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Applications of Spectrophotometry (Chapter 19). Red shift of l max with increasing conjugation CH 2 =CHCH 2 CH 2 CH=CH 2 l max =185 nm. CH 2 =CHCH=CH 2 l max =217 nm. vs. . Red shift of l max with # of rings Benzene l max =204 nm Naphthalene l max =286 nm. - PowerPoint PPT Presentation
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Applications of Spectrophotometry(Chapter 19)
Red shift of lmax with increasing conjugationCH2=CHCH2CH2CH=CH2 lmax =185 nm
CH2=CHCH=CH2 lmax =217 nm
vs.
Red shift of lmax with # of ringsBenzene lmax =204 nmNaphthalene lmax =286 nm
Polar solvents more likely to shift absorption maximaShifts of lmax with solvent polarityn®p* hypsochromic/blue shiftp®p* bathochromic/red shift
heptanemethanol
Hypsochromic shiftO
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
Determination of Concentration – Multicomponent System
CB
l1 l2
Abso
rban
ce
C+BAbso
rban
ce
Wavelength (nm)
Wavelength (nm)
l2l1
Determination of Concentration – Multicomponent System
CB
l1 l2
Abso
rban
ce
Wavelength (nm)
(1) Measure eB and eC at l1 and l2 (pure substances)
Determination of Concentration – Multicomponent System
C+BAbso
rban
ce
Wavelength (nm)l2l1
(2) Measure A at l1 and l2 (mixture)
Determination of Concentration – Multicomponent System
C+BAbso
rban
ce
Wavelength (nm)l2l1
A1 = xeB1 + yeC1
A2 = xeB2 + yeC2
x = molarity By = molarity C
Isobestic Points
Brocresol green
HIn (acid form) In- (base form)
When one absorbing species is converted to another, it is apparent in the absorption spectrum.
Isobestic Points
[OH-]pH[H+]
pH
I see an isobestic point!!!
Isobestic Points
The Total concentration of Bromocresol Green is constant throughout the reaction
HIn (acid form) In- (base form)
[Hin] + [In-] = F bromocresol green
A = b (exCx + eyCy)
Cx + Cy = C
But at the isosbestic point both molar absorptivities are the same!
ex + ey = e
Isobestic Points
HIn (acid form) In- (base 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
600 650 700 750 800 850 900 950 10000.0E+00
1.0E+03
2.0E+03
3.0E+03
4.0E+03
5.0E+03
6.0E+03
Hb02Hb
l (nm)
e (M
-1cm
-1)
400 500 600 700 800 900 10000.0E+00
2.0E+04
4.0E+04
6.0E+04
8.0E+04
1.0E+05
1.2E+05
1.4E+05
1.6E+05
1.8E+05
2.0E+05 Hb02Hb
l (nm)
e (M
-1cm
-1)
250 350 450 550 650 750 850 9500.0E+00
1.0E+05
2.0E+05
3.0E+05
4.0E+05
5.0E+05
6.0E+05
Hb02Hb
l (nm)
e (M
-1cm
-1)
Hb HbO2
O2
deoxyhemoglobin Oxyhemoglobin
100 % O2
0 % O2
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.
dr
fa Kk
kLR
RLK 1]][[
][
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
This expression can then be rearranged to find the ratio of bound to unbound (free) hormone
])[]([][
max RLBLRLKa -
])[(1])[(][][
][][
maxmax RLBK
RLBKLRL
freebound
da --
Measuring the equilibrium constant (The Scatchard Plot)
])[(1])[(][][
][][
maxmax RLBK
RLBKLRL
freebound
da --
The method of continuous variation (for isomolar solution)(Job’s method)
A + n B ABn
Call this C
Abso
rban
ce
Mole fraction (c = nb / na + nb)0 1
Allows for the determinationof the stoichiometry of thePredominant product.
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 + n B C
Start
Change
Equil.
M(1-x) Mx 0
-C -nC +C
M(1-x)-C Mx-nC C
M(1-x) is a concentrationT
A
VMVxM
-
1)1(
Cor
rect
ed A
bsor
banc
e
Mole fraction (c = nb / na)
0 1
A3X AX AX2
0.50.25 0.66