up at masses 141 and 113). This is exactly what we observe. All of the above fragments are observed in the mass spectrum and all of the predominant features of the mass spectrum are accounted for. We conclude that we have a good structure. Notice that this technique is incapable of distinguishing the possible o-, m- or p- positions of the chlorine substituent by fragment mass analysis alone, although it is possible that the relative stabilities of the ions may cause different relative intensities of fragment peaks.
2. For a drift length of 100 cm in a time-of-flight mass spectrometer, what is the difference in arrival time between ions of m/z = 44 and m/z = 43 when the accelerating voltage is 2800 V? At what frequency might you choose to have the instrument electronics sample the electron multiplier or multichannel detector signal in order to adequately resolve these masses?
Both ions are accelerated by a voltage of 2800V before entering the field-free drift tube. The kinetic energy gained by each is:
KE = qeV = (1)(1.602×10−19 C)(2800J / C) = 4.5×10−16 J
Note that they both have the same kinetic energy after being accelerated between plates held at a potential difference of 2800 V, since they both have the same charge. Their velocities are different because they have different masses and are determined by:
1 KE = mv 2 = 4.5×10−16 J 2 2KE v = m
Don’t forget to convert mass to kilograms:
g kg mol −26 m = (44 )( )( 23 ) = 7.3×10 kg 44 mol 1000g 6.022×10 −26 m43 = 7.1×10 kg
Now the velocities are:
2KE (2)(4.5×10−16 J ) v = = = 1.11×105 m / s 44 m 7.3×10−26 kg 5 v43 = 1.12×10 m / s
Since we know the velocities and the length of the drift tube, we can calculate the time it takes for each to reach the end of the tube and the detector.
t = (1m)(1.11×105 m / s) = 9.0×10−6 s 44 −6 t43 = 8.9×10 s
Then the difference in time of arrival is
−7 t44 − t43 = 1.0×10 s
The instrument then must sample the detector at least at a rate of
1 f = = 9.7 ×106 Hz 1.0×10−7 s
or about every 100 ns to have any chance of distinguishing the two masses. Actually, this is still probably not sufficient time resolution, since it would require precise coincidence of the detector sampling with the arrival time of the ions. What is typically done in practice is to divide the time into discrete bins of constant duration and integrate the detector current generated during each time interval. Therefore, let’s add some data points between the two masses by sampling every 10 ns (or a rate of 108 Hz), to avoid error in distinguishing the two masses due to time-base jitter. This may sound like an awfully fast rate, but it’s easily achievable with modern electronics.
3. Consider the following experimental data collected for separations on three columns identical in all ways except for the size of the stationary phase particles (3, 5 or 10 µm).
a) Which column would show the best resolution for two similar solutes? What aspect(s) of the graph above allowed you to make this decision?
Since resolution is proportional to N , we can expect the column with the highest efficiency (# theoretical plates) to have the best resolution. These columns are all the same length, so the one with the smallest plate height will have the highest efficiency. L N = H
The 3 µm particle column has the smallest plate height for all flow rates; thus, the 3 µm column should have the best resolution.
b) Use the van Deemter equation to explain the differences in these three plots.
The van Deemter “A” term, which is due to differences in analyte path distances, is independent of flow rate (u), and is relected in the vertical offsets of the three column plots. Larger particles produce a larger variation in possible path lengths; therefore, the 10 µm column has a larger plate height (smaller N) than the other columns at all flow rates.
The van Deemter “B” term is due to longitudinal diffusion and is inversely proportional to the flow rate. This term determines the downward curvature in the plot before the plot minima, and is roughly similar for all three columns. The lower the flow rate, the longer the retention time for the analyte on the column and the more time it has to diffuse away from the original injection volume. To minimize this broadening, use a higher flow rate.
The van Deemter “(Cm+Cs)” terms are linear in flow rate (u) and are attributed to “mass transport” between the mobile and stationary phases, which is determined by the attainment or non-attainment of partition equilibrium between the two phases. This determines the upward slopes of the plots after the minima. At low flow rates, the analyte can achieve equilibrium at each portion of the stationary phase before it is swept along in the mobile phase by the solvent flow. At higher flow rates, true equilibrium with the stationary phase can not be achieved before it is swept down the column. This non-equilibrium condition is reflected in an increase in plate height (decrease in N). Remember, to get the most effective separation we want to decrease H (increase N) as much as possible. The smaller particle size column (3 µm) has a larger stationary phase surface area per unit volume, and partition equilibrium is more easily achieved even at the highest flow rates. Therefore, the upward slope is smaller for the 3 µm column.
For the most efficient separation possible, we want to achieve the smallest plate height (largest N) possible. So, we should choose a flow rate fast enough to minimize longitudinal broadening, but slow enough to ensure partition equilibrium. The optimal flow rate is selected as that producing the minimum plate height.
c) What disadvantage in a chromatographic separation can result from a decrease in particle size?
A major problem with decreasing particle size to arbitrarily small dimensions is the corresponding increase in mobile phase pressure needed to elute the analyte in a reasonable time.
4. The data in the following table were extracted from the chromatogram of a two- component mixture of x and y.
Air x y Retention time (s) 5 25 30 Peak width, FWHM, (s) --- 2.5 2.0
a) Calculate the capacity factors k’ for x and y.
Capacity factors are derived from the retention times and the dead time (in this case reflected in the retention time of air).
x ' t R − tm 25 − 5 k x = = = 4 tm 5 y ' t R − tm 30 − 5 k y = = = 5 tm 5
b) Calculate the number of theoretical plates N for the column and the theoretical plate height H (assume a column 25 cm long).
Remember that each component will have its own column efficiency value:
2 2 t x 25 N = 5.54 R = 5.54 = 554 x x W 1 2.5 2
L 0.025m −4 H x = = = 4.5×10 m N x 554 2 2 t y 30 N = 5.54 R = 5.54 = 1247 y y W 1 2.0 2
L 0.025m −4 H y = = = 2.0×10 m N y 1247
c) Calculate the separation factor α. Is this a reasonable value for achieving separation of the components?
' k y 5 α = ' = = 1.25 k x 4
The separation factor alone doesn’t indicate the effectiveness of a separation. It doesn’t take into account the peak widths, so to get a good measure of the effectiveness of separating the two components we need to calculate the resolution of the two peaks.
d) Calculate the resolution. Is this sufficient to obtain fully resolved peaks? (NOTE: the equation for resolution given in class should have been in terms of the peak widths measured at baseline. If we are to stay consistent and refer to all peak widths in terms of full-width-at-half-maximum W1/2, then the correct equation for resolution should be:
t B − t A R = R R S A B 2 ln 2(W1/ 2 +W1/ 2 )
Since W1/ 2 = 2 2 ln 2 σ .
t y − t x 30 − 25 R = R R = = 1.1 S y x 2 ln 2(W1/ 2 +W1/ 2 ) 2 ln 2(2.0 + 2.5)
The resolution falls short of the critical value of 1.5, so we conclude that it is insufficient to obtain fully resolved peaks.
5. Consider the following chromatogram. The fast-eluting peaks and the slow-eluting peaks present a problem.
a) Describe the chromatographic nature of the problem.
This is a reflection of the Common Elution Problem, in which it becomes impossible to effectively separate nearby peaks at the fast-eluting and slow- eluting ends of the chromatogram simultaneously under the same column conditions. The first peaks are poorly resolved because of their short retention times, and the last peaks are retained too long, such that they are broadened to the point of being barely detectable.
b) Devise a way in gas chromatography to eliminate the problem.
We can manipulate the column conditions during the course of the run to retard the fast-eluting components to increase separation and then accelerate the slow-eluting components to increase peak intensities and minimize broadening. In GC the major parameter we have control over influencing retention time is temperature. Design a temperature ramp that begins at low temperature (somewhat lower than that at which the above chromatogram was obtained) and increases during the course of the run. Iterate runs under different temperature ramp parameters until the best possible mitigation of the common elution problem is achieved. It is not always possible to completely eliminate the problem.
c) Devise a way in liquid chromatography to eliminate the problem. Assume that the chromatogram has been acquired on a normal-phase chromatographic system.
In LC the major column parameter influencing retention time that we have control over is solvent strength. Since this was acquired on a normal-phase system, we can infer that the mobile phase was relatively nonpolar and the fast-eluting peaks were the most nonpolar components. We can increase the solvent strength by increasing the polarity of the mobile phase (introducing a second, more polar solvent to the mobile phase or increasing the proportion of that solvent). This will cause the most nonpolar analytes to be retained longer on the column, allowing them to be separated more effectively. Likewise, the stronger solvent strength will be able to compete with the polar stationary phase more effectively for the most polar components (slowest eluting), allowing them to be eluted fast enough to reduce peak broadening to a reasonable extent. So, we should try running with a higher proportion of polar solvent in the mobile phase mixture and examine the results. Gradient elution, in which the solvent polarity is changed during the course of a run, is an option if isocratic methods are insufficient. Again, it may not be possible to completely avoid the Common Elution Problem for all components in a mixture simultaneously.
6. Design an HPLC technique that would be appropriate for the separation of benzoic acid, benzene, chlorobenzene and phenol. Give the expected order of elution. What type of detector would be appropriate?
There are many possible answers; the following is just one of many.
The analytes range from nonpolar to fairly polar. We could use a reversed- phase stationary phase. To find the appropriate solvent strength, start with a fairly polar (aqueous perhaps) mobile phase, varying it by increasing the concentration of a nonpolar solvent until the solvent strength is sufficient to elute the least polar component (benzene) in a reasonably short retention time. The solvent strength could even be varied during the run time (gradient elution) to ensure reasonable retention times and acceptable peak separation.
The expected order of elution would be:
1. benzoic acid 2. phenol 3. chlorobenzene 4. benzene
Many different detection techniques could be employed.
All of the components likely absorb in the UV/vis because of the common chromophore benzene ring. The absorption spectra are probably too similar to distinguish between them spectroscopically. Remember, though, that this does not necessarily present a problem in chromatography as long as we have a reasonable expectation of the order of elution or can run pure standards of the individual components to verify retention times. We just need a detector that is sensitive to all components; it need not be selective between them, and UV/vis certainly fits the bill in this case.
Mass spectrometry is almost always a fine choice for detector. Parent or fragment ion masses may be monitored.
IR is another possibility.