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بسم الله الرحمن الرحيم. قَالُوا سُبْحَانَكَ لا عِلْمَ لَنَا إلاَّ مَا عَلَّمْتَنَا إنَّكَ أَنتَ العَلِيمُ الحَكِيمُ. Stability Indicating Methods. Stability Indicating Methods. - PowerPoint PPT Presentation
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اَن�َك� ْب�َح� ُس� اُل�وا اَن�َك� َق� ْب�َح� ُس� اُل�وا َق� � إَّال ُل�َن�ا ْل�َم� ِع� � َّال إَّال ُل�َن�ا ْل�َم� ِع� َّالإَن�َك� َت�َن�ا ِع�ْل�ْم� ا إَن�َك� َم� َت�َن�ا ِع�ْل�ْم� ا َم�
اُلَع�ْل�يَم� اُلَع�ْل�يَم� َأ�َنَت� َأ�َنَت�ِك�يَم� ِك�يَم�اُلَح� اُلَح�
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.