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Any method, that affords selective determination of the intact molecule in presence of its degradation products, is considered as stability indicating method.

1. The intact molecule is the active form of drug.2. Although degradations are usually inactive yet

sometimes they are harmful.3. Choice of suitable stability indicating method to

carry out stability study of certain drug.4. Registration of new pharmaceutical formulations

Most of official methods are not Most of official methods are not

stability indicating. In the case of the official stability indicating. In the case of the official

method the whole monograph must be applied, method the whole monograph must be applied,

including:including:

A.A. Tests for identityTests for identity

B.B. Limit tests for impurities and degradation productsLimit tests for impurities and degradation products

C.C. Last step is the analysisLast step is the analysis.

Despite the large number of possible reactions leading to drug decomposition, perhaps most of them can be classified as either:-

I) Hydrolysis of carboxylic acid derivative

II) Other kind of hydrolysis

Hydrolysis Of Carboxylic acid Derivatives

We are concerned in this part with compounds possessing acyl group

O II

R- C---

O II R---C---OR

Ester

O II R---C---NHR

Amide

O II R---C---SR

Thiol ester

O II R---C---Cl

Acid chloride

O O II II R---C---O—C---RAcid anhydride

O O II II R---C---NH---C---R Imides

R---CH-----C=O I I (CH2)n--O

Lactones

R---CH------C=O I I (CH2)n—NH

Lactams

O O II OH II

R--- C--/---X + H2O R---C---OH + HX H

Pharmaceutical Examples(I) Simple ester hydrolysis is exemplified by the

hydrolysis ofa) Procaine

CH2N O

O

CH2 CH2 N

Et

Et

OH

CH2N OH + HO

O

CH2 CH2 N

Et

Et

+H2OH

H2O

P-amino benzoic acid N-diethyl ethanolamine

b) Atropine

c) Aspirin hydrolyses to salicylic acid & acetic acid

NCH3OH + HO C

O

CH

C6H5

CH2OHNCH3 O C CH

C6H5

CH2OHO

O

COOH

C CH3

O

H OH

H OH

+H2OOH

COOH

OH C CH3

O

+H2O H2O

H2O

)2 (Hydrolysis of lactonese.g Pilocarpine

O

CH2C2H5N

NO

CH3

OH

C2H5

CH2

C

N

N

CH3

HOO

OH H

+H2O H2O

(3) amidea) Procainamide

CH2N NH CH2 CH2 N

OEt

Et

OH

+H2O

H

H2O

O

H 2 N O H

procamino benzoic acid

H2N - CH2 - CH2 - N

Et

EtN diethyl ethylendiamine

b) Nicotinamide

b) Chloramphenicol

N

C

O

NH2

OH HN

OH

O

Nicotinic acidpyridine 3-carboxylic acid

NH3

O2N CH

OH

CH NH

CH2OH

CH

Cl

Cl

C

OOH H

O2N CH

OH

CH NH2

CH2OHCH

Cl

Cl

C

O

OH

+H2OH2O

+H2OH2O

(4) Lactam ring

Penicillin contains both a Lactam ring & an amide function, although

both the Lactam & the amide can be hydrolyzed, it is observed that

the Lactam is more labile than the amide, apparently because of

strain induced by fusion of the four membered ring to 5- membered

ring.

Penicillin G

CH2 - C - NN

S

O

O H CH3

CH3COOH

benzyl penicillin

CH2 - C - NNH

SHO

CH3

CH3COOHHOOC

penicilloic acid deriviativeOH H

+H2O H2O

+H2O

5) Di-imide Barbituric acid derivative " phenobarbital"

NH2

H

C

O

C

O

C2H5

C6H5

OH

N

C=O

NC = O

NC2H5

C6H5

C

C

OH

O

65 3

24

1NH2

N- COOHH

C

O

C

O

C2H5

C6H5 H2O

1,2

H2O

1,6

II) Other kinds of hydrolysisa) Alkyl halides hydrolysis to the corresponding alcohol e.g. hydrolysis of chloramphenicol to the corresponding

dihydroxy derivative

O2N CH CHHN

CH2OH

C

HO

CH

OCL

CL

OH

H

H

OH

+2HCl

+2H2O HOHOO2N CH CHHN

CH2OH

C

HO

CH

OOH

OH

chloramphenicol

dihydroxy derivative

b) Hydrolysis as a reverse of condensation.Another kind of hydrolysis is the reverse of a

“condensation” reaction, in which an amine e.g., adds to the carbonyl (compounds with elimination of water).

c) Hydration (addition of water) rather than hydrolysis, but there is no fundamental difference.

HOOCC

CH

H

COOH

fumaric acid

HOC

CH

H

COOH

malic acid

H

COOH

CH2 – C – CH3 + H2O

NH

CH3 – C – CH3

O

HOhydrolysis

condensation

Primary amine

imine

+ NH3

Partial cleavage

+H2O

Stabilization of PharmaceuticalsWhen drug decomposition is the result of a

hydrolysis reaction, an obvious & effective means of stabilization is

1.To limit access of the drug to water. This is simply accomplished by using a solid

dosage form, as with aspirin, which is normally available in tablets, capsules & suppositories because of its instability in liquid formulation. Antibiotic is supplied in dry form, water is added before use.

2. Control of pH offers a mean for

So a compromise pH, at which the solubility and stability are both acceptable, must be selected.

3. Stability can always be increased by lowering

temperature. 4. Changes in solvent composition of formulation can

have significant effects. To decrease hydrolysis rate e.g. it may seem reasonable to replace water with an alcohol solvent; but one must be aware of possible complication (e.g. alcoholysis instead of hydrolysis as a decomposition reaction).

Controlling stability&

Affect solubility

OxidationOxidation– Oxidation reaction is a complementary one, its partner is reduction.– One can not happen without the other.– oxidation /reduction (redox) reactions involve the transfer of electrons.

(or one or more oxygen or hydrogen)

The oxidation state of carbon atom , & hence of a compound containing carbon atoms, is determined by the number of bonds from carbon to oxygen.

reduced form oxidized + ne

oxidationH

H - C - H

H

H

H - C - OH

H

HC = O

H

H-COH

O = C = O

O

X 1

reduction

2 3 4

Thus since oxygen is "added", we consider an aldehyde to be more oxidized than an alcohol, & So on.

Note that reduction can be thought of as the addition of hydrogen, accordingly we consider the addition of hydrogen to an olefin to be a reduction.

Another exampleoxidation of hydroquinone (1,4 dihydroxybenzene ) to p. benzoquinone.

In general oxidation is a loss of electrons & reduction is a gain of electrons.The rate of oxidation may be a function of the concentration of H + , i.e., of pH. Accordingly , many compounds are more resistant to oxidation at low pH values.

H2 + R.CH = CHR (red)

(ox)R CH2 CH2 -R

OHHO O O + 2H + + 2e

Oxidative Pathways ofPharmaceutical Interest

Autoxidation:-This term is used to refer to oxidations that takes place spontaneously under mild conditions.The majority of the reactions involved are free radical in type with organic peroxides often being intermediate or final products.Free radicals are atoms or molecules that have one or more unshared valence electrons ; e.g.

Free radicals are formed by: Thermal orPhotolytic hemolytic cleavage of covalent bond.Redox processes involving one electron transfer steps e.g.

Fe2+ + ROOH Fe3+ + RO + OH*

CH3 CH3 2CH3 *

The first 1st step is called initiation & takes a period of time called the induction period.

The length of the induction period depends on

The initiation step in a hydrocarbon reaction is written as follows.

The second step is propagation hydro peroxide formation occurs

The final step is terminationReactions occur that break the chainTermination may take place by coupling (two radicals combine to form a non radical),

the reaction

&the condition

R* + O2 ROO*ROO* + RH ROOH + R*

R H R* + *H

A few selected oxidativereactions of pharmaceutical interest

are illustrated here

Many drugs have been reported to be subjected to autoxidation including.

- apomorphine - Cyanocobalamin- ascorbic acid - epinephrine- chlorpromazine - ergometrine- phenothiazine derivative - heparin- hydrocortisone - kanamycin - morphine - neomycin- p. amino bezoic acid - penicillin - Phenylephrine - prednisolone- prednisone - procaine- riboflavin - streptomycin- sulfadiazine - terpenes- tetracyclines - thiamine - vitamins A, D & E - rancidity of oils & fats.

MorphineMorphine dimerizes when oxidized

OH

H3C-N

O OH

+ O2

morphine

O

H3C-N

O OH

+ HOO*

*

OH

H

H3C-N

O OH

morphine

H OH

O

H3C-N

O OH

* + H*

Then the morphine free radical couples with a morphine molecule ( at the free position ortho to phenolic oxygen) to give the dimer (bimorphine or

pseudomorphine) & H. Hydrogen peroxide is then produced. hydrogen peroxide can cause additional, oxidation to the N-oxide.

+H*

+ HOO*

H2O2

the dimer (bimorphine or pseudomorphine)

ChlorpromazineIt forms yellow green color initially

N

S

Cl

(CH2)3

CH3

CH3

N

Cp = Chlorpromazine

S

O

Cp Sulfoxide

S

O

O

Cp Sulfone

The marked effect of UV light on an oxidation reaction

5% solutions of 2 therapeutically useful phenothiazine salts (A) Chlorpromazine hydrochloride & (B) Prochlorperazine ethane

disulfonate were placed in Warburg respirometer to permit measurement of oxygen up take & were then exposed to a sun light. The solutions

became colored shortly after the light was turned on, & they continue to darken. As shown in the following figure.

Plot of oxygen – uptake data for chlorpromazine hydrochloride (A) & Prochlorperazine ethane disulfonate (B) illustrate induction period, the

linearity of uptake with time, the extreme light dependence of the oxidative degradation.

Induction period

Inhibition of Oxidation1.Protection from light:

The choice of suitable container is very important .

Note also that if a product is light sensitive, it is important that this fact should be stated on

the label.To exclude light, four main techniques are

availablea)Wrap – around label

b)Container coatings (some may incorporate UV absorbing material).

c)Various cartooning procedure.d)The use of so –called light resistant

containers.

2) Exclusion of oxygen:

Removing O2 from a formulation is an obvious

way to prevent oxidation.

This can be done in several ways:

a)Oxygen may be expelled from aqueous preparations by boiling the water beforehand, although some oxygen will redissolve during

the cooling process.

b)A better way is bubbling nitrogen through the

solvent to flush the O2 out of solution.

c)Another way is to flush the head space of the

container with N2 just prior sealing or capping.

3) addition of antioxidants:antioxidants are materials that act by being more readily

oxidized than the agents they are to protect.• Thus in a closed system (e.g. an ampoule) they may consume essentially all the oxygen present & there by

protect the drug.• Open system may require higher antioxidant concentration than closed system. antioxidants may also function by being

inhibitors of free radicals, i.e., they provide an H. to break the chain reaction.

• To enhance the effectiveness of the antioxidant approach it is sometimes useful to use more than one antioxidant.

• It has been found that a combination of two antioxidants along with a chelating agent to complex metals (thus

inhibiting their catalytic effect of the antioxidant).• Emulsions may need two antioxidant systems water soluble

one & an oil soluble one.• The concentration used generally vary between 0.01% 1%,

• 0.1% is a good place to start when formulating.

Chelating agents are:- citric acid

- phenylalanine- EDTA

- sorbitol - tartaric acid

Oil soluble antioxidants are: - Butylate dihydroxyanisole (BHA)

- ascorbyl palmitate- hydroquinone

- lecithin, propyl gallateWater soluble antioxidants are:

- Na bisulphite- cysteine hydrochloride

- thioglycolic acid- ascorbic acid

- Na sulphite

4- Control of pH:

- The principle of lowering pH was discussed

before.

- For purpose of pharmaceutical products, the

pH range of 3-4 is generally found most useful.

- This technique is not for use for drugs that

precipitate or decompose at lower pH values.

1) Functional group approach:The intact molecule

of the drug

via certain

functional group

Which does not persist

in any of the degradation

products

e.g. hydroxamic acid Method for the determination of penicillins Via -Lactam ring

Is

determined

e.g. hydroxamic acid Method for the determination of penicillins Via -Lactam ring

CH3

CH3N

ON

S

COOH

C26

R

O H

H2N - OH

reddish violet Complex of ferric

hydroxamate1 : 1

CH3

CH3

N

S

COOH

N

N - HOH

O

R

HO

H

Fe+++

2) Combination of

Separation

and

measurement

Operations

Separation may beCarried out either by

Solvent extraction

or

Chromatographic methods

A) Solvent extraction:Liquid – Liquid extraction It is the transfer of a solute From

another to

partition coefficient K = where Cu

Cl

Cu = concentration of the solute in the upper Phase

&

Cl = " " " " " " lower “

one liquid phase

depends on

the intact molecule from

Its degradation Products

Separability factor = whereKi

Kd

The feasibility of resolving

K's are the partition coefficient of the two substances.

If ~ 1 → no separation

much deviated from unity → good separation This can be accomplished by suitable choice

of solvent, pH and ionic strength

I) Choice of Solvents:I) Choice of Solvents:Sometimes the analyst may select both solvents but often the

sample is in aqueous solution. The difference in partition coefficient of both intact &

degradation products

a)The chemical nature of the 2nd solvent.b)The difference in the chemical structure of the intact molecule

& its degradation product.-The choice of the suitable solvent is very important for the

success of separation process.

Ki

Kd

will be markedly

influenced

by

II)II) Control of Ionic strength Control of Ionic strength::

Is made very high ↑the salt concentration of an aqueous solution

the solubility of non - electrolyte → will be decreased

- This reduction of solubility by an increase in ionic strength is " the Salting out effect"

- The ions of salt will reduce the availability of water molecules which act as a solvent for non electrolyte by tying up much water

as a hydration shell around the ions.- The salt also helps to break the emulsions that may be formed

when shaking the two phases together.- Salting out effect alters the apparent partition coefficient of many

substances.

III) Control of pH: III) Control of pH:

acids or

bases

Most of the drugs encountered by the

pharmaceutical analyst

are weak

The solubility of these substances depends upon their ionic formsThe ionic species are usually soluble in polar (specially aqueous)

solvents

If we have a mixture of A base

& an acid

In acid mediumcomplete extraction of the acid could be achieved while the base

will be retained as its salt in aqueous layer.

They are readily separated by solvent extraction

e.g.e.g. OH H

- COOH + - NH2

O

-C

H

N

amide linkage

H+

- COOH + - NH3+

OH-

- COO- + - NH2

Acetylsalicylic acid + antihistaminic in aqueous medium

H+

)HCl(

ether

acetylsalicylic acid in etherAntihistaminic as hydrochloride

salt in aqueous layer

Suppose, it is desired to retain weak acid, with pKa = 5 in aqueous layer while extracting a 2nd

basic substance.

[A-] pH = PKa + log ________

[HA]

at pH = 5 → [A-] = [HA]at pH = 6 → about 10% remains as [HA]

at pH = 7 → “ 1% remains as [HA]at pH = 8 → “0.1% remains as [HA]

So the pH should be at least 3 units more basic than pKa to ensure complete conversion to the anion

form.

b) Chromatographic Separation :a large number of stability indicating methods entail several

chromatographic techniques.

Zr4+ + eriochrome cyanine R colored product

Mineralization of F- bleaching of the color ZrF62-

- TLC (densitometry) .

- Column,

- GLC

- HPLC etc …

F C (CH2)3 -N

OH

Cl

O

Haloperidal safenare

A

F-

at 525nm

If it is possible to determine the degradation in presence of intact.

III- Determination via degration by quantitative conversion of intact molecule of drug into its

degradates e.g. :

penicillin

chlorazepate dipotassium .

In this case we determine the degradation products before and after degradation.

By simple subtraction, the concentration of the intact drug can be obtained e.g.

D (10) + Intact (90)degradation

10 (D) + 90 (D) = 100

I = 100 - 10 = 90

D1 (10) + D2 (10) + I (90)degradation

D1 (10) + D2 (10) + D1 (90) + D2 (90)

I = 100 (D1) - 10 (D1) = 90 via (D1) or I = 100 (D2) - 10 (D2) = 90 via (D2)

On the condition that 1) Method should be suitable for the determination of degradation

without any contribution from the intact or D2

2) Degradation is not affected during the process of degradation.3) degradation should be quantitative.

100

100

Cl

N - C

C = N

O

CH - COK - KOH

O

Clorazepate dipotassium

2N HClAt room

Cl

N - C

C = N

O

CH2

extracted beforehydrolysis CH2Cl2

N - desmethyl diazepam

with (benzene :

9 : 1

Temp½hr

6N HCl at 100=Co

for 1hr

NH2

C = OCl

* 2 amino - 5-chloro

benzophenone

- H2N – CH2 - COOH

Glycine in aqueous layer ninhydin + pyridine

bluish violet measured at 560nm

* Extracted from neutralized hydrolyzate with diethylether the extract is evaporated & the residue is dissolved in CHCl3 & its absorbance is measured at either 240nm or 380nm

Stability Indicating Method For The Determination of Chlorazepate dipotassium.

Via its degrading products

240nm

280nmA

IV- Determination of the intact molecule of the drug in presence of its degradation products, without

separation, by the use of suitable selective method :

a) Organic polarography :

It offers the desired specificity if the difference in E ½

and

the intact molecule

Its degradation productsis big enough to prevent overlapping. (0.4 )

Intact + Deginactive active (via degradation)active inactive (no problem)active active (the difference in E ½ should be at least 0.4 v )

current

Applied voltage polarogram.

b) Spectroscopic method :1) IR:

It has very few quantitative application in stability evaluation because of its limited sensitivity .

2) NMR:

specificity along with

simplicity

but it lacks as wellSensitivity

&precision

It offers

It depends on suitable choice of one resonance band of the intact molecule, which is free from any overlap by the others, specially the resonance bands of the degradation products.

NH2

O

ClNO2

COOH

CH2

NH2

+

Deg2

NO2

N

N

O

Cl

H

N-C-CH2-NH2

O2

ClNO2

N O

NO2

N O

NH2

Cl

H

Deg 1

Pathway of Degradation of Clonazepam

H OH

H2O

1H NMR Spectrum of 3-amino-4-(2-chlorophenyl)-6-nitrocarbostyril in DMSO.

1H NMR Spectrum of 10 mg Clonazepam and 10 mg Maleic acid in DMSO.

1H NMR Spectrum of 2-amino-2-chloro-5-nitrobenzophenone in DMSO.

It is the difference in absorbance

between

two equimolar solutions of the analyte in two different chemical forms

Whichexhibit twodifferentspectralcharacteristics

The criteria for a successful application of A method to the assay of substance in the presence of other absorbing degradation products are .

a) Reproducible changes of the spectrum of the analyte by addition of one or more reagents.

b) The absorbance of degradation products is not affected by reagent .

Difference Spectrophotometry ( A)

3) Spectrophotometric methods:

Direct spectrophotometry is unsuitable method for stability indication because of lack of selectivity.

- The difference absorption spectrum is a plot of the difference in absorbance

between two equimolar solutions of the analyte against .

differenceabsorbance

+

-

0

- The simplest technique for producing spectral changes of the analyte is the adjustment of pH by

addition :

ofacidbasebuffer

- The suitable analyte, producing spectral changes by changing the pH, are those containing ionizable

functional groups e.g.- phenols- amines

- aromatic carboxylic acids etc…

Difference absorption spectrum of B relative to

solution A

Phenylephrine hydrochloride in

(A) 0.1N NaOH

(B) 0.1N HCl

Phenolic group in alkaline medium generates an additional n (non bonding) electron that interacts with electrons in

the ring to produce a) Bathochromic shift of the max from 271 nm to 291

nm (in acid med).b) Hyperchromic effect .

The difference absorption spectrum generated automatically using double beam recording

spectrophotometer with the solution at pH 13 in a sample cell the solution at pH1 in a reference (blank) cell .

A = A alk - Aacid

A = a b c . A = (Aalk + Ax) – (Aacid + Ax) = Aalk – A acid

at 257 &278 nm

both solutions have identical absorbance & consequently exhibit zero difference absorbance it is

called isosbestic or isoabsorptive points.

OH

CH

OH

CH2 NH CH3

HClPhenyl epherine HCl

A substance whose spectrum

is unaffected bythe change in pH

may be determined byadding of suitable

reagent

Which induces

reproducible spectral changes

A = (Ai alk + Ad) – (Ai acid + Ad) = (Ai alk – Ai acid )

The A of unreacted substance = Zero The A of degradation product = Zero

Chlorpromazine in colored syrup formulations & in the

presence of its sulfoxide (decomposition product)

Peracetic

acid

Sulfoxide + sulfoxide originally presents

A at 343 nm concentration of intact chlorpromazine

Can be determinedBy adding of

In this case, one can not apply method III (degradation product) – because of the complete overlapping of the two spectra of the intact

& degradation products accordingly selective determination of degradation in the presence of intact molecule is impossible .

The selectivity of A method

The use of 0.1 N NaOH Are usually convenient

0.1 N HCl

It is possible to apply the A method :

Without altering the absorbance of degradation

products

Induce spectral changes of analyte

If the pK a values of

the analyte&

Degradation products

differ by more than 4

depends onthe correct choice of pH values which

The selection of 2 buffers

one at a pH, 2 units greater than

& the other at a pH2 unit

less than

The pKa of the analyte

ensures that the analyte is 99% in

& that the A is almost maximal

ionized or

molecularstate

Selective assay of

Chlordiazepoxide (pKa = 4.9)&

Demoxpam (pKa = 10.5) major hydrolysis product

At pH A at3 & 8 269nm

8 & 13 263 nm

In the presence of 2 amino 5 chlorobenzophenone (pKa = 1) (a minor hydrolysis product).

NH2

O

ClNO2

COOH

CH2

NH2

+

Deg2

NO2

N

N

O

Cl

H

N-C-CH2-NH2

O2

ClNO2

N O

NO2

N O

NH2

Cl

H

Deg 1

Pathway of Degradation of Clonzepam

Absorption spectra of 20 g. ml-1 Clonazepam, 10g. ml-1 Deg1 and 10g. ml-1 Deg2 in ethyl alcohol.

First derivative absorption spectra of 20 g.ml-1 Clonazepam, 10g. ml-1 Deg1, and 10g. ml-1 Deg2

in ethyl alcohol.

Difference absorption spectrum of 10 g.ml-1 Deg2 against Clonazepam as a blank.

Difference absorption spectra of (0.5-4 mg) Clonazepam after hydrolysis against the same

amount of Clonazepam without hydrolysis.

Derivative Spectrophotometry

It involves the conversion of normal absorption - 1st , 2nd , 3rd or 4th derivative

First Derivative spectrum (D1) is a plot of rate of change of absorbance with ,

The slope of the FundamentalSpectrum ,dA

d

A A2 – A1 = = slope 2 - 1

or dA 2 - 1

At d 2

Against

Against

Against

Against

2) Maximum positive slope in D0 which corresponds to a maximum in D1

4) Maximum negative slope in D0 which corresponds to a minimum in D1

3) max In D0 = cross-over point in D1 = minimum in D2

d2A d2

dA

d

*

*

*

*

***

At the maximum peak A = Zero A2 = A1

dA A2 – A1 ______ = _____________

d 2 - 1 dA

= _______

d

Slope =opposite

tan = ____________

adjacent

- At ascending part of the curve the slope has positive value were A2 > A1 (according to the direction of scanning).

- At descending part of the curve the slope negative value were A1 > A2 .

11stst derivative spectrum of an absorption band derivative spectrum of an absorption band

22ndnd derivative spectrum is characterised derivative spectrum is characterised

Is characterized by

maximum Cross over point at max of the absorption band

& a minimum

by

Two satellite maxima & An inverted band of which the minimum corresponds to

max and D0

2nd derivative (D2) spectrum is a plot of d2A

_________ V.S d2

- The max at 3 in D0 a minimum in D2 spectrum

At 3 = max

- odd numberderivatives = zero

D1, D3, etc ….

- even number derivatives = maxima

D2, D4, etc ….

This is because the derivative amplitude (D) i.e. the distance from a maximum to a

minimum, is inversely proportional to the fundamental spectral band widths (w), raised

to the power (n) of the Derivative order,

These spectral transformations confer two principal advantage

•Firstly, an even order spectrum is of narrower spectral band than its

fundamental spectrum. it shows better resolution of

overlapping bands.

Secondly, 2nd derivativespectrophotometry discriminatesin favour of substances of narrow

band-width against broad bandwidth.

A

d2A

d2

D0

D2

- For this reason the order of derivative spectra can increase

the detection &

Sensitivity

D ( 1w )

n

of minor spectral features.

Provided that Beer’s law is obeyed by the fundamental spectrum .

1) P.h., Peak height:By measuring the distance in

(cm) between a maximum or minimum & the base

line on the graph . 2) P.p, Peak Peak:

The distance in (cm) between maximum, and adjoining

minimum .3) Zero crossing :

The distance in cm between the maximum & zero

crossing of the derivative of the interfering band.

In quantitative Analysis:If Beer’s law is obeyed the following equation can be obtained

- All amplitude in the derivative spectrum are proportional to the concentration of the analyte

Direction of

Scanning

D2D2

Determination of carbimazole and methimazole by 1st & 3rd derivative spectrophotometry.

NCH3N

S

C2H5O

O

carbimazole

NCH3NH

Smethimazole

Stability Indicating method for the determination of carbamazepine in the presence of iminodibenzyl .

Carbamazepine (_______) (Tegretol).

N

CONH2

-antidepressant-Elevate mood

NH

Iminodibenzyl (-----)

Stability Indicating method for the determination of imipramine hydrochloride (Tofranil) in the presence of

iminodibenzyl .

CH - CN2 - CH2 - N . HCl

H

CH3

Derivative ratioDerivative ratioAnother method for resolving binary mixtures without

previous separation is the derivative ratio spectrophotometry, which was developed by Salinas et al

(Talanta, 37 (3) 1990, 347, 351).It is based on the use of the 1st derivative of the ratio

spectra. The absorption spectrum of the mixture is obtained & divided (amplitudes at each W.L) by the absorption

spectrum of a standard solution of one of the component. In this case the 1st derivative of ratio of one component & its

standard will be zero, Accordingly its interfering effect will be cancelled . After obtaining the 1st derivative ratio of the two component (one of them is cancelled). The concentration of

the other component is determined form a calibration graph. This method permits the use of the W.L. of greatest

sensitivity as the signal of measurement, either a maximum or a minimum. This method was then extended for the

determination of ternary mixtures .

Ratio = constant for

each wavelength.

Absorption spectra of disopyramide phosphate and its degradation product. Concentration of each is 75 g/ml.

Ratio spectra of disopyramide phosphate )12.5 – 87.5 g/ml (in methanol. Divisor

is 87.5 g/ml degradation product

First derivative of the ratio spectra of disopyramide phosphate (12.5 – 87.5 g/ml) in methanol. Divisor

is 87.5 g/ml degradation product

Theory of ratio subtraction method:The method depends on that, if you have a mixture of two drugs (X) and (Y)

with overlapping spectra and the spectrum of (Y) is extended than (X), the determination of (X) can be done by dividing the spectrum of the mixture by

a certain concentration of (Y) as a devisor (Y’). The division will give a new curve that represents

we subtract this constant, then multiply the new curve obtained after subtraction by (Y’) (the devisor), therefore we can obtain the original curve

of (X). This can be summarized as follows:

The constant can be determined directly from the curve

by the straight line which is parallel to the wavelength axis in the region where (Y) is extended.

X_______ + constant t

Y’

X + Y X Y X_________ = _______ + ______ = ______ + constant t

Y’ Y’ Y’ Y’

X _________ + constant t

Y’

X X _____ + constant t – constant t = _______

Y’ Y’

X_____ x Y’ = X

Y’

Absorption spectra of vincamine 20 g ml-1 (_____) degradation product 20 g ml-1 (------) and degradation

product 16g ml-1 (devisor) (__ _ __ _) using 0.1N hydrochloric acid as a solvent.

Division spectra of laboratory prepared mixtures of vincamine (X) and its degradation product (Y) using 16g

ml-1 of degradation product (Y’) as a divisor and 0.1N HCl as a solvent.

Division spectra of laboratory prepared mixtures of vincamine (X) and its degradation product (Y) using

16g ml-1 of degradation product (Y’) as a divisor and 0.1N HCl as a solvent after subtraction of the constant.

The obtained absorption spectra of vincamine in

lab.mixtures 8-32 g. ml-1.

The original absorption spectra of vincamine in cal.

curve from 8-40 g. ml-1.