Advanced photochemistry

Advanced Photohemistry
Course Code 5512 Credit Hours 3
Lectured by: Dr. Fateh Eltaboni
Assistant Professor of Physical Chemistry at the University of Benghazi
University of Benghazi
Faculty of Science
Chemistry Department
Advanced Photochemistry Lecture Notes (Dr Fateh Eltaboni) 1

2
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Photochemistry
Photochemical
principles; Rates of
intramolecular
processes and
Intermolecular
processes
Energy transfer
Photochemical
reactions and
their quantum
yields
Photochemistry
in nature
Special topics in
photochemistry.

What is
Photochemistry?
Its reaction of matter with light!
Mr. Photo

CHARACTERISTICS OF LIGHT
 Light = band of waves of different wavelengths
Increase in wavelengths (nm)
Increase in frequency (energy)

An artist’s colour wheel: complementary colours are opposite one another
on a diameter. The numbers correspond to wavelengths of light in nm.

(Light) Electromagnetic spectrum important for photochemistry:

 The dual nature of light: It means that, in some experiments, light behaves as a wave, In
other experiments, light behaves as a particle.
• Wavelength (): Distance between successive crests or troughs
• Frequency (): Number of crests or troughs that pass a point per second
• Speed of light (c) c =  
• Solar energy (Es): (Es = h) It’s the energy come from the sun or other radiant source.

PARTICLE CHARACTERISTICS OF LIGHT:
 Light = flux of discrete units (i.e quanta) called photons.
Fireworks:

ABSORBANCE, COLOUR AND TRANSPARENCY OF MATERIALS:
Transparent materials absorb no light in the visible wavelength range, and hence have no
energy states separated by between 1.65 and 3.10eV.
Examples:
SiO2 (glass, quartz);
Water soluble polymers like:
Poly(acrylic acid) (PAA)C(Diamond);

ABSORBANCE, COLOUR AND TRANSPARENCY OF MATERIALS:
Opaque materials: are absorb light in all the visible wavelength range, and hence
have many energy states separated by between 1.65 and 3.10eV.
Examples:
• C(graphite) also absorbs light at all wavelengths, but is black and not reflective like a
metal.
• Some metals (Cu, Fe…)
are coloured as well as reflective.

 Dyes and many other materials may be
coloured because they absorb some of the
wavelengths in the visible range.
UV Absorbance – Sunscreens:
 Absorbance in the ultraviolet (UV) range in
important for transparent materials. UV is higher
energy light, and UV absorbance can lead to
photochemical reactions and the formation of
highly reactive free radicals. As with ionizing
radiation these can lead to cell damage and (skin)
cancer.

Mechanism of UV-degradation of
polymer chain In presence of O2:
UV-light

 Sunscreens contain a mixture of compounds, like polymers, that absorb UV light
from the solar spectrum:

 The "Z" in Structure I is selected from a UV-chromophore:

ABSORPTION OF LIGHT BY ORGANIC
MOLECULES

• The absorption spectra of rigid hydrocarbons in nonpolar solvents may show
vibrational fi ne structure, but absorption spectra of other organic molecules in
solution tend to be broad, featureless bands with little or no vibrational structure
(Figure 2.8 ). This is due to the very large number of vibrational levels in organic
molecules and to blurring of any fi ne structure due to interaction between organic
molecules and solvent molecules.

Relation between wavelength and conjugation

• Using propenal as an example, we can examine the
conjugation between a C = C bond and a carbonyl
group. The electronic absorption spectrum of
propenal is shown in Figure 2.11 .
• The effect of conjugation on the π - and π * - orbitals
is analogous to the previous example with dienes
but the n orbital remain virtually unchanged as a
result of the conjugation. Figure 2.12 shows the
effect of conjugation on the lowest - energy π → π *
and n → π * transitions.

o Ordinary reaction (thermal or dark reaction):
o Its occurs by absorption of heat energy from outside in absence of light.
1. The reacting molecules are energised
2. Molecular collision is occurred
3. Product is formed
Steps of Ordinary reactions

o Photochemical reaction:
o Its a reaction which takes place by absorption of the visible and ultraviolet radiations
(200-800 nm)
o Photochemical reaction leads
to the heating of the
atmosphere during the
daytime by absorption of
ultraviolet radiation.
 Mechanism of photochemical reactions occurring during atmospheric: Photochemical change oc
only by absorption of photons.

o Photochemical reaction includes the absorption of visible radiation during
photosynthesis.
 Without photochemical processes, the Earth would be simply a warm, sterile, rock.
Schematic of photosynthesis in plants. The
carbohydrates produced are stored in or used
by the plant.
Overall equation for the type of
photosynthesis that occurs in plants

o Photochemistry:
o Its the branch of chemistry which deals with the study of the chemical reactions and
physical changes that result from interactions between matter and visible or ultraviolet
light.
o Demonstration of a Photochemical reaction

o Difference between photochemical and thermochemical reactions

The Photochemistry of
Alkenes

Electrocyclic Reactions

The Photochemistry of Carbonyl
CompoundsThe principal reaction types for ketone (n, π * ) excited states are:

α - CLEAVAGE REACTIONS (Decarbonylation )

• The efficiency of product formation in solution is also controlled by the stabilities
of the radicals. Stable radicals such as tertiary alkyl radicals or benzyl radicals
lead to efficient decarbonylation in solution.
• Because of steric factors involving bulky groups, tertiary radicals tend to
preferentially undergo disproportionation rather than radical combination and so
the quantum yield of the products formed by disproportionation exceeds that of
the radical combination product.

• If radicals are unable to undergo disproportination they must react by radical
combination. For example, irradiation of PhCH2COCHPh2 gives three radical
combination products in the statistically – expected ratio of 1 : 2 : 1:

• Cyclic ketones undergo decarbonylation on irradiation in the vapour phase
and in solution, provided they form stabilised radicals:

• Five - or six - membered saturated cyclic ketones can also react by another
pathway that does not involve decarbonylation.

• The photodecomposition is accelerated by using a sensitiser. If
such a sensitiser is added to the herbicide it will result in a less
environmentally- persistent herbicide.

INTRAMOLECULAR HYDROGEN -ABSTRACTION REACTIONS

THE ROLE OF CARBONYL COMPOUNDS IN POLYMER
CHEMISTRY

ENERGY TRANSITIONS:
• Molecules absorb radiation by increasing internal energy:
Internal energy  electronic, vibrational, & rotational states
• Energy requirements:
(1). Electronic transitions 
UV-vis ( 400-700 nm)
(2). Vibrational transitions
 Near-IR (700-2000 nm)
(3). Rotational transitions
 Far-IR-Mic (> 2000 nm)

Typical electronic transitions: Typical Molecular vibrations
Typical Molecular rotations

What Happens When Radiation Hits a Molecule?

Absorption and Emission:
• An electronic transition occurs when an electron changes from state to another.
• The orbital it is leaving must be partly filled (it contains at least one electron), and the
orbital it is entering must be partly unfilled
(it contains less than two electrons).

Absorption or Absorbance: An electron absorbs the energy of a photon, and
jumps from a lower into a higher energy orbital.
Emission: An electron jumps from a higher into a lower orbital, and releases (emits) a
photon of energy equal to the difference.

o Light absorption
o When light is passed through a medium, a part of it is absorbed. This absorbed portion
of light which causes photochemical reactions.
o Let a beam of monochromatic light pass through a thickness dx of the
medium. The intensity of radiation reduces from I to I-dI.
o The intensity of radiation can be defined as the number
of photons that pass across a unit area in unit time.
o Let us denote the number of incident photons by N and
the number absorbed in thickness dx by dN.
o The fraction of photons absorbed is then dN/N which is
proportional to thickness dx. That is,

α
o where b is proportionality constant called absorption coefficient.
o Let us set I = I0 at x = 0 and integrate. This gives
o Lambert first derived equation (1) and it is known as Lambert Law. Beer extended
this relation to solutions of compounds in transparent solvents. The equation (1)
then takes the form (2).
Lambert-Beer Law. This law forms the basis
of spectrophotometric methods of chemical analysis.
o where C = molar concentration; ∈ is a constant characteristic of the solute called the molar absorption
coefficient.

b = x
Beer-Lambert Law :
The attenuation of a beam
by absorbance is typically
represented in two ways:
1. Fraction or Percent
Transmission (%T)
2. Absorbance (A)

ABSORBANCE SPECTROMETRY
 Determines concentration of a substance in solution
 Measures light absorbed by solution at a specific wavelength
 Visible region: low energy electronic transition due to:
a. Compounds containing transition metals
b. Large aromatic structures & conjugated double bond systems
 UV region (200-400 nm):
a. Small conjugated ring systems
b. Carbonyl compounds

o Determination of absorbed intensity
o A photochemical reaction occurs by the absorption of photons of light by the molecules.
o Therefore, it is essential to determine the absorbed intensity of light for a study of the
rate of reaction.
Schematic diagram of the spectrophotometer used for measurement of light intensity
A sample UV-Vis spectrum.

Measurements at two wavelengths can
be used to find the individual
concentrations of two components A
and B in a mixture. For this analysis,
we write the total absorbance at a
given wavelength as:

1. Light beam from a suitable source (tungsten filament or mercury vapour lamp) is
rendered parallel by the lens.
2. The beam then passes through a ‘filter’ or monochrometer which yields light of one
wavelength only. The monochromatic light enters the reaction cell made of quartz.
3. The part of light that is not absorbed strikes the detector. Thus the intensity of light
is measured first with the empty cell and then the cell filled with the reaction sample.
4. The first reading gives the incident intensity, I0, and the second gives the
transmitted intensity, I. The difference, I0 – I = Ia, is the absorbed intensity.
o Steps of absorbance measurements

o The detector generally used for the measurement of intensity of transmitted light
is :
(a) a thermopile (b) photoelectric cell (c) a chemical actinometer.

o LAWSOF PHOTOCHEMISTRY
1) Grothus–Draper Law (qualitative aspect)
It is only the absorbed light radiations that are effective in producing a chemical reaction.
2) Stark-Einstein Law of Photochemical Equivalence
Stark and Einstein (1905) studied the quantitative aspect of photochemical reactions by
application of Quantum theory of light.
Illustration of Law of Photochemical
equivalence; absorption of one
photon decomposes one molecule.

o In a photochemical reaction, each molecule of the reacting substance absorbs
a single photon of radiation causing the reaction and is activated to form the
products.
o In practice, we use molar quantities. That is, one mole of A absorbs one mole of
photons or one einstein of energy, E. The value of E can be calculated by using the
expression given below:

o Primary and Secondary reactions:
The overall photochemical reaction may consist of :
(a) a primary reaction: proceeds by absorption of radiation.
(b) secondary reaction: is a thermal reaction which occurs subsequent to the primary
reaction.
For example, the decomposition of HBr occurs as follows :
o Evidently, the primary reaction only obeys the law of photochemical equivalence strictly. The
secondary reactions have no concern with the law.

o Quantum yield (or Quantum efficiency) (ᶲ)
o For a reaction that obeys strictly the Einstein law, one molecule decomposes per
photon, the quantum yield φ = 1.
o When two or more molecules are decomposed per photon, φ > 1 and the reaction
has a high quantum yield.
o If the number of molecules decomposed is less than one per photon, φ < 1 the
reaction has a low quantum yield.

o Causes of high quantum yield
The chief reasons for high quantum yield are :
(1) Reactions subsequent to the Primary reaction:
(2) A reaction chain forms many molecules per photon:
 A reaction chain steps:
1) Initiation……………….>
2) Self-propagating……>
3) Termination………….>
The number of HCl molecules formed
for a photon of light is very high. The
quantum yield of the
reaction varies from 104 to 106

o Causes of low quantum yield
(1) Deactivation of reacting molecules:

(2) Occurrence of reverse of primary reaction:
Dimerization of Anthracene:
o The quantum yield should be 2 but it is actually found to be 0.5. The low
quantum yield is explained as the reaction is accompanied by fluorescence
which deactivates the excited anthracene molecules. Furthermore, the
above reaction is reversible.

(3) Recombination of dissociated fragments:
Combination of H2 and Br2.
2
o The reaction (2) is extremely slow. The reactions (3), (4) and (5), depend directly or
indirectly on (2) and so are very slow. Therefore most of the Br atoms produced in the
primary process recombine to give back Br2 molecules.
o Thus the HBr molecules obtained per quantum is extremely small. The quantum yield of
the reaction is found to be 0.01 at ordinary temperature.

CALCULATION OF QUANTUM YIELD
By definition, the quantum yield, , of a photochemical reaction is expressed as :
Thus we can calculate quantum yield from :
(a) The amount of the reactant decomposed in a given time and
(b) The amount of radiation energy absorbed in the same time
o The radiation energy is absorbed by a chemical system as photons. Therefore we
should know the energy associated with a photon or a mole of photons.

The energy of photons; einstein
o We know that the energy of a photon (or quantum), ∈ , is given by the equation:
h = Planck’s constant (6.624 × 10– 27 erg-sec)
v = frequency of radiation
λ = wavelength of radiation
c = velocity of light (3 × 1010 cm sec– 1) o If λ is given in cm, the energy is expressed in ergs.
The energy, E, of an Avogadro number (N) of photons is referred to as one einstein:
Substituting the values of N (= 6.02 × 1023), h and c, in (2), we have:

If λ is expressed in Å units (1Å = 10– 8 cm),
Since 1 cal = 4.184 × 107 erg, energy in calories would be:
o It is evident from (3) that the numerical value of einstein varies inversely as the wavelength
of radiation.
o The higher the wavelength, the smaller will be the energy per einstein.
Electron-volt (eV) is another
commonly used energy unit, 1 eV
= 1.6 x 10-19 J.

SOLVED PROBLEM 1. Calculate the energy associated with (a) one photon;
(b) one einstein of radiation of wavelength 8000 Å. h = 6.62 × 10–27 erg-sec; c =
3 × 1010 cm sec–1.
SOLUTION
(a) Energy of a photon
(b) Energy per einstein

SOLVED PROBLEM 2. When a substance A was exposed to light, 0.002 mole of it
reacted in 20 minutes and 4 seconds. In the same time A absorbed 2.0 × 106 photons
of light per second. Calculate the quantum yield of the reaction. (Avogadro number
N = 6.02 × 1023)
SOLUTION
Number of molecules of A reacting = 0.002 × N = 0.002 × 6.02 × 1023
Number of photons absorbed per second = 2.0 × 106
Number of photons absorbed in 20 minutes and 4 seconds = 2.0 × 106 × 1204
Quantum yield

SOLVED PROBLEM 3. When irradiated with light of 5000 Å wavelength, 1 × 10–
4 mole of a substance is decomposed. How many photons are absorbed during the
reaction if its quantum efficiency is 10.00. (Avogadro number N = 6.02 × 1023)
SOLUTION
Quantum efficiency of the reaction = 10.00
No. of moles decomposed = 1 × 10–4
No. of molecules decomposed = 1 × 10–4 × 6.02 × 1023 = 6.02 x 1019
we know that,

SOLVED PROBLEM 4. When propionaldehyde is irradiated with light of λ =
3020 Å, it is decomposed to form carbon monoxide.
CH3CH2CHO + hv ⎯⎯→ CH3CH3 + CO
The quantum yield for the reaction is 0.54. The light energy absorbed is 15000 erg
mol in a given time. Find the amount of carbon monoxide formed in moles in the
same time.
SOLUTION

PHOTOSENSITIZED REACTIONS
o In many photochemical reactions the reactant molecule does not absorb the
radiation required for the reaction. Hence the reaction is not possible.
o In such cases the reaction may still occur if a foreign species such as mercury
vapour is present.
o A species which can both absorb and transfer radiant energy for activation of the
reactant molecule, is called a photosensitizer. The reaction so caused is called a
photosensitized reaction.
Hg + hv ⎯⎯→ Hg*
Hg* + A ⎯⎯→ A* +
Hg
Reaction between H2 and CO:
Hg + hv ⎯⎯→ Hg* Primary absorption
Hg* + H2 ⎯⎯→ 2H * + Hg Energy transfer
H + CO ⎯⎯→ HCO
Some glyoxal, CHO-CHO, is also formed by dimerization of formyl radicals, HCO.

EMISSION:
• An electron jumps from a higher
into a lower orbital, and releases
(emits) a photon of energy equal to
the difference (∆E = hv).
• Luminescence (emission of light)
works on the basis that if a
molecule is excited above its
ground state, then it must de-excite
somehow and release excess
energy as either non-radiative or
radiative energy.

 A good starting point for a discussion of
luminescence (fluorescence or phosphorescence)
principle is a simplified Jablonski diagram.
• The Jablonski diagram is employed to represent the
energy levels of a molecule. As depicted in JD, S0, S1 and
S2 represent ground, first and second singlet electronic
states, respectively, whereas T1 and T2 describe first and
second triplet electronic states, respectively.
• Each electronic energy level of a molecule also has
numerous vibrational (v) and rotational (n) energy
sublevels.
Jablonski diagram (JD)

(JD)
10-15 s
10-11 s
10-9 to 10-7 s
10-5 to 10-s
Absorbanceenergy

So
S1
VR
VR
F
A
T1
VR
P
DE
DE
Relationship between fluorescence, phosphorescence and absorption spectra
-spectrum: intensity
of luminescence vs
wavelength
Similarly, a molecule
can relax back to
a vibrationally excited
level of So from T1.
- have vibrational fine
structure in absorption
and in emission
molecule relaxes back
from the S1 state to a
vibrationally excited
level of the ground
state So.

• Mirror-image rule and Franck-Condon
factors. The absorption and emission
spectra are for anthracene. The
numbers 0, 1, and 2 refer to vibrational
energy levels.
• Phosphorescence and ISC are spin-
forbidden transitions. In contrast,
fluorescence and IC are spin-allowed
transitions.
“Stokes” shift
Absorption vs Emission

PHOTOPHYSICAL PROCESSES:
o If the absorbed radiation is not used to cause a chemical change, it is re-emitted
as light of longer wavelength. The three such photophysical processes which can
occur are :
(1) Fluorescence (2) Phosphorescence (3) Chemiluminescence
o Fluorescence
 Process at which certain molecules (or atoms) when exposed to light radiation of
short wavelength (high frequency), emit light of longer wavelength.
 Substance that exhibits fluorescence is called fluorophore.
 Florescence stops as soon as the incident radiation is cut off.
Examples:
(a) a solution of quinine sulphate on exposure to visible light, exhibits blue
fluorescence.
(b) a solution of chlorophyll in ether shows blood red fluorescence.

Molecular formulas of common dyes of fluorescent probes (fluoroph

o Fluorescence depends on:
1. pH
2. Solvent
3. Temperature
4. Salts
5. Concentration

Demonstration of a Solvent Sensitive Fluorophore
Corrected fluorescence emission spectra of DOS in
cyclohexane(CH), toluene (T), ethylacetate(EA) and
butanol(Bu). The dashed line shows the emission of
DOS from DPPC vesicles
Solvent Polarity

Fluorescent minerals, shown under ultraviolet light:

 The colour of fluorescence depends on the wavelength of light emitted.

Phosphorescence
 When a substance absorbs radiation of high frequency and emits light even after the
incident radiation is cut off, the process is called phosphorescence.
 The substance which shows phosphorescence is called phosphorescent substance.
 Phosphorescence is chiefly caused by ultraviolet and visible light. It is generally shown
by solids.
Examples:
(a) Sulphates of calcium, barium and strontium exhibit phosphorescence.
(b) Fluorescein in boric acid shows phosphorescence in the blue region at 570nm
wavelength.
o Phosphorescence could be designated as delayed fluorescence.
A + hv ⎯⎯→ A* ⎯slow⎯→ hv′

Phosphorescent powder under:
visible light, ultraviolet light, darkness

How much luminescence is observed depends on efficiency of the various
processes shown in the Jablonski Diagram.
- which compete with one another to deactivate the excited state
Consider fluorescence:
-amount of fluorescence observed is dependent on relative rates of
the following unimolecular processes:
process rate
hu1M 1M * absorption Ia
k F
1M * 1M + hu fluorescence k F [ 1M *]
k IC1M * 1M internal conversion k IC [ 1M *]
k ISC1M * 3 M * intersystem crossing k ISC [ 1M *]
Consider a molecule, M
(* denotes electronically excited)

Quantum Yield of Fluorescence
FO = number of photons emitted as fluorescence
total number of photons absorbed
rate
follow law of energy conversion
(i.e. what goes up must come down!)
rate of
absorption
rate of all deactivation processes
(i.e. total rate of deactivation)
FO rate of fluorescence
=
=
k F [ 1M *]
k F [ 1M *] + k IC [ 1M *] + k ISC [ 1M *]
(PHI)
rate

=
k F [ 1M *]
(k F + k IC + k ISC ) [ 1M *]
=
k F
k F + k IC + k ISC
Fo
since k n r = k IC + k ISC
rate of non-radiative deactivation
then
Fo
=
k F
k F + k nr

Now, since k = rate of all deactivation processes for the excited state
then
to =
1
k F + k nr
1
k
=
What does a lifetime mean?
This is a Relaxation Process
1 M * population relaxes back to the ground state lifetime to relaxation time
78.
Physically it corresponds (due to the exponential nature of the decay)
to the time taken for [ 1M *] to decay from [ 1M *]o
to 1/e [ 1M *]o
=
Fo
k F
k F
k F + k nr
x =
1
k F
Example: In water, the fluorescence quantum yield and observed fluorescence lifetime of tryptophan are φo =
0.20 and τ0 = 2.6 ns, respectively. The fluorescence rate constant kf is?
Solution: to =
Fo
k F
So, k F =
Fo
to
=
0.20
2.6 x 10-9 s
= 7.7 x 107 s-1

First, the sample is excited with a short light pulse from a laser using a
wavelength at which S absorbs strongly. Then, the exponential decay of the
fluorescence intensity after the pulse is monitored. From eqns 21.80 and
21.81, it follows that
Example: In water, the fluorescence quantum yield and observed fluorescence
lifetime of tryptophan are φf = 0.20 and τ0 = 2.6 ns, respectively. The
fluorescence rate constant kf is?

Fluorescence Lifetime (Lifetime of the Excited State)
-if use light of intensity Io
create an excited state population [ 1 M *]
(M absorbs the light to create 1 M *)
[ 1 M *] will remain constant (i.e. in a steady state) until light is removed
At the instant light is removed concentration of excited
states is [ 1 M *] o
t=0 time (t)t
[1M *] t
[1M *] o
Io
[1M *]
The excited state population [ 1M * ]o will
decay exponentially
- after some time, t, it’s [ 1M * ] t

The observed fluorescence lifetime can be measured by using a pulsed
laser technique:

This is a 1st order rate process & obeys 1st order kinetics.
Shape of decay can be described by:
[ 1M * ] t = [ 1M * ] o exp ( - k t )
[ 1M * ] t = [ 1M * ] o exp ( - t / t )
concentration of
excited states at
some time, t
initial concentration
of excited state
rate constant for process
time
Alternatively a physicist would use
where, t =
1
k
= “lifetime” of 1 M *
77.

Bimolecular Processes (Quenching)
-can expand kinetic scheme to include bimolecular processes.
Certain molecules, through collisions with M * can remove excitation energy
- called quenchers since they reduce the excitation state population
(in addition to that from unimolecular processes).
Examples of quenchers:
O2 ; CCl4 ; CH3NO2 ; Tl+
Mechanism can be expanded to include steps of the type:
1 M * + Q M + Q*
process
bimolecular
quenching
rate
k q [ 1 M *] [Q]
k q
bimolecular quenching constant
usually non fluorescent
• The shortening of the lifetime of the excited state by the presence of another species is called quenching

F
rate of fluorescence
=
=
k F [ 1M *]
k F [ 1M *] + k IC [ 1M *] + k ISC [ 1M *] + k q [ 1M *] [Q]
=
k F [ 1M *]
(k F + k IC + k ISC + k q [Q]) [ 1M *]
Quantum Yield in the Presence of Quencher (F )
F =
k F
(k F + k IC + k ISC + k q [Q])

Derivation of an expression which relates quantum yields in presence ( F )
& absence ( F o ) of quencher
k F
F o
F
=
k F
k F + k IC + k ISC
k F + k IC + k ISC
=
k F
(k F + k IC + k ISC + k q [Q])k F
k F + k IC + k ISC
X
k F + k IC + k ISC + k q [Q])
(k F + k IC + k ISC )
=
(k F + k IC + k ISC)
+

k q [Q])
(k F + k IC + k ISC)
1 +
F o
F
=
Such dynamic quenching can be described (simply) by Stern-Volmer Kinetics.
since t F
o
=
1
k F + k nr
1
k
=
lifetime in the absence of
quencher
quantum yield in
the absence of quencher
quantum yield in
the presence of quencher
t F
o is the lifetime in
the absence of quencher
quenching constant
given concentration
of quencher ([Q])
Stern-Volmer Equation
F o
F
= 1 + k q t F
o [Q]

FF
Fo
F
=
IF
Io
F
=
t F
t o
F
= 1 + kq tF
o [ Q ]
Obviously, a plot of
FF
Fo
F
IF
Io
F
or or
t F
t o
F
vs. [Q] is linear
with a slope = kq tF
o and intercept = 1
If we know tF
o can determine kq (the Bimolecular Quenching Constant)
- from the slope.
Since, I F  FF
(I F is the intensity of fluorescence and is what you measure “directly” via fluorescence
spectroscopy)
( y = c + m x )
In addition, for dynamic quenching:
FF
Fo
F
=
IF
Io
F
=
t F
t o
F
( t f data can be determined by a fluorescence lifetime spectrometer)

Example
(i) Show (graphically) that the quenching obeys Stern-Volmer kinetics.
84.
FF
Fo
F
=
IF
Io
F
=
t F
t o
F
= 1 + kq tF
o [ Q ]
(naphthalene)
When carbon tetrachloride is added to a dilute (10-5 M) solution of naphthalene in toluene, the
fluorescence, observed from the naphthalene, is quenched. The following table shows the effect
of varying amounts of carbon tetrachloride upon t F observed at 340 nm at 298 K.
[CCl4] / M t F / ns
0 18.0
0.0018 14.4
0.0049 10.8
0.0074 9.0
0.0110 7.2
t o
F
t F
vs. [Q] is linear

slope = 136 M-1
[CCl4]
/ M t F / ns to
F / t F
0 18.0 1.00
0.0018 14.4 1.25
0.0049 10.8 1.67
0.0074 9.0 2.00
0.0110 7.2 2.50
. : .
[CCl 4] / M
0.004 0.008 0.0120
1
2
3
to
F / t F
(ii) Estimate the value for k q, the bimolecular quenching rate constant, at 298K.
slope = k q to
F
k q = 136 M-1 / 18 x 10 –9 s
k q = 7.6 x 109 M-1 s-1

Since to
F of acetone is ~ 2 x10 -9 s
then k q = 7 M -1 / 2 x 10 -9 s
k q = 3.5 x 10 9 M-1 s-1 rate constant for the reaction
Since k q [ 1 M *] [Q] = rate of reaction
Can measure reaction rate
88.
0 0.1 0.2 0.3
1.0
2.0
Fo
F
FF
[dicyanoethylene] / M
slope = kq to
F = 7 M -1

e.g. photoaddition reaction
Stern-Volmer experiments can be used to study photochemical reactions:
fluorescent
-acetone fluorescence quenched by dicyanoethylene
C=O *
CN
CN
O
NC CN
+
k q
acetone dicyanoethylene dicyanooxetane
non-fluorescent
-can detect fluorescence from acetone to monitor progress of reaction

Energy Transfer
Energy exchange occurs through short distances ~ 2nm (collisions).
However, energy can be transferred over longer distances than collision diameters.
In general:
D* + A D + A*
-energy transfer can occur up to distances of 10 nm
FÖrster Transfer
-depends on overlap of the spectral bands of emission from donor
& absorption by the acceptor.
donor acceptor
M* + Q M + Q*
Fluorescence quenching - example of energy transfer.

• The excitation frequency of the DONOR must not excite the ACCEPTOR.!
• The emission spectrum of the DONOR must overlap the excitation spectrum of the
ACCEPTOR.!
• The emission spectrum of the DONOR cannot significantly overlap with emission of
ACCEPTOR. !
• DONOR and ACCEPTOR molecules must be in close proximity (typically 10–100 Å). !
Fluorescence Resonance Energy Transfer (FRET)

For example, 2 solutions in benzene:
(a) 10-3M in anthracene (b) 10-3M in perylene
-excitation of either solution within the chromphore’s excitation band
produces fluorescence from the fluorophore.
e.g. excitation of anthracene solution at 340 nm gives relatively
intense fluorescence between ~ 370-470 nm.
90.
(a) anthracene

Similarly, excitation of the perylene containing solution at 410 nm “see”
fluorescence from perylene (430-550 nm)
(b) perylene

Now, a solution containing both 10-3M anthracene and 10-3M perylene:
(i) excitation at 410 nm (where only perylene absorbs)
gives perylene fluorescence (unperturbed by presence of anthracene).
(ii) excitation at 340 nm (hardly absorbed at all by perylene) produces fluorescence
form both anthracene and perylene.
However, anthracene fluorescence quenched in mixture.
(c)
0
4 0
8 0
1 2 0
1 6 0
2 0 0
400 500
Wavelength / nm
Intensity
perylene
anthracene

EXCIMERS

EXCIPLEXES

FÖrster Transfer-form of non-radiative transfer
- depends upon spectral overlap of donor (anthracene) & absorption band of
acceptor (perylene)
Involves dipolar interactions:
-effectively electronic oscillations in D * induce oscillations in electron to be
promoted in A
(like resonance interactions between 2 tuning forks).
(iii) large eA for A
-also known as fluorescence resonance energy transfer (FRET).
Consequently, D * returns to D and A is promoted to A *
D * + A D + A *
This form of energy transfer is enhanced by:
(i) good overlap of fluorescence spectrum of D * & absorbance spectrum of A
(ii) large FF for D * (in absence of A)

Energy transfer Efficiency
E T =
Ro
6
Ro
6 + R 6
distance between donor & acceptor
- ET increases with decreasing distance (R)
Donor & acceptor must be within 10 nm (critical transfer distance)
-sometimes referred to as Spectroscopic Ruler Technique
-used extensively to measure distances in biological systems & across
science in general.
Fo
F
FF
E T = 1 -
for D in absence of acceptor
transfer
efficiency
for D in presence of acceptor
spectroscopic characteristic
of each donor/acceptor pair

Applications of FRET in Chemistry & Biology
Example: rhodopsin -protein (primary absorber of light contained in the eye)
-allows vision to take place
rhodopsin- consists of an opsin protein
molecule attached to 11- cis retinal
D A
-amino acid on the surface of rhodopsin was
labelled covalently with the energy
donor I AEDANS
Iodoacetyl amino ethyl amino naphthalene sulphonic acid (I AEDANS)

FF of label decreased from 0.75 to 0.68 due to quenching by the cis retinal acceptor
Ro is 5.4 nm for this pair from E T =
Ro
6
Ro
6 + R6
Can calculate the distance (R) between the surface of the rhodopsin protein
& cis retinal as 7.9 nm
AEDANS (D)
11- cis retinal (A)
opsin protein
rhodopsin
can calculate R
7.9 nm
ET = 1 –
0.75
0.68
= 0.093

• Example: Consider a study of the protein rhodopsin. When an amino acid on the surface
of rhodopsin was labelled covalently with the energy donor 1.5-I AEDANS (3), the
fluorescence quantum yield of the label decreased from 0.75 to 0.68 due to quenching
by the visual pigment 11-cis-retinal (4). Calculate the distance between the surface of
the protein and 11-cis-retinal (Ro = 5.4 nm for the 1.5-I AEDANS/11-cis-retinal pair )

PHOTOINDUCED
ELECTRON
TRANSFER ( PET )

Chemiluminescence:
o The emission of light as a result of chemical action is called chemiluminescence. The
reaction is referred to as a chemiluminescent reaction.
o Chemiluminescent reaction is the reverse of a photochemical reaction which proceeds
by absorption of light.
o The light emitted in a chemiluminescent reaction is also called ‘cold light’ because it is
produced at ordinary temperature.
o In a chemiluminescent reaction, the energy released in the reaction makes the product
molecule electronically excited. The excited molecule then gives up its excess energy
as visible light while reverting to ground state.

Chemiluminescence of fireflies and luminol
Examples:
(a) The glow of fireflies due to the aerial oxidation of luciferin (a protein) in the presence of
enzyme luciferase.
(b) The oxidation of 5-aminophthalic cyclic hydrazide (luminol) by hydrogen peroxide in
alkaline solution, producing bright green light.

Notes
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Advanced photochemistry

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