PH1002 Experiments 1, 2 and 3 Combined copy PDF

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PH1002: Physical Pharmacy I 1. Solubility Relationships of Drugs and their Metabolites

Name: Joseph O’Shea Student Number: 14311653 Date: 13/12/2014

Introduction:

In pharmaceutical science, it is of particular importance while developing/studying drugs that the solubility of the drug in the body is known. In this experiment, an investigation into the distribution of a drug between two phases was carried out. An aqueous phase and a lipid-like phase (chloroform) was used, as this simulates water and fatty tissue, respectively, which is found in our bodies. The aqueous phase solubility demonstrates how well the drug is disposed of in urine. The lipid phase solubility demonstrates how well the drug penetrates (dissolves in) the fatty tissue of the body. This is important as the cell membrane is composed of a lipid bilayer (liposome), therefore it is important to know if the drug will be biologically active i.e. if it will penetrate somatic cells and take effect. Solubility is the property of a solid, liquid, or gaseous chemical substance called solute to dissolve in a solid, liquid, or gaseous solvent to form a homogeneous solution of the solute in the solvent. The distribution of a chemical between the two phases of water and the lipidlike solvent chloroform shows how it might be distributed between fatty tissue and water. A compound’s biological activity is strongly influenced by how well it dissolves in body tissues which are usually fatty and how quickly it is removed in the urine. Knowledge of a material’s ability to penetrate fatty tissues is desirable, as it is believed that (all else being sterically and electronically equal) material is effective to the degree that it penetrates these tissues. The higher the polarity of a substance, the poorer its solubility in the fat/chloroform phase and the better its solubility in the aqueous phase. In this experiment, the lipid solubility and aqueous solubility of benzoic acid and its metabolite in the body, hippuric acid was investigated. Several titrations were carried out: HCl with Na2CO3, HCl with NaOH, Benzoic acid with NaOH and Hippuric acid with NaOH using methyl red as the indicator. The same acids were then placed in separatory funnels with chloroform, and left to allow the different layers to separate, and then the aqueous layer is titrated with NaOH using phenolphthalein as the indicator- allowing the amount of each acid in chloroform to be found.

Method:

1. Standardisation of NaOH and Determining Molarities of Benzoic and Hippuric AcidA standard solution of 0.05 M Na2CO3 was prepared by dissolving 1.32 g in 250 ml of water. A solution of HCl was titrated against 25 ml of Na2CO3 using methyl red as the indicator to standardise the HCl solution. (Colour change from red-orange to yellow). 25 ml of 0.02M NaOH was titrated with the standardised HCl using methyl red as the indicator. The molarity of the NaOH is then determined. 10 ml of 0.02 NaOH was titrated with Hippuric acid and Benzoic acid using methyl red as the indicator to determine the precise concentration of the Hippuric and Benzoic acid.

2. Investigating the Distribution between the Two Phases25 ml of each acid solution was placed into a separatory funnel with 25 ml of chloroform. The funnel was shaken (gently at first, and then vigorously for 10 minutes). The funnel was then left to stand and the layers allowed to separate completely. The aqueous layer was transferred to an Erlenmeyer flask and then titrated with NaOH using phenolphthalein as the indicator. Colour change: straw yellow to pink. The molarity of each acid was then calculated and the amount of each acid in chloroform was found by subtracting the number of moles found in the extracted aqueous solution from the number of moles in the original aqueous solution.

Results and Calculations:

Key Formula: M1 V1 / n1 = M2 V2 / n2 (M = Molarity of the substance, V = Volume used in titration (titre value), N = Stoichiometry of the reaction)



Na2CO3 solution:

1.32 g / 106 (Mw of Na2CO3) = 0.01245 mol in 1L 250 mL was made; therefore the solution is 4 times stronger. (0.01245)(4) = 0.0498 M solution of Na2CO3.



Standardising HCl:

2(0.0498)(25) = 25.1x x= [50(0.0498)] / 25.1 = 0.0992 M solution of HCl



Determining molarity of NaOH solution:

25x = 0.0992(6) x = 6(0.0992) / 25 = 0.0238 M NaOH solution



NaOH : Hippuric Acid:

0.0238(10) = M1 (9.9) M1 = 0.0238(10) / 9.9 = 0.024 M Hippuric Acid



NaOH : Benzoic Acid:

0.0238(10) = M1 (9.8) M1 = 0.0238(10) / 9.8 = 0.0243 M Benzoic Acid



Extracted Hippuric Acid : NaOH:

25x = 23.3(0.0238) x = 23.3(0.0238) / 25 =0.0222 M



Extracted Benzoic Acid : NaOH:

25x = 0.0238(4) x = 0.0238(4) / 25 = 0.0038 M

Hippuric Acid: Phase

Concentration/ M

Aqueous Chloroform

0.0222 M 0.0018 M

Concentration (mmoles /25 mL)

Benzoic Acid: Phase

Concentration/ M

Aqueous Chloroform

0.0038 M 0.0205 M

Concentration (mmoles /25 mL)

Conclusions: From the results, it is observed that the solubility alters when benzoic acid is metabolised into hippuric acid in the body. Benzoic acid is more lipophilic than hippuric acid, i.e. there are more moles of benzoic acid in the lipid phase (0.0205 M compared to 0.0018 M of hippuric acid). As expected, the metabolism process also changes hydrophilicity of the drug substances. Hippuric acid is more water soluble than benzoic acid (determined due to its higher concentration in the extracted aqueous phase). These results were expected. The drug as it enters the body is capable of penetrating fat tissue and when metabolised by the body, becomes water soluble so it can be easily excreted in the urine.

2. Kinetic Studies of an Oxidation Reaction

Name: Joseph O’Shea

Student Number: 14311653 Date: 14/12/2014

Introduction:

Kinetics is the study of the rates at which reactions take place. The rate of reaction is dependent on temperature and concentration of reactants. For the reaction: a A + b B + …  Products (P), the rate of reaction = -(1/a) x d[A]/dt = -(1/b) x d[B]/dt = d[P]/dt = k [A]α . [B]β (k is the reaction constant, α is the order of reaction regarding A and β is the order of reaction regarding B) The overall order of the reaction is α + β + … The Arrhenius equation shows how it is dependent on temperature: k = A e(-E / RT) (Where: A = Frequency Factor, E = Activation Energy, T = Temperature, R = Gas Constant) In this experiment, the reaction studied was the oxidation of iodide ion by peroxydisulphate ion: S2O82- + 2I-

→ 2SO4- + I2

Chemical kinetics is the study of the rates at which reactions occur (rates are dependent on concentration and temperature). Study of the kinetics of a chemical reaction may give information about the mechanism of the reaction. In this reaction, the rate of production of I2 is equal to the rate of disappearance of S2O82-. As the reaction between persulphate and iodide is quite slow and therefore thermodynamically favoured its kinetics can be measured. The rate of this reaction is given by: S 2O 8 2−¿ ¿ Rate = = k[S2O82-]x [I-]Y (k, x and y need to be calculated by measuring how the ¿ −d ¿ rate is affected by changing the concentrations of the reactants). The values of x and y are determined by observing how the rate is affected by changing the concentration and temperature. Firstly it in necessary to measure the initial rate of the reaction. The production of iodine is very fast and needs a sensitive measurement apparatus to determine the rate of its production. Persulphate and thiosulfate react in the ratio of 1:2.

2S2O32- + I2

→ S4O62- + 2I-

In this experiment, the effect of concentration and the effect of temperature on the rate of reaction were observed. When the iodine is produced in this experiment, is reacts with the thiosulphate and continues to until all the thiosulphate has been used up. Starch detects the presence of iodine, and turns the solution blue if iodine is present. The solution turns blue once all of the thiosulphate has been removed. The time taken for this to occur is recorded. The effect of concentration on the rate of reaction in this experiment is demonstrated by varying the concentration of the peroxydisulphate and the iodide. The times taken for the reactions to occur were recorded. Experiment B acts as the control in the experiment, having the constant amount of iodide and persulphate. Experiment A is used to demonstrate the effect of iodide concentration on the rate of reaction. Experiment C is used to demonstrate the effect of persulphate concentration on the rate of reaction. The effect of temperature on the rate of reaction in this experiment is demonstrated by recording the time taken for the reaction to occur at 0 ℃ , 30 ℃ and 40 ℃ .

Method: Effect of Concentration- 3 experiments were set up for this investigation: A, B and C. Each experiment required two tubes. The starch serves as an indicator because as I2 is produced

the mixture should turn blue-black. The Potassium Nitrate is added to maintain a constant ionic strength (note the extra volume in tube A to compensate for the lack of iodide). Experiment A- 5 ml of 0.05 M persulphate was added to Tube 1. 1 ml of 0.8 M iodide, 5 ml of starch, 1 ml of 0.01 M thiosulphate and 8 ml of 0.8 M potassium nitrate was added into Tube 2. The contents of Tube 2 were then poured into Tube 1. The time taken for the solution to turn blue was then recorded. Experiment B- 5 ml of 0.05 M persulphate was added to Tube 1. 2 ml of 0.8 M iodide, 5 ml of starch, 1 ml of 0.01 M thiosulphate and 7 ml of 0.8 M potassium nitrate was added into Tube 2. The contents of Tube 2 were then poured into Tube 1. The time taken for the solution to turn blue was then recorded. Experiment C- 2.5 ml of 0.05 M persulphate and 2.5 ml of water were added to Tube 1. 2 ml of 0.8 M iodide, 5 ml of starch, 1 ml of 0.01 M thiosulphate and 7 ml of 0.8 M potassium nitrate was added into Tube 2. The contents of Tube 2 were then poured into Tube 1. The time taken for the solution to turn blue was then recorded. Results: Experiment A 80 seconds

Experiment B 40 seconds

Experiment C 79 seconds

Conclusion- As concentration decreases, the rate of reaction decreases. This is evident from the results obtained, as when the concentration of the iodide and persulphate is decreased, the reaction takes almost 4 times longer to reach completion. By the same token, as the concentration increases, the rate of reaction increases. This is because when the concentration increases, there are more molecules- therefore more collisions occur, therefore more effective collisions occur, which increase the reaction rate.

Effect of Temperature5 ml of 0.05 M persulphate was added to Tube 1. 1 ml of 0.8 M iodide, 5 ml of starch, 1 ml of 0.01 M thiosulphate and 8 ml of 0.8 M potassium nitrate was added into Tube 2. The contents of Tube 2 were then poured into Tube 1. The tube was then placed in an ice bath at 0 ℃ turn blue was then recorded.

and the time taken for the solution to

The tube was placed in hot baths at 30 and 40 ℃ and the time taken for the solution to turn blue was then recorded. Results: 0 ℃

Room Temperature

30 ℃

40 ℃

120.5 seconds

80 seconds

60.5 seconds

43 seconds

Conclusion- As temperature increases, the rate of reaction increases. This is because the molecules gain more energy, and more collisions occur and therefore more effective collisions occur which bring the reaction to completion, and increase the reaction rate. Also, more of the molecule have the require activation energy.

Calculations: To calculate concentration-

Volume Used x Molarity Total Volume

Concentration [S2O82-] [I-] [S2O32-] Time difference

A 0.0125 0.04 5.0x10-4 80 seconds

B 0.0125 0.08 5.0x10-4 40 seconds

C 6.25x10-3 0.08 5.0x10-4 79 seconds

Rate constants- -d [S2O82-]/dt -d [S2O82-]= 2.5x10-4 From equation: S2O82- + 2I- → 2SO4- + I2 1 mole S2O82- = 1 mole I2 = 2 moles S2O321 ml 0.01 M S2O32- contains 0.00001 moles (1.0x10-5 moles) 1 ∴ 5.0x10-6 moles persulphate in 20 ml ( 2

x 1.0x10-5)

∴ 2.5x10-4 moles in 1000 ml (50 x 5.0x10-6) No. moles in 1 l = 2.5x10-4 moles/l 2.5x10-4/dt = k[S2O82-]X [I-]Y

Solving for X and Y using concentration:

Reaction A:

Reaction B:

Reaction C:

(2.5 × 10-4)/80 = 3.125 x 10-6

(2.5 × 10-4)/40 = 6.25× 10-6

(2.5 × 10-4)/103 = 3.16× 10-6

3.125 × 10-6 = k (0.0125) x (0.04) y

6.25 × 10-6 = k (0.0125) x (0.08) y

3.16× 10-6 = k (0.00625) x (0.08) y

Reactions A and B (6.25 × 10-6) / (3.125 × 10-6) = k (0.0125) x (0.08) y / k (0.0125) x (0.04) y 2 = 2y Y=1

Reactions B and C

(6.25 × 10-6) / (3.16× 10-6) = k (0.0125) x (0.08) y / k (0.00625) x (0.08) y 2 = 2x X=1

Solving for k using temperature (with X and Y already known): Rate = -d [S2O82-]/dt =k [S2O82-] x [I-] y X = 1, y = 1 from our previous calculation

Temperature

Time difference

T-1

0 oC

120.5

0.0083

Room Temperature

80

0.0125

30 oC

60.5

0.017

40 oC

43

0.023

0°C: (2.5 × 10-4)/120.5= k (0.0125)1(0.04)1 K=4.1 × 10-3 Lnk= – 5.50

Room temperature:

30°C

40°C

2.5 × 10-4/80= k (0.0125)1(0.04)1 K=6.25 × 10-3 Lnk= – 5.01

2.5 × 10-4/60.5= k (0.0125)1(0.04)1 K=0.0083 Lnk= – 4.79

2.5 × 10-4/43= k (0.0125)1(0.04)1 K=0.012 Lnk= – 4.42

Relationship between 1/t and lnk 0 0.01

0.01

0.01

0.01

0.01

0.02

0.02

0.02

0.02

0.02

-1

lnK

-2 lnk Linear (lnk)

-3

-4 -4.42 -4.79 -5.01

-5 -5.5 -6

1/T

Determining the value of E: Slope = -E / R Equation of trendline (from Excel): y = 70.424x – 6.0004 Slope = 70.424 70.424 = -E / 8.314 8.314 (70.424) (-1) = E E = -585.51 kJ / mole

3. Lipophilicity Constants of Sulfonamide Substituents

Name: Joseph O’Shea Student Number: 14311653 Date: 15/12/2014

Introduction:

The partition coefficient is the ratio of the concentrations of a solute in two immiscible or slightly miscible liquids, or in two solids, when it is in equilibrium across the interface between them. The partition coefficient is one of the most important factors controlling drug action in biological systems. It is often used in quantitative studies of structure-activity relationships of drugs. Lipophilicity means the tendency of the compound to partition between lipophilic organic phase (immiscible with water) and polar aqueous phase. The lipophilicity constant, π, is a constant that has been derived for the contribution of individual atoms or substituents to the partition coefficient and is defined as:

π = log PRX – log PRH = log

PRX P RH

PRH and PRX= the partition coefficients of the unsubstituted and substituted compounds respectively.

The chromatographic Rm value is used as an expression of the lipophilic character of molecules. The Rm value is derived from the Rf value which is obtained by means of reversedphase TLC, i.e. : 



Rm = log (Rf-1 – 1) for neutral substances K a+¿ ¿ H +¿ Rm = log (Rf-1 – 1) + log { for acids and bases ¿ ¿ ¿ ¿

The change in Rm for substituted and unsubstituted compounds is defined as: πx = RmRX – RmRH

In this experiment, the mobile phase was an aqueous buffer at pH 7.4. The sulphonamides were dissolved in acetone. The Rf values were calculated.

The following is known as sulphonamide:

Method: A TLC plate was set up using an aqueous buffer of pH 7.4 as the mobile phase. The sulphonamide derivatives were dissolved in acetone. The TLC plate was allowed to develop in a developing jar and the Rf values were calculated.

Results:

Compound Sulfalinamide

Sulfamethazine

R group H

pKa 10.45

Rf 0.73

7.7

0.38

Sulfamethoxypyridazine

7.05

0.41

Sulfathiazole

7.1

0.54

Calculations:

K a+¿ ¿ H +¿ -1 Rm = log (Rf – 1) + log { ¿ ¿ ¿ ¿ Ka = 10-pKa

pH = 7.4

[H+] = 10-pH

[H+] = 10-7.4 = 3.98 x 10-8

1. Calculate Rm value for each derivative:

Sulfalinamide: Rm = log (

1 −1¿ 0.73

Sulfamethazine: Rm = log (

1 −1¿ 0.38

Sulfamethoxypyridazine: Rm = log (

Sulfathiazole: Rm = log (

+ log

(10

+ log

1 −1 ¿ 0.41

−10.45

+10 7.4 10

(10

−7.7

−7.4

+10 7.4 10

+ log

)

−7.4

(10

)

} = -4.32 x 10 -1 }

−7.05

−7.1 −7.4 1 (10 +10 ) −1 ¿ + log } 7.4 0.54 10

= 3.89 x 10 -1

+10 7.4 10

−7.4

)

} = 6.68 x 10 -1

= 4.07 x 10 -1

2. Calculate the lipophilicity substituent constant of the three substituents above:

To do this, the Rm of sulfalinamide was subtracted from the Rm values of the other three substituents. For example, the π value for Sulfamethazine = 3.89 x 10-1 – (-4.32 x 10-1) = 0.821

The complete set of π Compound Sulfalinamide Sulfamethazine Sulfamethoxypyridazine Sulfathiazole

values are shown below: π 0.000 0.821 1.100 0.838

If the constant is positive, it is less hydrophilic than H.

3. Calculate the partition coefficients (and logP) of the three derivatives:

We are given that the partition coefficient of sulfalinamide is 0.15. Therefore the PRH = 0.15 Log PRH = -0.824 πx = log PRX – log PRH For Sulfamethazine; The complete set of values is shown in the table below. Compound Sulfalinamide Sulfamethazine Sulfamethoxypyridazine Sulfathiazole

log(PRX) 0.15 -0.003 0.276 0.014...