Drug Stability - National University

Dr. Adil Abdelrhman

Drug Stability

Stability of pharmaceutical products refers to the

capacity of the product or a given drug substance to

remain within established specifications of identity,

potency, and purity during a specified time period.

1. Chemical Stability of Drug Substances

Many drugs are susceptible to some form of chemical

decomposition when formulated in either liquid or even

solid dosage forms.

Such degradation not only leads to a loss of potency of

the drug but may, in some cases, cause changes in the

physical appearance of the dosage forms, for example,

discoloration following the photochemical

decomposition of the drug.

1. Identification of the product(s) formed provides a

better understanding of the mechanism(s) of these

chemical reactions as well as other valuable information.

2. Other reasons for quantitating drug loss include the

following:

Why study drug stability?

I. The drug may degrade to a toxic substance. Therefore, it is important to determine not only how much drug is lost with time but also what are its degradants. In some cases, the degradants may be of known toxicity.

II. Degradation of the drug may make the product esthetically unacceptable. Products are presumed to be adulterated if significant changes in, for instance, color or odor have occurred with time.

colorless pink

III. Even though a drug may be stabilized in its intended

formulation, the formulator must show that the drug is

also stable under the pH conditions found in the

gastrointestinal tract, if the drug is intended for oral

use.

Most drug substances are fairly stable at the neutral

pH values found in the small intestine (disregarding

enzymatic degradation) but can be unstable at pH

values found in the stomach. Examples of drugs that

are very acid-labile are various penicillins,

erythromycin and some of its analogs.

Pathways of Chemical Degradation

Drug substances used as pharmaceuticals have diverse

molecular structures and are, therefore, susceptible to

many and variable degradation pathways.

Possible degradation pathways include:

hydrolysis,

dehydration,

isomerization and racemization,

elimination,

oxidation,

photodegradation,

and complex interactions with excipients and other drugs.

It would be very useful if we could predict the chemical

instability of a drug based on its molecular structure.

This would help both in the design of stability studies

and, at the earliest stages of drug development, in

identifying ways in which problematic drugs could be

formulated to minimize chemical degradation.

1. Hydrolysis

Hydrolysis, in its widest sense, is the breaking of a

chemical bond due to the reaction of water.

Hydrolysis is often the main degradation pathway for

drug substances having ester and amide functional

groups within their structure.

1.1 Drugs susceptible to hydrolytic degradation

If the drug is a derivative of carboxylic acid or contains

functional groups based on this moiety, for example an

ester, amide, lactone, lactam , imide or carbamate

(Scheme 1) ,then we are dealing with a drug which is

liable to undergo hydrolytic degradation.

Scheme 1:Examples of chemical groups susceptible to hydrolysis

Drugs that contain ester linkages include acetylsalicylic

acid (aspirin),physostigmine, methyldopate, tetracaine

and procaine.

Ester hydrolysis is usually a bimolecular reaction

involving acyl–oxygen cleavage.

Acid-catalysed hydrolysis

The initial protonation on the carbonyl oxygen produces a

resonance stabilized cation; this increases the

electrophilicity of the carbonyl group, making it susceptible

to attack by the nucleophilic water. Proton transfer from the

water to the alcohol converts the latter into a better leaving

group (G).

Incidentally, this mechanism is the reverse of the

mechanism for formation of an ester from an acid

and an alcohol under acidic conditions (esterification).

Base-catalysed hydrolysis

This reaction is easier to follow; the nucleophile in this

case is the strongly basic OH ion, which attacks the

carbon of the carbonyl group directly.

Note that in base-catalysed hydrolysis the acid formed

by hydrolysis instantaneously reacts with the excess of

base to form the salt of the acid. The free acid may be

obtained, if desired, by acidification of the mixture.

Hydrolysis of the ester group of procaine

The hydrolysis of amides involves the cleavage of the amide linkage as for example, in the breakdown of the local anaesthetic cinchocaine

As examples of lactam ring hydrolysis we can consider

the decomposition of nitrazepam and chlordiazepoxide,

Other drugs, apart from the benzodiazepines, which are

susceptible to hydrolysis include the penicillins and

cephalosporins.

Hydrolysis of β-lactam cephalosporins

1.2 Controlling drug hydrolysis in solution

1.2.1 Optimization of formulation

Hydrolysis is frequently catalyzed by hydrogen ions

(specific acid-catalysis) or hydroxyl ions (specific

base-catalysis) and also by other acidic or basic

species that are commonly encountered as components

of buffers. This latter type of catalysis is referred to as

general acid–base catalysis.

Several methods are available to stabilise a solution of

a drug which is susceptible to acid–base catalyzed

hydrolysis.

1. The usual method is to determine the pH of

maximum stability from kinetic experiments at a

range of pH values and to formulate the product

at this pH.

2. Alteration of the dielectric constant by the addition of

nonaqueous solvents such as alcohol, glycerin or

propylene glycol may in many cases reduce hydrolysis.

3. Since only that portion of the drug which is in

solution will be hydrolyzed, it is possible to suppress

degradation by making the drug less soluble.

The stability of penicillin in procaine–penicillin

suspensions was significantly increased by reducing its

solubility by using additives such as citrates, dextrose,

sorbitol and gluconate.

4. Adding a compound that forms a complex with the

drug can increase stability.

The addition of caffeine to aqueous solutions of

benzocaine, procaine and tetracaine was shown to

decrease the base-catalysed hydrolysis of these local

anaesthetics in this way.

5. In many cases solubilisation of a drug by surfactants

protects against hydrolysis.

1.2.2 Modification of chemical structure of drug

The control of drug stability by modifying chemical

structure using appropriate substituents has been

suggested for drugs for which such a modification does

not reduce therapeutic efficacy.

AzithromycinClarithromycin

Erythromycin

2. Oxidation

Oxidation processes

Oxidation is a loss of electrons or an increase in the

oxidation state and reduction is a gain in electrons or a

decrease in the oxidation state. Redox reaction is an

electron transfer process:

where the reduced form loses n number of electrons

(e-).

Autoxidation (auto-oxidation) is a complex oxidation

mechanism that proceeds through a free radical chain

process.

It is a common degradation mechanism for

unsaturated fats, but a number of drugs containing

carbon-carbon double bonds also undergo

autoxidation. In general, the free radical chain process

consists of three steps

A free radical is formed in the initiation step, frequently

through the thermal or photochemical hemolytic

cleavage of an R-H bond.

The initiation is catalyzed by metal ions such as Cu 2+ ,

Ni 2+ , and Fe 3+ . Molecular oxygen is added to the free

radical in the propagation step and then in the rate-

determining step (RDS), the peroxyl radical extracts the

hydrogen atom from RH to form another R • radical.

The rate of the RDS depends mainly on the strength of

the C-H bond that is being cleaved. In the termination

step, the chain reaction is broken when two free

radicals react to form nonradical products.

The rancidification of fat, which gives a distinct rancid

smell (due to the formation of volatile aldehydes and

ketones), is an example of autoxidation:

2.1 Drugs susceptible to oxidation

Oxidation mechanisms for drug substances depend

on the chemical structure of the drug and the

presence of reactive oxygen species or other

oxidants.

Catechols such as methyldopa and epinephrine are readily oxidized to quinones.

Thiols such as 6-mercaptopurine and captopril are oxidized to disulfides.

Phenothiazines such as promethazine are oxidized via complex pathways and yield various products

Polyunsaturated molecules such as vitamin A, as well as other polyenes such as ergocalciferol, cholecalciferol

are susceptible to oxidation.

Polyene antibiotics, such as amphotericin B (II) which

contains seven conjugated double bonds (heptaene

moiety), are subject to attack by peroxyl radicals,

leading to aggregation and loss of activity.

The ether group in drugs such as econazole nitrate (III)

and miconazole nitrate (IV) is susceptible to oxidation.

The process involves removal of hydrogen from the C–H

bonds in the α-position to the oxygen to produce

radicals, which further degrade to α-hydroperoxides

and eventually to aldehydes, ketones,alcohols and

carboxylic acids.

2.2 Stabilization against oxidation

Various precautions should be taken during

manufacture and storage to minimize oxidation.

2.2.1 The oxygen in pharmaceutical containers should

be replaced with nitrogen or carbon dioxide; contact of

the drug with heavy-metal ions such as iron, cobalt or

nickel, which catalyze oxidation, should be avoided; and

storage should be at reduced temperatures.

are sterically hindered phenols that react with free

radicals, blocking the chain reaction.

2.2.2 Antioxidants

These hindered phenols are radical-trapping

antioxidants for oxy- and peroxy radicals. The phenoxy

radicals formed with their bulky substituents are

stabilized by steric hindrance and cannot attach drugs

or unsaturated fatty acids to maintain the chain

reaction.

The radical formed is stabilized by delocalization of the

unpaired electron around the phenol ring to form a

stable resonance hybrid (i.e. low-energy radical):

Reducing agents such as sodium metabisulfite

are compounds that have lower redox potentials and,

thus, are more readily oxidized than the drug they are

intended to protect.

2.2.3 Use of chelating agents

Chelating agents, such as disodium edetate, are

used to chelate and remove metal ions.

Chelating agents do not possess antioxidant activity as

such, but enhance the action of phenolic antioxidants

by reacting with catalyzing metal ions to make them

inactive.

2.2.4 Use of amber or coloured glass containers

Amber glass excludes light of wavelengths ‹470 nm and

so affords some protection to light-sensitive

compounds.

Special formulations, such as metered dose inhalers

used in the treatment of asthma, also offer protection

from light and oxygen since the drug is dissolved or

suspended in propellant and stored in a sealed

aluminium container.

Isomerisation is the process of conversion of a drug into

its optical or geometric isomers.

Since the various isomers of a drug are frequently of

different activity, such a conversion may be regarded as

a form of degradation, often resulting in a serious loss

of therapeutic activity.

3.Isomerisation and Racemization

The appreciable loss of activity of solutions of

adrenaline at low pH has been attributed to

racemisation – the conversion of the therapeutically

active form, in this case the levorotary form, into its less

active isomer.

In acidic conditions the tetracyclines undergo

epimerisation at carbon atom 4 to form an equilibrium

mixture of tetracycline and the epimer, 4-epi-

tetracycline. The 4-epi-tetracycline is toxic and its

content in medicines is restricted to not more than 3%.

Cis–trans isomerisation may be a cause of loss

of potency of a drug if the two geometric

isomers have different therapeutic activities.

Vitamin A (all-trans-retinol) is enzymatically

oxidised to the aldehyde and then isomerised

to yield 11-cis-retinal ,which has a decreased

activity compared with the all-trans molecule.

Racemization and epimerization, which are

reversible conversions between optical isomers,

have been reported for many drug substances.

Pilocarpine undergoes epimerization by base

catalysis

Primary photochemical reaction occurs when the

wavelength of the incident light is within the

wavelength range of absorption of the drug

(usually within the ultraviolet range, unless the

drug is coloured), so that the drug molecule

itself absorbs radiation and degrades.

4. Photochemical decomposition

Photodegradation may also occur with drugs

that do not directly absorb the incident

radiation, as a consequence of absorption of

radiation by excipients in the formulation

(photosensitisers) which transfer the

absorbed energy to the drug, causing it to

degrade.

Although it is difficult to predict which drugs

are likely to be prone to photodegradation,

there are certain chemical functions that are

expected to introduce photoreactivity, including

carbonyl, nitroaromatic and N-oxide functions,

aryl halides, alkenes, polyenes and sulfides.

The phenothiazine chlorpromazine (CLP) is

rapidly decomposed under the action of

ultraviolet light, the decomposition being

accompanied by discoloration of the solutions

Stabilization against photochemical decomposition

1. Pharmaceutical products can be adequately

protected from photo induced decomposition

by the use of coloured glass containers and

storage in the dark. Amber glass excludes light

of wavelength ‹ 470 nm and so affords

considerable protection of compounds sensitive

to ultraviolet light.

2. Coating tablets with a polymer film containing

ultraviolet absorbers has been suggested as an

additional method for protection from light.

Polymerisation is the process by which two or more

identical drug molecules combine together to form a

complex molecule.

5. Polymerization

It has been demonstrated that a polymerisation

process occurs during the storage of concentrated

aqueous solutions of aminopenicillins, such as

ampicillin sodium and also formaldehyde.

The reactive β-lactam bond of the ampicillin molecule

is opened by reaction with the side-chain of a second

ampicillin molecule and a dimer is formed.

• Sugars such as glucose and lactose are known to

undergo dehydration to form 5-(hydroxymethyl)

furural.

• Erythromycin is susceptible to acid catalyzed

dehydration,

• Prostaglandins E1 and E2 undergo dehydration

followed by isomerization

6. Dehydration

Dehydration of glucose

Scheme 28

Dehydration of erythromycin

Scheme 29

Dehydration and isomerization of prostaglandin E2

Scheme 30

Drug substances having a carboxylic acid group

are sometimes susceptible to decarboxylation,

4-Aminosalicylic acid is a good example

7. Decarboxylation and Elimination

Quite often, reactions can occur between the

drug and one or more additives. Similarly, two

drugs might be formulated in the same product

and react with each other.

8. Drug-Excipient and Drug-Drug Interactions

8.1 Reactions of Bisulfite, an Antioxidant

It was reported that epinephrine, a catecholamine,

undergoes displacement of its hydroxy group by

bisulfite, .

Dexamethasone 21-phosphate, an α/β-unsaturated

ketone, is known to undergo addition by bisulfite.

Reducing sugars readily react with primary

amines, including those of amino acids, through

the Maillard reaction. Drug substances with

primary or secondary amine groups undergo this

addition/rearrangement reaction, also called

the browning reaction because of the resulting

discoloration.

8.2 Reaction of Amines with Reducing Sugars

In the presence of drug substances with hydroxy groups,

aspirin undergoes a reversible transacylation reaction

to form salicylic acid, while acetylating the drug

substance. For example, codeine and sulfadiazine are

acetylated by aspirin.

8.3 Transesterification Reactions

Another example of transesterification is the reaction of

benzocaine with polyvinyl acetate phthalate.

2. Kinetics of chemical decomposition in solution

Kinetics deals with the study of the rate at which

processes occur and mechanism of chemical reactions

KineticsMotion ormovement

Velocity, rate orrate of change

It involves the study of rate of change and the way in

which this rate is influenced by the concentration of

reactants, products, and other chemical species that

may be present, and by factors such as solvents,

pressure, and temperature.

• It gives an in light into the mechanism of changes involved

• Allows a prediction of the degree of change that will occur after a given time has elapsed.

WHY DO WE STUDY ABOUT KINETICS?

RATES

• the speed or velocity of a reaction with which a

reactant or reactants undergoes a change.

• It is determined by the change in the

concentration of the reactants or products as a

function of time.

• The rate may be determined by the slowest or rate determining step.

RATES AND ORDERS OF REACTIONS

Before we can predict the shelf-life of a dosage form it

is essential to determine the kinetics of the breakdown

of the drug under carefully controlled conditions.

2.1 REACTION RATE AND HALF-LIFE (t1/2)

For a hypothetical reaction,

aA Product

the rate of reaction of A into product (P) is defined by the

derivative

−dA/dt

where the minus sign means the concentration (or

amount) of A is decreasing with time. For P, the rate of

production is defined by

+dP/dt

Here the concentration (or amount; amount = concentration × volume) of P always increaseswith time. According to the law of mass action, the rate of chemical reaction is proportional to the product of the molar concentration of the reactants each carried to the power equal to the number of moles of the substance undergoing reaction.

−dA/dt α [A]a

Therefore,

−dA/dt = k[A]a or

dA/dt = −k[A]a

where k is the proportionality constant, also known

as the rate constant, [A] is the molar concentration

of A as a function of time, and a is the number of

moles of A undergoing reaction.

• Half-life (t1/2) is defined as the time required for

half of the initial concentration (or amount)

of reactants to form products.

• Shelf-life is defined as the time for the original

potency (i.e., 100%) of the active drug to be reduced to

95% (t 95 ) or, more frequently, 90% (t 90 ), although

more stringent time limits may apply if the degradation

products are toxic.

2.2 ORDER OF REACTION

• The order of a chemical reaction refers to the

way in which the concentration of the reactant

influences the rate.

• Most commonly, zero-order and first-order reactions

are encountered in pharmacy and will be discussed in

details below. However, the concept of second order will

also be introduced.

2.2.1 Zero-Order Reactions

Again, for a hypothetical reaction,

A → P

The rate equation for a zero-order reaction is

defined as

dA/dt = −k[A]0

Since [A]0= 1, the rate equation can be simply

rewritten as

dA/dt = −k0

The rate equation can be integrated as

and solved to give the integrated rate equation

for zero-order kinetics:

At − A0 = −k0t

or

A0 − At = k0t

where At is the amount of A at any time t, A0 is

the initial amount of A, and k0 is the zero-order

rate constant with the unit of concentration (or

mass)/time.

The rate constant is obtained by plotting the change in [A] over time

Zero-order plot of [A] versus time. [A]0 is the y-intercept.

The half-life (t1/2) of a zero-order reaction

can be deduced by the fact that at t1/2

At = A0/2

Substituting into the integrated equation for

zero-order reactions will give

A0/2 = A0 − k0 t1/2

or

A0 − A0/2 = k0 t1/2 and A0/2 = k0 t1/2

Therefore, the half-life of zero-order reactions

is defined by

t1/2 = 0.5A0/k0 or A0/2 k0

Therefore, the half-life of zero-order reactions is directly

proportional to the initial concentration of the

reactants.

Example

Drug X degrades by a zero-order process with a

rate constant of 0.05 mg ml-1 year−1 at room

temperature. If a 1% weight/volume (w/v) solution is

prepared and stored at room

temperature:

1. What concentration will remain after 18 months?

2. What is the half-life of the drug?

Answers

A0 = 1% w/v = 10 mg/ml; t =18 months = 1.5 year;

k0 = 0.05 mg ml−1 year −1

1. At = A0 – k0t = 10 – (0.05 × 1.5) = 9.25 mg/ml

2. t1/2 = 0.5 A0 /k0 = (0.5 × 10)/0.05 =100 years

2.2.2 First-Order Reactions

The rate equation for first-order kinetics is

given by

dA/dt = −k1[A] 1

or simply

dA/dt = −k1 [A]

The rate equation can be integrated to give

which is solved to give the integrated rate equation for

first order kinetics:

ln At − ln A0 = −k1t

which can be rewritten as

ln At = ln A0 − k1t or At = A0 e−k

1t

Converting into base 10 log,

log = ln/2.303

log At = log A0 – k1t/2.303

The half-life of a first-order reaction is defined as the

time (t1/2) when At = A0/2.

Substituting this into the integrated equation:

log A0/2 = log A0 – k1t1/2/2.303

which will give

log 2 = k1t1/2/2.303

or

t1/2 = 0.693/k1

and

k1 = 0.693/ t1/2

Note that the half-life of first-order reactions is

independent of the initial concentration of the

reactants.

The units of k1 will be (1/time) or time-1.

Plot of [A] and [P] versus time

First-order plot of ln [A] versus time. ln [A]0 is the y-intercept.

– /2.303

The characteristic of a first-order reaction is such that

over the same time period, the fraction of unchanged

drug remaining will always be the same as from

At = A0 e−k

1t,

At/A0 = e−k1

t

which is a constant.

Example

Homatropine is an ester that undergoes hydrolysis in

aqueous solutions. Samples were collected and the

homatropine concentration at various time points

calculated:

Apparent (or Pseudo) Zero-Order Reactions for Suspensions

Suspensions (e.g., Amoxil®, Mylanta®, and Maalox®)

are dosage forms in which the concentration of a drug

exceeds the solubility.

Suspension formulations, therefore, have solid

particles suspended in a solution of drug. The

decomposition of drugs in suspensions depends

on the concentration of the drug in solution as

shown below.

Asolid → Asoln → Product

The rate of decomposition of A in solution, therefore, is

given by the first-order expression

dA/dt = −k[A]soln

Since there is excess solid drug present, and it

dissolves continuously to replace the portion of drug

in solution, which is being converted to product, the

concentration of drug in solution (k[A]soln) remains

constant, and we can write

k[A]soln = k′0

which converts the decomposition process an

apparent zero-order process, with k′0 being the

apparent zero-order rate constant, although

the actual decomposition of the drug from the

solution may be first order. The above rate

equation, therefore, can be rewritten as

dA/dt = −k′0

and the integrated form of the apparent zero-order rate

equation is expressed as

At = A0 − k′0 t

Example

An aqueous suspension of drug X contains 200

mg of drug X per teaspoon (5.0 ml). The

solubility of drug X at 25°C is 1 g/350 ml in

water, and the first-order rate constant for

degradation of drug X in solution at 25°C is

3.9 × 10−6 sec−1 .

Calculate (a) the zero-order rate constant and

(b) the shelf life of the liquid preparation.

Answers

(a) From k′0 = k[A]soln =

(3.9 × 10−6 sec−1)(1 g/350 ml) =

1.11 × 10−8 g/ml sec−1

(b) t90 = 0.1[A0]/ k′0 =

(0.1)(0.04 g/ml)/ 1.11 × 10−8 g/ml sec−1=

3.6 × 105 sec = 4.2 days

2.2.3 Second-Order Reactions

The rate of a second-order reaction is proportional to the

concentration of two reactants:

where k2 is the second-order rate constant and [A] and

[B] are the reactant concentrations.

A simpler form of second-order reaction is obtained if

[A]=[B] or if two molecules of A react:

Example

Ethyl acetate undergoes hydrolysis in aqueous alkaline

solutions containing equal concentrations of both the

ester and sodium hydroxide or 0.020 M. Samples were

collected and the ethyl acetate concentration was

determined at two time points:

The initial concentrations of both reactants are the same

and, thus, can be used to calculate the second-order rate

constant.

Since the initial concentrations of both the ester and the

sodium hydroxide are identical, can be used to calculate

the half-life:

EXAMPLE Calculation of first-order rate constant and

half-life

The following data were obtained for the hydrolysis of homatropine in 0.226 mol dm-3 HCl at 90°C:Show that the hydrolysis follows first-order kinetics and calculate (a) the rate constant, and (b) the half-life.

Answer(a) The reaction will be first-order if a plotof the logarithm of the amount of homatropineremaining against time is linear.

The following figure shows a linear plot with aslope = -(1.96 - 1.55)/(20 - 2) = -2.278 x 10-2 h-1

Slope = K1/2.303

Therefore,

K1 = 5.25 x 10-2 h-1

b) From this equation

t0.5 = 0.693/K1 = 13.2 h

The half-life of the reaction is 13.2 h.

First-order plot for hydrolysis of homatropine in hydrochloric acid (0.226 mol dm-3) at 90°C

2.1.3 Determination of the order of reaction

• The most obvious method of determining the order of

a reaction is to determine the amount of drug

decomposed after various intervals and to substitute

the data into the integrated equations for zero-, first-

and second-order reactions.

• The equation giving the most consistent value of k for

a series of time intervals is that corresponding most

closely to the order of the reaction.

For zero order, At = A0 – k0t.950 = 1000 − k0(2)900 = 1000 − k0 (4)

Solving both the equations shows that k0 is the same (i.e., 25 mg/ml/h). Therefore, this is anexample of zero-order reaction kinetics. The same conclusion could also be drawn by looking atthe data as the amount lost every 2 hours being constant

The best way to determine the order of a reaction is to

plot the data according to the equation for zero-order,

first-order ,or second-order reactions:

If a linear zero-order plot is obtained, then the reaction

follows zero-order kinetics. If a nonlinear plot is

observed, then re-plot the data according to a first-

order equation. If a linear plot is observed, then the

reaction is first-order; if not, then try a second-order

plot, and so on.

Fitting data to the standard rate equations may,

however, produce misleading results if a fractional

order of reaction applies.

An alternative method of determining the order of

reaction, which avoids this problem, is based

on the following equations :

where [A]0 is the initial reactant (drug) concentration

and n is the reaction order., t 1/2 is determined at two

different [A]0 values:

The half-life of the reaction is determined for a series of

initial drug concentrations, C0 , and the order, n, is

calculated from the slope of plots of log t0.5 as a

function of log C0 .

Example Determination of the order of reaction

Example

The kinetics of decomposition of a drug in aqueous

solution were studied using a series of solutions of

different initial drug concentrations, C0 . For each

solution the time taken for half the drug to decompose

(that is, t0.5) was determined with the following results:

Determine the order of reaction and calculate the rate

constant.

Answer

Application of the equation requires values for log C0

and log t0.5; thus

A plot of log t0.5 against log C0 is linear with slope (1 - n)

= -1.01. Hence n = 2.01, i.e. the reaction is second-

order.

The intercept of the graph (that is, the value of log t0.5

at log C0 = 0) is 2.60, and

k = 2.51 × 10−3 (mol dm −3) −1 min −1

3.Factors influencing drug stability

•Liquids

•Semisolids

•Solids

3.1 Liquid dosage forms1. pH:

Degradation rates of drug substances are generally affected

by pH because most degradation pathways are catalyzed by

hydronium and/or hydroxide ions. Water itself is also a

critical reactant. If the critical path in a reaction involves a

proton transfer or abstraction step, other acids and bases

present in solution (usually buffer species) can affect

the reaction rate.

These reactions will also be pH-dependent because the

fraction of any species present in its acid or base form

will be dependent on its dissociation constant and the

solution pH.

Also, for ionizable drugs, the fraction of drug present in

any particular form will depend on the solution pH.

Therefore, if the reactivity of the drug depends on its

form, its reactivity will be pH-dependent.

When a reaction dependent on hydronium and hydroxide ion activity is performed at constant pH, it usually follows pseudo-first-order kinetics, which can be described by a first-order rate constant kobs. A reaction in which hydronium ion, hydroxide ion, and watercatalysis are observed can be described by

where kobs is the sum of the specific rate constants

and activities for each parallel pathway, and a H+

and a OH- are the activities of hydronium and

hydroxide ion, respectively. This equation is for the

case when the drug itself is neutral in the pH range

of study, that is, where ionization of the drug does

not have to be taken into account. The pH-rate

profiles for drugs meeting this criterion are relatively

simple, as shown in the following figures

Simple pH-rate profiles for drug degradation

If the contributions of the first and second terms in the

equation are larger than that of the third term, the pH-rate

profile shown in (panel 1) of the figure is seen. If the second

and third terms are dominant, then the profile illustrated in

(panel 2) is observed. If the first and third terms are

dominant, then a V-type pH-rate profile (panel 3) occurs. If

all terms contribute significantly, the U-shaped pH-rate

profile shown in (panel 4) is observed.

2. Temperature:

• Increase in temperature usually causes a very

pronounced increase in the hydrolysis rate of drugs in

solution. This effect is used as the basis for drug

stability testing.

The equation which describes the effect of temperature

on decomposition is the Arrhenius equation:

log k = log A – Ea /(2.303RT)

and

where Ea is the activation energy (The minimum

amount of energy needed for a reaction to proceed), A

is the frequency factor, R is the gas constant

(8.314 J mol –1 K –1 ) and T is the temperature in kelvins.

The Arrhenius equation predicts that a plot of the log

rate constant, k, against the reciprocal of the

temperature should be linear with a gradient of

– Ea/2.303R.

• Therefore, assuming that there is no change in the

order of reaction with temperature, we can measure

rates of reaction at high temperatures (where the reaction

occurs relatively rapidly) and extrapolate the Arrhenius

plots to estimate the rate constant at room temperature

(where reaction occurs at a very slow rate)

• This method therefore provides a means of speeding up

the measurements of drug stability during preformulation.

A typical Arrhenius plot showing the determination of a rate constant at room temperature by extrapolation of data at high temperatures.

For a given reaction under a given reaction condition

(i.e., reaction media, pH, etc.), the following equations

can be used to calculate k-values at different

temperatures:

Thus, if Ea is known and the rate constant (k1 ) at a

given temperature is T1 , then it is possible to calculate

k 2 at temperature T2

Application of the Arrhenius Equation

EXAMPLE

The following values were determined for the specific acid-catalytic constants for an anti-inflammatory drug:

Determine graphically (a) the rate constant for

acid-catalysis at 25°C, (b) the activation energy.

Answer

According to the Arrhenius equation, a plot of

log k against 1/T has a gradient of -Ea/2.303R.

From the graph:

(a) At 1/T = 3.356x 10-3 K-1, log k = -5.85.

Therefore, k at 25°C = 1.41 x10-6 (mol.dm-3)-1 s-1.

(b) Gradient = -5.91x103 K.

Therefore Ea = 113 kJ mol -1.

Calculating the rate constant

The first-order rate constant for the hydrolysis of sulfacetamide at 120°C is 9x10-6s-1 and the activation energy is 94 kJ mol-1. Calculate the rate constant at

25°C.

Removing the logarithms,

k120/k25 = 9.55 x103.

Therefore, k25 = 9 x 10-6/9.55 x103

=9.4x10-10 s-1.

Calculation of Ea , t1/2 , and t90

Atropine is an ester that is hydrolyzed in aqueous

solutions. The following data for atropine hydrolysis

was obtained for an aqueous atropine solution:

Using the data in the table, calculate A, Ea , t1/2 , and t90

at room temperature (25 °C).

We start by plotting the data according to

Then we need to convert the data from the table to the

Y (lnk) and X (1/T) values:

Calculating the shelf-life

The initial concentration of active principle in

an aqueous preparation was 5x10-3 g cm-3.

After 20 months the concentration was shown by

analysis to be 4.2x10-3 g cm-3. The drug is known to be

ineffective after it has decomposed to 70% of its original

concentration.

Assuming that decomposition follows first order

kinetics, calculate the expiry date of the drug

preparation.

Substituting into the first-order equation

70% of the initial concentration= 3.5x10-3 g cm-3

3. Ionic strength:

We often need to add electrolytes to drug solutions, for

example to control their tonicity. Consequently we must

pay particular attention to any effect they may have on

stability.

• The equation which describes the influence of

electrolyte on the rate constant is the Brønsted–

Bjerrum equation:

log k = log k0 + 2AzAzB√μ

where zA and zB are the charge numbers of the two

interacting ions, A is a constant for a given solvent and

temperature and μ is the ionic strength.

• The Brønsted–Bjerrum equation predicts that a plot

of log k against μ1/2 should be linear for a reaction in

the presence of different concentrations of the same

electrolyte with a gradient of 2AzAzB

• The gradient will be positive (i.e. the reaction rate will be

increased by electrolyte addition) when reaction is between

ions of similar charge, for example, the acid-catalysed

hydrolysis of a cationic drug ion.

• The gradient will be negative (i.e. the reaction

rate will be decreased by electrolyte addition) when the

reaction is between ions of opposite charge, for

example, the base-catalysed hydrolysis of positively

charged drug species.

The variation of rate constant, k, with square root of ionic strength, μ, for reaction betweena: ions of similar charge,b: ion and uncharged molecule andc: ions of opposite charge.

4. Oxygen:

Since molecular oxygen is involved in many oxidation

schemes, we could use oxygen as a challenge to find out

whether a particular drug is likely to be affected by

oxidative breakdown. We would do this by storing solutions

of the drug in ampoules purged with oxygen and then

comparing their rate of breakdown with similar solutions

stored under nitrogen.

• Drugs which have a higher rate of decomposition

when exposed to oxygen can be stabilized by replacing

the oxygen in the storage container with nitrogen or

carbon dioxide. These drugs should also be kept out of

contact with heavy metals and should be stabilized with

antioxidants.

5. Solvent effects:

Since we are considering the hydrolysis of drugs it might

seem that an obvious way to reduce the breakdown

would be to replace some or all of the water in the

system with a solvent such as alcohol or propylene

glycol.

The equation that describes the effect of the dielectric

constant, ε , on the rate of hydrolysis is:

log k = log kε=∞ – KzAzB/ε

where K is a constant for a particular reaction at a given

temperature, zA and zB are the charge numbers of the two

interacting ions and kε=∞ is the rate constant in a theoretical

solvent of infinite dielectric constant.

• This equation predicts that a plot of log k against the

reciprocal of the dielectric constant of the solvent

should be linear with a gradient –KzAzB. The intercept

when 1/ε = 0 (i.e. when ε = ∞) is equal to the logarithm

of the rate constant, kε=∞ , in a theoretical solvent of

infinite dielectric constant

The variation of rate constant with reciprocal of dielectric constant for reaction between a , ions of opposite charge, b, ion and uncharged molecule and c, ions of similar charge.

• The gradient will be negative when the charges on the

drug ion and the interacting species are the same. This

means that if we replace the water with a solvent of lower

dielectric constant then we will achieve the desired effect of

reducing the reaction rate.

• The gradient will be positive if the drug ion and the

interacting ion are of opposite signs and therefore the

choice of a non-polar solvent will only result in an increase

of decomposition.

6. Light:

Photolabile drugs are usually stored in containers

which exclude ultraviolet light, since exposure to

light in this wavelength range is the most usual

cause of photodegradation . Amber glass is

particularly effective in this respect because it

excludes light of wavelength of less than about 470

nm. As an added precaution, it is always advisable to

store photolabile drugs in the dark.

7.Surfactants:

The presence of surfactants in micellar form

has a modifying effect on the rate of hydrolysis

of drugs. The magnitude of the effect depends

on the difference in the rate constant when the

drug is in aqueous solution and when it is

solubilised within the micelle, and also on the

extent of solubilisation.

4. Stability testing andprediction of shelf-life

It is clearly most important to be able to ensure

that a particular formulation when packaged in a

specific container will remain within its physical,

chemical, microbiological, therapeutic and

toxicological specifications on storage for a

specified time.

A recently agreed stability-testing requirement for a

Registration Application within the three areas of the

EC, Japan and the USA exemplifies the core stability

data package required for new drug substances and

associated drug products.

Current Stability Testing Concepts

Testing under temperature plus humidity

conditions instead of isothermal conditions

Both drug substance and drug product to be

subjected to similar test conditions

Zone concept for determination of world-wide storage conditions,

based on calculation of mean kinetic temperature

Long-term

Zone IZone II

Zone IIIZone IV

- 25°C/60%RH

- 30°C/35%RH- 30°C/65%RH

AcceleratedFor all zones - 40°C/75% RH

IntermediateZone I and II - 30°C/65%RH

Consideration for product-wise storage conditions

General ProductsStudy Storage Condition Minimum time

period at Submission

Long-term Testing 30 ± 2°C/65 ± 5% RH 12 months

AcceleratedTesting

40 ± 2°C/75 ± 5% RH 3 months

Where "significant change" occurs due to accelerated testing, onlylong-term testing should be conducted.

Study Storage condition Minimum time period covered by data at submission

Long term 30 ± 2°C/40 ± 5% RH

12 months

Accelerated 40 ± 2°C/not more than 25% RH

6 months

Liquid Products Packed in Semi-permeable containers

Drug Products Intended for Storage in a Refrigerator

Study Storage condition Minimum time period covered by data at submission

Long term 5°C ± 3°C 12 months

Accelerated If available, 25 ± 2°C/60± 5% RH, otherwise 30 ± 2°C/65 ± 5% RH

6 months

Drug Products Intended for Storage in a Freezer

Study Storage condition Minimum time period covered by data at submission

Long term

- 20°C ± 5°C 12 months