Drug Stability Lecture Notes PDF

Introduction to Pharmaceutics 30:721:301 Professor Guofeng You, Ph.D. Department of Pharmaceutics Room 218 848-445-6349 [email protected] Stability/ Chemical Kinetics  Definition of Stability Stability according...

Introduction to Pharmaceutics 30:721:301 Professor Guofeng You, Ph.D. Department of Pharmaceutics Room 218 848-445-6349 [email protected] Stability/ Chemical Kinetics  Definition of Stability Stability according to USP is: ”extent to which a product retains,within specified limits, & throughout its period of storage and use (i.e. its shelf life), the same properties & characteristics that it possessed at the time of its manufacture”  Responsibility of the Pharmacist USP recommendations Watch for & comply with expiration dates, rotate stock, & use older products first Observe products for instability Properly handle drugs and drug products that require extemporaneous preparation Package products using recommended containers & closures Educate patients about correct storage & use of products  Example 1 : Chlorpromazine Chlorpromazine HCl: solubility 1g / 2.5 ml water Chlorpromazine base : insoluble in water If you raise the pH of an aq. solution of Chlorpromazine HCl, some of the salt form of the drug would be converted into the unionized free form. If the conc. of the chlorpromazine base exceeds its water solubility, what will happen?  Example 2: Effect of temperature The solubility of most drugs decreases as the temp. of the solution decreases Refrigeration is often recommended for drug solutions to increase their chemical stability and retard microbial growth but this may cause problems with precipitation Ex. of parenteral drugs which cannot be refrigerated because of ppt. effects : fluorouracil, cisplatin, metronidazole & brands of aminophylline.  Example 3: Efflorescent powders Powders contain water of hydration that may be released when powders are triturated or stored in low relative humidity Examples: atropine sulfate, caffeine, cocaine, codeine, sodium acetate, terpin hydrate, morphine acetate, scopolamine hydrobromide, etc.  Example 4: Hygroscopic & Deliquescent Powders Hygroscopic drugs absorb moisture from the air. “Deliquescent” refers to hygroscopic powders that can absorb enough moisture to dissolve & form a solution Examples: ammonium bromide, ephedrine sulfate, phenobarbital Na, potassium acetate, lithium bromide, calcium chloride, Ca bromide, pepsin, etc.  Example 5: Eutectic mixtures Two or more substances that liquefy when mixed at room temperature Examples:acetaminophen, aspirin, benzocaine, camphor, chloral hydrate, lidocaine, menthol, phenacetin, prilocaine, salicylic acid, thymol, resorcinol, phenylsalicylate (salol), aminopyrine, etc.  Example 6: Solvent effects When a drug is dissolved in a solvent, and a second solvent, one in which the drug is poorly soluble, is added, the drug may precipitate Example: Digoxin has a water solubility of 0.08 mg/ml. Digoxin Injection is available in a cosolvent system containing 40% propylene glycol & 10% alcohol. If it is diluted with an aqueous injectable solution, precipitation of digoxin may occur. Example 7: Polymorphic Conversions Drugs which exist in different crystalline structure in the solid sate, although they are identical chemically in liquid and gas states Example: Cocoa Butter has several polymorphic forms with MPs 18, 24, 28-31, & 340C. It is very easily overheated and it then solidifies with one of the lower MP polymorphs which may make a suppository melt at room temp.Hence, you need to heat Cocoa Butter slowly and not go over 340C.  Useful References Remington: The Science & Practice of Pharmacy The Merck Index Handbook of Chemistry & Physics Trissel’s Handbook of Injectable Drugs United States Pharmacopoea Trissel’s Stability of Compounded Formulations  Chemical Kinetics Important since involves kinetics of stability & decomposition All human clinical pharmacokinetics also involves same fundamental principles and equations Concentration, temperature, light & catalysts play an important role  Rates and Orders of Reaction The rate , velocity, or speed of a reaction is given by dc/dt, where dc is the increase or decrease of concentration over an infinitesimal time interval, dt. According to Law of Mass Action, rate of reaction is proportional to product of molar conc. of reactants raised to the power usually equal to # of molecules a & b of substances A & B undergoing reaction (see next slide)  In the reaction: aA + bB + cC……= Products The rate of reaction is: Rate = - 1/a. d(A)/dt = - 1/b. d(B)/dt = k (A)a (B)b….. Where K is the RATE CONSTANT & (a + b) is overall ORDER of the reaction  Actual example Reaction of ethyl acetate and sodium hydroxide in aqueous solution: CH3COOC2H5 + NaOH → CH3COONa + C2H5OH Rate is - d[CH3COOC2H5]/dt = - d[NaOH]/dt = k[CH3COOC2H5]1[NaOH]1  Order of Reaction?? a =1 first order with respect to ethyl acetate b = 1 first order with respect to NaOH solution BUT a + b = 2 So overall the reaction is SECOND ORDER  Pseudo First Order If water and NaOH in great excess in the reaction Hence concentration of ethyl acetate changes during reaction, but that of the other reactants stays nearly constant Reaction rate = - d[CH3COOC2H5]/dt =k[CH3COOC2H5]1[NaOH]1 = k’ [CH3COOC2H5] and k’ = k [NaOH] Rate depends only on first power of conc. of ethyl acetate hence PSEUDO FIRST ORDER  Units of Basic Rate Constants For zero order reactions: k= - dA/dt = moles/liter = moles = moles liter-1 sec-1 second liter sec For first order k = -dA/dt x 1/A = moles/liter = 1 = sec-1 sec-moles/liter sec  Units of Basic Rate Constants For second order reaction: k = - dA/dt x 1/A2 = moles/liter sec (moles/liter)2 = liter = liter second-1mole-1 mole-sec  Zero order Reactions Example: loss of color of a drug (measured using spectroscopy) followed zero order kinetics Means that change of absorbance A with time is - dA/dt = k0. Why minus?? If you integrate this with limits of t=0 at A0 & t=t when A= At:  Zero order Reactions  dA = - k0  dt and integrate between limits A0 when t=0 and At when t=t. Hence At - A0 = -k0 t Or At = A0 - k0t OR C =C0 -kt  Zero order Reactions C =C0 -kt If you plot C vs. t you get a straight line, slope is -k & rate is INDEPENDENT of C -K C t Zero order Reactions: Half Life Half life or half period is time for half of the original conc. to disappear It is the time for A to go to 0.5A At - A0 = -k0 t or C =C0 -kt Hence t1/2 = 0.5A0 k0 Units are in HOURS  Apparent Zero Order: Suspensions Suspensions are a zero order case where conc in solution depends on drug’s solubility As drug decomposes in solution, more drug is released from suspended particles, keeping the overall conc. in solution relatively constant. Reservoir of solid drug in suspension is important to maintaining the drug’s equilibrium solubility Common case in pharmaceutical formulations  Apparent Zero Order: Suspensions The first order expression becomes: -d[A]/dt = k[A] When A is constant: k[A] = k0 So - d[A]/ dt = k0 a zero order equa. Actually referred to as APPARENT ZERO ORDER Aspirin suspension: an example A prescription for a liquid aspirin preparation is called for. It is to contain 325 mg/5ml or 6.5g/100ml. The solubility of aspirin at 250C is 0.33 g/100ml; therefore, the preparation will definitely be a suspension. The other ingredients in the prescription cause the product to have a pH of 6.0. The first order rate constant for aspirin degradation in this solution is 4.5 x 10-6 sec-1. Calculate the zero order rate constant. Determine the shelf life of this prescription, assuming that the product is satisfactory until the time at which it has decomposed to 90% of its original concentration at 250C.  ANSWER k0 = k x [aspirin in solution] since k0=k[A] = [4.5 x 10-6 sec-1] x [0.33 g/100ml] = 1.5 x 10-6 g/100ml sec-1 C =C0 -kt t90 = 0.10[A]0/ k0 = 0.10 [6.5 g/100ml] 1.5 x 10-6 g/100ml sec-1 4.3 x 105 sec = 5.0 days  First Order Reactions Example is decomposition of drug X in solution follows: -dc/dt = kc Note rate DEPENDENT on concentration c Integrating equation between conc c0 at t=0 and c at time t:  dc/c = -k dt and…….  First Order Reactions ln c - ln c0 = -k (t-0) ln c = ln c0 - kt Converting into common logs: Log c = log c0 - kt/2.303 Or k = 2.303 log c0/c OR c=c0e-kt t  First Order Reactions OR c = c0 10-kt/2.303 1st order equation also written as K = 2.303 log a t (a-x) Where a replaces c0 and (a-x) replaces c. First Order Reactions : half life Period of time for drug to decompose to half its original concentration For first order: K = 2.303 log a t (a-x) t1/2 = 2.303/k log 500/250 = 2.303/k. log2 t1/2 = 0.693/k  Example A solution of a drug contained 500 units/ml when prepared.It was analyzed after a period of 40 days and was found to contain 300 units/ml. Assuming the decomposition is first-order, at what time will the drug have decomposed to one half its original concentration?  Answer K = 2.303 log a C0 t (a-x) C k = [2.303/40]. log 500/300 = 0.0128 day-1 t = 2.303/0.0128. log 500/250 = 54.3 days  Next Example The catalytic decomposition of hydrogen peroxide may be followed by measuring the volume of oxygen liberated in a glass burette. From such an experiment, it was found that the conc. of hydrogen peroxide remaining after 65 mins, expressed as the volume in ml of gas evolved, was 9.60 from an initial conc of 57.90. A) calculate k B) how much hydrogen peroxide remained undecomposed after 25 minutes?  Answers A) K = 2.303 log a t (a-x) k = 2.303/65. Log 57.90/9.60 = 0.0277 min-1 B) 0.0277 = 2.303/25. Log 57.90/c c = 29.01  Second Order Reactions Usually involves bimolecular reactions, which occur when 2 molecules come together: A + B = Products - dA/dt = - dB/dt = k[A][B] or if a and b are initial conc of A and B and x is conc of each reacting in time t: dx/dt = k(a-x) (b-x) and if a=b: dx/dt = k(a-x)2  Second Order Reactions Integration gives: k = 2.303 log b(a-x) t(a-x) a(b-x) Half life is t1/2 = 1/ak  Determination of Order Substitution method: try data in equations and see which fit best Graphic method/ Linear Regression. c vs. time & get straight line = zero order; log c vs. time = first order ; and 1/c vs. time = second order. Half life method. Each order has different equations so see which fits best  Influence of temperature The speed of reactions increased by temperature, catalysts,solvents and light Arrhenius equation describes the effect of temperature: K(rate constant) = Ae -Ea/RT or log k = logA - Ea/2.203. 1/RT [Equation1] where Ea is energy of activation, A =Arrhenius constant, R is gas constant, and T is temperature Decomposition of medicinal agents Pharmaceutical decomposition can be classified as: hydrolysis,oxidation,isomerization, epimerization, & photolysis Affects stability in liquid, solid and semisolid products  Drug Classes susceptible to oxidation Catecholamines (epinephrine) Phenolics (morphine, phenylephrine) Phenothiazines (chlorpromazine) Steroids & tricyclic antidepressants Thiols (captopril) Amphotericin B, tetracycline, furosemide, etc. Summary 1. Order of reaction: 0, 1st, 2nd 2. Concentration dependence of each order - dc/dt = k0 -dc/dt = kc dc/dt = k(a-c)2 3. How to determine the order c vs. time: zero order log c vs. time: first order 1/c vs. time: second order Based on half life 4. Application