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Complexation & Protein Binding
Learning topics
• Complexation mechanisms• Methods for determination of stoichimiometric ratio and
stability constant • Protein binding• Analysis of protein binding• Scatchard method
ComplexationDefinition: usually non-covalent interactions between two or
more compounds that are capable of independent existence.
Mechanisms: classic donor-acceptor (Lewis acid-base), but any intermolecular forces may be considered:
van der Waals< induced dipolar <dipolar<hydrogen binding
Physico-chemical properties that may be altered:• Chemical (pH), instrumental (UV/IR spectra, conductance) • Formulation (solubility, partitioning, stability, drug delivery by
ion-exchange resins e.g. Delsym, Tussionex, Kayexalate, Renagel, Cholestyramine)
• Pharmacokinetic – permeability (bioavailability), protein binding (distribution)
Cation exchanger resins in drug delivery
Types of complex formation
• Metal ion complexes (metal ion coordination with ligands): monodendate, multidendate (chelate)
• Organic complexes (due to electrostatic, charge transfer, hydrogen binding, van der Waals forces or other hydrophobic interaction e.g. п- п stacking): molecular complexes, donor-acceptor complexes, charge-transfer complexes
• Inclusion / Occlusion complexes (a certain type of molecular complexes with unique properties)
Metal ion Complexes
Monodentate (Inorganic): • Inner-sphere complex: [Cu(NH3)4]2+ .2Cl- , Cu2+ (metal ion,
unfilled orbitals), 4 (coordination number), Cl- (counter ion), NH3 (ligand, electron pairs), orbital configuration dsp2 (square planar)
• Outer-sphere complex: [Fe(CN)6]3-, orbital configuration d2sp3 (octahedral)
• Biologic Examples: iron complexes (hemoglobin, cytochrome c, myoglobin), platinum complexes (cisplatin, carboplatin), copper complexes (hemocyanin, superoxide dismutase, cytochrome oxidase), cobalt (vitamin B12), zinc complexes (insulin, carbonic anhydrase, carboxypeptidase)
Bidendate: EDA in aminophiline, carboxylate
Multidentate (Chelate): EDTA (hexadentate)
Outer sphere vs. inner sphere
Organic complexes
• Charge transfer (donor-acceptor) complexes (complete or partial charge transfer, develop new band in spectra e.g. trinitrobenzene and benzene п bonding)
• Molecular complex (No charge transfer, contribution of weak van der Waals, dipole-dipole, hydrogen binding)
Inclusion/Occlusion Complexes
the molecular size matters1. Channel lattice type: starch for iodine2. Layer type: Bentonite, graphite3. Clathrate (cage-like lattice): hydroquinone cage for
warfarin4. Molecular sieves: zeolites (aluminosilicate) for
organic solvent, H2O
Mono and Macromolecular Inclusion/Occlusion
ComplexesCyclodextrins: cyclic oligosaccarides 6, 7 or 8
glucopyranose (α, β, γ) naturally by Bacillus macerans amylase from starch sized from 5-8A; applied for:
drug solubilization, flavor stabilization, long-acting deodorants, changing from liquid to solid state, prevention of incompatibilities, improvement of organoleptic properties, enantiomer separation, and bioavailability enhancement.
QuestionDue to low aqueous solubility and nephrotoxicity of β-CD, what kinds of synthetic CDs were developed ?
Complexation parametersstability constant,
stoichimiometric ratio
CD
CDK
CDCnD
nn
a
n
Analysis of stability constantsphase-solubility profile
• By Higuchi and Connors:
)1
.(][
))(
(
)(
)(
)(
0
000
000
0
000
0
0
0
n
a
n
a
na
nn
a
n
n
n
nn
a
n
SK
SnKCSS
n
SSCS
n
SS
K
CD
CDK
n
SSCCDCC
n
SSCD
SD
CDnSS
CD
CDK
CDCnD
Example
• Please calculate intrinsic solubility of PABA and stability constant of the β-CD complexes.
• Assumption 1:1 complex
Concentrations (mM)
β-CD PABA0.50 1.741.00 2.231.50 2.722.00 3.212.50 3.703.00 4.194.00 5.175.00 6.16
13
0
0
CD-
0
000
30.41)98.01(*10*25.1
98.0
)1(
25.1
25.198.0
C vs.
)1
.(][
MslopeS
slopeK
mMS
xy
S
SK
SKCSS
a
PABA
a
a
6.195)1)106.4105.5(101(104.6
106.4105.5
eine]PABA][Caff[
caffeine]-PABA[
1105.5107.3
101102.8
complexin PABA
complexin caffeinen
)105.5 ,10(2.8 C
)105.5 ,10(1 B
)104.6 (0,A
2-2-2-2-
2-2-
2-2-
2-2-
2-2-
2-2-
-2
K
Analysis of complex formation Solubility
Excess PABA
Dissolved PABA
Caffeine
Method of continuous variation
and Job’s method
• A- maximum deviation from linear changes of an additive property (e.g. dielectric constant) at a specific mole ratio (n)
• B- plot of absorbance difference (as the additive property) against mole fraction of ligand is used to calculate n• Job’s method: if the magnitude of absorbance is proportional only to the concentration of the complex, then:
nlog[A]log[M]logK]log[MA
[M][A]
][MAK
MAnAM
n
nn
n
Analysis of stability constantsSpectroscopy
Due to charge transfer (e.g. additional UV absorption for iodine in electron donor solvents)
0
0
0
00
Donor0
Acceptor0
00
D
1.
ε.K
1
ε
1
A
A
D1
DAεA
complex), ofty absorptiviε(molar
.[complex]A
),(CD
),(CA
[complex])1(A
[complex]K
:Equation HildebrandBenesi
K
K
D
Analysis of stability constantspotentiometric titration
][logK][][1.5n if
] [logK][][0.5n if
)1 ( ][2log][
1][][MA 1n f
]][[
][
][][][
][2][
ion] metal[
ligand] bound[
][
][
2H)COOCHCu(NHCOOCH2NHCu
22
1
22
22
21
2
2
222
1
222232 _
ApMAMA
ApMMA
nifApA
Mi
AM
MAKK
MAMAM
MAMAn
AMA
MAKMAAMA
AM
MAKMAAM
-
ExamplePlease determine average number of ligand (n) and stability exponent if addition of 0.001 mole Cu2+ to 0.03 M glycine HCl (pKa=9.7) in 100mL sample causes shifts of 1.7 mL (pH=4) and 5 mL (pH=8) as titrated with 0.3 M NaOH.
82.1030.752.3log
30.7)100
1000*10*7.1*3.003.0log(47.9p[A]
52.3)100
1000*10*5*3.003.0log(87.9p[A]
NaOH])[]log([]HAlog[][
][
]][OH[
1.5)nat p[A]logK 0.5,nat p[A](logK
4)(pH 5.0001.0
3.0*10*7.1n
8)(pH 5.1001.0
3.0*10*5n
ion metal of moles
releasedproton of oles
ion] metal[
ligand] bound[
3
3
3
2121
3
3
initialHApHpKapHpKaAp
HA
AKa
KK
mn
Analysis of stability constants Distribution (Question)
• Question : I2+I-→I3-
H20 replaced by 0.125 M KI → Co (I2) = 0.1890, Cw (I2) = 0.0283.
K (stability constant) ?Kiodine (o/w) = 625
1972MK
0.096M0.0280.125free w,[KI]
0.028M0.00030.0283complexed w,]2[I
M4103.034625
0.189free w,]2[I
free[KI]free]2[I
[complex]K
I2
I2
H2O
CS2
Protein BindingDrugs almost bind to albumin (acidic) and α1-acid glycoprotein (basic)
To act as drug reservoir and sustain disposition / therapeutic or toxic effects
affinity) site binding(K, ].[DK1
].[DK.νr
sites) binding ofnumber ,(ν ].[DK1
].[DKν.
][P
[PD]r
occupancy) fractional (r, ].[DK1
].[DK
][P
[PD]r
:absorption isothermalLangmuir
].[D].K[P]).[DK[PD].(1
[PD])]].([P.[DK[PD]
[PD]][P][P
]].[D[P
[PD]K
PDDP
fi
fii
fa
fa
t
fa
fa
t
fatfa
tfa
tf
ffa
Protein bindingAffinity and capacity
Langmuir isothermPB capacity & affinity
Double reciprocal vs. Scatchard
].[Pν.K][DK][D
][D
:ionconcentrat and natureprotein unkown if
r.Kν.K][D
r : Scatchard
].[Dν.K].[Dr.Kr
ν
1
][D
1.
ν.K
1
r
1 reciprocal Double
].[DK1
].[DKν.r
tabaf
b
aaf
fafa
fa
fa
fa
Protein bindingcalculations
Fraction of drug bound, β
Example
• Please calculate affinity and capacity indices of protein binding of a new anti-HIV drug
0.00 0.50 1.00 1.50 2.00 2.500
0.05
0.1
0.15
f(x) = − 0.060022266 x + 0.159872829R² = 0.996707969872418
r
r /
[Df]
(L
/nm
ole
)
65.20.06
0.159
06.0K
r.Kν.K][D
r
a
aaf
Protein BindingMethod of Analysis
• Dialysis (cut-off 12-14 KDa): Equilibrium vs. Dynamic mode• Ultrafiltration (cut-off 10-30 KDa)• Ultracentrifugation (20000-50000 rpm, Ficoll or cesium
chloride gradient may be required)• Gel filtration chromatography (Sephadex columns)• Capillary Electrophoresis• Spectroscopy methods (qualitative)
Equilibrium Dialysis
Dynamic Dialysis
I
II
t
t
t
t
S
S100100%PB
).dt,100
%PB-k.(1
][D
]d[D:II
k.dt,][D
]d[D :I
constantty permeabiliapparent k
Homework
• Martin’s Physical Pharmacy:• Problem 11.7• Problem 11.13• Problem 11.15
Chemical kinetics and drug stability
Ali Tamaddon, Ph.DDepartment of Pharmaceutics
Shiraz Faculty of Pharmacy
Topics
• Reaction rate and order• Zero, first and second order reactions• Half-life and shelf-life calculations• Complex reactions (reversible, parallel, series) • Michaelis-Menten nonlinear kinetic• Parameters influencing reaction rate (temp, pH, ionic
strength, solvent, dielectric constant)• Collision theory and Arrhenius equation• Stability testing methods (shelf-life estimation)
Kineticimplications in pharmaceutics
• Degradation reactions• Storage conditions, shelf-life, expiration date• Stability testing• Pharmacokinetics• …
Reaction rate
• k: reaction rate constant in elementary reactions
ba BAkrate
dt
Bd
bdt
Ad
arate
productbBaA
][][
][.
1][.
1
Reaction order• Reaction order: zero, first, second, etc.• depending on molecularity (number of molecules, ions,
atoms reacting simultaneously) in elementary reactions,
• but not in complex reactions
Zero order reactions
0
00.90
0
01/20
00
00
00
0
0
0
1.0 then t9.0C
5.0 then t5.0C
.
)0.(
.
sec)./(
k
CCif
k
CCif
t
x
t
CCk
tkCC
tkCC
dtkdC
lmolesk
kdt
dCrate
First order reaction
110.90
111/20
0
10
303.2
.
0
.0
10
10
1
11
1
105.09.0ln then t9.0C
693.05.0ln then t5.0C
)(log
302.2log
302.2303.2
.loglog
10.
.
.lnln
)0.(lnln
.
)(sec
.
1
1
kkCif
kkCif
xa
a
tC
C
tk
tkCC
CC
eCC
tkCC
tkCC
dtkC
dC
k
Ckdt
dCrate
tk
tk
First order reaction
Second order reaction1st type
200.90
201/20
20
20
22
112
22
.9
1 then t9.0C
.
1 then t5.0C
.11
)0.()1
(1
.
)sec..(
.
kCCif
kCCif
tkCC
tkCC
dtkC
dC
molelk
Ckdt
dCrate
Example• Please determine rate-constant
and half-life of a second-order reaction if the initial concentration of 0.05 M changed by 0.0088 M over 35 min.
min9.163.
1t
min./122.0
35).0088.005.0(*05.0
0088.0
)..(
.).(
,
..
.11
201/2
2
2
2
00
20
0
20
kC
molelk
txaa
xk
tkxaa
x
xCCCa
tkCC
CC
tkCC
Second order reaction2nd type
tkABA
B
A
B
tkABBA
AB
tkBA
AB
AB
tkxB
B
xA
A
AB
dtkxBxA
dx
xBxAkdt
dx
CCkrate BA
.).][]([][
][ln
][
][ln
.).][]([]][[
]][[ln
.]][[
]][[ln.
)][]([
1
)0.()]][
][ln()
][
][.[ln(
)][]([
1
.)]).([]([
)]).([].([
..
2000
0
2000
0
20
0
00
20
0
0
0
00
200
002
2
Pseudo-order reactionsApparent rate constants
• Pseudo-zero order (example):• Aspirin hydrolysis in aqueous suspension (6.5%) follows
pseudo zero-order reaction at pH=6 and room temperature. What is the shelf-life? (Saspirin=0.33%), K1=4.5*10-6 sec-1
• Pseudo-first order:
daysk
C
lgCkkrate
Ckdt
dCrate
hydrolysisAA
app
solid
51.0
t
sec./105.1.
.
0
00.9
510
1
solution saturated
].[
then[A][B]
]][[
1
2
AKrate
if
BAkrate
productBA
app
Determination of reaction order
• Substitution method• Graphical method ((a-x) vs. t, ln(a-x) vs. t, 1/(a-x) vs. t) • Half-life method:
1)log(
loglog
loglog)1(log.
1
1
2
)2(2/1)1(2/1
2/1
)1(2/1
a
a
ttn
kantak
tn
Complex ReactionsReversible
• Example: epimerization of tetracycline, 32% iso-7-chlorotetracycline + 68% iso-7-chloro-4-epi-tetracycline in equilibrium, please calculate kf and kr if slope = 0.01 min-1
tkk
AA
AA
A
AA
A
B
kr
kfK
BkrAkfdt
Ad
BA
rf
eq
eq
eq
eq
eq
eq
303.2
)(log
].[].[][
0
0
1
1
min016.0
min007.0
01.0303.2
1.2
01.0303.2
1.232
68
kf
kr
krkr
krkfkr
kfK
Complex reactionparallel vs. series
• Parallel (e.g. general acid-base catalyzed reactions)
• Series or consecutive (a complex calculations!)
)1.(.
]).[(].[].[][
0
21
2121
2
1
ktii eA
K
kB
kkK
AkkAkAkdt
Ad
BA
BA
).()(
11.(
).()(
.
21
21
1
1221
02
12
101
0
21
tktk
tktk
tk
ekekkk
AB
eekk
kAB
eAA
BBA
Complex reactionMichaelis-Menten (Enzyme
catalyzed)
][
]][[][
][
])[]]([[
][][]])[[]([
][][][
]])[[]([][
1
32
321
32
1
31
2
Sk
SEES
ES
ESES
k
kkK
ESkESkSESEk
ESkESkdt
ESd
SESEkdt
ESd
PSESE
M
T
TM
T
T
kk
k
][
1.
1
V
1
:Burk-Lineweaver
)][
][.(
][
]][[.].[
max
33
SV
K
V
SK
SVV
Sk
SEkESkV
m
M
m
M
M
T
Line-weaver Burk equation
V (μg/l.min) 0.0350 0.0415 0.0450 0.0490 0.0505
C (M) 0.0025 0.0050 0.0100 0.0167 0.0333
Question: Please calculate KM
KineticFactors: A-Temperature
• Based on collision theory:
)11
.(2.303R
E-log
2.303RT
E-logAlogk
Equation) (.
.
..
on)Distributi (.
..
12
a
1
2
a
TTk
k
ArrheniuseAk
ePZk
NePZrate
BoltzmaneNN
NZPrate
RT
Ea
RT
Ea
tRT
Ea
RT
Ea
ti
i
Example
• Please determine the activation energy for the below data:
0.002 0.00220.00240.00260.0028 0.0030.544
0.545
0.546
0.547
0.548
0.549
0.55
0.551
0.552
f(x) = − 8.95655653247871 x + 0.571367059892067
1/T
Logk+
2
Absolute T, K k, year-1
450 0.0356
400 0.0354
361 0.0352
Accelerated Stability Testing
Arrhenius method• Shelf-life calculation based on logK in the elevated
temperature and extrapolation to normal condition
• Example: Please calculate shelf-life for the last slide corresponding example
Log (k+2)=0.541, then k=0.035t0.9=0.105/k=0.105/0.035=3 years
Accelerated Stability Testing
Amirjahed method
Q10 and shelf-life estimation
• Example: Reconstituted ampicillin suspension is stable for 14 days when stored in the refrigerator, please calculate shelf-life at room temperature
T
T
T
TTT
TTR
Ea
TR
Ea
TR
Ea
T
T
Q
TtTTt
Qk
kQ
e
eA
eAQ
k
kQ
)()(
.
.
190190
1010)(
)1
)10(
1(
1.
)10(
1.
10
)10(10
days 875.016
14)5()25(
164
9090
10
20
1010
T
T
T
Q
CtCt
KineticFactors: B-Medium Polarity
• Solvent polarity: polar (high solubility parameter) solvents generally accelerate reactions if the product is more polar than the reactants and vice versa.
PolarityDielectric constant
• Dielectric constant (ε): charge storation i.e. ε= C/C0
• Water (78.5)• Methanol (31.5)• Ethanol (24.3)• Acetone (19.1)• PEG400 (12.5)• Ether (4.3)• Ethyl acetate (3.0)
KineticFactors: C-Dielectric
Constant• Opposite sign ions vs. similar sign ions:• By increasing dielectric constant:
– K decreases for opposite signs– K increases for similar signs
• For reaction of ion and neutral molecules:– K increases with decreasing dielectric constant
KineticFactors: D-ionic strength
• Primary salt effect (non-catalytic salts):
• If zA and zB have the same sign, then zAzB is positive and the rate constant increases with ionic strength.
• 2. If zA and zB have different signs, zAzB is negative and the rate constant decreases with ionic strength.
• 3. If one of the reactants is uncharged, zAzB is zero and the rate constant is
independent of the ionic strength.
BA
ii
BA
BA
BA
BAZZ
ZZ
ZZZZ
ZZkk
Z
BABAkk
kk
BA
BAK
PBABA
BA
BA
BABA
02.1loglog
zc2
1μ
Eq.) Huckel(Debye 51.0log
)loglog(logloglog
.
.]][[
][
0
j
1
2ii
2
0
0
)(
)(
KineticFactors: E-pH/Buffer
• Specific acid-base catalysis:
][][0 OHkHkkk OHHobs
KineticFactors: E-pH/Buffer
• General acid-base catalysis: rate constant furthermore depends on the acidity constant and the concentration of buffer reagents (kA[acid], kB[Base]).
Questions
• 15-4• 15-19• 15-22
Q15.4
• Zero order rate constant for degradation of the colorant in a multisulfa preparation. Please determine Ea, K25C, and the predicated life for changing absorbance from initial value of 0.470 to 0.225.°C k 1/T ln K
40 0.00011 0.003194888 -9.1150350 0.00028 0.003095975 -8.1807260 0.00082 0.003003003 -7.1062170 0.00196 0.002915452 -6.23481
ln k= 24.2 - 10428.2 (1/T) -Ea/R=10428.2Ea = 1.9872 * 10428.2 = 20754 cal /moleT=298 °K, then k= 2.05231E-05At 25°C, 0.225=0.470-2.05 * 10-5tt= 11937.8h
Q15.19
• The following data are for the decomposition of 0.056 M glucose at 140 C, please calculate K0 and KH
kobs (hr-1) [H3O+]0.00366 0.0108
0.0058 0.01970.00818 0.02950.01076 0.03940.01217 0.0492
kobs=k0+kH*[H3O+]k0=0.001(hr-
1)kH=0.227 (hr-1)
0 0.01 0.02 0.03 0.04 0.05 0.060
0.002
0.004
0.006
0.008
0.01
0.012
0.014
f(x) = 0.2276388274645 x + 0.0013485740478R² = 0.992288609004898
[H3O+]
k
Q15.22• The hydrolysis of the prostaglandin in aqueous solution
was studied by varying the buffer. The dependence of kobs and pH is given as follows, do the buffer system influence hydrolysis? Please compute KH+ and KOH
-.
Buffer pHKobs * 107 (sec-1) log Kobs
HCl 1.15 3360-
3.47366
Formate 2.99 35-
5.45593
Formate 3.21 22.1-
5.65561
Formate 3.22 21.6-
5.66555
Phosphate 6.51 18.6-
5.73049
Phosphate 6.57 21.4-
5.66959
Phosphate 7.21 84.5-
5.07314
Carbonate 9.22 8350-
3.07831if specific acid-base catalysis:kobs=k0+kH*[H3O+]+kOH*[OH-]at low pH, then log Kobs = log kH
- pHintercept= -2.25 kH= 5.62 * 10-3
at high pH, then kobs=KOH.Kw/[H3O+]
log Kobs= log kOH.kw +pHintercept=-12.11
if kw=12.63 * 10
-12 then KOH=6.1 0
0 1 2 3 4 5 6 7 8 9 10
-7-6-5-4-3-2-10
pHL
og
k