Applications of
Spectrophotometry
(Chapter 19)
Red shift of lmax with increasing conjugation
CH2=CHCH2CH2CH=CH2 lmax =185 nm
CH2=CHCH=CH2 lmax =217 nm
vs.
Red shift of lmax with # of rings
Benzene lmax =204 nm
Naphthalene lmax =286 nm
Polar solvents more likely to shift absorption maxima
Shifts of lmax with solvent polarity
n* hypsochromic/blue shift
* bathochromic/red shift
O
Hypsochromic shift
methanol
heptane
Generally, extending conjugation leads
to red shift
“particle in a box” QM theory; bigger box
Substituents attached to a chromophore that cause a red shift are called
“auxochromes”
Strain has an effect…
lmax
253
239
256
248
Generally, extending conjugation leads
to red shift
“particle in a box” QM theory; bigger box
Absorbance
Determination of Concentration – Multicomponent System
C
B
l2
Wavelength (nm)
Absorbance
l1
C+B
l1
l2
Wavelength (nm)
Absorbance
Determination of Concentration – Multicomponent System
C
B
l1
l2
Wavelength (nm)
(1) Measure eB and eC at l1 and l2 (pure substances)
Determination of Concentration – Multicomponent System
Absorbance
(2) Measure A at l1 and l2 (mixture)
C+B
l1
l2
Wavelength (nm)
Determination of Concentration – Multicomponent System
A1 = xeB1 + yeC1
A2 = xeB2 + yeC2
Absorbance
x = molarity B
y = molarity C
C+B
l1
l2
Wavelength (nm)
Isobestic Points
HIn (acid form)
In- (base form)
Brocresol green
When one absorbing species is converted to another, it is apparent in the
absorption spectrum.
Isobestic Points
pH
I see an
isobestic
point!!!
pH
[H+]
[OH-]
Isobestic Points
In- (base form)
HIn (acid form)
The Total concentration of Bromocresol Green is constant throughout the reaction
[Hin] + [In-] = F bromocresol green
Cx + Cy = C
A = b (exCx + eyCy)
But at the isosbestic point both molar absorptivities are the same!
e x + ey = e
Isobestic Points
In- (base form)
HIn (acid form)
ex + ey = e
Therefore, the absorbance does not depend on the extent of reaction
(i.e. on the particular concentrations of x and y)
A = b (exCx + eyCy) = b e (Cx + Cy) = b e C
An isobestic point is good evidence
that only two principal species
are present in a reaction.
Isobestic Points - Application
Oximetry
6.0E+05
5.0E+05
Hb
deoxyhemoglobin
HbO2
Oxyhemoglobin
e (M-1cm-1)
O2
4.0E+05
Hb02
3.0E+05
Hb
2.0E+05
1.0E+05
0.0E+00
250
450
650
850
l (nm)
2.0E+05
Hb02
1.8E+05
Hb
6.0E+03
1.6E+05
5.0E+03
1.2E+05
e (M-1cm-1)
e (M-1cm-1)
1.4E+05
1.0E+05
8.0E+04
6.0E+04
Hb02
Hb
4.0E+03
3.0E+03
2.0E+03
100 % O2
4.0E+04
1.0E+03
2.0E+04
0.0E+00
400
600
800
l (nm)
1000
0 % O2
0.0E+00
600
700
800
l (nm)
900
1000
Measuring the equilibrium constant
(The Scatchard Plot)
Biochemistry example: The cellular action of a hormone begins when the hormone (L)
Binds to it’s receptor protein (R) in a tight and specific way.
The binding thus gives rise to conformational changes which change the biological activity
of the receptor (an enzyme, an enzyme regulator, an ion channel,
or a regulator of gene expression.
R
+ L
RL
The binding depends of the concentration of the concentration of the components.
Ka 
[ RL ]
[ R ][ L ]

kf
kr

1
Kd
Ka is the association constant and Kd is the dissociation constant
Measuring the equilibrium constant
(The Scatchard Plot)
When binding has reached equilibrium, the total number of binding sites, Bmax = [R] + [RL]
whereas, the number of unbound sites would be [R] = Bmax – [RL].
The equilibrium constant would then be
Ka 
[ RL ]
[ L ]( B max  [ RL ])
This expression can then be rearranged to find the ratio of bound to unbound (free) hormone
[ bound ]
[ free ]

[ RL ]
[L]
 K a ( B max  [ RL ]) 
1
Kd
( B max  [ RL ])
Measuring the equilibrium constant
(The Scatchard Plot)
[ bound ]
[ free ]

[ RL ]
[L]
 K a ( B max  [ RL ]) 
1
Kd
( B max  [ RL ])
The method of continuous variation (for isomolar solution)
(Job’s method)
A
+
nB
ABn
Absorbance
Call this C
Allows for the determination
of the stoichiometry of the
Predominant product.
0
Mole fraction (c = nb / na + nb)
1
The method of continuous variation (for isomolar solution)
(Job’s method)
Requirements:
The system must conform to Beer's law.
Only single equilibrium
Equimolar solutions MA = MB
Constant volume 1 = VT = VA + VB
K reasonably greater than 1
pH and ionic strength must be maintained constant
The method of continuous variation (for isomolar solution)
(Job’s method)
A
Start
Change
Equil.
+
nB
C
M(1-x)
Mx
0
-C
-nC
+C
M(1-x)-C
M(1-x) is a concentration
Mx-nC
M (1  x )
1
C

MV
VT
A
Corrected Absorbance
A3 X
0
0.25
AX
0.5
AX2
0.66
Mole fraction (c = nb / na)
1