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ASU / Emergency Medicine / EM 103 / 107 148 class notes

107 148 class notes

107 148 class notes

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School: Arizona State University
Department: Emergency Medicine
Course: Emergency Medical Technician
Professor: N. cutsumbis
Term: Winter 2016
Tags: Chemistry
Cost: 25
Name: Pharmaceutical Analysis .week 10
Description: Pharmaceutical Analysis A Textbook for Pharmacy Students and Pharmaceutical Chemists
Uploaded: 03/13/2017
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• Atwhat rateisthe drug released fromits formulation sothatit can be absorbed bythe body?




• What is the stability of a drug in the formulation and hence the shelf-life of the product?




• What is the percentage of the stated content of a drug present in a formulation?



Pharmaceutical Analysis A Textbook for Pharmacy Students and Pharmaceutical ChemistsSenior Commissioning Editor: Pauline Graham Senior Development Editor: Ailsa Laing Project Manager: Sruthi Viswam If you want to learn more check out supply chain management refers to a set of approaches and techniques firms employ to efficiently and effectively integrate their manufacturers, warehouses, transportation intermediaries, stores, and
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Designer: Miles Hitchen Illustration Manager: Jennifer RosePharmaceutical Analysis A Textbook for Pharmacy Students and Pharmaceutical Chemists THIRD EDITION David G. Watson BSc PhD PGCE Senior Lecturer in Pharmaceutical Sciences, Strathclyde Institute of Pharmacy and Biomedical Sciences, University of Strathclyde, Glasgow, UK With a contribution by RuAngelie Edrada-Ebel BSc MSc DrRerNat AMRSC Lecturer, Strathclyde Institute of Pharmacy and Biomedical Sciences, University of Strathclyde, Glasgow, UK EDINBURGH LONDON NEW YORK OXFORD PHILADELPHIA ST LOUIS SYDNEY TORONTO 2012# 2012 Elsevier Ltd. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). First edition 1999 Second edition 2005 Third edition 2012 ISBN 978-0-7020-4621-6 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging in Publication Data A catalog record for this book is available from the Library of Congress Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. With respect to any drug or pharmaceutical products identified, readers are advised to check the most current information provided (i) on procedures featured or (ii) by the manufacturer of each product to be administered, to verify the recommended dose or formula, the method and duration of administration, and contraindications. It is the responsibility of practitioners, relying on their own experience and knowledge of their patients, to make diagnoses, to determine dosages and the best treatment for each individual patient, and to take all appropriate safety precautions. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. The Publisher's policy is to use paper manufactured from sustainable forestsPrinted in China Preface to the third edition The fundamental principles of obtaining reliable analytical data do not change, but some instrumental methods have changed greatly in the 12 years since the first edition – none more so than mass spectrometry. The chapter on mass spectrometry has been completely rewritten to reflect these extensive changes in instrumentation. High-pressure liquid chromatography remains at the heart of pharmaceutical analysis and its most up-to-date version, ultra-high-pressure liquid chromatography, is becoming more widely adopted since it can offer faster analyses. Chapter 12 has been revised to reflect recent developments in liquid chromatography. Chapter 8 on nuclear magnetic resonance spectroscopy (NMR) has been extensively revised by Dr Edrada-Ebel, who is a more frequent practitioner of the art than I. Since NMR is the most complex analytical technique, a chapter written by an expert seems most appropriate. Changes have also been made to Chapters 1, 3 and 11. The promised advance in techniques with potential for use in direct process control such as near infra red spectroscopy (NIRS) has been slow, so this section has not been revised. It is to be hoped that perhaps for the next edition I will be able to write an account of how NIRS process control has revolutionised pharmaceutical manufacture. For myself, as primarily a mass spectrometry specialist, the last 10 years has seen the most fantastic period of instrumental development; however, mass spectrometry remains primarily a research tool. I hope that in the course of this revision the book has remained as a succinct account of the most frequently used techniques in the field of pharmaceutical quality control. Dave Watson, University of StrathclydeIntentionally left as blankContents 1. Control of the quality of analytical methods 1 Introduction 1 Control of errors in analysis 2 Accuracy and precision 5 Validation of analytical procedures 7 Standard operating procedure (SOP) for the assay of paracetamol tablets 10 Compound random errors 11 Reporting of results 13 Other terms used in the control of analytical procedures 14 Basic calculations in pharmaceutical analysis 19 Additional problems 24 2. Physical and chemical properties of drug molecules 26 Introduction 26 Calculation of pH value of aqueous solutions of strong and weak acids and bases 27 Acidic and basic strength and pKa 29 Henderson–Hasselbalch equation 29 Ionisation of drug molecules 31 Buffers 33 Salt hydrolysis 36 Activity, ionic strength and dielectric constant 37 Partition coefficient 38 Drug stability 41 Stereochemistry of drugs 43 Measurement of optical rotation 49 Profiles of physico-chemical properties of some drug molecules 50 Additional problems 57 3. Titrimetric and chemical analysis methods 60 Keypoints 60 Introduction 61 Instrumentation and reagents 61 Direct acid/base titrations in the aqueous phase 62 Titrations of the salts of weak bases in mixed aqueous/non-aqueous media 65 Indirect titrations in the aqueous phase 66 Non-aqueous titrations 68 Argentimetric titrations 70 Compleximetric titrations 70 Redox titrations 71viii Pharmaceutical analysisIodometric titrations 73 Ion pair titrations 75 Diazotisation titrations 76 Potentiometric titrations 77 Karl Fischer titration (coulometric end-point detection) 81 Automation of wet chemical methods 82 Applications of FIA in pharmaceutical analysis 84 Additional problems 87 4. Ultraviolet and visible spectroscopy 90 Keypoints 90 Introduction 91 Factors governing absorption of radiation in the UV/visible region 92 Beer–Lambert Law 94 Instrumentation 95 Diode array instruments 96 Instrument calibration 96 UV spectra of some representative drug molecules 98 Use of UV/visible spectrophotometry to determine pKa values 102 Applications of UV/visible spectroscopy to pharmaceutical quantitative analysis 103 Difference spectrophotometry 107 Derivative spectra 109 Applications of UV/visible spectroscopy in preformulation and formulation 112 Additional problems 113 5. Infrared spectrophotometry 115 Keypoints 115 Introduction 116 Factors determining intensity and energy level of absorption in IR spectra 117 Instrumentation 118 Sample preparation 120 Application of IR spectrophotometry in structure elucidation 123 Examples of IR spectra of drug molecules 124 IR spectrophotometry as a fingerprint technique 127 Infrared spectrophotometry as a method for identifying polymorphs 130 Near-infrared analysis (NIRA) 130 Keypoints 130 Introduction 131 Examples of NIRA applications 131 Additional problems 135 6. Atomic spectrophotometry 138 Atomic emission spectrophotometry (AES) 138 Keypoints 138 Introduction 138 Instrumentation 139 Contents ixExamples of quantitation by AES 140 Interferences in AES analysis 142 Assays based on the method of standard additions 143 Atomic absorption spectrophotometry (AAS) 145 Keypoints 145 Introduction 145 Instrumentation 146 Examples of assays using AAS 146 Some examples of limit tests employing AAS 148 Inductively coupled plasma emission spectroscopy 150 7. Molecular emission spectroscopy 152 Fluorescence spectrophotometry 152 Keypoints 152 Introduction 153 Instrumentation 154 Molecules which exhibit fluorescence 154 Factors interfering with fluorescence intensity 155 Applications of fluorescence spectrophotometry in pharmaceutical analysis 156 Raman spectroscopy 159 Keypoints 159 Introduction 160 Instrumentation 161 Applications 161 8. Nuclear magnetic resonance spectroscopy 165 Keypoints 165 Introduction 166 Instrumentation 167 Proton (1H) 168 Application of NMR to structure confirmation in some drug molecules 188 Carbon NMR 192 Two-dimensional NMR spectra 194 Application of NMR to quantitative analysis 199 Other specialised applications of NMR 200 9. Mass spectrometry 204 Keypoints 204 Introduction 205 Ion generation 205 Other ionisation methods 214 Ion separation techniques 215 A more detailed consideration of mass spectra 219 Molecular fragmentation patterns 220 Gas chromatography–mass spectrometry (GC–MS) 231 Applications of GC–MS with EI 232 x Pharmaceutical analysisTandem mass spectrometry 236 High-resolution mass spectrometry 244 Mass spectrometry of proteins 246 Mass spectrometry in drug discovery 248 10. Chromatographic theory 252 Introduction 252 Void volume and capacity factor 252 Calculation of column efficiency 253 Origins of band broadening in HPLC 254 Parameters used in evaluating column performance 258 Data acquisition 262 Report generation 263 11. Gas chromatography 265 Keypoints 265 Introduction 266 Instrumentation 266 Selectivity of liquid stationary phases 272 Use of derivatisation in GC 280 Summary of parameters governing capillary GC performance 283 GC detectors 284 Applications of GC in quantitative analysis 284 Determination of manufacturing and degradation residues by GC 291 Determination of residual solvents 294 Solid-phase microextraction (SPME) 297 Applications of GC in bioanalysis 298 Additional problems 299 12. High-performance liquid chromatography 301 Keypoints 302 Introduction 302 Instrumentation 302 Stationary and mobile phases 303 Structural factors which govern rate of elution of compounds from HPLC columns 306 More advanced consideration of solvent selectivity in reverse-phase chromatography 313 Effect of temperature on HPLC 316 Summary of stationary phases used in HPLC 317 A more advanced consideration of reverse-phase stationary phases 320 Summary of detectors used in HPLC 322 Performance of a diode array detector 323 Applications of HPLC to the quantitative analysis of drugs in formulations 327 Assays involving more specialised HPLC techniques 340 Additional problems 355 Contents xi13. Thin-layer chromatography 358 Keypoints 358 Introduction 359 Instrumentation 359 TLC chromatogram 360 Stationary phases 361 Elutropic series and mobile phases 361 Modification of TLC adsorbant 365 Detection of compounds on TLC plates following development 366 Applications of TLC analysis 367 High-performance TLC (HPTLC) 372 14. High-performance capillary electrophoresis 376 Keypoints 376 Introduction 377 Instrumentation 380 Control of separation 382 Applications of CE in pharmaceutical analysis 384 Use of additives in the running buffer 388 Additional problems 396 15. Extraction methods in pharmaceutical analysis 398 Keypoints 398 Introduction 399 Commonly used excipients in formulations 399 Tablets and capsules 399 Solvent extraction methods 400 Microdialysis extraction 404 Solid-phase extraction (SPE) 404 Keypoints 404 Introduction 405 Methodology 405 Types of adsorbants used in SPE 407 Typical extraction methodologies using lipophilic silica gels 408 Adaptation of SPE for automated on-line extraction prior to HPLC analysis 413 Recent developments in solid-phase and on-line extraction 414 Index 417 Intentionally left as blank1 Control of the quality of analytical methods Introduction 1 Control of errors in analysis 2 Accuracy and precision 5 Validation of analytical procedures 7 The analytical procedure 7 Levels of precision 7 Repeatability 8 Intermediate precision 10 Reproducibility 10 Accuracy 10 Standard operating procedure (SOP) for the assay of paracetamol tablets 10 Compound random errors 11 Reporting of results 13 Introduction Other terms used in the control of analytical procedures 14 System suitability 14 Analytical blank 14 Calibration 14 Limit of detection 14 Limit of quantification 16 Linearity 16 Range 17 Robustness 18 Selectivity 18 Sensitivity 18 Weighing by difference 19 Basic calculations in pharmaceutical analysis 19 Percentage volume/volume (%v/v) 20 Percentage weight in volume (%w/v) 20 Dilutions 20 Preparation of standard stock solutions 21 Percentage weight/weight (%w/w) 22 Parts per million (ppm) calculations 23 Working between weights and molarity 23 Additional problems 24 Pharmaceutical analysis procedures may be used to answer any of the questions outlined inBox 1.1. The quality of a productmay deviate fromthe standard required butin carrying out an analysis one also has to be certain that the quality of the analysis itself is of the standard required. Quality control is integral to all modern industrial processes and the pharmaceutical industry is no exception. Testing a pharmaceutical product involves chemical, physical and sometimes microbiological analyses. It has been estimated that £10 billion is spent each year on analyses in the UK alone and such analytical processes can be found in industries as diverse as those producing food, beverages, cosmetics, detergents, metals, paints, water, agrochemicals, biotechnological products and pharmaceuticals. With such large amounts of money being spent on analytical quality control, great importance must be placed on providing accurate and precise analyses (Box 1.2). Thusitis appropriateto begin a book onthetopic of pharmaceutical analysis by considering, at a basic level, the criteria which are used to judge the quality of an analysis. The terms used in defining analytical quality form a rather elegant vocabulary that can be used to describe quality in many fields, and in writing this book the author would hope to describe each topic under consideration with accuracy, precision and, most importantly, reproducibility, so that the information included in it can be readily assimilated and2 Pharmaceutical analysisBox 1.1 Questions pharmaceutical analysis methods are used to answer • Is the identity of the drug in the formulated product correct? • What is the percentage of the stated content of a drug present in a formulation? • Does this formulation contain solely the active ingredient or are additional impurities present? • What is the stability of a drug in the formulation and hence the shelf-life of the product? • Atwhat rateisthe drug released fromits formulation sothatit can be absorbed bythe body? • Do the identity and purity of a pure drug substance to be used in the preparation of a formulation meet specification? • Do the identity and purity of excipients to be used in the preparation of a formulation meet specification? • What are the concentrations of specified impurities in the pure drug substance? • What is the concentration of the drug in a sample of tissue or biological fluid? • What are the pKa value(s), partition coefficients, solubilities and stability of a drug substance under development? Box 1.2 International Conference on Harmonisation (ICH) guidelines The requirements for control of the quality of methods of analysis (validation) have been addressed by the International Conference on Harmonisation of Technical Requirements For Registration of Pharmaceuticals for Human Use, or, more briefly, the ICH (www.ich. org). The ICH was initiated in Brussels in 1990 and brought together representatives of regulatory agencies and industry associations of Europe, Japan and the USA. The purpose of the organisation was to standardise the requirements for medicines regulation throughout the world. The standardisation of the validation of analytical procedures is one area that the ICH has addressed. The ICH indicated that the most important analytical procedures that require validation are: • Identification tests • Quantitative tests for impurities • Limit tests for the control of impurities • Quantitative tests of the active moiety in samples of drug substance or drug product or other selected component(s) in the drug product. reproduced where required by the reader. The following sections provide an introduction to the control of analytical quality. More detailed treatment of the topic is given in the reference cited at the end of the chapter.1 Control of errors in analysis A quantitative analysis is not a great deal of use unless there is some estimation of how prone to error the analytical procedure is. Simply accepting the analytical result could lead to rejection or acceptance of a product on the basis of a faulty analysis. For this reason it is usual to make several repeat measurements of the same sample in order to determine the degree of agreement between them. There are three types of errors which may occur in the course of an analysis: gross, systematic and random. Gross errors are easily recognised since they involve a major breakdown in the analytical process such as samples being spilt, wrong dilutions being prepared or instruments breaking down or being used in the wrong way. If a gross error occurs the results are rejected and the analysis is repeated from the beginning. Random and systematic errors can be distinguished in the following example: Control of the quality of analytical methods 3A batch of paracetamol tablets are stated to contain 500 mg of paracetamol per tablet; for the purpose of this example it is presumed that 100% of the stated content is the correct answer. Four students carry out a spectrophotometric analysis of an extract from the tablets and obtain the following percentages of stated content for the repeat analysis of paracetamol in the tablets: Student 1: 99.5%, 99.9%, 100.2%, 99.4%, 100.5% Student 2: 95.6%, 96.1%, 95.2%, 95.1%, 96.1% Student 3: 93.5%, 98.3%, 92.5%, 102.5%, 97.6% Student 4: 94.4%, 100.2%, 104.5%, 97.4%, 102.1% The means of these results can be simply calculated according to the formula: x  ¼ X i xi n [Equation 1] Fig. 1.1 Diagrammatic representation of accuracy and precision for analysis of paracetamol in tablet form. where x  is the arithmetic mean, xi is the individual value and n is the number of measurements. These results can be seen diagrammatically in Figure 1.1. Student 1 has obtained a set of results which are all clustered close to 100% of the stated content and with a mean for the five measurements very close to the correct answer. In this case the measurements made were both precise and accurate, and obviously the steps in the assay have been controlled very carefully. Student 2has obtaineda set of resultswhich are closelyclustered, but give amean which is lower than the correct answer. Thus, although this assay is precise, it is not completely accurate. Such a set of results indicates that the analyst has not produced random errors, which would produce a large scatter in the results, but has produced an analysis containing a systematic error. Such errors might include repeated inaccuracy in the measurement of a volume or failure to zero the spectrophotometer correctly prior to taking the set of readings. The analysis has been mainly well controlled except for probably one step, which has caused the inaccuracy and thus the assay is precisely inaccurate. mean Student 1: Precise  and accurate mean  Student 2: Precise  and inaccurate Student 3: Imprecise and inaccurate Student 4: Imprecise  and accurate mean mean  90 95 100 105 Paracetamol % of stated content 4 Pharmaceutical analysisStudent 3 has obtained a set of results which are widely scattered and hence imprecise, and which give a mean which is lower than the correct answer. Thus the analysis contains random errors or, possibly, looking at the spread of the results, three defined errors which have been produced randomly. The analysis was thus poorly controlled and it would require more work than that required in the case of student 2 to eliminate the errors. In such a simple analysis the random results might simply be produced by, for instance, a poor pipetting technique, where volumes both higher and lower than that required were measured. Student 4 has obtained a set of results which are widely scattered yet a mean which is close to the correct answer. It is probably only chance that separates the results of student 4 from those of student 3 and, although the answer obtained is accurate, it would not be wise to trust it to always be so. The best assay was carried out by student 1, and student 2 produced an assay that might be improved with a little work. In practice it might be rather difficult to tell whether student 1 or student 2 had carried out the best analysis, since it is rare, unless the sample is a pure analytical standard, that the exact content of a sample is known. In order to determine whether student 1 or 2 had carried out the best assay it might be necessary to get other analysts to obtain similar sets of precise results in order to be absolutely sure of the correct answer. The factors leading to imprecision and inaccuracy in assay results are outlined in Box 1.3. Box 1.3 Some factors giving rise to imprecision and inaccuracy in an assay • Incorrect weighing and transfer of analytes and standards • Inefficient extraction of the analyte from a matrix, e.g. tablets • Incorrect use of pipettes, burettes or volumetric flasks for volume measurement • Measurement carried out using improperly calibrated instrumentation • Failure to use an analytical blank • Selection of assay conditions that cause degradation of the analyte • Failure to allow for or to remove interference by excipients in the measurement of an analyte Self-test 1.1 Suggest how the following might give rise to errors in an analytical procedure: (i) Analysis of a sucrose-based elixir using a pipette to measure aliquots of the elixir for analysis. (ii) Weighing out 2 mg of an analytical standard on a four-place analytical balance which weighs a minimum of 0.1 mg. (iii) Use of an analytical standard that absorbs moisture from the atmosphere. (iv) Incomplete powdering of coated tablets prior to extraction. (v) Extraction of an ointment with a solvent in which it is poorly soluble. (vi) Use of a burette that has not been rinsed free of traces of grease. of the analyte; (vi) Distortion of meniscus making reading of the burette inaccurate. of moisture absorption is uncertain; (iv) Poor recovery of the analyte; (v) Poor recovery 2.5%; (iii) The degree   0.05 mg, which in relation to 2 mg is   there is an uncertainty of : (i) Viscosity leads to incomplete drainage of the pipette; (ii) In any weighing Answers Accuracy and precision Control of the quality of analytical methods 5The most fundamental requirements of an analysis are that it should be accurate and precise. It is presumed, although it cannot be proven, that a series of measurements (y) of the same sample will be normally distributed about a mean (m), i.e. they fall into a Gaussian pattern as shown in Figure 1.2. The distance s shown in Figure 1.2 appears to be nearly 0.5 of the width of distribution; however, because the function of the curve is exponential it tends to zero and does not actually meet the x axis until infinity, where there is an infinitesimal probability that there may be a value for x. For practical purposes approximately 68% of a series of measurements should fall within the distance s either side of the mean and 95% of the measurements should lie with 2s of the mean. The aim in an analysis is to make s as small a percentage of the value of m as possible. The value of s can be estimated using Equation 2: s Pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi [Equation 2] s ¼ where: s ¼ standard deviation n ¼ number of samples ðxi   x Þ2 ðn   1Þ xi ¼ values obtained for each measurement x  ¼ mean of the measurements Sometimes n rather than n – 1 is used in the equation but, particularly for small samples, it tends to produce an underestimate ofs. For a small number of values it is simple to work out s using a calculator and the above equation. Most calculators have a function which enables calculation ofsdirectly andsestimated usingthe above equationis usuallylabelled as sn – 1. For instance, if the example of results obtained by student 1, where the mean is calculated to be 99.9%, are substituted into equation 2, the following calculation results: vuut ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s ¼ vuut ð99:5   99:9Þ2 þ ð99:9   99:9Þ2 þ ð100:2   99:9Þ2 þ ð99:4   99:9Þ2 þ ð100:5   99:9Þ2 ð5   1Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 0:4Þ2 þ ð0Þ2 þ ð0:3Þ2 þ ð 0:5Þ2 þ ð0:6Þ2 ¼ 4 s ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:16 þ 0 þ 0:09 þ 0:25 þ 0:36 4 ¼ s ffiffiffiffiffiffiffiffiffi 0:86 4 ¼ ffiffiffiffiffiffiffiffiffiffiffi 0:215 p ¼ 0:46 s ¼ 0.46% of stated content Fig. 1.2 The Gaussian distribution. y x 68% μ – 2σ μ – σ μ μ + σ μ + 2σ 6 Pharmaceutical analysisThe calculated value for s provides a formal expression of the scatter in the results from the analysis rather than the visual judgement used in Figure 1.1. From the figure obtained for the standard deviation (SD), we can say that 68% of the results of the analysis will lie within the range 99.9   0.46% (  s) or within the range 99.44–100.36%. If we re-examine the figures obtained by student 1, it can be seen that 60% of the results fall within this range, with two outside the range, including one only very slightly below the range. The range based on   s defines the 68% confidence limits; for 95% confidence   2s must be used, i.e. 95% of the results of student 1 lie within 99.9   0.92% or 98.98–100.82%. It can be seen that this range includes all the results obtained by student 1. The precision of an analysis is often expressed as the   relative standard deviation (  RSD) (Equation 3). RSD ¼ sx   100% [Equation 3] The confidence limits in this case are often not quoted but, since it is the SD that is an estimate of s which is being used, they are usually 68%. The advantage of expressing precision in this way is that it eliminates any units and expresses the precision as a percentage of the mean. The results obtained from the assay of paracetamol tablets are shown in Table 1.1. Table 1.1 Results obtained for the analysis of paracetamol tablets by four analysts Student Mean (% of stated content) S (% of stated content)   RSD (68% confidence) 1 99.9 0.5   0.5% 2 95.6 0.5   0.5% 3 96.9 4.0   4.4% 4 99.7 4.0   4.0% Self-test 1.2 Four analysts obtain the following data for a spectrophotometric analysis of an injection containing the local anaesthetic bupivacaine. The stated content of the injection is 0.25% weight in volume (w/v). Analyst 1: 0.245% w/v, 0.234% w/v, 0.263% w/v, 0.261% w/v, 0.233% w/v. Analyst 2: 0.236% w/v, 0.268% w/v, 0.247% w/v, 0.275% w/v, 0.285% w/v. Analyst 3: 0.248% w/v, 0.247% w/v, 0.248% w/v, 0.249% w/v, 0.253% w/v. Analyst 4: 0.230% w/v, 0.233% w/v, 0.227% w/v, 0.230% w/v, 0.229% w/v. Calculate the mean percentage of stated content and RSD for each set of results at the 68% confidence level. Assuming the content really is as stated on the label, comment on the accuracy and precision of each set of results. Calculate the precision of each assay with regard to 95% confidence limits. 1.8. ¼   0.9%: inaccurate and precise. At 95% confidence RSD   : 91.9% Analyst 4 1.8%; ¼   0.9%: accurate and precise. At 95% confidence RSD   : 99.6% Analyst 3 15.4%;   ¼ 7.7%: inaccurate and imprecise. At 95% confidence RSD   : 104.9 Analyst 2 11.6%;   ¼ 5.8%: accurate but imprecise. At 95% confidence RSD   : 98.9% Analyst 1 : Answers Control of the quality of analytical methods 7Validation of analytical procedures The International Conference on Harmonisation (ICH) has adopted the following terms for defining how the quality of an assay is controlled. The analytical procedure The analytical procedure provides an exact description of how the analysis is carried out. It should describe in detail the steps necessary to perform each analytical test. The full method should describe: (i) the quality and source of the reference standard for the compound being analysed (ii) the procedures used for preparing solutions of the reference standard (iii) the quality of any reagents or solvents used in the assay and their method of preparation (iv) the procedures and settings used for the operation of any equipment required in the assay (v) the methodology used for calibration of the assay and methodology used for the processing of the sample prior to analysis. In fact it is difficult to be comprehensive in this short account, since the description of a fully validated method is a lengthy document. Levels of precision The ICH guidelines define precision as follows: “the precision of an analytical procedure expresses the closeness of agreement (degree of scatter) between a series of measurements obtained from multiple sampling of the same homogeneous sample under the prescribed conditions... The precision of an analytical procedure is usually expressed as the variance, standard deviation or coefficient of variation of a series of measurements.” This is broadly what was described in more detail above for the assay of paracetamol tablets. There is no absolute guideline for how good precision should be for the active ingredient in a formulation but, in general, a precision of <   1.0% is desirable. The precision achievable depends on the nature of the sample being analysed. The RSDs achievable in the analysis of trace impurities in a bulk drug or drugs in biological fluids may be considerably greater than   1.0% because of the increased likelihood of losses when very low concentrations of analyte are being extracted and analysed. The precision ofthe assay of a particular sample,inthe firstinstance,is generally obtained by repeating the assay procedure a minimum of five times starting from five separate aliquots of sample (e.g. five weights of tablet powder or five volumes of elixir) giving a total of 25 measurements. Repetition of the sample extraction gives a measure of any variation in recovery during extraction from the formulation matrix. One difficulty in defining the precision of an assay is in indicating which steps in the assay should be examined. Initially an assay will be characterised in detail but thereafter, in re-determining precision (e.g. in order to establish repeatability and intermediate precision), certain elements in the assay may be taken for granted. For example, the same standard calibration solution may be used for several days provided its stability to storage has been established. Similarlythere needsto be alimited number of samples extracted for assay, provided it has been established that the recovery of the sample upon extraction does not vary greatly. According to the ICH guidelines, precision may be considered at three levels: repeatability, intermediate precision and reproducibility. 8 Pharmaceutical analysisRepeatability Repeatability expresses the precision obtained under the same operating conditions over a short interval of time. Repeatability can also be termed intra-assay precision. It is likely that the assay would be repeated by the same person using a single instrument. Within repeatability it is convenient to separate the sample preparation method from the instrument performance. Figure 1.3 shows the levels of precision including some of the parameters which govern the system precision of a high-pressure liquid chromatography (HPLC) instrument. It would be expected that the system precision of a well-maintained instrument would be better than the overall repeatibility where sample extraction and dilution steps are prone to greater variation than the instrumental analysis step. An excellent detailed summary of levels of precision is provided by Ermer. For example Figure 1.4 shows the results obtained from five repeat injections of a mixture of the steroids prednisone (P) and hydrocortisone (H) into a HPLC using a manual loop injector. The mixture was prepared by pipetting 5 ml of a 1 mg/ml stock solution of each steroid into a 100 ml volumetric flask and making up to volume with water. The precision obtained for the areas of the hydrocortisone peak is   0.3%; the injection process in HPLC is generally very precise and one might expect even better precision from an automated injection system, thus this aspect of system precision is working well. The precision obtained for the prednisone peak is   0.6%; not quite as good, but this is not to do with the injector but is due to a small impurity peak (I) which Reproducibility Between laboratory transfer Round robin trials Long term performance Intermediate precision Instrument columns Chemicals Reference standards Operators Time Repeatability Sample extraction method Weighing Solution preparation Dilution Glassware pH adjustment System precision Injection precision Integration settings Flow rate Mobile phase composition Column performance Fig. 1.3 Parameters involved in three levels of precision. 0.321 240 nm Detector stability Control of the quality of analytical methods 9Fig. 1.4 Repeat injection of a mixture of prednisone (0.05 mg/ml) and hydrocortisone (0.05 mg/ ml) into a high-pressure liquid chromatography (HPLC) system. P H I P H I P H I P H I P H I runs closely after the prednisone causing some slight variation in the way the prednisone peak is integrated. The integration aspect of the system precision is not working quite as well when challenged with a difficulty in the integration method but the effect is really only minor. For the repeat, analysis is carried out on a subsequent day on a solution freshly prepared by the same method. The injection precision for hydrocortisone was   0.2%. The variation for the means of the areas of hydrocortisone peaks obtained on the 2 days was   0.8%; this indicates that there was small variation in the sample preparation (repeatability) between the 2 days since the variation in injection precision is   0.2%–  0.3%. Usually it is expected that instrument precision is better than the precision of sample preparation if a robust type of instrument is used in order to carry out the analysis. Table 1.2 shows the results for the repeat absorbance measurement of the same sample with a UV spectrophotometer in comparison with the results obtained from measurement of five samples by a two-stage dilution from a stock solution. In both cases the precision is good but, as would be expected, the better precision is obtained where it is only instrumental precision that is being assessed. Of course instruments can malfunction and produce poor precision, for example a spectrophotometer might be nearing the end of the useful lifetime of its lamp and this would result in poor precision due to unstable readings. A target level for system precision is generally accepted to be <   1.0%. This should be easily achievable in a correctly functioning HPLC system but might be more difficult in, for instance, a gas chromatography assay where the injection Table 1.2 Comparison of precision obtained from the repeat measurement of the absorbance of a single sample compared with the measurement of the absorbance of five separately prepared dilutions of the same sample Sample Absorbance readings RSD% Repeat measurement 0.842, 0.844, 0.842, 0.845, 0.841  0.14 Repeat dilution/measurement 0.838, 0.840, 0.842, 0.845, 0.847  0.42 10 Pharmaceutical analysisprocess is less well controlled. The tolerance levels may be set at much higher RSD in trace analyses where instruments are operated at higher levels of sensitivity and achieving 100% recovery of the analyte may be difficult. Intermediate precision Intermediate precision expresses within-laboratory variation of precision when the analysis is carried out by different analysts, on different days and with different equipment. Obviously a laboratory will want to cut down the possibility for such variations being large and thus it will standardise on particular items of equipment, particular methods of data handling, and make sure that all their analysts are trained to the same standard. Reproducibility Reproducibility expresses the precision between laboratories. Such a trial would be carried out when a method was being transferred from one part of a company to another. The data obtained during such method transfer does not usually form part of the marketing dossier submitted in order to obtain a product licence. For new methodologies a popular method for surveying the performance of a method is to carry out a round robin trial, where many laboratories are asked to carry out qualitative and quantitative analysis of a sample where the composition is only known to those organising the trial. Accuracy As described above, methods may be precise without being accurate. The determination of accuracy in the assay of an unformulated drug substance is relatively straightforward. The simplest method is to compare the substance being analysed with a reference standard analysed by the same procedure. The reference standard is a highly characterised form of the drug which has been subjected to extensive analysis including a test for elemental composition. The methods for determining the accuracy of an assay of a formulated drug are less straightforward. The analytical procedure may be applied to: a drug formulation prepared on a small scale so that the amount of drug in the formulation is more precisely controlled than in a bulk process; a placebo formulation spiked with a known amount of drug or the formulated drug spiked with a known amount of drug. The accuracy of the method may also be assessed by comparison of the method with a previously established reference method such as a pharmacopoeial method. Accuracy should be reported as percent recovery in relation to the known amount of analyte added to the sample or as the difference between the known amount and the amount determined by analysis. In general, at least five determinations, at 80, 100 and 120% of the label claim for drug in the formulated product, should be carried out in order to determine accuracy. Standard operating procedure (SOP) for the assay of paracetamol tablets The terms defined above are perhaps best illustrated by using the example of the simple assay that we mentioned before. The assay in Box 1.4 is laid out in the style of a standard operating procedure (SOP). This particular section of the operating procedure Control of the quality of analytical methods 11Box 1.4 Extract from a standard operating procedure for the analysis of paracetamol tablets 8. Assay procedure: 8.1 Use a calibrated balance 8.2 Weigh 20 tablets 8.3 Powder the 20 paracetamol tablets and weigh accurately by difference a quantity of tablet powder equivalent to 125   10 mg of paracetamol 8.4 Shake the tablet powder sample with ca 150 ml of acetic acid (0.05 M) for 10 min in a 500 ml volumetric flask and then adjust the volume to 500 ml with more acetic acid (0.05 M). 8.5 Filter ca 100 ml of the solution into a conical flask and then transfer five separate 5 ml aliquots of the filtrate to 100 ml volumetric flasks and adjust the volumes to 100 ml with acetic acid (0.05 M) 8.6 Take two readings of each dilution using a UV spectrophotometer and using the procedure specified in Section 9 describes the assay itself, but there would also be other sections in the procedure dealing with safety issues, the preparation and storage of the solutions used for extraction and dilution, the glassware required and a specification of the instrumentation to be used. The assay described in Box 1.4 assesses the precision of some of the operations within the assay. If a single analyst were to assess the repeatability of the assay, instructions might be issued to the effect that the assay as described was to be repeated five times in sequence, i.e. completing one assay before commencing another. If between-day repeatability were to be assessed, the process used for determining the repeatability would be repeated on two separate days. If the within laboratory reproducibility were to be assessed two or more analysts would be assigned to carry out the repeatability procedure. In arriving at an SOP such as the one described in Box 1.4, there should be some justification in leaving out certain steps in the complete assay. For instance, weighing is often the most precise step in the process and thus repeat weighings of samples of tablet powder would not be necessary to guarantee precision; the precision of the extraction might be more open to question. Each of the sections within an assay would have other SOPs associated with them, governing, for instance, the correct use and care of balances, as listed in Box 1.5. Compound random errors Systematic errors in analysis can usually be eliminated, but true random errors are due to operations in an assay which are not completely controlled. A common type of random error arises from the acceptance of manufacturers’ tolerances for glassware. Table 1.3 gives the RSD values specified for certain items of grades A and B glassware. An estimate of compound random errors is obtained from the square root of the sum of the squares of the RSDs attributed to each component or operation in the analysis. If the analysis of paracetamol described in Box 1.4 is considered, then, assuming the items of glassware are used correctly, the errors involved in the dilution steps can be simply estimated from the tolerances given for the pipette and volumetric flasks. 12 Pharmaceutical analysisBox 1.5 Procedure for the use of a calibrated balance SOP/001A/01 This balance is a high-grade analytical balance. The balance is sited in a vibration-free area and disturbance by draughts should be avoided. It carries out internal calibration but as a double check it is checked with certified check weights. Any deviation of the check weight values from those expected indicates need for servicing of the balance. Check weight calibration should be carried out once a week according to the instructions in SOP/001C/01. Caution: The logbook (form SOP/001 AR/01) must be filled in. Any spillages on the balance must be cleaned up immediately and recorded in the log. This balance is to be used only for analytical grade weighings. Operation 1. When carrying out weighing of amounts < 50 mg use tweezers to handle the weighing vessel. 2. Make sure the door of the balance is shut. Switch on the balance and allow it to undergo its internal calibration procedure. When it is ready the digital read-out will be 0.0000. Wait 30 s to ensure that the reading has stabilised. 3. Introduce the weighing vessel onto the balance pan. Close the door. Wait 30 s to ensure that the reading has stabilised and then send the reading to the printer. 4. If the tare is used in the weighing procedure, press the tare button and wait until the balance reads 0.0000. Wait 30 s to ensure that the reading has stabilised. If it drifts, which under normal circumstances it should not, press the tare button again and wait for a stable reading. 5. Remove the weighing vessel from the balance, introduce the sample into the vessel and put it back onto the balance pan. Close the door and note the reading. 6. Remove the sample and adjust the sample size to bring it closer to the required amount. Re-introduce the sample onto the balance pan. Close the door and note the reading. 7. Repeat step 5 until the target weight is reached. When the required weight is reached wait 30 s to ensure that the reading has stabilised. Send the reading to the printer. N.B. An unstable reading may indicate that moisture is being lost or gained and that the sample must be weighed in a capped vessel. Date of issue: 6/10/95 Signature: Table 1.3 Manufacturers’ tolerances on some items of glassware Item of glassware Grade A Grade B 1 ml bulb pipette   0.7%   1.5% 5 ml bulb pipette   0.3%   0.6% 100 ml volumetric flask   0.08%   0.15% 500 ml volumetric flask   0.05%   0.1% full 25 ml burette   0.2%   0.4% The British Standards Institution (BS) tolerances for the grade A glassware used in the assay are as follows: 500ml volumetric flask 500 ml   0:05% 100ml volumetric flask 100 ml   0:08% 5ml one mark pipette 5 ml   0:3% Standard deviation of error from glassware ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:052 þ 0:082 þ 0:32 p¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi 0:0989 p ¼ 0:31% Control of the quality of analytical methods 13Thus it can be seen that the compound error from the glassware differs little from the largest errorinthe process. Of coursethe glassware errors can be eliminated by calibration of the glassware prior to use, but, in general, analysts will accept manufacturers’ tolerances. The tolerated random error from glassware could be readily eliminated; other random errors, such as variation in the extraction efficiency, are more difficult to control. Self-test 1.3 Estimate the compound random error in the following assay with respect to the dilution steps described and calculate the error as SD of the w/v percentage of the injection, assuming it is exactly 2% w/v. A 2% w/v injection was diluted twice using grade A 5 ml bulb pipettes and grade A 100 ml volumetric flasks as follows: Dilution 1: 5 to 100 ml Dilution 2: 5 to 100 ml The uncertainty in the spectrophotometric reading was   0.2%. 0.01% w/v   0.48% and  : Answer Reporting of results In calculating an answer from the data obtained in an analysis it is important to not indicate a higher level of precision than was actually possible in the assay. As mentioned in the previous section, when considering the accuracy of glassware used with the assumption that it complied with the BS grade A standard, it was obvious that there was some uncertainty in any figure < 1%. It might be possible to improve on this degree of precision by calibrating glassware; however, any improvement in precision in the real world would take time and hence have cost implications. Thus for the purposes of most analyses, and for the purposes of the calculations in this book, it would seem sensible to report four significant figures, i.e. to 0.1%. In the process of carrying out calculations, five figures can be retained and rounded up to four figures at the end of the calculation. Since in pharmaceutical analyses the percentage of the stated content of a drug in a formulation may be reported as being between 90 and 99.9%, if the first significant figure is 9, then at the end of the calculation a more realistic estimate of precision is given by rounding the answer up to three significant figures. The SD or RSD reported with the answer should reflect the number of significant figures given; since there is usually uncertainty in figures < 1% of the answer, the RSD should not be reported below 0.1%. Taking this into consideration the correct and incorrect ways of reporting some answers are given in Table 1.4. Table 1.4 Significant figures in the reporting of analytical results Answer   S Incorrect RSD Answer   S Correct RSD % of stated content ¼ 99.2   0.22 0.22 % of stated content ¼ 99.2   0.2 0.2 % of stated content ¼ 101.15   0.35 0.35 % of stated content ¼ 101.2   0.4 0.4 0.2534   0.00443%w/v 1.75 0.2534   0.0044%w/v 1.7 1.0051   0.0063%w/w 0.63 1.005   0.006%w/w 0.6 1.784   0.1242 mg/ml 6.962 1.784   0.124 mg/ml 7.0 14 Pharmaceutical analysisOther terms used in the control of analytical procedures System suitability System suitability should not be confused with method validation. System suitability tests are most often applied to analytical instrumentation. They are designed to evaluate the components of the analytical system in order to show that the performance of the system meets the standards required by the method. Method validation is performed once at the end of method development, whereas system suitability tests are performed on a system periodically to determine whether or not it is still working properly and is capable of carrying out the analysis. System suitability relates to the performance of the equipment. In selecting equipment, the four Qs rule can be applied:2 (i) Design qualification (fit for purpose). What is the equipment required to do? (ii) Installation qualification. Does the equipment work in the way that the manufacturer claims? (iii) Operational qualification. Does the equipment work for the analyst’s particular application? (iv) Performance qualification. Does the instrument continue to perform to the standard required? In routine use it is point 4 that is checked, and, for a given procedure, an analyst will use several tests routinely, in order to monitor instrument performance, e.g. the resolution test during chromatography. The system suitability tests that would routinely be carried out on a HPLC are described in Chapter 10. Analytical blank This consists of all the reagents or solvents used in an analysis without any of the analyte being present. A true analytical blank should reflect all the operations to which the analyte in a real sample is subjected. It is used, for example, in checking that reagents or indicators do not contribute to the volume of titrant required for a titration, including zeroing spectrophotometers or in checking for chromatographic interference. Calibration The calibration of a method involves comparison of the value or values of a particular parameter measured by the system under strictly defined conditions with pre-set standard values. Examplesinclude: calibration ofthe wavelength and absorbance scales of a UV/visible spectrophotometer (Ch. 4), calibration of the wavelength scale of an IR spectrometer (Ch. 5) and construction of chromatographic calibration curves (Ch. 12). Limit of detection This is the smallest amount of an analyte which can be detected by a particular method. It is formally defined as follows: x   xB ¼ 3sB where x is the signal from the sample, xB is the signal from the analytical blank and sB is the SD of the reading for the analytical blank. In other words, the criterion for a reading reflecting the presence of an analyte in a sample is that the difference between Control of the quality of analytical methods 15the reading taken and the reading for the blank should be three times the SD of the blank reading. The SD of the signal from the sample can be disregarded since the sample and the blank should have been prepared in the same manner so that it and the sample produce a similar SD in their readings. A true limit of detection should reflect all the processes to which the analyte in a real assay is subjected and not be a simple dilution of a pure standard for the analyte until it can no longer be detected. The definition of limit of detection (LOD) above applies mostly to spectrophotometric readings. In the case of chromatographic separations there is usually a constant background reading called the baseline. According to the EP the limit of detection is where the height of a chromatographic peak is three times the height of the fluctuation of the baseline noise over a distance equal to 20 times width of the peak at half height. Chromatographic software will generally calculate this for the operator; however, manual estimation can be carried out as shown in Figure 1.5 where the LOD is reached where 2H/h is < 3. Fig. 1.5 Estimation of limit of detection (LOD) and limit of quantification (LOQ) for a chromatographic system. e cnadnuba evitaleRe cnadnuba evitaleR100 LOQ 90 80 70 H=5 60 50 40 30 H=1 20 10 0 1 2 3 4 5 6 7 8 9 10 Time (min) 100 LOD 90 80 H=1.5 70 60 50 H=1 40 30 20 10 0 1 2 3 4 5 6 7 8 9 10 Time (min) 16 Pharmaceutical analysisLimit of quantification The limit of quantification is defined as the smallest amount of analyte which can be quantified reliably, i.e. with an RSD for repeat measurement of <  20%. The limit of quantification is defined as: x–xB ¼ 10sB. In this case the analyte should give a peak at more than ten times the standard deviation of the chromatographic baseline during chromatographic analysis. As shown in Figure 1.5 where the limit of quantification (LOQ) is reached where 2H/h is <10. Self-test 1.4 In which of the following cases has the limit of detection been reached? Signal from sample Sample SD Signal from analytical blank Analytical blank SD 1. Abs 0.0063 0.0003 0.0045 0.0003 2. Abs 0.0075 0.0017 0.0046 0.0018 3. 0.335 ng/ml 0.045 ng/ml 0.045 ng/ml 0.037 ng/ml : 2 Answer Linearity Most analytical methods are based on processes where the method produces a response that is linear and which increases or decreases linearly with analyte concentration. The equation of a straight line takes the form: y ¼ a þ bx where a is the intercept of the straight line with the y axis and b is the slope of the line. Taking a simple example, a three-point calibration curve is constructed through readings of absorbance against procaine concentration (Table 1.5). The best fit of a straight line through these values can be determined by determining a and b from the following equations: X ðxi   x Þðyi   y Þ b ¼ i X i ðxi   x Þ2 a ¼ y   bx  where xi is the individual value for x, x is the mean value of xi, yi is the individual value for y and y  is the mean of yi. Table 1.5 Data used for the construction of a calibration curve for the spectrophotometric determination of procaine Procaine concentration mg/100 ml Absorbance reading 0.8 0.604 1.0 0.763 1.2 0.931 From the data in Table 1.4: x  ¼ 0:8 þ 1:0 þ 1:2 3 ¼ 1:0 y  ¼ 0:604 þ 0:763 þ 0:931 3 ¼ 0:766 Control of the quality of analytical methods 17b ¼ ð0:8   1:0Þð0:604   0:766Þþð1:0   1:0Þð0:763   0:766Þþð1:2   1:0Þð0:931   0:766Þ ð0:8   1:0Þ2 þ ð1:0   1:0Þ2 þ ð1:2   1:0Þ2 ¼ 0:0324 þ 0 þ 0:033 0:04 þ 0:04 ¼ 0:818 a ¼ 0:766   0:818   1:0 ¼  0:052 Thus the equation for the best fit is: y ¼ 0:818x   0:052 The statistical measure of the quality of fit of the line through the data is the correlation coefficient r. A correlation coefficient of > 0.99 is regarded as indicating linearity. The correlation coefficient is determined from the following equation: X i fðxi   x Þðyi   y Þg ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ¼ r X i ½ðxi   x Þ2 X i ½ðyi   y Þ2  Substituting the values from Table 1.4: r ¼ ð0:8   1:0Þð0:604   0:766Þþð1:0   1:0Þð0:763   0:766Þþð1:2   1:0Þð0:931   0:766Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ½ð0:8   1:0Þ2 þ ð1:0   1:0Þ2 þ ð1:2   1:0Þ2 ½ð0:604   0:766Þ2 þ ð0:763   0:766Þ2 þ ð0:931   0:766Þ2  r ¼ 0:0324 þ 0 þ 0:033 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:08   0:0534 p ¼ 1:00 Thus, to three significant figures, the straight line fit through the values in Table 1.4 is perfect. For a fuller treatment of the mathematical determination and significance of a correlation coefficient see Miller and Miller.1 The equation for the correlation coefficient is very useful in that it can be applied to correlations between curves of any shape and thus it can be used for spectral comparisons, such as those carried out between diode array spectra obtained during HPLC (Ch. 12). Range The term range can be applied to instrument performance (dynamic range) but, when applied to the performance of an assay, it refers to the interval between the upper and lower concentration of an analyte for which an acceptable level of precision and accuracy has been established. Typical ranges are: 80–120% of the stated amount for a finished product; 70–130% of the expected concentration, e.g. for content of single tablets (the range may be even wider for some products, such as doses delivered by a metered dose inhaler) and 0–110% for dissolution tests where the drug is released from the dosage form over a time period. 18 Pharmaceutical analysisRobustness Robustness is evaluated in order to determine how resistant the precision and accuracy of an assay is to small variations in the method. The types of parameters which are assessed in order to determine the robustness of a method include: the stability of analytical solutions; the length of the extraction time; the effect of variations in the pH of a HPLC mobile phase; the effect of small variations in mobile phase composition; the effect of changing chromatographic columns; the effect of temperature and flow rate during chromatography. Selectivity The selectivity of a method is a measure of how capable it is of measuring the analyte alone in the presence of other compounds contained in the sample. The most selective analytical methods involve a chromatographic separation. Detection methods can be ranked according to their selectivity. A simple comparison is between fluorescence spectrophotometry and UV spectrophotometry. There are many more compounds which exhibit UV absorption than there are those which exhibit strong fluorescence; thus fluorescence spectrophotometry is a more selective method. Because selective methods are based on more complex principles than non-selective methods, they may be less robust, e.g. fluorescence spectrophotometry is more affected by changes in the analytical method than UV spectrophotometry. Sensitivity The sensitivity of method indicates how responsive it is to a small change in the concentration of an analyte. It can be viewed as the slope of a response curve and may be a function of the method itself or of the way in which the instrumentation has been calibrated. In Figure 1.6 the method having a linear response y ¼ 2.5x is five times more sensitive than the method exhibiting a linear response y ¼ 0.5x. Sensitivity and the limit of detection of a method are often confused. The limit of detection is due to a combination of range and sensitivity. e snopseR60 50 40 30 20 10 0 y = 2.500x y = 0.500x 0 5 10 15 20 25 Concentration Response Response Fig. 1.6 Sensitivity as a function of the slope of a response curve. Weighing by difference Control of the quality of analytical methods 19Weighing by difference is used to minimise weighing errors in an analytical procedure. The sample is weighed in a suitable vessel, e.g. a glass weighing boat with a spout, and then transferred immediately to the vessel in which it is going to be analysed or dissolved. The weighing vessel is then reweighed and the difference between the weights before and after transfer gives the weight of the sample. This method of weighing minimises errors due to, for example, the absorption of moisture onto the surface of the vessel. It also means that there is not a requirement for complete transfer of the sample that is to be analysed. The points listed in Boxes 1.6 and 1.7 indicate how pharmaceutical preparations may come to be out of specification. Box 1.6 Sources of impurities in pharmaceutical manufacture • During the course of the manufacture of a pure drug substance, impurities may arise as follows: (i) Present in the synthetic starting materials (ii) Result from residual amounts of chemical intermediates used in the synthetic process and from unintended side reactions (iii) Result from reagents, solvents and catalysts used in manufacture. • The process used to produce the formulated drug substance may introduce impurities as follows: (i) Particulate matter from the atmosphere, machines and devices used in the manufacturing process and from containers (ii) Impurities that are present in the excipients used in the formulation (iii) Cross contamination may occur from other processes carried out using the same equipment, e.g. from mixers (iv) Microbial contamination may occur (v) The drug may react with the excipients used in the formulation (vi) Impurities may be introduced from packaging, e.g. polymeric monomers. Box 1.7 Processes leading to the deviation of the actual content from the stated content of a drug in a formulation • Incomplete mixing of drug with formulation excipients prior to compression into tablets or filling into capsules • Physical instability of the dosage form: tablets that disintegrate too readily; creams or suspensions that separate and over- or undercompression of tablets, leading to deviation from the required weight • Chemical breakdown of the drug resulting from its reaction with air, water, light, excipients in a formulation or with packaging materials • Partitioning of the drug into packaging materials. Basic calculations in pharmaceutical analysis The data from an analysis can be obtained using computer-based methods. However, in order to have some idea of the accuracy of an answer, it is necessary to be able to carry out calculations in the traditional manner. There are various units used to 20 Pharmaceutical analysisexpress amounts and concentrations in pharmaceutical analysis and examples of these units will be considered below. Percentage volume/volume (%v/v) Percentage volume/volume (%v/v) is most often encountered in relation to the composition of mobile phases used in HPLC. Thus, when 30 ml of methanol is mixed with 70 ml of water, a 30:70 v/v mixture is formed. Since some shrinking in volume occurs when two liquids are mixed, %v/v may only be approximate. Some chromatographers prefer to prepare mixtures of solvents by weighing them rather than by volume measurement and in this case the solvent mixture can be expressed as % weight in weight (%w/w). Percentage weight in volume (%w/v) Percentage weight in volume (%w/v) is normally used to express the content of active ingredient in liquid formulations such as injections, infusions and eyedrops. The density of the solvent in this case is irrelevant; thus, a 1 g/100 ml solution of a drug is 1%w/v whether it is dissolved in ethanol or water. Self-test 1.5 Convert the following concentrations to % w/v: (i) 0.1 g/100 ml (ii) 1 mg/ml (iii) 0.1 g/ml (iv) 100 mg/ml (iv) 0.01% w/v (iii) 10% w/v (ii) 0.1% w/v : (i) 0.1% w/v Answers Dilutions In order for an extract from a formulation or a solution of a pure drug substance to be measured it must be diluted so that it falls within the working range of the instrument used to make the measurement. Thus an understanding of dilution factors is fundamental to calculations based on analytical data. Calculation example 1.1 An infusion stated to contain 0.95% w/v NaCl was diluted so that its Na content could be determined by flame photometry. The following dilutions were carried out: (i) 10 ml of the sample was diluted to 250 ml with water. (ii) 10 ml of the diluted sample was diluted to 200 ml with water. The sample was found to contain 0.74 mg/100 ml of Na. Atomic weights: Na ¼ 23 Cl ¼ 35.5 (Continued) Calculation example 1.1 (Continued) Calculate: The % w/v of NaCl present in the infusion. The % of stated content of NaCl. Control of the quality of analytical methods 21Dilution factor ¼ 10 to 250 (  25), 10 to 200 (  20). Total dilution ¼ 25   20 ¼ 500. Therefore, in the original injection, conc. of Na ¼ 0.74   500 mg/100 ml ¼ 370 mg/100 ml. Conc. of NaCl in the injection ¼ 370   (58.5/23) ¼ 941 mg/100 ml ¼ 0.941 g/100 ml ¼ 0.941%w/v. % of stated content ¼ (0.941/0.95)   100 ¼ 99.1%. In the sample there are 941 mg/100 ml. Calculation example 1.2 A 2 ml volume of eyedrops containing the local anaesthetic proxymetacaine. HCl is diluted to 100 ml and then 5 ml of the dilution is diluted to 200 ml. The diluted sample was measured by UV spectrophotometry and was found to contain 0.512 mg/100 ml of the drug. Calculate the %w/v of the drug in the eyedrops. Dilution factors 2 to 100 (  50), 5 to 200 (  40). Total dilution 40   50 ¼ 2000. Original concentration ¼ 2000   0.512 ¼ 1024 mg/100 ml ¼ 1.024 g/100 ml ¼ 1.024% w/v. Self-test 1.6 A 5 ml sample of an injection of a steroid phosphate was diluted to 100 ml. Then 10 ml of the diluted injection was diluted to 100 ml and this dilution was further diluted 10 to100 ml. From measurement by UV the diluted sample was found to contain 0.249 mg/100 ml of the steroid. What was the original concentration of the injection in %w/v and in mg/ml? : 0.498% w/v, 4.98 mg/ml Answers Self-test 1.7 A sample of an infusion was diluted 5 ml to 200 ml and then 10 ml to 200 ml. It was then analysed and was found to contain sodium at 0.789 mg/100 ml. Calculate the concentration of sodium in the original sample in %w/v. The sample was composed of a mixture of sodium lactate and sodium bicarbonate in equimolar amounts. Calculate the amount of sodium lactate and sodium bicarbonate in mg in 10 ml of the sample (Na ¼ 23, lactate ¼ 89, bicarbonate ¼ 61). 115.3 mg/10 ml ¼ 153.7 mg/10 ml Na bicarbonate ¼ : 0.6312% w/v, Na lactate Answers Preparation of standard stock solutions In preparing a stock solution of a standard using a standard four-place balance, assuming there is no lack of availability of standard, it is best to weigh at least 100 mg of material, since an error of 0.1 mg in weight is only 0.1% of the weight taken. 22 Pharmaceutical analysisCalculation example 1.3 Assuming that you wish to avoid pipetting less than a 5 ml volume, starting from a 102.1 mg/100 ml stock solution, how would you prepare the following concentrations for a calibration series? 0.2042 mg/100 ml, 0.4084 mg/100 ml, 0.6126 mg/100 ml, 0.8168 mg/100 ml, 1.021 mg/ml When a large dilution (>10) is required it is best to carry it out in two stages. In this case an initial dilution of 20 ml to 100 ml is carried out, producing a 10.21 mg/100 ml solution. Then 250 ml volumetric flasks can be used to carry out the following dilutions. 5 to 250 (  50), 10 to 250 (  25), 15 to 250 (  16.67), 20 to 250 (  12.5), and 25 to 250 (  10), giving the dilution series above. N.B. Once the volume required for the lowest concentration (5 ml) has been determined, then it can be multiplied   2,   3,   4 and   5 to give the series. Self-test 1.8 A stock solution containing 125.6 mg of standard in 250 ml is prepared. Suggest how the following dilution series could be prepared (pipettes 5 ml or greater must be used). 0.1005 mg/100 ml, 0.2010 mg/100 ml, 0.3015 mg/100 ml, 0.4020 mg/100 ml, 0.5025 mg/100 ml 1, 5, 10, 15, 20 and 25 ml pipettes used to transfer into 100 ml volumetric flasks. : Number of possible answers, e.g. dilution 1, 10 ml to 250 ml. Then, from dilution Answer Percentage weight/weight (%w/w) Percentage weight/weight (%w/w) is a common measure used to express the concentration of active ingredient in a formulation such as a cream or to express the content of a minor impurity in a drug substance. Thus, a cream containing 10 mg (0.01 g) of drug per gram is a: (0.01/1)   100 ¼ 1% w/w formulation. Equally, if a drug contains 0.5 mg (0.0005 g) per gram of an impurity, the impurity is at a concentration of (0.0005/1)   100 ¼ 0.05% w/w. It is generally accepted that, for a drug, all impurities above 0.05% w/w should be characterised, i.e. their structures should be known and their toxicities should be assessed. In determining impurities in a drug, 1 g of the drug might be dissolved in 100 ml of solvent. If an analysis was carried out and the drug solution was found to contain 3 mg/100 ml of an impurity, then the %w/w referring back to the original weight of drug substance would be: (0.003/1)   100 ¼ 0.3% w/w Control of the quality of analytical methods 23Self-test 1.9 (i) 0.1521 g of a corticosteroid drug was dissolved in 100 ml of methanol/water (1:1 v/v). The sample was analysed by HPLC and was found to contain a known impurity in a concentration of 0.354 mg/100 ml. What is the %w/w of the impurity in the drug? (ii) 0.5321 g of a b-adrenergic blocking drug is dissolved in 50 ml of methanol/0.1% w/v acetic acid (20:80 v/v) and the sample is then diluted by taking 10 ml of the solution and diluting it to 50 ml. The diluted sample was analysed by HPLC and was found to contain 0.153 mg/100 ml of an impurity. What is the %w/w of the impurity in the drug? (ii) 0.0719% w/w : (i) 0.233% w/w Answers Parts per million (ppm) calculations Parts per million (ppm) on a w/w basis is 1 mg/g (1 mg/kg). It is a common measure used for impurities in drug substances, particularly heavy metals and solvents. 1 ppm is also 0.0001%w/w. Calculation example 1.4 The potassium content of an intravenous infusion containing sodium chloride was determined. The infusion was found to contain 0.9092% w/v NaCl. The undiluted infusion was measured for potassium content in comparison with a potassium chloride standard. The potassium content of the undiluted infusion was found to be 0.141 mg/ 100 ml. Calculate the potassium content in the sodium chloride in ppm: 0.9092%w/v ¼ 0.9092 g/100 ml Potassium content ¼ 0.141 mg/100 ml ¼ 141 mg/100 ml. Relative to the sodium chloride, there are 141 mg/0.9092 g ¼ 141/0.9092 ppm ¼ 155 ppm. Self-test 1.10 125.1 mg of streptomycin sulphate are dissolved in 10 ml of water. A GC headspace analysis is carried out in order to determine the methanol content of the drug. A peak for methanol is produced which has 73.2% of the area of a peak for a methanol standard containing 0.532 mg/100 ml of methanol in water analysed under exactly the same conditions. What is the methanol content of the streptomycin sulphate in ppm and %w/w? : 311.3 ppm, 0.03113%w/w. Answer Working between weights and molarity Weights are much easier to appreciate than molar concentrations but sometimes, particularly in bioanalytical methods, molar concentrations are used. Definitions Molar: molecular weight in g/l (mg/ml) mMolar: molecular weight in mg/l (mg/ml) mMolar: molecular weight in mg/l (ng/ml) nMolar: molecular weight in ng/l (pg/ml). 24 Pharmaceutical analysisCalculation example 1.5 The metabolism of paracetamol (molecular weight 151.2 amu) by liver microsomes was studied by preparing a 100 mM solution of paracetamol in 1 ml of incubation mixture. A 30.12 mg/100 ml solution of paracetamol in buffer was prepared. What volume of paracetamol solution had to be added to incubation mixture prior to making up the volume to 1 ml: 100 mM ¼ 100   151.2 ¼ 15120 mg/l ¼ 15.12 mg/ml. 30.12 mg/100 ml ¼ 30120 mg/100 ml ¼ 301.2 mg/ml ¼ 0.3012 mg/ml. Volume of paracetamol solution required ¼ 15.12/0.3012 ¼ 50.2 ml. Self-test 1.11 (i) Calculate the concentrations in mg/ml and mg/ml of a 10 mM solution of kanamycin (molecular weight 484.5). (ii) A solution containing diclofenac sodium (molecular weight 318.1) at a concentration of 79.5 mg/100 ml in buffer was prepared. What volume of this solution was required in order to carry out a microsomal incubation containing a 25 mM concentration of the drug in 1 ml? l m (ii) 10 l m g/ m : (i) 0.004845 mg/ml, 0.004845 Answers Additional problems All answers to be given to four significant figures. e.g. 1% w/v to four significant figures is 1.000 % w/v. 1. Convert the following concentrations to %w/v. (i) 10 mg/ml (ii) 100 mg/l (iii) 0.025 g/ml (iv) 250 mg/ml (v) 20 mg/ml. (v) 2.000% w/v (iv) 0.02500% w/v (iii) 2.500% w/v (ii) 0.01000% w/v : (i) 1.000% w/v Answers 2. An infusion which was stated to contain 0.5000% w/v KCl was diluted so that its K content could be determined by flame photometry. The following dilutions were carried out: (i) 10 ml of the sample was diluted to 200 ml with water. (ii) 10 ml of the diluted sample was diluted to 200 ml with water. The sample was found to contain 0.6670 mg/100 ml of K. (Atomic weights: K ¼ 39.1 Cl ¼ 35). Calculate: (i) The % w/v of KCl present in the infusion (ii) The % of stated content of KCl. (ii) 101.8% : (i) 0.5090% w/v Answers (Continued) Control of the quality of analytical methods 25Additional problems (Continued) 3. Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5 g Potassium Chloride 1.5 g Sodium Citrate 2.9 g Anhydrous Glucose 20.0 g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7,2H2O) molecular weight 294.1 and sodium chloride (NaCl) molecular weight 58.5 Atomic weight of Na ¼ 23 The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml Determine the % of stated content of Na in the sample. : 104.5% Answer 4. A 5 ml volume of eyedrops containing the mydriatic drug cyclopentolate. HCl is diluted to 100 ml and then 20 ml of the dilution is diluted to 100 ml. The diluted sample was measured by UV spectrophotometry and was found to contain 20.20 mg/100 ml of the drug. Calculate the % w/v of the drug in the eyedrops. : 2.020% w/v Answer 5. 0.5% w/v of an injection is to be used as an anaesthetic for a 2-week-old baby weighing 3.4 kg. The recommended dose for a bolus injection is 0.5 mg/kg. The injection must be given in 1 ml. Calculate the amount of water for injection that must be drawn into the syringe with the appropriate volume of injection. : 0.6600 ml Answer 6. 0.0641 g of a semi-synthetic alkaloid was dissolved in 25 ml of 1% w/v acetic acid and was analysed directly by HPLC. The solution was found to contain 0.142 mg/100 ml of an impurity. What is the level of impurity in % w/w and ppm. : 0.05538% w/w, 553.8 ppm. Answers 7. The level of ethyl acetate is determined in colchicine by headspace gas chromatography. A solution containing 4.361 g/100 ml of colchicine in water was prepared. An aqueous standard containing 0.5912 mg/100 ml of ethyl acetate was also prepared. Headspace analysis of 2 ml volumes of the two solutions produced GC peaks for ethyl acetate with the following areas: Colchicine solution: Peak area 13457 Ethyl acetate solution: Peak area 14689 Calculate the ethyl acetate content in the colchicine sample in ppm. : 124.2 ppm Answer References 1. Miller JC, Miller JN. Statistics and Chemometrics for Analytical Chemistry. 6th ed. Chichester: Ellis Horwood; 2010. 2. Burgess C. Good spectroscopic practice: science and compliance. Spectroscopy Europe 1994;6:10–3. Further reading Ermer J, Miller JHM, editors. Method Validation in Pharmaceutical Analysis. Weinheim: Wiley-VCH; 2005. Jones D. Pharmaceutical Statistics. London: Pharmaceutical Press; 2002. 2 Physical and chemical properties of drug molecules Introduction 26 Calculation of pH value of aqueous solutions of strong and weak acids and bases 27 Dissociation of water 27 Strong acids and bases 27 Weak acids and bases 28 Acidic and basic strength and pKa 29 Henderson–Hasselbalch equation 29 Ionisation of drug molecules 31 Buffers 33 Salt hydrolysis 36 Activity, ionic strength and dielectric constant 37 Introduction Partition coefficient 38 Effect of pH on partitioning 40 Drug stability 41 Zero-order degradation 42 First-order degradation 42 Stereochemistry of drugs 43 Geometrical isomerism 43 Chirality and optical isomerism 44 Diastereoisomers 47 Measurement of optical rotation 49 Profiles of physico-chemical properties of some drug molecules 50 Procaine 50 Paracetamol 51 Aspirin 52 Benzylpenicillin 52 5-Fluorouracil 53 Acebutolol 54 Sulfadiazine 54 Isoprenaline 55 Prednisolone 55 Guanethidine 56 Pyridostigmine bromide 56 Additional problems 57 The physical properties of organic molecules, such as pKa and partition coefficient, are dealt with extensively in pharmacy courses1,2 but do not feature greatly in analytical chemistry courses. It is often surprising that analytical chemists cannot distinguish between, for instance, basic, weakly basic, acidic, weakly acidic and neutral nitrogen functions. The physical properties of drug molecules, along with simple chemical derivatisation and degradation reactions, play an important part in the design of analytical methods. Drug molecules can be complex, containing multiple functional groups that in combination produce the overall properties of the molecule. This chapter will serve as a starting point for understanding the chemical and physico-chemical behaviour of drug molecules, which influence the development of analytical methods. The latter part of the chapter focuses on some typical drugs that are representative of a class of drug molecules and lists their physical properties and the properties of their functional groups in so far as they are known.Physical and chemical properties of drug molecules 27Calculation of pH value of aqueous solutions of strong and weak acids and bases Dissociation of water The pH of a solution is defined as log [Hþ], where [Hþ] is the concentration of hydrogen ions in solution. In pure water the concentration of hydrogen ions is governed by the equilibrium: H2O  KaHþ þ HO  Ka is the dissociation constant for the equilibrium, is known as Kw in the case of the dissociation of water and is determined by the following expression: Kw ¼ Hþ ½   HO  ½   ½   H2O ¼ Hþ ½   HO  ½  ¼ 10 14 Since the concentration of water does not change appreciably as a result of ionisation, its concentration can be regarded as not having an effect on the equilibrium and it can be omitted from the equation, and this means that in pure water: Hþ ½   ¼ HO  ½   ¼ 10 7 The pH of water is thus given by log 10 7¼7.00. Strong acids and bases If an acid is introduced into an aqueous solution the [Hþ] increases. If the pH of an aqueous solution is known, the [Hþ] is given by the expression 10 pH, e.g. [Hþ] in pH 4 solution¼10 4 M¼0.0001 M. Since [Hþ][OH ]¼10 14 for water, the concentration of [OH ] in this solution is 10 10 M. A strong acid is completely ionised in water and [Hþ] is equal to its molarity, e.g. 0.1 M HCl contains 0.1 M Hþ (10 1 Hþ) and has a pH of log 0.1¼1. For a solution of a strong base such as 0.1 M NaOH, [OH ]¼0.1 M and [0.1][Hþ]¼10 14; therefore, [Hþ]¼10 13 M and the pH of the solution¼13. Although the pH range is regarded as being between 0 and 14, it does extend above and below these values, e.g. 10 M HCl, in theory, has a pH of  1. Self-test 2.1 Calculate the pH of the following solutions: (i) 0.05 M HCl (ii) 0.05 M NaOH (iii) 0.05 M H2SO4. contains 0.1 M H 4 SO2 : (i) 1.3; (ii) 12.7; (iii) 1.0, since 0.05 M H Answers þ 28 Pharmaceutical analysisWeak acids and bases Weak acids are not completely ionised in aqueous solution and are in equilibrium with the undissociated acid, as is the case for water, which is a very weak acid. The dissociation constant Ka is given by the expression below: HA  KaA  þ Hþ Ka ¼ A  ½   Hþ ½   ½   HA For instance, in a 0.1 M solution of acetic acid (Ka¼1.75 10 5), the equilibrium can be written as follows: C0H3COOH ð0:1 xÞ  KaCH3COO  x þHþ x Ka ¼ CH3COO  ½   Hþ ½   ½   CH3COOH The pH can be calculated as follows: 1:75   10 5 ¼ x2 ð0:1   xÞ Since the dissociation of the acetic acid does not greatly change the concentration of the un-ionised acid, the above expression can be approximated to: 1:75   10 5 ¼ x2 0:1 1:75   10 6 p ¼ 0:00132 M x ¼ Hþ ½  ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pH ¼ 29 In comparison, the pH of 0.1 M HCl is 1. The calculation of the pH of a weak base can be considered in the same way. For instance, in a 0.1 M solution of ammonia (Kb¼1.8 10 5), the equilibrium can be written as follows: NH3 þ H2O ð0:1 xÞ  KbNH4þ x þ HO  x If the concentration/activity of water is regarded as being 1, then the equilibrium constant is given by the following expression:     HO  ½   Kb ¼ NHþ4 ½   NH3 1:8   10 5 ¼ x2 ð0:1   xÞ Physical and chemical properties of drug molecules 29The concentration of NH3 can be regarded as being unchanged by a small amount of ionisation and the expression can be written as: 1:8   10 6 p ¼ 0:0013 M x ¼ HO  ½  ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:0013 ¼ 10 14 Hþ ½  ¼ Kw pH ¼ 11:1 0:0013 ¼ 7:7   10 12 M In comparison, the pH of 0.1 M NaOH is 13. Self-test 2.2 Calculate the pH of the following solutions: (i) 0.1 M formic acid (Ka¼1.77 10 4) (ii) 0.05 M phenol (Ka¼1.3 10 10) (iii) 0.15 M ethylamine (Kb¼5.6 10 4) : (i) 2.4; (ii) 5.6; (iii) 11.9 Answers Acidic and basic strength and pKa The pKa value of a compound is defined as: pKa¼ log Ka. A pKa value can be assigned to both acids and bases. For an acid, the higher the [Hþ] the stronger the acid, e.g.: CH3COOH  KaCH3COO  þ Hþ In the case of a base, it is the protonated form of the base that acts as a proton donor, e.g.: NHþ4  KaNH3 þ H  In this case, the lower the [Hþ] the stronger the base. If pKa is used as a measure of acidic or basic strength: for an acid, the smaller the pKa value the stronger the acid; for a base the larger the pKa value the stronger the base. Henderson–Hasselbalch equation Ka ¼ A  ½   Hþ ½   ½   HA Can be rearranged substituting pH for log [Hþ] and pKa for log Ka to give: pH ¼ pKa þ log A  ½   ½   HA 30 Pharmaceutical analysisFor example, when acetic acid (pKa 4.76) is in solution at pH 4.76, the Henderson–Hasselbalch equation can be written as follows: pH ¼ 4:76 þ log CH3COO  ½   ½   CH3COOH From this relationship for acetic acid it is possible to determine the degree of ionisation of acetic acid at a given pH. Thus, when the pH¼4.76, then: 4:76 ¼ 4:76 þ log CH3COO  ½   ½   CH3COOH log CH3COO  ½   ½   CH3COOH ¼ 0 CH3COO  ½   ½   CH3COOH ¼ 100 ¼ 1 Self-test 2.3 Determine the percentage of ionisation of acetic acid at (i) pH 3.76 and (ii) pH 5.76. : (i) 9.09%; (ii) 90.0% Answers Acetic acid is 50% ionised at pH 4.76. In the case of a weak acid, it is the protonated form of the acid that is un-ionised, and, as the pH falls, the acid becomes less ionised. For a base, the Henderson–Hasselbalch equation is written as follows: BHþ   B þ Hþ pH ¼ pKa þ log ½   B BHþ ½   For example, when ammonia (pKa 9.25) is in a solution at pH 9.25, the Henderson–Hasselbalch equation can be written as follows: 9:25 ¼ 9:25 þ log ½   NH3     NHþ4 log ½   NH3     ¼ 0 NHþ4 ½   NH3     ¼ 100 ¼ 1 NHþ4 Ammonia is 50% ionised at pH 9.25. In this case, it is the protonated form of the base that is ionised, and, as the pH falls, the base becomes more ionised. Physical and chemical properties of drug molecules 31Self-test 2.4 Calculate the percentage of ionisation of ammonia at (i) pH 8.25 and (ii) pH 10.25. : (i) 90.9%; (ii) 9.09% Answers An alternative way of writing the expression giving the percentage of ionisation for an acid or base of a particular pKa value at a particular pH value is: Acid: % ionisation ¼ 10pH pKa 1 þ 10pH pKa   100 Base: % ionisation ¼ 10pKa pH 1 þ 10pKa pH   100 Ionisation of drug molecules The ionisation of drug molecules is important with regard to their absorption into the circulation and their distribution to different tissues within the body. The pKa value of a drug is also important with regard to its formulation into a medicine and to the design of analytical methods for its determination. Calculation example 2.1 Calculate the percentage of ionisation of the drugs shown in Figure 2.1 at pH 7.0. Fig. 2.1 Ionisation of a basic and an acidic drug. CHOCH CH ) N (CH 2 2 3 2 CHOCH CH N+H (CH3) 2 2 2 pKa 9.0  Diphenhydramine Ionised form iC4H9 CH3 CH3 CHCOOH CHCOO_ iC4H9 pKa 4.4 Ibuprofen Ionised form 32 Pharmaceutical analysisDiphenhydramine This drug contains one basic nitrogen and, at pH 7.0, its percentage of ionisation can be calculated as follows: % Ionisation diphenhydramine ¼ 109:0 7:0 1 þ 109:0 7:0   100 ¼ 102:0 1 þ 102:0   100 ¼ 100 101  100 ¼ 99:0% Ibuprofen This drug contains one acidic group and, at pH 7.0, its percentage of ionisation can be calculated as follows: % Ionisation ibuprofen ¼ 107:0 4:4 1 þ 107:0 4:4   100 ¼ 102:6 1 þ 102:6   100 ¼ 398 399  100 ¼ 99:8% Self-test 2.5 Calculate the percentage of ionisation of the following drugs, which contain 1 group that ionises in the pH range 0–14, at the pH values of (i) 4 and (ii) 9. CH OCH CHCH NHCH CH3 3 NH.CO CH3 N C4H9 Base pKa 8.1 2 2 CH3 Base pKa 9.6 CH CONH 2 2 Bupivacaine Atenolol H CH O3 CH COOH 2 Acid pKa 4.5 O O N N N H CH3 CO F Acid pKa 8.0  5-fluorouracil CI Indomethacin (i) 24.0%, (ii) 99.99%; 5-fluorouracil (i) 0.01%, (ii) 90.9% : Bupivacaine (i) 99.99%, (ii) 11.2%; Atenolol (i) 99.99%, (ii) 79.9%; Indomethacin Answers Physical and chemical properties of drug molecules 33Buffers Buffers can be prepared from any weak acid or base and are used to maintain the pH of a solution in a narrow range. This is important in living systems; for example, human plasma is buffered at pH 7.4 by a carbonic acid/bicarbonate buffer system. Buffers are used in a number of areas of analytical chemistry, such as the preparation of mobile phases for chromatography and the extraction of drugs from aqueous solution. The simplest type of buffer is composed of a weak acid or base in combination with a strong base or acid. A common buffer system is the sodium acetate/acetic acid buffer system. The most direct way of preparing this buffer is by the addition of sodium hydroxide to a solution of acetic acid until the required pH is reached. The most effective range for a buffer is 1 pH unit either side of the pKa value of the weak acid or base used in the buffer. The pKa value of acetic acid is 4.76; thus its effective buffer range is 3.76–5.76. Calculation example 2.2 1 litre of 0.1 M sodium acetate buffer with a pH of 4.0 is required. Molecular weight of acetic acid¼60; therefore, in a litre of 0.1 M buffer there will be 6 g of acetic acid. To prepare the buffer, 6 g of acetic acid are weighed and made up to ca 500 ml with water. The pH of the acetic acid solution is adjusted to 4.0 by addition of 2 M sodium hydroxide solution, using a pH meter to monitor the pH. The solution is then made up to 1 litre with water. Calculate the concentration of acetate and acetic acid in the buffer at pH 4.0. Using the Henderson–Hasselbalch equation: 4:00 ¼ 4:76 þ log CH3COO  ½   ½   CH3COOH log CH3COO  ½   ½   CH3COOH ¼  0:76 CH3COO  ½   ½   CH3COOH ¼ 10 0:76 1 ¼ 0:171 The buffer is composed of 1 part acetic acid and 0.17 part acetate. The buffer was prepared from 0.1 moles of acetic acid; after adjustment to pH 4.0, the amounts of acetic acid and acetate present are as follows: CH3COOH ¼ 11:17   0:1 moles ¼ 0:085 moles CH3COO  ¼ 0:17 1:17   0:1 moles ¼ 0:015 moles Since the acetic acid and acetate are dissolved in 1 litre of water, the buffer is composed of 0.085 M CH3COOH and 0.015 M CH3COO . Although the concentrations of acetate and acetic acid vary with pH, such a buffer would be known as a 0.1 M sodium acetate buffer. Note: The 0.1 M buffer should not be prepared by adding acetic acid to a solution of 0.1 M NaOH, since a lot of acetic acid would be required to adjust the pH to 4.0; in fact, 0.1 moles of NaOH would require 0.57 moles of acetic acid to produce a buffer with pH 4.0 and a strength of 0.57 M. 34 Pharmaceutical analysisAn alternative way of producing 1 litre of 0.1 M acetate buffer would be to mix 850 ml of a 0.1 M solution of acetic acid with 150 ml of a 0.1 M solution of sodium acetate. Self-test 2.6 1 litre of a 0.1 ammonium chloride buffer with a pH of 9.0 is required. Ammonia has a pKa value of 9.25. If a precise molarity is required this buffer is best prepared from ammonium chloride. The pH of the ammonium chloride may be adjusted to pH 9.0 by addition of a solution of sodium hydroxide (assuming the presence of sodium is not a problem), to avoid adding an excessive amount of the weak base ammonia. A total of 5.35 g (0.1 moles) of ammonium chloride are weighed and dissolved in ca 500 ml of water. The pH is then adjusted to pH 9.0 by addition of 5 M NaOH. The solution is then made up to 1 litre with water. Calculate the concentrations of NH4þ and NH3 in the buffer at pH 9.0 and indicate an alternative method for preparing the buffer. In this case: pH ¼ pKa þ log ½   NH3     NHþ4 Cl 4 a 0.1 M ammonia solution with 640 ml of 0.1 M NH M NH 064 : 0 and 3 : 0.036 M NH Answer þ44 The buffer could be prepared by mixing 360 ml of Some weak acids and bases have more than one buffer range; for example, phosphoric acid has three ionisable protons with three different pKa values and can be used to prepare buffers to cover three different pH ranges. The ionic species involved in the ranges covered by phosphate buffer are: H2PO 4 =H3PO4 HPO2  4 =H2PO 4 PO3  4 =HPO2  4 pH 1:13 3:13 pH 6:2 8:2 pH 11:3 13:3 The buffering ranges of a weak electrolyte are only discrete if the pKa values of its acidic and/or basic groups are separated by more than 2 pH units. Some acids have ionisable groups with pKa values less than 2 pH units apart, so they produce buffers with wide ranges. For example, succinic acid, which has pKa values of 4.19 and 5.57, can be considered to have a continuous buffering range between pH 3.19 and 6.57. Self-test 2.7 Determine the buffer range(s) for the following compounds: (i) Carbonic acid pKa 6.38, 10.32 (ii) Boric acid pKa 9.14, 12.74 (iii) Glycine 2.34, 9.60 (iv) Citric acid 3.06, 4.74, 5.4. 6.4   (iv) Continuous buffering range 2.06 10.6;   3.34, 8.6   13.74; (iii) 1.34   10.14, 11.74   is lost from solution; (ii) 8.14 2 which CO 11.32. The lower range is not useful because of the ease with   7.38, 9.32   : (i) 5.38 Answers Physical and chemical properties of drug molecules 35Sometimes a salt of a weak acid with weak base is used in a chromatographic mobile phase to, apparently, set the pH at a defined level, e.g. ammonium acetate or ammonium carbonate. These salts are marginally more effective than a salt of strong acid with a strong base at preventing a change in pH but they are not truly buffers. Such salts have buffering ranges ca 1 pH unit either side of the pKa values of the weak acid and weak base composing them. For example, the pH of a solution of ammonium acetate is ca 7.0 but it does not function effectively as a buffer unless the pH either rises to ca 8.25 or falls to ca 5.76. A buffer is most effective where its molarity is greater than the molarity of the acid or base it is buffering against. Calculation example 2.3 If 10 ml of 0.05 M HCl is added to 100 ml of 0.2 M sodium acetate buffer pH 4.5, the resultant pH can be calculated as follows: Molarity volume¼mmoles Number of mmoles of acetateþacetic acid in 100 ml of buffer¼0.2 100¼20 mmoles. Using the Henderson–Hasselbalch equation: 4:5 ¼ 4:76 þ log CH3COO  ½   ½   CH3COOH log CH3COO  ½   ½   CH3COOH ¼  0:26 CH3COO  ½   ½   CH3COOH ¼ 10 0:26 1 ¼ 0:551 The buffer contains 1 part CH3COOH and 0.55 parts CH3COO . CH3COOH ¼ 11:55   20 ¼ 12:9mmoles;CH3COO  ¼ 0:55 1:55   20 ¼ 7:1mmoles When HCl is added the following reaction occurs: CH3COO  þ HCl ! CH3COOH þ Cl  The amount of HCl added is 10 0.05¼0.5 mmoles Therefore, after addition of HCl, the amount of acetate remaining is: 7.1 0.5¼6.6 mmoles The amount of acetic acid now present is: 12.9þ0.5¼13.4 mmoles Therefore, the pH of the buffer after addition of HCl is determined as follows (amounts may be substituted in the equation instead of concentrations since CH3COO  and CH3COOH are present in the same volume): pH ¼ 4:76 þ log 6:6 13:4 ¼ 4:76   0:31 ¼ 4:45 (Continued) 36 Pharmaceutical analysisCalculation example 2.3 (Continued) The new pH of the buffer is 4.45. The molarity of the buffer has also changed, since the total amount of CH3COO  and CH3COOH is now contained in 110 ml instead of 100 ml, giving a new molarity of 0:2   100 110 ¼ 0:182 M If 10 ml of 0.05 M HCl were added to 100 ml of water, the pH would be determined as follows:     Self-test 2.8  log 0:05   10 110 ¼ 2:34 10 ml of 0.1 M HCl is added to 20 ml of a 0.5 M sodium acetate buffer with a pH of 4.3. Calculate: the pH of the buffer after addition of the HCl, the molarity of the buffer after addition of the HCl, the resultant pH if the HCl had been added to 20 ml of water. would give a pH of 1.48 0.33. Addition of 10 ml of 0.1 M HCl to 20 ml of water ¼ 4.04, new molarity ¼ : pH Answer Salt hydrolysis When the salt of a strong acid and a strong base is dissolved in water it produces a pH of ca 7.0. When salts of a weak acid and a strong base or of a strong acid and a weak base are dissolved in water, they will produce, respectively, alkaline and acidic solutions. When sodium acetate is dissolved in water, the acetate ion behaves as a base, removing protons from solution. For a weak electrolyte in water, Kb Ka¼Kw. If a 0.1 M solution of sodium acetate in water is considered: CH3COO  þ H2O ð0:1 xÞ  KbCH3COO x H þ HO x Ka CH ð Þ 3COOH ¼ 10 14 Kb ðCH3COO Þ ¼ Kw 1:75   10 5 ¼ 5:7   10 10 Regarding the change in the concentration of water as not affecting the equilibrium and regarding the [CH3COO ] as being relatively unchanged by hydrolysis. 5:7   10 10 ¼ x2 0:1 5:7   10 11 p ¼ 7:6   10 6 HO  ½  ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Hþ ½  ¼ 10 14 7:6   10 6 ¼ 1:33   10 9 pH ¼ 8:9 Physical and chemical properties of drug molecules 37Self-test 2.9 Calculate the pH of a 0.1 M solution of NH4Cl. Here salt hydrolysis increases [Hþ] and the equilibrium in this case is: NH4þ ð Þ 0:1 x The Ka for this reaction is 5.6 10 10. 5.13 ¼ : pH Answer  KaNH3 x þ Hþ x Activity, ionic strength and dielectric constant The activity of ions in a solution is governed by the dielectric constant of the medium they are dissolved in and by the total concentration of ions in solution. For solutions of electrolytes in water with concentrations<0.5 M, the activity of the ions present in solution is usually approximated to their individual concentrations. The mean activity coefficient for an ion in solution is defined as: g  ¼ activity concentration Although activity is regarded as 1 in dilute solutions this is still an approximation. The activity of an electrolyte solution in water can be estimated from the following equation: logg  ¼  0:509ð Þ zþz ffiffiI p where I ¼ 12Xmiz2i where  0.509 is a constant related to the dielectric constant of the solvent used to prepare the electrolyte solution and to temperature; z is the charge on a particular ion; I is the ionic strength of the solution; and m is the molality (moles per kg of solvent) of a particular ion in solution. Using this equation, the activity of Hþ in 0.1 M HCl can be calculated to be 0.69. Thus, the true pH of 0.1 M HCl is calculated as follows: pH ¼  log 0:1   0:69 ¼ 1:2 Calculation example 2.4 g  ¼ activity concentration log g  ¼  0:509j j zþz ffiffiI p where I ¼ 12Xmizi2 (Continued) 38 Pharmaceutical analysisCalculation example 2.4 (Continued)  0.509 is a constant related to the dielectric constant of the solvent used to prepare the electrolyte solution and to temperature. z is the charge on a particular ion. I is the ionic strength of the solution. m is the molality (moles per kg of solvent) of a particular ion in solution. For dilute solutions the molality can be approximated to the molarity Using this equation, the activity of Hþ in 0.1 M HCl can be calculated: where I ¼ 12X0:1   12 þ 0:1   12 ¼ 0:1 log g  ¼  0:509 1j j   1 ffiffiffiffiffiffi 0:1 p ¼  0:509   0:32 ¼  0:16 g  ¼ 0:69 Thus, the true pH of 0.1 M HCl is calculated as follows: pH ¼  log 0:1   0:69 ¼ 1:2 The ionic strength of buffers varies according to their pH. For instance: 0.1 M sodium phosphate buffer at pH 7.2 is composed of 50 mM H2PO4  þ50 mM HPO42  þ150 mM Naþ, having an ionic strength of 0.2. 0.1 M sodium phosphate buffer at pH 8.2 is composed of 9.1 mM H2PO4 , 90.9 mM HPO42  and 190.9 mM Naþ, having an ionic strength of 0.282. This slight difference between the pH determined from activity and from concentration is usually ignored. However, from the equation used to calculate the activity coefficientit can be seenthatthe activity decreases withincreasing ionic strength. In addition, the constant ( 0.509 for water) increases with decreasing dielectric constant, e.g. water has a dielectric constant of 78.5 and methanol a dielectric constant of 32.6. Addition of methanol to an aqueous solution of an acid or buffer will cause an increase in the pH of the solution through decreasing the activity of all of the ions in solution including Hþ. This effect should be noted with regard to the preparation of high-pressure liquid chromatography (HPLC) mobile phases, which are often composed of mixtures of buffers and organic solvents. Ionic strength is important with regard to the preparation of running buffers for capillary electrophoresis, where the greater the ionic strength of the buffer the higher the current through the capillary (Ch. 14). Partition coefficient An understanding of partition coefficient and the effect of pH on partition coefficient is useful in relation to the extraction and chromatography of drugs. The partition coefficient for a compound (P) can be simply defined as follows: P ¼ Co Cw where Co is the concentration of the substance in an organic phase and Cw is the concentration of the substance in water. The greater P the more a substance has an affinity for organic media. The value of P for a given substance of course depends on the particular organic solvent used Physical and chemical properties of drug molecules 39to make the measurement. Many measurements have been made of partitioning between n-octanol and water, since n-octanol, to some extent, resembles biological membranes and is also quite a good model for reverse-phase chromatographic partitioning. P is often quoted as a log P-value, e.g. a log P of 1 is equivalent to P¼10. Where P¼10 for a particular compound partitioning into a particular organic solvent, and partitioning is carried out between equal volumes of the organic solvent and water, then ten parts of the compound will be present in the organic layer for each part present in the water layer. Calculation example 2.5 A neutral compound has a partition coefficient of 5 between ether and water. What percentage of the compound would be extracted from 10 ml of water if (i) 30 ml of ether were used to extract the compound or (ii) three 10 ml volumes of ether were used in succession to extract the compound? (i) Between water and an equal volume of ether, 5 parts of the drug would be in the ether layer compared with 1 part in the water layer. Where three volumes (30 ml) of ether were used to one volume (10 ml) of water, the distribution would be 15 parts of drug in the ether layer to 1 part in the water layer. Percentage extracted ¼ 1516   100 ¼ 93:75% (ii) First extraction 5 parts in the ether layer and 1 part in the water layer (total six parts) At each extraction 56 of the material in the water layer is extracted. Percentage extracted ¼ 56   100 ¼ 83:3% Percentage of drug remaining in water layer¼16.7% Second extraction 5 6 of the 16.7% remaining in the water layer is extracted. Percentage extracted ¼ 56   16:7 ¼ 13:9% Percentage of drug remaining in water layer¼2.8% Third extraction 5 6 of the 2.8% remaining in the water layer is extracted. Percentage extracted ¼ 56   2:8 ¼ 2:3% Percentage of drug remaining in water layer¼0.5% Total percentage of drug extracted¼83.3þ13.9þ2.3¼99.5% Self-test 2.10 A drug has a partition coefficient of 12 between chloroform and water. Calculate the percentage of a drug that would be extracted from 10 ml of water with (i) 30 ml of chloroform; (ii) 3 10 ml of chloroform. : (i) 97.3%; (ii) 99.95% Answers 40 Pharmaceutical analysisEffect of pH on partitioning Many drugs contain ionisable groups and their partition coefficient at a given pH may be difficult to predict if more than one ionised group is involved. However, often one group in a molecule may be much more ionised than another at a particular pH, thus governing its partitioning. It is possible to derive from the Henderson– Hasselbalch equation expressions for the variation in the partitioning of organic acids and bases into organic solvent with respect to the pH of the solution that they are dissolved in. From the Henderson–Hasselbalch equation: For acid: Papp ¼ P 1 þ 10pH pKa For bases: Papp ¼ P 1 þ 10pKa pH Papp is the apparent partition coefficient, which varies with pH. Thus it can be seen that, when a compound, acid or base, is 50% ionised (i.e. pH¼pKa), its partition coefficient is half that of the drug in the un-ionised state: Papp ¼ P 1 þ 100 ¼ P2 As a general rule, for the efficient extraction of a base into an organic medium from an aqueous medium, the pKa of the aqueous medium should be at least 1 pH unit higher than the pKa value of the base, and, in the same situation for an acid, the pH should be 1 pH unit lower than the pKa value of the acid. Calculation example 2.6 A linctus formulation contains the following components: Base A pKa 6.7, P (chloroform/0.1 M NaOH)¼100 5 mg/ml Base B pKa 9.7, P (chloroform/0.1 M NaOH)¼10 30 mg/ml Benzoic acid pKa 4.2 P (chloroform /0.1 M HCl)¼50 5 mg/ml In order to selectively extract base A, 5 ml of the linctus is mixed with 15 ml of phosphate buffer pH 6.7 and extracted once with 60 ml of chloroform. Calculate the percentage and the weight of each component extracted. Base A At pH 6.7: Papp ¼ 100=ð1 þ 106:7 6:7Þ ¼ 100=2 ¼ 50 If 20 ml of aqueous buffer phase was extracted with 20 ml of chloroform, there would be 1 part of the base in the aqueous phase to 50 parts in the chloroform layer. Since 60 ml of chloroform is used in the extraction, there will be 1 part of the base remaining in the aqueous phase and 150 parts in the chloroform layer. Percentage extracted¼(150/151) 100¼99.3% Base A is present at 5 mg/ml in the elixir. Amount of base A in 5 ml of elixir¼5 5 mg¼25 mg Amount of base A extracted¼25 (150/151)¼24.8 mg (Continued) Calculation example 2.6 (Continued) Base B At pH 6.7: Physical and chemical properties of drug molecules 41Papp ¼ 10=ð1 þ 109:7 6:7Þ ¼ 10=1001 ¼ 0:01 If 20 ml of aqueous buffer phase was extracted with 20 ml of chloroform, there would be 1 part of the base in the aqueous phase to 0.01 parts in the chloroform layer. Since 60 ml of chloroform is used in the extraction, then 1 part of the base will remain in the aqueous phase while there will be 0.03 parts in the chloroform layer. Percentage extracted¼(0.03/1.03) 100¼3.0% Base B is present at 30 mg/ml in the elixir Amount of base B in 5 ml of elixir¼5 30 mg¼150 mg Amount of base B extracted¼0.03 150¼4.5 mg Benzoic acid At pH 6.7 for an acid: Papp ¼ 50=ð1 þ 106:7 4:2Þ ¼ 50=317 ¼ 0:158 If 20 ml of aqueous buffer phase was extracted with 20 ml of chloroform, there would be 1 part of the preservative in the aqueous phase to 0.158 parts in the chloroform layer. Since 60 ml of chloroform is used in the extraction, 1 part of the benzoic acid will remain in the aqueous phase while there will be 0.474 parts in the chloroform layer. Percentage extracted¼(0.474/1.474) 100¼32.2% Benzoic acid is present at 5 mg/ml in the elixir Amount of benzoic acid in 5 ml of elixir¼5 5 mg¼25 mg Amount of benzoic acid extracted¼(0.474/1.474) 25¼8.0 mg The extract is not completely free of the other ingredients in the formulation. If back extraction of the extract with an equal volume of pH 7.7 buffer is carried out, approximately 1% of extracted base A will be removed but the amount of base B and benzoic acid will be reduced to<0.5 mg. Self-test 2.11 A cough mixture contains the following components: (i) Base 1 pKa¼9.0, P (CHCl3/0.1 M NaOH)¼1000 0 mg/5 ml (ii) Base 2 pKa¼9.7, P (CHCl3/0.1 M NaOH)¼10 30 mg/5 ml (iii) Acidic preservative pKa¼4.3, P (CHCl3/0.1 M HCl)¼10 5 mg/5 ml 5 ml of the cough mixture is mixed with 15 ml of phosphate buffer pH 7.0 and extracted with 60 ml of chloroform. Calculate the weight of each component extracted. : (i) Base 1, 29.0 mg; (ii) Base 2,1.69 mg; (iii) Preservative, 0.28 mg Answers Drug stability The purpose of drug stability testing is to provide evidence on how the quality of a drug substance or drug product changes with time under the influence of environmental factors such as temperature, humidity and light. The aim of the testing is to establish a re-test period for the drug substance or a shelf-life for the drug product and 42 Pharmaceutical analysisrecommended storage conditions. ICH guidelines divide the world into four climate zones, I–IV, and stability testing to establish the shelf-life of a product should be carried out in accordance with the climatic conditions in which the drug product is to be sold. Many drugs are quite stable, but functional groups such as esters and lactam rings, which occur in some drugs, are susceptible to hydrolysis, and functional groups such as catechols and phenols are quite readily oxidised. The most common types of degradation which occur in pure and formulated drugs obey zero- or first-order kinetics. Zero-order degradation In zero-order kinetics the rate of degradation is independent of the concentration of the reactants. Thus, if the rate constant for the zero-order degradation of a substance is 0.01 moles h 1, then, after 10 h, 0.1 moles of the substance will have degraded. This type of degradation is typical of hydrolysis of drugs in suspensions or in tablets, where the drug is initially in the solid state and gradually dissolves at more or less the same rate as the drug in solution is degraded, i.e. the equilibrium concentration in free solution remains constant. First-order degradation First-order kinetics of drug degradation have been widely studied. This type of degradation would be typical of the hydrolysis of a drug in solution. Such reactions are pseudo first order, since the concentration of water is usually in such large excess that it is regarded as constant even though it does participate in the reaction. In first-order kinetics the rate constant k has the units h 1 or s 1 and the rate of the reaction for a drug is governed by the expression:   d½   Adt ¼ k A½   where A is the concentration of the drug, which will change as degradation proceeds. This expression can be written as: dx dt¼ k að Þ   x which can also be written as Zs o dx ð Þ a   x ¼ Zt o kdt where x is the amount of degraded product and a is the starting concentration of the drug. From this expression, by integration and rearrangement, the following expression arises: t ¼ 1k ln a a   x The half-life of the drug (the time taken for 50% of a sample drug to degrade, i.e. where x is a/2) is thus given by the following expression: t0:5 ¼ 1k ln aa=2 ¼ 1k ln2 Physical and chemical properties of drug molecules 43Thus, for aspirin, which has a rate constant of 0.0133 h 1 for the hydrolysis of its ester group at 25 C and pH 7.0, the half-life can be calculated as follows: t0:5 ¼ 0:693 0:0133 ¼ 52:1 h The shelf-life of a drug, the time required for 10% degradation (where x is 0.1 a), is given by the following expression: t0:9 ¼ 1kln a 0:9a ¼ 1kln 1:11 Calculation example 2.7 In a high-pressure liquid chromatography assay of aspirin tablets, ten extracts are made and the extracts are diluted with mobile-phase solution, which consists of acetonitrile/0.1 M sodium acetate buffer pH 4.5 (10:90) and analysed sequentially. If the rate constant for the degradation of aspirin in the mobile phase is 0.0101 h 1 at room temperature, how long can the analyst store the solutions at room temperature before the degradation of the analyte is greater than 0.5%? In this case we are interested in t(0.995): In tð Þ 0:995 ¼ ð Þ a=0:995a 0:0101 ¼ 0:5 h Thus, in order for degradation to be<0.5%, the solutions would have to be analysed within 30 min of their being prepared. Self-test 2.12 Determine the half-lives of the following drugs, which can undergo hydrolysis of their ester functions in solution, under the conditions specified. (i) Atropine at 40  C and pH 7.0 where k¼2.27 10 4 h 1. (ii) Procaine at 37  C and pH 8.0 where k¼1.04 10 2 h 1. (iii) Benzocaine at 30  C and pH 9.0 where k¼2.27 10 3 h 1. : (i) 127.2 d; (ii) 66.6 h; (iii) 12.7 d Answers Stereochemistry of drugs The physiological properties of a drug are governed to a great extent by its stereochemistry. In recent years it has emerged that, in some instances, even optical isomers of a drug can have very different physiological effects. Since stereochemistry is concerned with the way in which a drug is orientated in space, this is something that is difficultto visualise ona flat piece of paper andthe assignment of absoluteconfiguration to a drug some people find confusing. The three types of isomerism encountered in drug molecules are geometrical isomerism, optical isomerism and diastereoisomerism. Geometrical isomerism Drugs which have a geometrical isomer are relatively uncommon. 44 Pharmaceutical analysisN C C H CH2N(CH3)2 N C C (CH ) 3 2NCH2 H Fig. 2.2 Geometrical isomers. Zimeldine Geometrical isomer An example of a drug with a geometrical isomer is the antidepressant zimeldine (Fig. 2.2). The lack of free rotation about the double bond ensures that the stereochemistry of this drug and and its isomer is different. Zimeldine is the only drug used. Other drugs which could also have geometric isomers of this type include amitriptyline and triprolidine. Chirality and optical isomerism Optical isomerism of drug molecules is widespread. Many drug molecules only contain one or two chiral centres. A simple example is the naturally occurring neurotransmitter adrenaline. When a compound has no symmetry about a particular carbon atom, the carbon atom is said to be a chiral centre. When a compound contains one or more chiral centres, it is able to rotate plane-polarised light to the right (þ) or the left ( ). A chiral centre arises when a carbon atom has four structurally different groups attached to it. Adrenaline can exist as two enantiomers that are mirror images of each other (Fig. 2.3) and are thus non-superimposible. In Figure 2.3 the wedge-shaped bonds indicate bonds above the plane of the paper, the dotted bonds indicate bonds pointing down into the paper and unbroken lines indicate bonds in the same plane as the paper. In common with all pairs of enantiomers, the adrenaline enantiomers have identical physical and chemical properties; the only difference in their properties is that the enantiomers rotate plane-polarised light in opposite directions. However, the two enantiomers of adrenaline do have different biological properties; the ( ) enantiomer exerts a much stronger effect, for instance, in increasing heart rate. It is not possible simply by looking at a structure drawn on paper to say which way it will rotate plane-polarised light – this can only be determined by experiment. In order to describe the configuration about a chiral centre, a set of precedence rules was developed: Fig. 2.3 HO HO OH 1 C 3 H 2 CH2NHCH3 CH3NHCH2 2 OH 1 C 3 H OH OH R and S enantiomers of adrenaline. A B Physical and chemical properties of drug molecules 45(i) The group of lowest priority attached to the chiral carbon, often hydrogen, is placed behind the plane of the paper with all the other groups pointing forwards. (ii) The priorities are assigned to the atoms immediately attached to the chiral centre in order of decreasing atomic mass. For example: Br > Cl > S > F > O > N > C > H (iii) If two atoms attached to the chiral centre are of the same precedence, the priority is assigned on the basis of the atoms attached to these atoms, for example: C — C– Cl > —C–S > — C–O > — C– N > —C–C > — —C –C > — C–H (iv) If required, the third atom in a chain may be considered, for example: —CCCl > —CCO > —CCN > —CCC > —CCH Using these rules we can assign the absolute configurations for adrenaline structures A and B. Placing the group of lowest priority behind the paper, in this case H. For structure A we find moving in a clockwise direction (clockwise¼R): —O > —CN > —CC Thus the absolute configuration of A is R and it follows that its mirror image B must be S (the order of precedence moves anti-clockwise). To relate the (þ) (dextrorotatory) and ( ) (laevorotatory) forms of a molecule to an absolute (R or S) configuration is complex and requires preparation of a crystal of the compound suitable for analysis by X-ray crystallography. In contrast, the direction in which a molecule rotates plane-polarised light is easily determined using a polarimeter. X-ray crystallography of the enantiomers of adrenaline has shown that the ( ) form has the R configuration and the (þ) form has S configuration. It should be noted that in older literature the terms d and l are used to denote (þ) and ( ) respectively and D and L are used to denote R and S respectively. A mixture containing equal amounts of (þ) and ( ) adrenaline, or indeed enantiomers of any drug, is known as a racemic mixture and of course will not rotate plane-polarised light. The physical separation of enantiomers in a racemic mixture into their pure (þ) and ( ) forms is often technically difficult. Ampicillin (Fig. 2.4) provides a more complex example than adrenaline with regard to assignment of absolute configuration since it contains four chiral centres. Fig. 2.4 NH2 1 CCONH H H CH3 H 2 O Ampicillin S 3 N 4 CH3 COOH H 46 Pharmaceutical analysisAssignment can be made as follows: Chiral centre 1: —N > —CO > —CC configuration R Chiral centre 2: —N > —CS > —CO configuration R Chiral centre 3: —S > —N > —CN configuration R Chiral centre 4: In this case, since the molecule is drawn with the hydrogen pointing forward, it is best to determine the configuration from the molecule as drawn and then assign the opposite configuration. —N > —CS > —CO configuration R as drawn, therefore configuration S with H behind the plane of the paper The structures of drugs as drawn on paper do not always lend themselves to ready assignment of absolute configuration and sometimes a certain amount of thinking in three dimensions is required in order to draw the structures in a form where the absolute configuration can be assigned. In drugs such as steroids, penicillins and morphine alkaloids, which are all based on natural products, chirality is built into the molecules as a result of the action of the stereoselective enzymes present in the plant or micro-organism producing them. However, there are many synthetic drugs where chiral centres are part of the structure. Many of these drugs are used in the form of racemates, since there are technical difficulties in carrying out stereospecific chemical synthesis or in resolving mixtures of enantiomers resulting from non-stereoselective synthesis. In the past, many such racemic mixtures have been used as drugs without regard for the fact that effectively a drug that is only 50% pure is being administered. The so-called ‘inactive’ enantiomer may in fact be antagonistic to the active form or it may have different physiological effects.3,4 The most notable example, which gave rise to much of the medicines legislation in the past 30 years, is the case of thalidomide, which contains one chiral centre and was administered as a racemate in order to alleviate morning sickness during pregnancy. The active enantiomer produced the intended therapeutic effects whilst the ‘inactive’ enantiomer was responsible for producing birth defects in the children of mothers who took the racemic drug. In fact, administering a pure, therapeutically active enantiomer would not have helped in this case since thalidomide becomes racemised within the body. Current legislation requires a manufacturer seeking to license a new drug in the form of a racemate to justify the use of the racemate as opposed to a pure enantiomer. Physical and chemical properties of drug molecules 47Self-test 2.13 Assign absolute configurations to the chiral centres in the following drugs. CH3 HO NH2 CCOOH HSCH2 C H 1 H CO N HO CH 2 2 HOOC Captopril S2 S ; captopril 1 R ) DOPA  : ( Answers Diastereoisomers H  (–) DOPA Fig. 2.5 The stereoisomers Where more than one chiral centre is present in a molecule, there is the possibility of diastereoisomers, e.g. captopril. Another example of a synthetic drug with two chiral centres is labetalol. The number of diastereoisomers arising from n chiral centres is 2n 1, i.e. 2 in the case of labetalol. In the structure shown in Figure 2.5, chiral centres 1 and 2 in structure A have the configurations R and S respectively; the enantiomer of this structure (B) has the S and R configurations in centres 1 and 2. In addition, there is a pair of enantiomers, C and D, that are diastereoisomers of the structures A and B, and which have the configurations 1R2R and 1S2S. R S R S CH3 of labetalol. H2NOC OH CH3 OH CONH2 C HO C CH2NH H H CH2 H2C C OH C HNH2C H H 1 2 2 1 A B SS R R H2NOC CH3 CH3 HCONH2 HOCCH2NH C H OH CH2 H2C HNH2C C H H C OH OH 1 2 2 1 C D 48 Pharmaceutical analysisThe diastereoisomers in a mixture can usually be separated by ordinary chromatographic methods. In the case of labetalol, two peaks would be seen in a chromatographic trace obtained from a non-chiral phase – one due to the 1R 2S, 1S 2R pair of enantiomers and the other to the 1R 2R and 1S 2S pair of enantiomers. The commercial drug is in fact administered as a mixture of all four isomers, and the BP monograph for labetalol checks the ratio of the two chromatographic peaks produced by the two enantiomeric pairs, of diastereoisomers. In order to separate the enantiomeric pairs, a chiral chromatography column would be required, and separation on a chiral column produces four peaks (Ch. 12). Self-test 2.14 Indicate the configuration of the pairs of enantiomeric diastereoisomers which compose the drug isoxsuprine. CH OH 3 CH3 HO NH CH2 O CH CH CH 12 3 Isoxsuprine S3 R2 S and 1 R3 S2 R ; 1 S3 S2 R and 1 R3 R2 S ; 1 R3 S2 S and 1 S3 R2 R ; 1 S3 S2 S and 1 R3 R2 R : 1 Answer An example of a pair of diastereoisomers used separately as drugs is betamethasone and dexamethasone, shown in Figure 2.6. With a total of eight chiral centres within the betamethasone and dexamethasone structures, there is the possibility of 28¼256 isomers of the structure and these divide into 128 enantiomeric pairs. Both dexamethasone and betamethasone could have corresponding enantiomers but because they are largely natural products, made by stereospecific enzymes, their optical isomers do not exist. Semi-synthetic steroids are largely derived via microbial fermentation from naturally occurring plant sterols, originally obtained from the Mexican yam. The partially degraded natural products are then subjected to a number of chemical synthetic steps to produce the required steroid. In betamethasone and dexamethasone, the only stereoisomerism is at the synthetically substituted 16 position. In the case of betamethasone, the methyl group is CH2OH CH2OH Fig. 2.6 HO O C 17 H 9 8 F H O OH 16 HO CH3 H O C 17 H 8 9 F H 16 O OHCH3 H An example of diastereoisomers used separately as drug molecules. Betamethasone Dexamethasone Physical and chemical properties of drug molecules 49in the b-position, i.e. as one looks down on the molecule; it is closer than the hydrogen at position 16 (not necessarily projecting vertically out of the plane of the paper). In dexamethasone, the methyl group is in the a-position – further away than the hydrogen at 16 (not necessarily projecting vertically down into the paper). Of the two steroids, betamethasone has the slightly stronger anti-inflammatory potency. Another stereochemical term is cis and trans and this refers to the relative orientation of two substituents. In betamethasone, the hydroxyl group at 17 and the methyl group at 16 are trans – on opposite sides of the ring; in dexamethasone, the hydroxyl group at 17 and the methyl group at 16 are cis – on the same side of the ring. Otherwise the relative orientation of the substituents is the same in both drugs, e.g. fluorine at 9 and hydrogen at 8 are trans to each other. Measurement of optical rotation Figure 2.7 shows a schematic diagram of a polarimeter. Light can be viewed as normally oscillating throughout 360  at 90  to its direction of travel. The light source is usually a sodium lamp; the polarising material can be a crystal of Iceland spar, a Nicol prism or a polymeric material such as polaroid. When circularly polarised light is passed through the polariser its oscillations are confined to one plane. In the absence of an optically active material, the instrument is set so that no light is able to pass through the analyser. When the polarised light is passed through an optically active medium, the plane in which the light is oscillating becomes tilted and light is able to pass through the analyser. The angle of rotation can be measured by correcting for the tilt by rotating the analyser until light again does not pass through it. The angle that the second polariser has to be rotated through to prevent, once again, the passage of light through it gives the measured rotation a. The standard value for the rotation produced by an optically active compound is [a], the specific rotation of a substance where: ½a  ¼ 100a 1c where a is the measured rotation, l is the pathlength of the cell in which the measurement is made in dm and c is the concentration of the sample solution in g/100 ml. The observed optical rotation is dependent on both the wavelength of the light and the temperature. The sodium D line (589 nm) is usually used to make measurements. The solvent in which the sample is dissolved may also greatly affect the [a] of a Fig. 2.7 Schematic diagram of a polarimeter. Circularly polarised light Plane-polarised light Tilted plane-polarised light   Polariser Analyser Detector Sample usually in  10 cm (1 dm) cell 50 Pharmaceutical analysissubstance. Values for [a] are usually quoted with details of the concentration of the solution used for measurement, the solvent, the temperature and the type of light used. For example, the specific rotation for ( ) adrenaline is given as: ½a 25D ¼  51 ðc ¼ 2; 0:5M HClÞ The optical rotation was obtained at 25  using the sodium D line with a 2 g 100 ml 1 solution in 0.5 M HCl. If an enantiomer is chemically pure it is possible to determine its degree of enantiomeric purity by measuring its optical rotation relative to a standard value, e.g. if an enantiomeric mixture contains 1% of enantiomer A and 99% of enantiomer B, [a] will be reduced by 2% compared with the value for optically pure B. Examples of the measurement of optical rotation as a quality control check are found in the BP monographs for Timolol Maleate, Tobramycin and Phenylephrine Hydrochloride. Self-test 2.15 Optical rotation measurements were made using the sodium D line at 25  C in a 1 dm cell and the readings obtained were as follows: (i) Phenylephrine HCl 2.6% w/v in 0.1 M HCl; a¼ 0.98  (ii) Timolol maleate 9.8% w/v in 1 M HCl; a¼ 0.59 Calculate [a] for these drugs and express it in the conventional form. ] a : (i) [ Answers 25 ] a (2.6, 0.1 M HCl); (ii) [   37.7 ¼  D 25 (9.8, 1 M HCl)   6.02 ¼  D Profiles of physico-chemical properties of some drug molecules Procaine C B A COOCH2CH2N(C2H5) H2N 2 Procaine Drug type: local anaesthetic. Functional groups: • A – Tertiary aliphatic amine, pKa 9.0. • B – Ester, neutral. • C – Aromatic amine, very weak base, pKa ca 2. Half-life in water: 26 days at pH 7.0, 37 C. Fig. 2.8 Additional information: Procaine is formulated in injections and thus susceptible to aqueous-phase hydrolysis; in simple solution its degradation is first order (Fig. 2.9). Closely related drug molecules: Proxymetacaine, benzocaine, tetracaine (amethocaine) butacaine, propoxycaine, procainamide (for stability of an amide group see paracetamol), bupivacaine, lidocaine (lignocaine), prilocaine. Physical and chemical properties of drug molecules 51Fig. 2.9 Hydrolysis of procaine. H+/OH– COOCH2CH2N(C2H5) H2N 2 Procaine Self-test 2.16 Calculate: H2N COOH + HOCH2CH2N(C2H5)2 (i) The percentage ionisation of procaine at pH 7.0 (the ionisation of group C at this pH will be negligible). (ii) The rate constant for its hydrolysis at pH 7.0 and 37  C. 10   : (i) 99.01%; (ii) 1.11 Answers 3 h 1   Paracetamol Fig. 2.10 Drug type: analgesic. Functional groups: • A – Amide group, neutral. • B – Phenolic hydroxy group, very weak acid, pKa 9.5. A NH.CO.CH3 OH B Paracetamol Half-life in water: 21.8 years at pH 6 and 25 C. Most amides are very stable to hydrolysis (Fig. 2.11). Fig. 2.11 Hydrolysis of paracetamol. NH.CO.CH3 H+/OH– NH2 + CH3COOH OH OH 52 Pharmaceutical analysisOther drugs containing an amide group: Bupivacaine, procainamide, lidocaine, beclamide, acebutolol. Aspirin A COOH Aspirin B O.CO.CH3 Fig. 2.12 Drug type: analgesic. Functional groups: • A – Carboxylic acid, weak acid, pKa 3.5. • B – Phenolic ester, particularly unstable. Half-life in water: 52 h at pH 7.0 and 25 C; ca 40 days at pH 2.5 and 25 C. Additional information: The rate of HO  catalysed hydrolysis of esters is>the rate of Hþ hydrolysis of esters. Solid aspirin absorbs water from the atmosphere and then hydrolyses (Fig. 2.13). COOH O.CO.CH3 H+ /OH– COOH OH + CH3COOH Fig. 2.13 Hydrolysis of aspirin. Partition coefficient of un-ionised compound at acidic pH: octanol/water ca 631. Other drugs containing phenolic ester group: metipranolol, vitamin E, benorilate, dipivefrin. Benzylpenicillin A B Fig. 2.14 CH2.CO.NH N S CH3 CH3 O C COOH Benzylpenicillin D Drug type: antibiotic. Physical and chemical properties of drug molecules 53Functional groups: • A – Amide, neutral, relatively stable, see paracetamol. • B – Thioether, neutral, can be oxidised at high levels of oxygen stress. • C – Lactam ring, neutral, particularly susceptible to hydrolysis (Fig. 2.15). H+/OH– Fig. 2.15 Hydrolysis of benzylpenicillin. CH2.CO.NH N O CH2.CO.NH S CH3 CH3 COOH S CH3 HOOC N CH3 COOH H Half-life: 38 days at pH 6.75 and 30 C; ampicillin: 39 days at pH 6.5 and 25 C. Additional information: Consideration of the stability of these compounds is particularly important since they may be formulated as oral suspensions. • D – Carboxylic acid, weak acid pKa 2.8 (the strength of the acid is increased by the adjacent lactam ring). Compounds with similar properties: amoxicillin, cloxacillin, carbenicillin, flucloxacillin, phenoxymethylpenicillin, cefalexin, cefuroxime. 5-Fluorouracil Fig. 2.16 5-Fluorouracil and its ionised form. O A B H H N N O F H O N N F O– Predominant form at pH 9.0 Drug type: anticancer Functional groups: • A – Ureide nitrogen, acidic, pKa 7.0. • B – Ureide nitrogen, very weakly acidic, pKa 13.0. Additional information: The molecule is quite stable. Partition coefficient of un-ionised molecule: octanol/water ca 0.13. Compounds containing an acidic nitrogen in a heterocyclic ring: phenobarbital, amylobarbital, butobarbital, bemegride, uracil, phenytoin, theophylline, theobromine. 54 Pharmaceutical analysisAcebutolol A NH.COCH2CH2CH3 COCH3 OCH2CHCH2NH.CH(CH3)2 Fig. 2.17 B C Acebutolol OH Drug type: b-adrenergic blocker. Functional groups: • A – Amide, neutral, susceptible to hydrolysis under strongly acidic conditions (see paracetamol). • B – Aromatic ether group, potentially oxidisable under stress conditions. • C – Secondary amine group, pKa 9.4. Partition coefficient of un-ionised compound: octanol/water ca 483. A Fig. 2.18 N B C O N O– Sulfadiazine and its N A NH S O NH2 N N S O Ionised form  NH2 ionised form. Sulfadiazine Drug type: antibacterial. Functional groups: • A – Diazine ring nitrogens, very weakly basic, pKa<2. • B – Sulfonamide nitrogen, weak acid, pKa 6.5. • C – Weakly basic aromatic amine, pKa<2. Partition coefficient of un-ionised compound: octanol/water ca 0.55 (log P octanol/ water at pH 7.5¼ 1.3). Closely related compounds: sulfadoxine, sulfamerazine, sulfametopyrazine, sulfquinoxaline, sulfachloropyridazine, sulfamethoxazole, sulfathiazole. Self-test 2.17 Calculate the % ionisation of sulfadiazine at pH 7.5. at pH 7.5 : 90.9%. The weakly basic groups in the molecule are not ionised to any degree Answer Isoprenaline Fig. 2.19 Oxidation of isoprenaline. Physical and chemical properties of drug molecules 55C B HO OH CH3 A C HO O CHCH2NHCH O2 CH3 O Drug type: sympathomimetic. Functional groups: • A – Secondary amine, base, pKa 8.6. • B – Benzyl alcohol group, neutral. OH CH3 CHCH2NHCH CH3 • C – Catechol group, weakly acidic, pKa values ca 10 and 12. Additional information: As indicated in the reaction above, the catechol group is very readily oxidised upon exposure to light or air. Oxidation results in such compounds turning brown, solutions of isoprenaline and related compounds must contain antioxidants as preservatives. Partition coefficient of the un-ionised drug: The compound is highly water soluble and cannot be extracted to any great extent into an organic solvent since it is ionised to some extent at all pH values. Closely related compounds: dopamine, adrenaline, noradrenaline, terbutaline, DOPA. Prednisolone Fig. 2.20 C OH B O D HO OH E A Drug type: corticosteroid. Functional groups: • A and D – Ketone groups, neutral. O Prednisolone • B and C – Secondary and primary alcohol groups, neutral. • E – Tertiary alcohol group, neutral, prone to elimination by dehydration at high temperatures. Where the hydroxyl group at E is converted to an ester such as 56 Pharmaceutical analysisvalerate, e.g. betamethasone valerate, thermal elimination of the ester can occur quite readily. Another decomposition reaction of the valerate esters is the intramolecular transfer of the ester group from E to C (see the BP tests for Betamethasone Valerate products). Partition coefficient: Octanol/water ca 70. The drug does not ionise so that the partition coefficient is unaffected by pH. Despite being neutral compound the hydroxyl groups in prednisolone give it a water solubility of ca 0.5 mg/ml. Closely related compounds: dexamethasone, betamethasone, triamcinolone, hydrocortisone, betamethasone valerate, betamethasone dipropionate. Guanethidine A A + Fig. 2.21 B NH B NH2 Guanethidine and its + N CH2CH2NHC.NH2 N CH2CH2NHC.NH2 H ionised form. Drug type: anti-hypertensive. Functional groups: Predominant form of guanethidine at pH 7.0; charge on A is delocalised  over all three nitrogens • A – Guanidine group; one of the strongest nitrogen bases, pKa 11.4. • B – Tertiary amine, pKa 8.3. Compounds containing guanidine group: arginine, creatinine, bethanidine, streptomycin, phenformin, metformin, chlorhexidine. Pyridostigmine bromide A CH3 Br – + N B Fig. 2.22 O.CO.N(CH3)2 Drug type: anti-cholinergic. Functional groups: Pyridostigmine bromide • A – Salt of strongly basic quaternary ammonium ion. Quaternary ammonium ions are charged at all pH values. • B – Carbamate group, the nitrogen is neutral as in an amide but the carbamate group is less stable than an amide, having a stability similar to that of a phenolic ester. Compounds containing a quaternary ammonium group: atracarium besylate, bretylium tosylate, clindium bromide, glycopyrronium bromide. Physical and chemical properties of drug molecules 57Additional problems 1. Calculate the pH of the following solutions assuming that the concentration and the activity of the solutions are the same: (i) 0.05 M HCl. (ii) 0.1 M chloroacetic acid (Ka¼1.4 10 3). (iii) 0.1 M phosphoric acid (First Ka 7.5 10 3). (iv) 0.1 M fumaric acid (Ka 9.3 10 4). (v) 0.1 M di-isopropylamine base (Kb¼9.09 10 4). (vi) 0.1 M imidazole base (Kb¼1.6 10 7). : (i) 1.3; (ii) 1.93; (iii) 1.56; (iv) 2.02; (v) 12.0; (vi) 10.1 Answers 2. Calculate the pH of the following salt solutions: (i) 0.1 M sodium formate (Ka formic acid¼1.77 10 4) (ii) 0.1 M sodium fusidate (Ka fusidic acid¼4.0 10 6). (iii) 0.1 M ephedrine hydrochloride (Ka of ephedrine¼2.5 10 10). : (i) 8.4; (ii) 9.2; (iii) 5.3 Answers 3. Calculate what volumes of the salt solutions specified would be required to prepare 1 l of the following buffers: (i) 0.1 M phosphoric acid/0.1 M sodium dihydrogen phosphate pH 2.5 (pKa H3PO4 2.13) (ii) 0.1 M sodium dihydrogen phosphate/0.1 M disodium hydrogen phosphate pH 8.0 pKa H2PO 4 7:21     (iii) 0.1 M sodium bicarbonate/0.1 M sodium carbonate pH 9.5 (pKa HCO 3 10.32). : (i) 299 ml/701 ml; (ii) 139.5 ml/860.5 ml; (iii) 868.5 ml/131.5 ml. Answers 4. Indicate the percentage of ionisation of the functional groups specified in the following drugs at pH 7.0 (Fig. 2.23). Fig. 2.23 O CH3 N Cl H A pKa 8.8 N B pKa 9.2 CHCH2CH2N(CH3)2 O NN CH3 Theophylline N A pKa 6.0  Chlorpheniramine : theophylline 1.56% (acidic nitrogen); chlorpheniramine A 9.09% B 99.4% Answers 5. Calculate the percentage of the following compounds that would be extracted under the conditions specified: (i) A solution of basic drug pKa 9.2 in an oral liquid is mixed with a buffer having a pH of 7.2 and is extracted with an equal volume of chloroform (the partition coefficient of the un-ionised base into chloroform is 500). (ii) an acidic drug with a pKa 4.2 is extracted from a solution of pH 4.5 with an equal volume of chloroform (the partition coefficient of the un-ionised acid into chloroform is 300). : (i) 83.2%; (ii) 99.0% Answers (Continued) 58 Pharmaceutical analysisAdditional problems (Continued) 6. Extracts containing benzylpenicillin were prepared for analysis in buffer at pH 6.5 at 25 C; the rate constant for the hydrolysis of benzylpenicillin under these conditions is 1.7 10 7 s. What is the maximum length of time the solutions can be stored before analysis so that no more than 1% decomposition occurs. : 16.4 h Answers 7. Determine the absolute configurations of the chiral centres in menthol and phenbutrazate. List the configurations of the pairs of enantiomeric diastereoisomers of menthol (Fig. 2.24). 2 3 CH3 O 1 2 Fig. 2.24 HO1 N CH3 CH3 CH3 Menthol CH2CH2COO CCH2CH3 H 3 Phenbutrazate Phenbutrazate as for menthol. . R3 S2 R /1 S3 R2 S ; 1 R3 R2 R /1 S3 S2 S ; 1 S3 R2 R /1 R3 S2 S ; 1 S2 S2 R /1 R3 R2 S Menthol: 1 S3 S2 S ; Phenbutrazate 1 R3 R2 S : Menthol 1 Answers 8. Name the functional groups indicated in these molecules and state whether they are acidic, basic or neutral. A OH B COOH E HOH2C OH Fig. 2.25 O O NNS HO NH N CHCH2NHtC4H9 F C D Sulfasalazine Salbutamol L M OPO32- O G NH2 CH H CONH N S HO OH K O O I Ampicillin COOH J Hydrocortisone phosphate J carboxylic acid; K ketone; L secondary alcohol; M phosphate ester. (benzyl) alcohol; F secondary amine; G primary amine; H amide; l lactam ring; : A phenol; B carboxylic acid; C sulphonamide; D pyridine ring; E secondary Answers Physical and chemical properties of drug molecules 59References 1. Stenlake JB. Foundations of Molecular Pharmacology. Athlone Press; 1979. 2. Florence AT, Attwood D. Physicochemical Principals of Pharmacy. 2nd ed. Macmillan Press; 1988. 3. Ariens EJ. Trends Pharmacol Sci 1986;7:200–5. 4. Lehmann PA. Trends Pharmacol Sci 1986;7:281–5. Useful weblinks http://www.sirius-analytical.com The website provides some additional information on pKa and partition coefficient determination. 3 Titrimetric and chemical analysis methods Keypoints 60 Introduction 61 Instrumentation and reagents 61 Glassware 61 Primary standards and standard solutions 61 Direct acid/base titrations in the aqueous phase 62 Strong acid/strong base titrations 62 Weak acid/strong base and weak base/strong acid titrations 63 Titrations of the salts of weak bases in mixed aqueous/non-aqueous media 65 Indirect titrations in the aqueous phase 66 Estimation of esters by back titration 66 Saponification value 66 Estimation of alcohols and hydroxyl values by reaction with acetic anhydride (AA) 67 Non-aqueous titrations 68 Theory 68 KEYPOINTS Principles Non-aqueous titration of weak bases 69 Non-aqueous titration of weak acids 70 Argentimetric titrations 70 Compleximetric titrations 70 Redox titrations 71 Theory 71 Iodometric titrations 73 Direct titrations 73 Iodine displacement titrations 74 Iodine-absorbing substances in penicillins 75 Ion pair titrations 75 Titrations using indicator dyes 76 Titrations using iodide as a lipophilic anion 76 Diazotisation titrations 76 Potentiometric titrations 77 Potentiometric end-point detection 77 Use of potentiometric titration to determine pKa values 80 Karl Fischer titration (coulometric end-point detection) 81 Automation of wet chemical methods 82 Automatic titration 82 Flow injection analysis 83 Applications of FIA in pharmaceutical analysis 84 Determination of chloroxine 84 Determination of captopril 85 Determination of non-steroidal anti inflammatory drugs 85 Determination of promethazine 85 Determination of chlorocresol 85 Limit test for heavy metals 85 Use of segmented flow in determination of partition coefficients 86 Automated dissolution testing 86 Additional problems 87 An analyte is chemically reacted with a standard solution of a reagent of precisely known concentration or with a concentration that can be precisely determined. The amount of a standard solution required to completely react with all of the sample is used to estimate the purity of the sample. (Continued)Titrimetric and chemical analysis methods 61KEYPOINTS (Continued) Applications • Provide standard pharmacopoeial methods for the assay of unformulated drugs and excipients and some formulated drugs, e.g. those that lack a strong chromophore. • Used for standardisations of raw materials and intermediates used in drug synthesis in industry. Suppliers of raw materials may provide these materials at a specified purity which has been assayed titrimetrically to a pharmacopoeial standard. • Certain specialist titrations, such as the Karl Fischer titration used to estimate water content, are widely used in the pharmaceutical industry. Advantages • Capable of a higher degree of precision and accuracy than instrumental methods of analysis, with precisions of ca   0.1% being achievable. • The methods are generally robust. • Analyses can be automated. • Cheap to perform and do not require specialised apparatus. • They are absolute methods and are not dependent on the calibration of an instrument. Limitations • Non-selective. • Time-consuming if not automated and require a greater level of operator skill than routine instrumental methods. • Require large amounts of sample and reagents. • Reactions of standard solutions with the analyte should be rapid and complete. Introduction Titrimetric methods are still widely used in pharmaceutical analysis because of their robustness, cheapness and capability for high precision. The only requirement of an analytical method that they lack is specificity. This chapter covers the theoretical basis of most of the commonly used methods; the practical aspects of titrations have been covered thoroughly by other textbooks.1,2 Instrumentation and reagents Glassware The manufacturers’ tolerances for the volumes of a number of items of glassware are given in Chapter 1. The larger the volume measure the smaller the tolerance percentage is of the nominal volume. Thus, for a Grade A 1 ml pipette the volume is within   0.7% of the nominal volume, whereas for the 5 ml pipette the volume is within   0.3% of the nominal volume. If greater accuracy than those guaranteed by the tolerances is required, then the glassware has to be calibrated by repeated weighing of the volume of water contained or delivered by the item of glassware. This exercise is also useful for judging how good one’s ability to use a pipette is, since weighing of the volumes of water dispensed correctly several times from the same pipette should give weights that agree closely. Primary standards and standard solutions Primary standards are stable chemical compounds that are available in high purity and which can be used to standardise the standard solutions used in titrations. Titrants such as sodium hydroxide or hydrochloric acid cannot be considered as primary 62 Pharmaceutical analysisTable 3.1 Primary standards and their uses Primary standard Uses Potassium hydrogen phthalate Standardisation of sodium hydroxide solution Potassium hydrogen phthalate Standardisation of acetous perchloric acid Potassium iodate Standardisation of sodium thiosulphate solution through generation of iodine Anhydrous sodium carbonate Standardisation of hydrochloric acid Zinc metal Standardisation of EDTA solution EDTA, Ethylenediamine tetracetic acid. standards since their purity is quite variable. So, for instance, sodium hydroxide standard solution may be standardised against potassium hydrogen phthalate, which is available in high purity. The standardised sodium hydroxide solution (secondary standard) may then be used to standardise a standard solution of hydrochloric acid. Table 3.1 lists some commonly used primary standards and their uses. Direct acid/base titrations in the aqueous phase Strong acid/strong base titrations Figure 3.1 shows the titration curve obtained from the titration of a strong acid with a strong base. The pH remains low until just before the equivalence point, when it rises rapidly to a high value. In many titrations a coloured indicator is used, although electrochemical methods of end-point detection are also used. An indicator is a weak acid or base that changes colour between its ionised and un-ionised forms; the useful range for an indicator is 1 pH either side of its pKa value. For example, phenolphthalein (PP) pKa 9.4 (colour changes between pH 8.4 and pH 10.4) undergoes a structural rearrangement as a proton is removed from one of its phenol groups when the pH rises, and this causes the colour change (Fig. 3.2). Methyl orange (MO) pKa 3.7 (colour changes between pH 2.7 and pH 4.7) undergoes a similar pH-dependent structural change. Both these indicators fall within the range of the inflection of the strong acid/strong base titration curve. PP pH 14 12 10 8 Fig. 3.1 The change in pH as 25 ml of 1 M HCl are titrated with 1 M NaOH. 6 4 MO 2 0 0 10 20 30 40 50 ml of 1 M NaOH added Fig. 3.2 The structural rearrangement responsible for the colour change in phenolphthalein. OH OH Titrimetric and chemical analysis methods 63OH –H+ O O O Colourless +H+ Phenolphthalein pKa 9.4 COOH Red There are only a few direct strong acid/strong base titrations carried out in pharmacopoeial assays. Strong acid/strong base titrations are used in pharmacopoeial assays of: perchloric acid, hydrochloric acid, sulphuric acid and thiamine hydrochloride. Weak acid/strong base and weak base/strong acid titrations On addition of a small volume of the strong acid or strong base to a solution of the weak base or weak acid, the pH rises or falls rapidly to about 1 pH unit below or above the pKa value of the acid or base. Often a water-miscible organic solvent such as ethanol is used to dissolve the analyte prior to the addition of the aqueous titrant. Figure 3.3 shows a plot of pH when 1 M NaOH is added to 25 ml of a 1 M solution of the weak acid aspirin. In the case of aspirin, the choice of indicator is restricted by where the inflection in its titration curve lies; PP is suitable as an indicator whereas MO is not. In the example of the titration of quinine with hydrochloric acid (Fig. 3.4), MO is a suitable indicator because it falls within the inflection of the titration curve whereas PP is not suitable. Fig. 3.3 Titration curve for 25 ml of a 1.0 M solution of aspirin (pKa 3.5) titrated with 1.0 M NaOH. PP pH 14 12 10 8 6 4 MO 2 0 0 10 20 30 40 50 ml of 1 M NaOH added 64 Pharmaceutical analysis14 12 10  8 pH  6  4  2  0 PP MO Fig. 3.4 25 ml of a 1.0 M solution of quinine (pKa 8.05) titrated with 1.0 M hydrochloric acid. 0 10 20 30 40 50 ml of HCl added Some acids or bases can donate or accept more than one proton, i.e. 1 mole of analyte is equivalent to more than 1 mole of titrant. If the pKa values of any acidic or basic groups differ by more than ca 4, then the compound will have more than one inflection in its titration curve. Sodium carbonate is a salt of carbonic acid and it can accept two protons. The pKa values of carbonate and bicarbonate are sufficiently different (pKa 10.32 and 6.38) for there to be two inflections in the titration curve. The two stages in the titration are: CO2  3 þ Hþ ! HCO3 HCO 3 þ Hþ ! H2CO3 Self-test 3.1 Which of these indicators could be used in the titration of aspirin and which could be used in the titration of quinine? (i) Bromophenol blue pKa 4.0 (ii) Methyl red pKa 5.1 (iii) Cresol red pKa 8.3 (iv) Chlorophenol blue pKa 6.0. Aspirin: (iii) and (iv). Quinine: (i) and (ii) Answers: In a titration of sodium carbonate, the first inflection is indicated by PP and the whole titration by MO (Fig. 3.5). Self-test 3.2 A sample containing 25.14 g of neutral salts, glucose and a sodium carbonate/bicarbonate buffer was dissolved in 100 ml of water. A 25 ml aliquot of the resultant solution required 20.35 ml of 0.0987 M HCl when titrated to the PP end-point. A second 25 ml aliquot was titrated to the MO end-point and required 56.75 ml of the acid. Calculate the percentage of Na2CO3 (molecular weight 106) and NaHCO3 (molecular weight 84) in the sample. 3.39% and 2.12%, respectively Answers: Fig. 3.5 Titration of 1 M sodium carbonate with 1 M HCl. Titrimetric and chemical analysis methods 6514 12 PP pH 10  8  6  4  2  0 Self-test 3.3 MO 0 10 20 30 40 50 ml HCl added How many inflections do the following substances have in their titration curves when titrated with a strong base? Draw the predominant forms of these substances which would exist at pH 14. COOH CHCOOH COOH HOOCCH Oxalic acid pKa 1.19, 4.21 CH COOH 2 Fumaric acid pKa 3.03, 4.47 COOH HO C CH CHNH CI 2 2+ – COOH CH COOH 2 Citric acid pKa 3.06, 4.74, 5.4 Phenylalanine pKa 1.83, 9.13 oxalic acid 1, fumaric acid 1, citric acid 1, phenylalanine 2 Answers: Weak acid/strong base titration is used in the pharmacopoeial assays of: benzoic acid, citric acid, chlorambucil injection, mustine injection, nicotinic acid tablets and undecanoic acid. Titrations of the salts of weak bases in mixed aqueous/non-aqueous media Non-aqueous titrations, which are described below, are still used for the analysis of acids and salts of weak bases. However, in many instances it is simpler to titrate weak bases as their salts in a mixed non-aqueous/aqueous medium using potentiometric end-point detection. The protonated base behaves as a weak acid when titrated with sodium hydroxide. RNHþ3 þ HO  ! RNH2 þ H2O The advantage of adding a water-miscible solvent such as methanol to the titration is twofold. Firstly, the addition of the organic solvent effectively lowers the pKa value 66 Pharmaceutical analysis11 EP1EP1 Fig. 3.6 Potentiometric titration of lidocaine (lignocaine) hydrochloride with 0.1 M pH 9 Methanol:water (30:70) 7 Methanol:water (70:30) 5 0 2 4 6 8 10 12 ml 0.1 M NaOH NaOH. The samples were dissolved in methanol/ water (30:70, 50 ml) and methanol/water (70:30, 50 ml). of the base, since the ionised form of the base is less stable in a mixed solvent system where the dielectric constant is lower, and, secondly, the organic solvent keeps the base in solution as it is converted to its free base form during the titration. An example of this can be seen for the titration of lidocaine (lignocaine) hydrochloride in methanol/water mixtures (Fig. 3.6), where the size of the inflection in the titration curve increases when moving from 30% methanol to 70% methanol. This is a very convenient procedure for many organic bases. Indirect titrations in the aqueous phase These can be of the strong acid/strong base, weak acid/strong base or weak base/strong acid type. The more common examples are weak acid/strong base. Estimation of esters by back titration Excess of sodium hydroxide is added to the ester. The following reaction occurs: RCOOR0 þ XSNaOH ! RCOONa þ R0 OH The XSNaOH is back titrated with HCl using PP as an indicator. This procedure is used in pharmacopoeial assays of: benzyl benzoate, dimethyl phthalate, ethyl oleate, methyl salicylate, cetostearyl alcohol, emulsifying wax, castor oil, arachis oil, cod liver oil and coconut oil. Saponification value The assay of fixed oils provides a special case of ester hydrolysis since they are triesters of glycerol. The saponification value for a fixed oil is the number of mg of potassium hydroxide (KOH) equivalent to 1 g of oil. A high value means rancidity, a low value possible adulteration with mineral oil. Almost all edible oils have a saponification value between 188 and 196. Hydrolysis of the fixed oil is carried out with ethanolic KOH. This procedure is used in the pharmacopoeial assays of: castor oil, cod liver oil, cotton seed oil, almond oil and sesame seed oil. Acid values are also determined for fixed oils. The acid value for a substance is the number of mg of KOH required to neutralise 1 g of the test substance when it is titrated Titrimetric and chemical analysis methods 67with 0.1 M ethanolic KOH to a PP end-point. This value is quoted for many fixed oils in order to eliminate rancid oils, which contain large amounts of free fatty acid. Typically acid values for fixed oils are in the range of 1–2. Calculation example 3.1 The following data were obtained for a sample of cod liver oil: Weight of oil taken for analysis ¼ 2.398 g Ethanolic KOH (molecular weight 56.1) used in determination ¼ 0.986 M Amount of ethanolic KOH used for hydrolysis and in blank titration ¼ 25 ml Amount of 0.470 M HCl required to neutralise excess KOH ¼ 35.2 ml Amount of 0.470 M HCl required in the titration of blank ¼ 52.3 ml Calculation Amount of KOH used initially ¼ 52.3   0.47 ¼ 24.6 mmole Amount of HCl required to neutralise excess KOH ¼ 35.20   0.470 ¼ 16.5 mmole Amount of KOH used in hydrolysis ¼ 24.6 – 16.5 ¼ 8.1 mmole   molecular weight ¼ mg Amount of KOH used in the hydrolysis ¼ 8.1   56.1 ¼ 454.0 mg Amount of KOH/g of fixed oil used in the hydrolysis ¼ 454/2.398 ¼ 189.3 mg Therefore saponification value ¼ 189.3. Self-test 3.4 Calculate the saponification value of a sample of castor oil from the following data: • Weight of oil taken for analysis ¼ 2.535 g • Ethanolic KOH used in the hydrolysis ¼ 1.03 M • Amount of KOH used in hydrolysis ¼ 25 ml • Amount of 0.514 M HCl required to neutralise excess KOH ¼ 34.2 ml • Amount of 0.514 M HCl required in the titration of blank ¼ 50.2 ml. : 182 Answer Estimation of alcohols and hydroxyl values by reaction with acetic anhydride (AA) Alcohols can be determined by reaction with excess acetic anhydride (AA) (Fig. 3.7). This is a useful titrimetric method because the alcohol group is difficult to estimate by any other means. Fig. 3.7 Estimation of benzyl alcohol by reaction with acetic anhydride. 2OH +CH (CH3CO)2O CH2OCOCH3 + CH3COOH (CH3CO) + XS 2O

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