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CH 429 NMR by xiaopangnv

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									Jarrett Long

Quantitative Analysis with NMR



INTRODUCTION

       In this laboratory experiment, the amount of various compounds in an unknown sample

will be determined through the use of NMR techniques. Nuclear Magnetic Resonance

spectrometry can be used for quantitative analysis through the method of internal standards, or

by using relative peak proportions. These peaks are characteristic of NMR spectrometry due to

the underlying principles of the technique. The basics are that a magnetic field is produced which

can excite protons at various frequencies. The signal strength, which arises from excitations and

positioned by their decay times (with the help of basic Fourier transform techniques, converting

frequency to time), is proportional to the number of magnetic nuclei. This fact allows for the

integration of characteristic peaks normalized to an internal standard in order to determine the

relative amount of the analyte compound. Donald Hollis’ 1963 laboratory experiment “The

Quantiative Analysis of Aspirin, phenacetin, and caffeine Mixtures by [NMR] Spectrometry”

and is used as a guide, reference, and quality control check.



METHOD

       In order to determine the amount of three compounds (Aspirin, Phenacetin, and

Caffeine), three known standards are created each containing approximately 20 mg of one of the

compounds. These are then dissolved in approximately 1.58 g NMR solution made up of 9 g

(approximately 6 ml) of CDCl3 and 0.1 g of dibromomethane (DBM). The peak characteristic of

DBM is very strong and is an excellent internal standard. However, having an accurate NMR
solution is key since all the calculations involve the mass of DBM in each solution. After the

known standards are created a test (quality control) sample is created. This involves

approximately 10-25 mg of each of the compounds dissolved in 1.5 g of NMR solution. The

exact figures are summarized on the attached Table 1.

       For the unknown sample, NMR solution is poured into the vial to dissolve the unknown

which is then transferred to a NMR tube. By comparing the integrated peaks of the unknown to

the known amounts in the standards, the mass of each compound in the unknown is easily

attained through the formula:

                           Area 2 ( X )          Area1 ( DBM ) / Mass1 ( DBM )
Mass2 ( X )                                   
                Area 2 ( DBM ) / Mass2 ( DBM )      Area1 ( X ) / Mass1 ( X )                (Eq. 1)

where the subscript 1 refers to the known sample run and the subscript 2 refers to the

unknown/analyte sample.

       The machine used here is a Varian 500 MHz NMR spectrometer. The basic outlay of the

technique is to carefully insert the NMR test tubes one by one and take readings of magnetic

resonance absorptions to gather various information. The procedure involves tuning the magnetic

field in such a way that the decay times of the excited protons are in an exponential fashion with

high consistency. If the field is tuned incorrectly, the protons no longer travel with any type of

regularity and thus are hard to analyze. The tuning involves a computer program and adjusting

variables such as the strength of the field in the X, Y, Z directions as well as power output and

others. Also, the same integration width and positions is also important.
CALIBRATION

       By knowing the amounts of compound in the quality control, a simple test run can be ran

to check the accuracy of the method. In this test run, the deviation from calculated amount to the

known quality control of Aspirin (A), Phenacetin (P), and Caffeine (C) were calculated to be A:

0.8%, P: 0.5%, C: 0.9% with uncertainty of about 1% on the calculated amount. This is reflected

in Hollis’ laboratory write up where he reports errors of A: 1.1%, P: 2.2%, C: 3.2%, rotationally

leading to a highly-accurate quantitative determination of the unknown in this experiment. The

exact data for the quality control test run is available in Table 2. A sample calculation for the

quality control is presented as Sample Calculation 1.



ERROR ANALYSIS

       The error budget is presented as Table 1 in the TABLES section.

       One idea to describe further is how the determinate errors cancel each other out in the

case of the Unknown sample. This is because for the estimated amount of error in the Standards

which would predictably lower the Analyte amount, there was also an amount of unknown that

was lost in the transferring. So in order to account for the Standards error the -1.5% of the

calculated amount would balance out with the +1.5% needed to account for lost unknown.

However, since little unknown was lost then only the -1.5% applies.

       The final error is  0.3 mg on the determination of each of the compounds, along with a

relative error on integral ratios of about 2% or .3 mg. So in calculating the total error in the

summation of the masses, the square root of the sum of the squares is taken and calculated to be

0.7 mg. This calculation is demonstrated as Sample Calculation 2.
RESULTS

        The full results are attached on the front page Summary Results. Briefly, the amounts

determined in the Unknown sample was A: 18.0  0.3 mg, P: 15.0  0.3 mg, and C: 17.4  0.3

mg. Thus, the total mass by NMR determination is 51.4  0.7 mg. Comparing this against the

total mass by weight of 52  2 mg (the mass of vial with compound minus the mass of vial

without compound (15.129 g - 15.077 g)) seems to optimistically confirm positive results with

little precision error.



DISCUSSION

        The results are in fact fairly ambitious showing little precision error and little accuracy

error. Considering that the majority of the error comes from the concentration of the standards,

one can see why there might be little error. By doing this experiment by weight instead of

volume allows for precise results. The accuracy of the machine itself is fairly substantial,

especially comparing it against the quality control. This comparison yielded even better results

than Hollis’ experiment, and that alone could be due to increased NMR technology in the past 50

years. Any time a <1% error can be achieved as with the quality control check, it must be

thought that the technology itself is well perfected upon.



SOURCES

Hollis, Donald P. "Quantitative Analysis of Aspirin, Phenacetin, and Caffeine Mixtures by

Nuclear Magnetic Resosnance Spectrometry." Analytical Chemistry (1963): 1682-684. Print.
TABLES

Table 1 - ERROR BUDGET

Error Source          Indeterminate   Random     Determinate

Conc. DBM              0.02 mg

Conc. Standards        0.25 mg                  -1.5%

Error on Integral                      0.3 mg
Ratios
Transfer of                                      +1.5%

Unknown

Error in Total Mass    2 mg
by Weight




FIGURES AND GRAPHS
DATA

Data Table 1 - 3 STANDARDS w/ Amount of DBM In Each

Compound         Amount Present (mg)      Peak Location (ppm)       Peak Intensity

A: Aspirin       18.6                     2.363                     135.17

P: Phenacetin    20.3                     1.4                       148.87

C: Caffeine      22.6                     3.58                      149.19

DBM              18.5                     4.94                      100




Data Table 2 - QUALITY CONTROL

Compound         Amount Present (mg)      Peak Location (ppm)       Peak Intensity

A: Aspirin       18.3                     2.34                      177.27

P: Phenacetin    24.4                     1.40                      148.87

Phenacetin

C: Caffeine      11.3                     3.59                      76.01

DBM              18.6                     4.94                      100




SAMPLE CALCULATIONS

Sample Calculation 1 - Relative Error of Calculated vs. Known Amounts in Quality Control
Sample Calculation 2 - Error in Total Mass by NMR Determination




QUESTIONS

11.1.

A) TMS - 0 Hz, cyclohexane - 346, t-butanol - 408, acetone - 467, dioxane - 950, benzene - 1787

B) Double all the frequencies, 346  692 Hz, etc.

C)



11.2. No, if the 7.2ppm correlates to the aromatic H and the 2.35 ppm correlates to the methyl H,

then there should be a 5/3 ratio of aromatic/methyl, but instead there is a 2:1 abundance, meaning

there are extra benzene rings.



11.5. 441 g/mol.



11.6. A) 0.500005 in lower, 0.499995 in upper.

B) 0.500008 in lower, 0.499992 in upper for 200 Mhz; 0.500002 in lower, 0.499998 in upper for

500 Mhz
C) Nope



17.2. 10,000 and 40,000




17.15.

								
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