Clinical Pharmacy Practical Notes (PC 810) 2015/2016. Student's name: Student's number: Date Lesson Mark Signature. Medicinal Chemistry

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1 Student's name: Student's number: Date Lesson Mark Signature 1

2 Clinical Pharmacy PC810 Practical Sessions Section Date Subject 1 13/2/2016 Assay of Lugol's solution 2 20/2/2016 Assay of vitamin C + Assay of Captopril 3 27/2/2016 Assay of Ca gluconate-ascorbic acid Mixture 4 5/3/2016 Case Study about CNS 5 12/3/2016 Case Study about CVS 6 19/3/2016 Assay Exam - 26/3/2016 Midterm exam 7 2/4/2016 Assay of Aspirin and Amidopyrine 8 9/4/2016 Assay of Phenazone 9 16/4/2016 Case Study about Analgesics 10 23/4/2016 Case Study about Hormones - 30/4/2016 Spring Vacation 11 7/5/2016 Case Study Exam + Sheet Exam 2

3 Subject index Subject page Intended Learning Outcomes of courses (ILO's) 4 Assay of Aspirin (Acetyl salicylic acid) 6 Assay of Amidopyrine 11 Assay of Antipyrine(Phenazone) 16 Assay of lugol's solution 19 Assay of Vitamin C 27 Assay of Calcium gluconate-ascorbic acid 32 Assay of Captopril 35 Case studies for Opioid Analgesics & Non-Steroidal Anti-inflammatory Drugs (NSAIDs) 39 Case studies for Centrally Acting Drugs (CNS) 47 Case studies for Cardiovascular system Drugs 53 (CVS) Case studies for Hormonal Drugs 58 3

4 Intended Learning Outcomes of courses (ILO's) Upon completion of the course the student should be able to: 1. Handle laboratory chemicals in a correct and safe way to obtain optimum results without harmful effects. 2. Perform titrimetric techniques for determination of some pharmaceutical compounds, handling data in term of graphical presentation. 3. Recognize the chemical structure, nomenclature, pharmacophoric moieties and consequently the structure activity relationships in each class of drugs. 4. Be familiar with the analysis of drugs and the laboratory techniques used for that. 5. Analyze pharmaceutical dosage forms and determine their active constituents by different analytical methods. 6. Identifying the drug impunities and different related substances and giving possible explanation of their existence. 4

5 PART 1 5

6 Assay of some Nonsteroidal Anti-Inflammatory Drugs (NSAIDs) 1- Aspirin (Acetyl salicylic acid) M. Wt. = 180 Action and uses: Aspirin is indicated for the relief of minor aches and mild to moderate pain, for arthritis and related arthritic conditions to reduce the risk of transient ischemic attacks for myocardial infarction prophylaxis and as a platelet aggregation inhibitor. Methods of determination: [1] Acid-base titration Principle: Hydrolysis of aspirin by heating with known excess of standard NaOH then the excess unused NaOH is back titrated against standard acid using Phenolphthalein indicator. Procedure: To a Rivo tablet in a glass conical flask, add 30 ml 0.15 M NaOH and reflux for 10 min. Add 10 ml H2O then titrate the excess unused NaOH against 0.15 M HCl using Phenolphthalein as indicator. 6

7 Results and calculations for acid-base titration:- 1 Mole Aspirin 2 Mole NaOH 2 Mole HCl 1 NaOH Aspirin 2 E.P = ml M HCl 1 ml M HCl M.wt of Aspirin x Molar ratio 1000 Molarity Molarity Molarity HCl % Purity (30-(Ep )) f NaOH 100 actor weight taken 7

8 Results and calculations for acid-base titration:- 8

9 [2] Bromometric determination Principle Aspirin is hydrolyzed with NaOH to liberate equivalent amount of sodium salicylate that is converted to salicylic acid upon acidification with HCl. The produced salicylic acid was determined by indirect bromometry; where known excess of standard Br2 is added to produce tribromophenol. The excess unused bromine was reacted with potassium iodide to give equivalent amount of iodine which can be titrated against sodium thiosulphate solution using starch as indicator. Sodium tetrathionate 9

10 Results and calculations for bromometric determination:- 3 moles Br 1 mole Salicylic acid 1 mole Aspirin 2 Br Aspirin E.P.= ml Na S O ml M Br 2 M.wt of Aspirin Molar ratio x Molarity mole Br 1mole I 2 moles Na S O Na 2S2O 3 Br 2 2 Ep Molarity Na S O Molarity Weight taken % purity (25-( )) factor Dilution factor x Br 2 10

11 2) Amidopyrine 2, 3-Dimethyl-4-dimethylamino-1-phenyl-5-pyrazolone [1] Acid-base titration. Principle: Being basic in nature, amidopyrine can be titrated against standard H2SO4 giving amidopyrine sulfate. Procedure: Transfer10 ml amidopyrine sample to a conical flask and titrate against 0.05M H2SO4 using 10 drops ph.ph indicator. 11

12 Results and calculations (Acid-base titration):- 12

13 [2] Potassium Permanganate method. Principle: Amidopyrine is peroxidized with known excess of standard KMnO4 solution in alkaline medium (to decrease the oxidation potential of KMnO4 just to form peroxide). The excess unused KMnO4 can be determined by adding 10% KI solution to liberate equivalent amount of I2 which is titrated against standard Na2S2O3. Procedure: Transfer 10 ml sample to glass conical stoppered flask, add 25 ml 0.1M NaOH, 25 ml 0.1 M KMnO4 solution, 10 ml 10% KI solution and 10 ml dilute H2SO4. Titrate the liberated iodine against Na2S2O3 using 1 ml starch solution as indicator. 13

14 Results and calculations (Potassium Permanganate method):- 2 KMnO 3O (x 2) 4 4 Amidopyrine 2O (x 3) 4 KMnO 6O 3 Amidopyrine 3 So KMnO 4 Amidopyrine 4 Also 2 KMnO 5 I 10 Na S O E.P.= ml Na S O ml M KMnO 4 M.wt of Amidopyrine Molar ratio x Molarity Na S O KMnO EP MolarityNa S O 100 )) 5 Molarity volume taken % Concentration (25-( factor KMnO 4 14

15 [3] Picrate method. Principle: It depends on the precipitation of amidopyrine as picrate salt using known excess of picric acid, filter to remove the precipitate and the excess unused picric acid is determined by titration against standard NaOH solution. 15

16 3)Antipyrine(phenazone) Principle: Antipyrine can be easily iodinated giving 4-iodoantipyrine using known excess standard I2. The reaction is reversible and can be forced to one direction by neutralizing the liberated HI with KHCO3. 16

17 Procedure: 1- Transfer 10 ml (Otocalm ear drop) into 200 ml measuring flask and complete to the mark with water. 2- Transfer 5 ml of the dilution to glass conical stoppered flask, add 1 ml KHCO3 solution followed by 25 ml M I2. 3- Stand for 20 min., then add 1ml acetic acid and 1ml CHCl3 (To dissolve the 4- iodo derivative otherwise it adsorbs I2 so decreases the excess unused). 4-Titrate against 0.01 M Na2S2O3 using I ml starch solution indicator. 17

18 Result and Calculation 18

19 Assay of Iodine solutions Physical properties of iodine Iodine is Sublimable solid Sparingly soluble in water (1: 4000) Slightly soluble in alcohol (1: 20) Freely soluble in concentrated solutions of iodides (KI) forming KI3 (potassium tri-iodide) Chemical properties of iodine Iodine is oxidizing agent (but weaker than It gives blue color with starch It gives pink color in chloroform chlorine and bromine) Official iodine preparations 1) Strong iodine solution (Lugol's Slution) Iodine 5.0 gm KI 10.0 gm Purified water to ml It is used in treatment of hypothyroidism 2) Weak iodine solution (Tincture iodine) Iodine 2.5 gm KI 2.5 gm Purified water 5.0 ml Ethanol (90%) to ml It is used as antimicrobial agent 19

20 3) Anticorn paint Iodine 2.5 gm KI 2.5 gm Salicylic acid 3.0 gm Ethanol (90%) 70.0 ml Purified water to ml It is used in treatment of warts & tinea infections Role of KI; To increase the solubility of iodine by the formation of KI3 (triiodide) complex as the solubility of iodine in water is 1: 4000 (very sparingly soluble) & in alcohol 1: 20 (slightly soluble). To decrease the volatilization of iodine. Role of alcohol; It increase the solubilization of iodine so the amount of KI used in tincture iodine decreased & become equal to the amount of iodine. 20

21 Assay of Lugol's solution 1) For I2 content By direct titration of known volume of Lugol s solution # Na2S2O3 using starch as indicator. I2 + 2 Na2S2O3 2 NaI + Na2S4O6 1mole of Na2S2O3 = 1/2 mole of I2 Procedure for I2; 10 ml sample (in GSCF) + 10 ml H2O # M/50 Na2S2O3 using 1 ml starch as indicator (EP1) Calculations 1mole of Na2S2O3 = 1/2 mole of I2 1cc of M 50 N a S O M. Wt of I 2 x m olar ratio 1000 x M olarity x x gm I 2 % Conc. = E P x x Volu m e taken 2) For total iodide & I2 content A- Andrew's method It depends on the direct titration of the sample # KIO3 solution in highly acidic medium (not less than 4N HCl) using CHCl3 as indicator. In dil. HCl, I2 is only formed but by using conc. HCl, I2 will react with it producing ICl (iodonium monochloride). HCl + ICl H + [ICl2] - 21

22 High acidity is needed, not less than 4N HCl, to increase the stability of ICl through formation of a stable complex (hydroiododichloride complex). 2x KIO3 + 5KI + 6HCl 3I2 + 6KCl + 3H2O 3x KIO3 + 2I2 + 6HCl 5ICl + KCl + 3H2O 2KIO3 + 10KI + 12HCl 6I2 + 12KCl + 6H2O (1) 3KIO3 + 6I2 + 18HCl 15ICl + 3KCl + 9H2O (2) Summation of the equations (1) & (2) yields 5KIO3 + 10KI + 30HCl 15ICl + 15KCl + 15H2O Dividing by 5 KIO3 + 2KI + 6HCl 3ICl + 3KCl + 3H2O 1 mole KIO3 2 mole KI Disadvantages of Andrew s method High acidity (not less than 4 N HCl) prevent the use of starch as indicator due to its hydrolysis in high acidic medium. So CHCl3 is used as indicator & EP. is determined by the disappearance of the violet color of I2 from CHCl3, this need strong shaking to liberate I2, hence loss of iodine. B-Lang's method It is a modification of Andrew s method, for stabilization of cationic iodine (I + ) by using KCN where ICN is more stable than ICl. The medium is not highly acidic so starch can be used as indicator. This method is more sensitive & not require vigorous shaking, hence no loss of iodine. 2x KIO3 + 5KI + 6HCl 3I2 + 6KCl + 3H2O 22

23 3x KIO3 + 2I2 + 5KCN + 6HCl 5ICN + 6KCl + 3H2O 2KIO3 + 10KI + 12HCl 6I2 + 12KCl + 6H2O (1) 3KIO3 +6I2 +15KCN +18HCl 15ICN + 18KCl +9H2O (2) Summation of the equations (1) & (2) yields 5KIO3 +10KI +15KCN +30HCl 15ICN +30KCl +15H2O Dividing by 5 KIO3 + 2KI + 3KCN + 6HCl 3ICN + 6KCl + 3H2O 1 mole KIO3 2 mole KI Calculations M M.wt of KI x Molar ratio x Molaraty 1 cc of KIO x 2 x 100 = gm KI 1000 N.B) EP I & EP KI + I EP KI = EP - EP I 2Na S O & KIO 2I 1KIO 2I 4Na S O M M & as we use KIO & Na S O EP KI = EP EP 1 4 2EP 100 % Conc. of KI = (EP - 1 ) x x Disadvantage of Lang s method The liberation of HCN gas which is toxic. KCN + HCL HCN + KCl Procedure for Andrew s method 10 ml sample (in GCSF) + 5 ml Conc. HCl # M/100 KIO3 using 1 ml CHCl3 as indicator EP2 (indicator color changes from pink in CHCl3 to colorless). 23

24 Comparison between Andrew's & lang's methods Item Andrew's method Lang's method Medium Strong acidic Slightly acidic Titrant KIO3 KIO3 Stable form of iodonium ion Hydroiododichloride (H + [ICl2 - ]) (Iodonium cyanide) (ICN) Indicator CHCl3 Starch Disadvantages Sensitivity 1) High acidity hydrolysis of starch not used as ind. 2) CHCl3 need vigorous shaking loss of I2 Less sensitive Liberation of HCN (toxic) More sensitive 24

25 Results and calculations 25

26 Problems 1) In the determination of 10.0 mls Lugol's solution by Lang's method, 8.0 mls M/50 KIO3 were consumed. Find the concentration of KI in the solution if the concentration of iodine = gm% (1cc M/100 Na2S2O3 = gm I2 & M.wt of KI = 166). 26

27 2) In the determination of 10.0 mls Lugol's solution by Andrew's method, 15.0 mls M/100 KIO3 were consumed. Find the concentration of KI in the solution if the concentration of I2 = gm% (1cc M/50 Na2S2O3 = gm I2 & M.wt of KI = 166). 27

28 3) In the determination of 10.0 mls Lugol's solution by Lang's method, 15.0 mls M/20 KIO3 were consumed. Find the concentration of iodine in the solution if the concentration of KI = 1.66 gm% (1cc M/10 Na2S2O3 = gm I2 & M.wt of KI = 166). 28

29 4) In the determination of 5.0 mls Lugol's solution by Andrew's method, 15.0 mls M/40 KIO3 were consumed. Find the concentration of iodine in the solution if the concentration of KI = 1.66 gm% (1cc M/10 Na2S2O3 = gm I2 & M.wt of KI = 166). 29

30 Assay of Vitamin C Action and uses:- Vitamin C or L-ascorbate is an essential nutrient and is required for a range of essential metabolic reactions. The pharmacophore of vitamin C is the ascorbate ion. In living organisms, ascorbate is an antioxidant, since it protects the body against oxidative stress and is a cofactor in several vital enzymatic reactions. There are several methods for assay of vitamin C:- 1) Iodimetric determination Principle It depends on the reducing power of ascorbic acid where it could be oxidized by iodine in acidic medium into dehydroascorbic acid. The method could be carried out either:- a) direct (i.e. direct titration with standard iodine solution using starch solution as indicator which is added at the start) b) Indirect (adding known excess iodine and the excess unused is back titrated with Na2S2O3 using starch solution as indicator which is added when the color become straw yellow) 30

31 2) By Andrew's method Principle: The method depends on the titration of vitamin C solution in strong acidic medium against KIO3 using chloroform as indicator. 5C6H8O6 + 2KIO3 + 2HCl 5C6H6O6 + I2 + 2KCl + 6H2O (1) KIO3 + 2I2 + 6HCl 5ICl + KCl + 3H2O (2) Multiply equation 2 by 2 10C6H8O6 + 4KIO3 + 4HCl 10 C6H6O6 + 2I2 + 4KCl + 12H2O (3) Summation of equation 2 and 3 10C6H8O6+5KIO3+ 10HCl 10C6H6O6 + 5ICl + 5KCl + 15H20 (4) Dividing equation 4 by 5 give the final equation 2C6H8O6+ KIO3+ 2 HCl 2 C6H6O6 + ICl + KCl + 3 H20 (5) So 2 mole C6H8O6 1 KIO3 Notes: 1. In dilute HCl, iodine is the only formed but by using conc. HCl, I2 will react with it ICl (iodinium monochloride). 2. High acidity is needed, not less than 4 N HCl, to increase the stability of ICl through formation of a stable complex (hydroiodochloride complex) HCl + ICl H + [ICl] - 3. At E.P. CHCl3 layer becomes colorless, while aqueous layer is faint yellow. 31

32 Disadvantages of this method: The high acidic medium prevents the use of starch as indicator as it will be hydrolyzed in this acidic medium, so CHCl3 is used instead, but it requires vigorous shaking that may be associated with loss of I2. 3) Oxidation with cerric salts Principle: The method depends on the direct titration of vitamin C with cerric ammonium sulfate (which act as oxidizing agent) using diphenylamine as indicator. C6H8O6 + 2 Ce +4 C6H6O6 + 2 Ce H + 32

33 Assay of Vitamin C effervescent tablets (Andrew's method) Procedure: Dissolve the effervescent vitamin C tablet in exactly 400 ml H2O. Transfer 10 ml of the solution to a glass stoppered conical flask, add 5 ml conc. HCl and 1 ml CHCl3. Titrate against 0.01 M KIO3 with strong shaking till the disappearance of violet color from the chloroformic layer. Results and calculations:- 1 mole KIO 2 mole vitamin C 3 E.P.= ml KIO 3 1 ml M KIO 3 M.wt of Vitamin C Molar ratio x Molarity % Purity E.P. Factor x Dilution fact or x Weight taken Total volume after dilution N. B: Dilution factor Volume taken for assay 33

34 Results and calculations:- 34

35 Assay of Calcium gluconate-ascorbic acid Mixture 1) Assay of ascorbic acid (Direct iodimetry) Transfer 10 ml sample into a glass stoppered conical flask. Add 25 ml 20% H2SO4 and 1 ml starch solution indicator. Titrate Against 0.02 M I2 till blue color. Results and calculations (Direct iodimetry):- 1 mole I 1 mole vitamin C 2 E.P.= ml I 2 1 ml M I 2 M.wt of Vitamin C Molar ratio x Molarity % Purity E.P. Factor x Dilution factor x Weight taken Total volume after dilution NB. : Dilution factor Volume taken for assay 35

36 2) For Calcium gluconate (C12H22CaO14, H2O 448.4) Action and uses: Calcium gluconate is a common calcium source used in the treatment of hypocalcemia. It is also used to counteract an overdose of magnesium sulfate, often administered to pregnant women experiencing premature labor to slow or stop contractions. Excess magnesium sulfate can cause respiratory depression, for which calcium gluconate would be the antidote. Principle (Complexometry) Depends on direct titration of sample solution against standard EDTA solution in presence of ammonia buffer using E.B.T. indicator. Procedure Take 10 ml of sample to a glass conical flask. Add 2 ml of ammonia buffer and few specks of EBT indicator. Titrate against 0.05 M EDTA till blue color is obtained. 36

37 Results and calculations (For Calcium gluconate-vit.c Mixture):- 37

38 Determination of Captopril (M.wt 217.3) Action and use: Captopril is Angiotensin-converting enzyme inhibitor & so used for treatment of hypertension. Principle of Argentometric assay (Mohr's method) Direct titration of captopril with silver nitrate using K.chromate as an indicator. The reaction stoichiometry was found to be 1: 1 (CPT: AgNO3), according to the following scheme: RSH + AgNO3 RSAg + HNO3 2AgN03 + CrO4-2 Ag2CrO4 +2 NO3 - Procedure Take 10 ml of sample to a glass conical flask. Add 0.5 ml of K.chromate indicator. Titrate against 0.01 M AgNO3 till first tinge brown. 38

39 Results and calculations 39

40 Results and calculations 40