Chromatographic and spectroscopic methods

Analysis of a Pharmacopoeia controlled chemical substance using chromatographic and spectroscopic methods.

Introduction

There are several powerful pharmaceutical analysis tools to identify and quantify active substances in samples namely infrared spectroscopy, ultraviolet spectroscopy, high pressure liquid chromatography (HPLC), mass spectroscopy and nuclei magnetic resonance (NMR). Using chromatographic and spectroscopic techniques can identify and quantify the active ingredient, identify the presence of impurity in a substance, and quality control of drugs. Each analytical method has its own pro and cons. Thus, hypernated technique such as LC-MS can improve the quality of pharmaceutical analysis work. The objective of these experiments is to analyze whether the two batches of paracetamol powder meet the BP specification using different chromatographic and spectroscopic methods. Besides that, another objective is to determine which techniques suitable analysis methods are in order to confirm the compliance of the two batches.

1. Assay of Paracetamol BP by HPLC

Background

High performance liquid chromatography (HPLC) is one type of column chromatography. It is a technique used to separate components of mixtures which can be used to quantify substances in a drug, identify compounds and examine specific impurities.

In the HPLC, there are 2 phases namely stationary phase and mobile phase. The stationary phase is actually the chromatographic holding material which does not move while the mobile phase in HPLC is a liquid used to carry the samples and is forced through the column that contains the stationary phase via pressure. The separation of components is based on the difference in the degree of attraction between the stationary and mobile phase. This indicates that the components in a sample might have different retention time as the analytes have different physical or chemical interactions with the stationary phase when they cross the column.

Method:

Refer to laboratory manual page 5 and 6.

Results:

Preparation of different concentration of standard solutions, samples and impurity:

Table 1: Weight and concentration of standard paracetamol, sample A, sample B and impurity X

Standard

Sample A

Sample B

Impurity X

Weight of sample transferred

0.2563g

0.1260g

0.1246g

0.1016g

Concentration of samples

1025.2 µg/ml

504 µg/ml

498.4 µg/ml

1016 µg/ml

1. Preparation of paracetamol standards

a. 1025.2g/ml x 10/100 dilution factor (intermediate) = 102.52µg/ml

b. Prepare different concentration of standards from intermediate solution

i. 102.52µg/ml x 5/100 dilution factor = 5.126µg/ml

ii. 102.52µg/ml x 10/100 dilution factor = 10.252µg/ml

iii. 102.52µg/ml x 15/100 dilution factor = 15.378µg/ml

iv. 102.52µg/ml x 20/100 dilution factor = 20.504µg/ml

v. 102.52µg/ml x 25/100 dilution factor = 25.63µg/ml

2. Preparation of sample A

a. Sample A

i. 504 µg/ml x 1/25 dilution factor = 20.16 µg/ml

b. Sample B

i. 498.4 µg/ml x 1/25 dilution factor =19.936 µg/ml

3. Preparation of Impurities X

a. 1016 µg/ml of impurity X x 10/100 dilution factor =101.6 µg/ml

b. 101.6 µg/ml of impurity X x 20/100 dilution factor =20.32 µg/ml

Table 2: Peak areas of different concentrations of Paracetamol standard solutions that were obtained from the automated HPLC system

Concentration (µg/ml)

Peak Area

5.126

409406

10.252

655987

15.378

1266084

20.504

1660219

25.63

2098600

Graph 1: Calibration curve was obtained by plotting peak areas against the concentration of the Paracetamol standards. The equation of this linear was y = 85498x - 96727.

Table 3: Calculation of total paracetamol in sample A using the calibration curve that obtained from the automated HPLC system

Sample

A1

A2

Peak

1618464

1442096

Concentration in sample solution

y = 85498x - 96727

1618464 = 85498x - 96727

x = 20.06µg/ml

y = 85498x - 96727

1442096 = 85498x - 96727

x = 17.998 µg/ml

Concentration in original extract

20.06 µg/ml x 25

= 501.5µg/ml

17.998 µg/ml x 25

= 449.96µg/ml

Amount of paracetamol found in sample A original extract

Mean of amount of paracetamol found in sample A original extract

[(125.375+112.49)/2] mg

= 118.93 mg

Concentration of paracetamol in sample A (%w/w)

In 126mg of sample A contained 116.05mg of paracetamol

Thus, the concentration of paracetamol is 94.39%w/w

Table 4: Calculation of total paracetamol in sample B using the calibration curve that obtained from the automated HPLC system

Sample

B1

B2

Peak

978175

1293462

Concentration in sample solution

y = 85498x - 96727

978175 = 85498x - 96727

x = 12.57 µg/ml

y = 85498x - 96727

1293462 = 85498x - 96727

x = 16.26µg/ml

Concentration in original extract

12.57 µg/ml x 25

= 314.3µg/ml

16.26 µg/ml x 25

= 406.5µg/ml

Amount of paracetamol found in sample A original extract

Mean of amount of paracetamol found in sample A original extract

[(78.58+101.63)/2] mg

= 90.10mg

Concentration of paracetamol in sample B (%w/w)

In 124.6mg of sample B contained 87.938mg of paracetamol

Thus, the concentration of paracetamol is 72.31%w/w

Table 5: Calculations of total amount of paracetamol that were found in both sample A1 and sample B1 by using a single point calibration. The peak areas of both samples were obtained from manual HPLC system.

Sample

A1

B1

Peak

18650416

15056728

Concentration in sample solution (Given that the peak area of standard paracetamol = 19377184)

x = 19.735 µg/ml

x = 15.932 µg/ml

Concentration in original extract

19.735 µg/ml x 25 = 493.375 µg/ml

15.932 µg/ml x 25

= 398.31µg/ml

Amount of paracetamol found in sample A original extract

Concentration of paracetamol (%w/w)

In 126mg of sample A contained 123.344mg of paracetamol

Thus, the concentration of paracetamol is 97.89%w/w

In 124.6mg of sample B contained 99.577mg of paracetamol

Thus, the concentration of paracetamol is 79.92%w/w

Table 6: Calculation amount of impurity X in sample B.

Sample

B1

Peak

2633917

Concentration in sample solution

(Given that the peak area of standard Impurity X = 11019760)

x = 4.857µg/ml

Concentration in original extract

4.857 µg/ml x 25

= 121.42 µg/ml

Amount of Impurity X found in sample B1 original extract

Concentration of Impurities X in sample B (%w/w)

In 124.6mg of sample B contained 30.355mg of Impurity X

Thus, the concentration of Impurity X is 24.36%w/w

TLC analysis of Paracetamol

Background

Thin Layer Chromatography is a simple and efficient method to separate the substance into different components in the sample. Every chromatography has similar principle of action. Thus, same as HPLC, TLC also contains both stationary phase which is usually silica gel, aluminium oxide or cellulose and mobile phase which consists of liquid solvent. However, compared with HPLC, TLC is cheaper screening test and always used prior to other expensive analytical tests.

The mobile phase in TLC is raised up in the plate via capillary action. Since each molecule has different attraction and solubility against the stationary and mobile phase, some of the components can carry further up the plate compared with others. Therefore, they will have different Rf value. Rf value can be counted by the distance travelled by the analyte divided by distance travelled by the solvent itself. By comparing the Rf value of a unknown compound with a known compare, it is possible to identify a compound in a mixture. Usually, the TLC plate contains fluoresces in order to visualize those uncoloured dots which appear on the plate.

Method:

Refer to laboratory manual page 7 and 8

Result:

Distance moved by solvent from the origin = 7.9cm

Table 7: Distance travelled by analytes in each solutions and Rf value of analytes in the solutions were calculated by a/S where a is the distance travelled by the analyte and S is the total distance moved by the solvent from the origin. From this table, it indicated that in sample A and B contained paracetamol while impurity X was also found in sample B as sample B had 2 spots which had similar Rf values as paracetamol and impurity X.

Samples

Distance moved by analytes

Rf value

Standard paracetamol solution

4.3

0.5443

Impurity X

5

0.6329

Sample A

4

0.5063

Sample B

1) 4

2) 4.9

1) 0.5063

2) 0.6203

Assay of Paracetamol BP by UV Spetrophotometry

Background

UV spectrophotometry is an instrument which is used to measure the intensity of UV light transmits through the sample which is filled in a cuvette. The transmitted light (I) over the intensity of UV light before the light passes through the sample (I%) is called transmittance. Another more common used parameter that used in UV spectrophotometer is called absorbance. The relationship between absorbance and transmittance is A= -log (%T), where A is absorbance while %T is the transmittance.

UV spectroscopy is a universal method to quantify the substances which occur in the sample. There is an assumption of application of the Beer-Lambert law which is as below:

A=A(1%, 1cm) x c(g/100ml) x l (1cm)

where A is the absorbance, A(1%,1cm) is the slope wavelength dependent absorptivity coefficient, c is the concentration of the sample and l is the pathlength.

Method:

Refer to laboratory manual page 9 and 10.

Results:

Table 8: Preparation of standard and sample solutions

Weight of sample transferred

Concentration of stock solution (%w/v)

Dilution factor

Concentration of final solution (%w/v)

Standard 1

0.1047g

0.1047

100

0.001047

Standard 2

0.1046g

0.1046

100

0.001046

Sample A1

0.1477g

0.0739

100

0.0007391

Sample A2

0.1500g

0.0750

100

0.000750

Sample B1

0.1502g

0.0751

100

0.000751

Sample B2

0.1482g

0.0741

100

0.000741

Table 9: Absorbance of standard 1 and standard 2 at 258nm wavelength

Sample

Absorbance

Standard 1

0.746

Standard 2

0.740

Calculation of experimental A(1%,1cm) at the wavelength of 258nm

λmax value = 258nm

According to Beer Lambert law A = A (1%,1cm) x c (g/100ml) x 1(cm)

Table 10: Calculation of experimental A (1%, 1cm) at 258nm wavelength.

Standard 1

Standard 2

Calculation

Experimental A(1%,1cm)

mean A(1%,1cm) = (712.51 + 707.46)/2

= 709.99

Table 11: Absorbance of samples at wavelength of 258nm

Sample

Absorbance

Sample A1

0.539

Sample A2

0.539

Sample B1

0.558

Sample B2

0.559

Table 12: Calculations of amount of paracetamol in both sample A and sample B

Samples

Concentration

(final solution)

(%w/v)

Concentration

(stock solution)

(%w/v)

Content of Paracetamol (g)

Mean content of Paracetamol (g)

Sample A1

Dilution factor = 100

Concentration

= 7.592 x 10-4

= 0.07592%w/v

0.1518g

0.1518

Sample A2

Dilution factor = 100

Concentration

= 7.592 x 10-4

= 0.07592%w/v

0.1518g

Sample B1

Dilution factor = 100

Concentration

= 7.859 x 10-4

= 0.07859%w/v

0.1572g

Sample B2

Dilution factor = 100

Concentration

= 7.873 x 10-4

= 0.07873%w/v

0.1575g

0.1489g sample A contained 0.1518g of paracetamol,

Concentration of paracetamol in sample A =101.95%w/w

In 0.1492g of sample B, there is 0.1574g of paracetamol.

Concentration of paracetamol in sample B = 105.50%w/w.

Question:

Propose reasons for differences between the calculated A(1%,1cm) and that reported BP 2003.

There is a difference between the calculated A(1%,1cm) and that reported BP 2003. This is due to the standard paracetamol solutions used in this experiment may not as accurate as standard paracetamol that used in BP 2003. Besides that, the personnel that conducted in this experiment was not well trained while personnel that conduct the test for BP is well trained. Besides the personnel factor, instrument that used for BP has higher quality compared with the instrument that had been used in the experiment.

Compare your assay value with that obtained in Experiment 2 (HPLC analysis of paracetamol)

The value of paracetamol content that found in sample A and sample B using HPLC and UV spectroscopy were different. In sample A, the concentration of paracetamol in sample A was 97.89% using HPLC while 101.95% using UV spectroscopy. Since sample A was purely paracetamol, the difference between both values may due to random error such as sloppy in making up solutions. Besides that, the big difference between the values of sample B gained by HPLC (79.92%) and UV spectroscopy (105.50%) was due to the presence of impurity in sample B which can also absorb the UV light at the given wavelength.

Identification of compounds using reference spectra

Background

Infrared spectroscopy is a useful technique to identify certain functional groups in the chemical structure of a compound. Moreover, it is a quick and relatively cheap method. Besides that, it can produce unique IR spectrum for a given compound which can serve as the fingerprint of the compound. As a result, the IR spectroscopy usually is used to identify a compound by comparing the fingerprint of the compound with the standard fingerprints of different compounds which can be found in the British Pharmacopoeia.

IR spectroscopy is based on the principle of absorption of different frequency of light due to different energies involved in bond vibrations. When a particular frequency of light is being absorbed means that the energy of the light is transferred to the compound that is being analysed. This means that the absorption of IR radiation is correlated with the bonding of the compound.

Table 13: Comparing experimental resolution with the BP specification

Wavenumber

Experimental resolution

BP specification resolution

2869.9/2849.2

25.9852

>18

1589/1583

21.1687

>12

The result showed that the IR instrument used in this experiment which met the BP specification was in good condition.

Table 14: Functional group analysis of compound in Sample A using the absorption value that obtained from the IR spectrum (Appendix I)

Experimental Absorption (cm-1)

Approximate Absorption (cm-1)

Functional group

Band characteristics

3326.1

3100 -3500

NH

Sharp band,

3162.0

3100 -3500

H-bonded OH

Broad band, stretching

2879.6

3000-3100

C-H (aromatic)

Weak, stretching

1653.5

1640-1700

C=O (conj)

Strong

1516.2

1550-1600

C=C (aromatic)

Moderate and stretch

Table 15: Functional group analysis of compound in Sample B using the absorption value that obtained from the IR spectrum

Experimental Absorption (cm-1)

Approximate Absorption (cm-1)

Functional group

Band characteristics

3326.9

3100 -3500

NH

Sharp band,

3162.6

3100 -3500

H-bonded OH

Broad band, stretching

3032.7

3000-3100

C-H (aromatic)

Weak, stretching

1654.0

1640-1700

C=O (conj)

Strong

1514.8

1550-1600

C=C (aromatic)

Moderate and stretch

Two IR spectrums were obtained and were compared with the BP IR fingerprint. (Appendix I) Both samples had similar fingerprint at 4000 -1300 cm-1 region which is particularly useful to identify the presence of specific functional groups. However, after comparing the fingerprint of the whole spectrums of sample A and sample B, sample A was further identified as paracetamol as its fingerprint identical with the standard paracetamol in British Pharmacopoeia. On the other hand, sample B may be compound that has similar functional group as paracetamol or it may contain paracetamol with additional of other impurity.

Nuclear Magnetic Resonance and Mass Spectral interpretation

Background

Nuclear Magnetic Resonance (NMR) is a powerful tool to elucidate the chemical structure of a compound. The NMR fingerprint of each compound is produced based on the proton the posses spinning property. Each proton has its own energy. When electromagnetic radiation of correct frequency occurs, there will be resonance absorption. And this resonance absorption is detected in NMR. Based on the chemical shifting and splitting patterns of the fingerprint of the compound, the chemical structure of the compound can be identified. However, in order to find the exact chemical structure, mass spectrophotometer is essential. Mass spectrophotometer (MS) is an instrument that has both quantitative and qualitative uses. In this case, MS is used to identify the molecular weight of a compound by producing a mass spectrum. Thus, with combination of both NMR and MS spectrum, a chemical structure of a compound can be figured out.

Results:

Table 16: Interpretation of standard paracetamol 1H NMR spectra

ppm value

Integration

Coupling pattern

Functional group

1.982

3.00 (3H)

Singlet

Methyl-CH3

3.434

0.73 (1H)

Singlet

Solvent- CD3OD

6.664-6.702

2.18 (2H)

Doublet

Aromatic-CH

7.323-7.361

2.17 (2H)

Doublet

Aromatic-CH

9.120

1.02 (1H)

Singlet

NH

9.630

1.04 (1H)

Singlet

OH

Table 17: Interpretation of 13C NMR spectrum for standard paracetamol. The number of carbon atoms is interpreted by the relative intensities of the peak. At the ppm value of 115.06 and 120.92, the intensities were 2 times larger compared with others. Thus, there were 2 carbon atoms at 115.06 and 120.92 ppm respectively. Besides that, the number of H atoms attach to the C atoms was determined by JMOD technique.

ppm value

Number of carbon atom

Number of H atoms attach to the C atoms (Odd/Even)

Functional group

23.78

1C

Odd

H3C13-C

39.51

-

Even

Solvent - CD3OD

115.06

2C

Odd

ArC13-H

120.92

2C

Odd

ArC13-H

131.07

1C

Odd

ArC13-N

153.19

1C

Odd

ArC13-O

167.63

1C

Odd

ArC13=O

Table 18: Interpretation of impurity standard 1H NMR spectra

ppm value

Integration

Coupling pattern

Functional group

4.489-4.358

2.81 (2H)

Singlet

Ar-NH

6.413-6.450

2.58 (2H)

Doublet

Aromatic-CH2

6.477-6.506

2.56 (2H)

Doublet

Aromatic-CH2

8.350-8.441

1.38 (1H)

Singlet

Ar-OH

Table 19: Interpretation of 13C NMR spectrum for standard impurity. The number of carbon atoms is interpreted by the relative intensities of the peak. At the ppm value of 115.349 and 115.618, the intensities were 2 times larger compared with another 2 peaks. Thus, it can be concluded that there were 2 carbon atoms at 115.349 and 115.618 ppm respectively. Besides that, the number of H atoms attach to the C atoms was determined by JMOD technique which is based on the influence of hydrogen atom on the 13C relaxation time.

ppm value

Number of carbon atom

Number of H atoms attach to the C atoms (Odd/Even)

Functional group

115.349

2C

Odd

ArC13-H

115.618

2C

Odd

ArC13-H

140.66

1C

Even

ArC13-C

148.31

1C

Even

ArC13-C

Table 20: Interpretation of Sample B 1H NMR spectra

ppm value

Integration

Coupling pattern

Functional group

Derived from

1.984

3.00 (3H)

Singlet

Methyl-CH3

Paracetamol

3.420

0.73 (1H)

Singlet

Solvent- CD3OD

Solvent

4.367

0.72 (1H)

Singlet

Ar-NH

Impurity

6.419-6.443

0.743 (1H)

Doublet

Aromatic-CH2

Impurity

6.469-6.492

0.698 (1H)

Doublet

Aromatic-CH2

Impurity

6.665-6.703

2.162 (2H)

Doublet

Aromatic-CH (HB)

Paracetamol

7.325-7.363

2.13 (2H)

Doublet

Aromatic-CH (HA)

Paracetamol

8.324

0.35 (1H)

Singlet

Ar-OH

Impurity

9.121

1.11 (1H)

Singlet

NH

Paracetamol

9.631

1.06 (1H)

Singlet

OH

Paracetamol

Table 21: Interpretation of 13C NMR spectrum for Sample B

ppm value

Number of carbon atom

Number of H atoms attach to the C atoms (Odd/Even)

Functional group

Derived from

23.78

1C

Odd

H3C13-C

Paracetamol

39.51

-

Even

CD3OD

Paracetamol

115.05

2C

Odd

ArC13-H

Paracetamol

115.317

2C

Odd

ArC13-H

Impurity

115.32

2C

Odd

ArC13-H

Impurity

120.90

2C

Odd

ArC13-H

Paracetamol

131.06

1C

Odd

ArC13-N

Paracetamol

140.66

1C

Even

ArC13-C

Impurity

148.30

1C

Even

ArC13-C

Impurity

153.19

1C

Odd

ArC13-O

Paracetamol

167.63

1C

Odd

ArC13=O

Paracetamol

Table 22: Interpretation of Electron Spray Mass Spectra. From the spectrum, the possible molecular weight of impurity was either 97 or 109.

m/Z

Ion source

Real molecular weight of the compound

Compound

174.05225

Sodium ion

174.05225 - 23 = 151.05225

Paracetamol

152.07047

Proton

152.07047 - 1 = 151.0747

Paracetamol

110.05995

Sodium ion

110.05995 - 23 = 97.05995

Impurity

Proton

110.05995 - 1 = 109.05995

The 1H NMR spectrum and 13C NMR spectrum of Sample B showed that paracetamol and impurity were found in Sample B. This was showed by comparing the spectrum of Sample B with the 1H NMR spectrum and 13C NMR spectrum of standard paracetamol and standard impurity. With the aid of mass spectrophotometer, the molecular structure of impurity can be identified after knowing the molecular weight of the impurity. The impurity that was found in Sample B was para-aminophenol and the chemical structure was as below:-

Relative amount of impurity present can be calculated using the integration ratio between pure compound and impurity.

P:I = 2.16:0.74

2.16 x 151:0.74 x 109 = 326.16:80.7

Percentage of impurity = 80.7/(326.16 + 80.7) x 100%

= 19.8%

Thus, the relative amount of impurity present in sample B was 19.8%.

Discussion

HPLC is used to identify and to quantify not only paracetamol but also the presence of impurity in the samples in this experiment. There were two approaches to find the concentrations of paracetamol content in the samples. The first one was by using a 5 points calibration curve while the other one was using one external calibration point. From the result of both methods, the content of paracetamol in both samples did not meet the BP specification (99%-101%). Besides that, since there was another peak in sample B which had identical retention time as impurity X standard, it can be concluded that there was impurity X in the sample B which was then calculated as 24.36%. However, HPLC can identify neither what the impurity is nor how chemical structure is.

In the TLC experiment, we can more confirm that both samples contained paracetamol while only sample B had impurity X within. Although TLC is a quick, inexpensive and simple analysis method, it is not suitable for quantify the concentration of paracetamol and impurity in samples. It can only be used to identify the presence of paracetamol and impurity in samples.

In UV Spectrophotometry experiment, the results obtained from sample A and sample B were used to quantify the content. This is because UV spectroscopy is a quantitation technique which has not been specific enough to characterize adequately the compounds in the samples. In this experiment, both samples A and sample B had similar concentration of compound that can absorb UV light at wavelength of 258nm. The result of UV indicated that sample B may have 101.74%w/w of paracetamol. However, the result of HPLC indicated that there was only 79.92%w/w of paracetamol in sample B which was contrast with the result of UV spectroscopy. Moreover, the previous HPLC and TLC results indicated that sample B contained impurity. Thus, the high absorbance in sample B may due to the presence of impurity which can absorb at the same wavelength as paracetamol. From this experiment, it showed that UV spectroscopy is not a good technique to identify the presence of impurity as it has low selectivity where other compound may interfere the assay.

Infrared spectroscopy on the other hand is useful to ensure the identity a substance as no two substances have exactly the same IR spectra. In this experiment, the IR spectrum for sample A was identical to the standard paracetamol spectrum in BP while sample B had similar fingerprint with standard paracetamol spectrum but there were small difference in the fingerprint region. Since IR spectroscopy is the most specific technique to confirm the identity of drug substances, it showed that sample B might contain both paracetamol and impurities or it may contain substance that has similar chemical structure as paracetamol. From the previous HPLC and TLC result, it showed that sample B contain paracetamol. Thus, in this case, there results of IR spectroscopy further confirm that sample B contain impurity and had failed in the test. IR spectroscopy is quite useful in detecting impurity in a substance as the presence of impurity alters the fingerprint of the compound. However, IR cannot be used to quantify the concentration of each substance or impurity in a sample.

NMR is an important instrument for structure elucidation. In this experiment, NMR and MS were used to identify the structure of the impurity. Moreover, it can be used to determine the relative amount of impurity in sample B. However, compared with IR spectroscopy, it is more complicated, expensive and takes longer time to generate the result.

The result in HPLC and UV showed that neither sample A nor sample B passes the specification of British Pharmacopoeia. As both samples had readings which were out of the range of 99% to 101% of N-(4-hydroxylphenyl) acetamide in HPLC and UV spectroscopy analysis.

Conclusion

The experiments showed that although sample A did not contain impurity X, it was rejected as falling outside its BP specification. Besides that, all analysis tests which had been done in this experiment suggested that there was impurity in sample B. Thus, sample B had also fails to meet the criteria in BP specification. As a conclusion, 2 batches of paracetamol powder had been rejected.

Please be aware that the free essay that you were just reading was not written by us. This essay, and all of the others available to view on the website, were provided to us by students in exchange for services that we offer. This relationship helps our students to get an even better deal while also contributing to the biggest free essay resource in the UK!