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Practice Problem Set #1 – Photochemistry Fundamentals 1. Is there such a thing as a T0 state? 2. What is the approximate bond dissociation energy of a C–Br bond? Using Planck’s equation, calculate the maximum wavelength of light capable of photolyzing a brominated organic molecule. However, when most bromides are irradiated at this wavelength, no decomposition is observed. Why? 3. In general, diamagnetic molecules are better reducing agents in their excited states than in their ground states. Explain, and support your answer using an orbital energy diagram. 4. Label each of the following processes (#1 to 10) in the Jablonski diagram below with the following letters : (A) “allowed” absorption (B) “forbidden” absorption (C) fluorescence

(D) phosphorescence (E) internal conversion/vibrational relaxation (F) intersystem crossing

S2 5

3

1

T2 S1

6

7

8

4

T1

2 9

10

S0

5. Explain why “λphos < λfluor” would be impossible. Use a Jablonski diagram to illustrate your answer. 6. What is Kasha’s Rule? Exceptions to this rule exist – search the literature for an example, and then illustrate the process using a Jablonski diagram. 7. A fluorescent object and a phosphorescent object are both lit with an appropriate UV lamp. How could you distinguish between the objects? Explain why your method works.

Practice Problem Set #1 – Photochemistry Fundamentals 1. Is there such a thing as a T0 state? Yes: paramagnetic molecules, such as oxygen, have triplet ground states.

2. What is the approximate bond dissociation energy of a C–Br bond? Using Planck’s equation, calculate the maximum wavelength of light capable of photolyzing a brominated organic molecule. However, when most bromides are irradiated at this wavelength, no decomposition is observed. Why? From the energy bar in the course notes, the C–Br BDE is about 68 kcal/mol.

28600 λ (in nm) 28600 28600 λ= = = 420 nm E 68

E(in kcal / mol) =

Therefore the wavelength of light containing enough energy to photolyze brominated compounds is 420 nm (in the visible region; specifically, it’s violet light). However, usually brominated compounds do not decompose readily when exposed to visible light. This is a consequence of the First Law of Photochemistry: only photons that are absorbed lead to excited states. No observed photodecomposition implies that most bromides are transparent at 420 nm (which is reasonable, since they are usually colourless).

3. In general, diamagnetic molecules are better reducing agents in their excited states than in their ground states. Explain, and support your answer using an orbital energy diagram.

By definition, a reducing agent is an electron donor. The ionization energy (IE) corresponds to the amount of energy that must be supplied to an atom or molecule to remove a single electron from the highest occupied orbital: X + IE → X+ + e– The smaller the value of IE, the easier it is to remove an electron – thus, the value of IE reflects the reducing power of a species.

For a diamagnetic molecule (with a singlet ground state), after the absorption of a photon, the electron is now higher in energy in the excited state, by a factor of ∆E = hν. Therefore, the ionization energy of the excited state is now decreased by exactly this amount: IEexcited state = IEground state – ∆E As a result, less energy is required to ionize the excited state, and consequently, the excited state can be considered to be a better reducing agent than the ground state. (Note that this explanation also works for oxidizing agents!).

4. Label each of the following processes (#1 to 10) in the Jablonski diagram below with the following letters : (A) “allowed” absorption (B) “forbidden” absorption (C) fluorescence

(D) phosphorescence (E) internal conversion/vibrational relaxation (F) intersystem crossing

S2

F

E1

5

3C

T2 S1

E6 4

C2

C

8A T1

D B9

S0

7

10 F

5. Explain why “λphos < λfluor” would be impossible. Use a Jablonski diagram to illustrate your answer. Phosphorescence emission must always occur at longer wavelengths than fluorescence emission (from the same molecule) because the emitted photon will always be of lower energy. In the intersystem crossing process to the triplet state, some energy is lost in order to flip the spin of the promoted electron. [Alternatively, we can explain this by invoking Hund’s rule: the parallel spins of the two electrons in the triplet state result in maximum multiplicity and therefore represent a lower energy configuration]. S1

T1 ∆E (S 1 –>S0) ∆E (T1 –>S0)

S0

Therefore, the T1 state lies at a lower energy than the S1 state. According to the Jablonski diagram, this means that ∆E(S1!S0) > ∆E(T1!S0), and since wavelength is inversely proportional to energy, the smaller energy gap for the T1!S0 transition, phosphorescence occurs at longer wavelengths.

6. What is Kasha’s Rule? Exceptions to this rule exist – search the literature for an example, and then illustrate the process using a Jablonski diagram. S2

Kasha’s Rule states that vibrational relaxation from upper singlet excited states to the first singlet excited state is faster than all other deactivation processes. As a consequence, all observed photochemistry occurs from S1. One famous exception to this rule is azulene: this polycyclic aromatic molecule is capable of exhibiting S1 fluorescence from the S2 state. (the reasons behind this exception are beyond the scope of this course, but if you are interested to learn more, then you should take Dr. Scaiano’s Photochemistry course!)

S0

7. A fluorescent object and a phosphorescent object are both lit with an appropriate UV lamp. How could you distinguish between the objects? Explain why your method works. Fluorescence is a spin-allowed process and is therefore FAST (usually lasting a few nanoseconds). Phosphorescence is a spin-forbidden process and is therefore SLOW (usually lasting anywhere from milliseconds to several minutes). Thus, in order to distinguish between the two objects, all we need to do is illuminate the objects for a brief time and then turn off the UV lamp: the fluorescent object stops fluorescing nearly instantly (meaning, light emission only occurs as long as the excitation source is present), but the phosphorescent object will continue to emit light for some time.

Practice Problem Set #2 – Absorption 1. A protein has a molar mass of 52542 g/mol. If the absorption of a 1.00% m/v aqueous solution of this protein at 280 nm is 7.22, what is the value of ε at this wavelength? 2. The absorption spectrum of a compound is mistakenly taken in a 1.5 cm path length cuvette instead of a 1.0 cm cuvette. If the measured absorption at a particular wavelength is 0.50, what absorption would have been obtained had the correct cuvette been used? 3. Two substances A and B are dissolved together in a solvent to form a homogenous solution. A has extinction coefficients εAλ1 and εAλ2 at wavelengths λ1 and λ2, respectively. B has extinction coefficients εBλ1 and εBλ1 at the same wavelengths. Using the absorption of the solution measured at these two wavelengths Abs(λ1) and Abs(λ2), derive two general equations for the concentrations of each substance. (e.g. CA = ? and CB = ?) 4. A compound C5H8O2 has a pKa of 4.8. In its UV-vis absorption spectra, this molecule has a λmax at 270 nm when dissolved in methanol, and a λmax at 290 nm when dissolved in hexane. Propose a reasonable structure for the molecule, and support your reasoning. Explain the observed shift in the spectra and give the name for this phenomenon. 5. Two aromatic compounds, A and B, are isolated from a reaction product mixture. Both molecules have a molecular formula of C9H10. In their UV-vis spectra, A has a λmax at 279 nm and B has a λmax at 300 nm. Can you propose two reasonable structures for A and B? 6. You’re studying the photochemistry of cyclohexanone. a) What would be the first singlet excited state of cyclohexanone? Give its abbreviation and draw an energy diagram for the relevant orbitals involved. b) When you irradiate a dilute solution of cyclohexanone (solvent = cyclohexane) by LFP, you observe the formation of the following radical products. What is the possible origin of these products?

7. The UV-vis absorption spectrum of the molecule at right is shown below. What electronic transition in the molecule is responsible for the absorption at right in the spectrum? Give its abbreviation and draw an energy diagram for the relevant orbitals involved.

O

N H

H N

O

A piece of white cloth is stained with a solution of this molecule, a known industrial dye. What colour would the dyed cloth appear to be? The cloth is then treated with a solution of bleach and the apparent colour disappears. What chemical change to the molecule is responsible for this loss of colour?

8. A biochemist is studying thionine (T), a ligand that forms a complex with trypsine (S). The figure below is the Job plot for the interaction of the two molecules. Calculate the value of the stoichiometric factor « n ». 0.7

S + nT → (STn) ∆Absorption

0.6 0.5 0.4 0.3 0.2 0

0.2 0.4 0.6 0.8 Mole fraction of thionine

1

Suggested Problems from Pavia, Chapter 7: #4, #5, and #6.

Practice Problem Set #2 – Absorption 1. A protein has a molar mass of 52542 g/mol. If the absorption of a 1.00% m/v aqueous solution of this protein at 280 nm is 7.22, what is the value of ε at this wavelength? mol 1000 mL 1g = 1.903 × 10 −4 mol / L × × 1L 100 mL 52542 g Abs 7.22 = 3.79 ×10 4 M −1cm−1 ε 280 = = −4 [protein]×  1.903 × 10 mol / L × 1 cm

[protein] =

2. The absorption spectrum of a compound is mistakenly taken in a 1.5 cm path length cuvette instead of a 1.0 cm cuvette. If the measured absorption at a particular wavelength is 0.50, what absorption would have been obtained had the correct cuvette been used?

Abs1 1 = Abs2  2 0.50 1.5 = Abs2 1.0 Abs2 = 0.33 3. Two substances A and B are dissolved together in a solvent to form a homogenous solution. A has extinction coefficients εAλ1 and εAλ2 at wavelengths λ1 and λ2, respectively. B has extinction coefficients εBλ1 and εBλ1 at the same wavelengths. Using the absorption of the solution measured at these two wavelengths Abs(λ1) and Abs(λ2), derive two general equations for the concentrations of each substance. (e.g. CA = ? and CB = ?)

Abs at λ1 = Abs1 = Abs1A + Abs2B = ε1ACA  + ε1BCB  Abs at λ2 = Abs2 = Abs2A + Abs2B A

= ε2 CA  + ε2BCB  These are two equations with two unknowns: solving them gives us CA and CB: (see added scanned sheets for extended solution)

CA =

Abs1εB2 − Abs2 ε1B (ε1Aε2B − ε2Aε1B )

CB =

Abs2ε1A − Abs1ε2A (ε1Aε2B −ε A2ε1B )

4. A compound C5H8O2 has a pKa of 4.8. In its UV-vis absorption spectra, this molecule has a λmax at 270 nm when dissolved in methanol, and a λmax at 290 nm when dissolved in hexane. Propose a reasonable structure for the molecule, and support your reasoning. Explain the observed shift in the spectra and give the name for this phenomenon. A pKa of 4.8 suggests that this is a carboxylic acid. A possible structure: O OH

The fact that the absorption peak undergoes a blueshift when added to a polar protic solvent suggests that hydrogen-bonding with the solvent lowers the energy of the ground state (S0) with respect to the excited state (S1). Consequently, the ∆ES energy gap increases, and the corresponding S0!S1 transition shifts to shorter wavelengths. This is an example of solvatochromism.

5. Two aromatic compounds, A and B, are isolated from a reaction product mixture. Both molecules have a molecular formula of C9H10. In their UV-vis spectra, A has a λmax at 279 nm and B has a λmax at 300 nm. Can you propose two reasonable structures for A and B? Aromatic component is likely a phenyl ring, –C6H5. Using the remaining atoms, we get two likely structures:

Looking at these structures, we see that the molecule on the left contains a carbon-carbon double bond that is conjugated with the aromatic ring. The molecule on the right shows no such conjugation (the two π systems are independent of one another). Since conjugation decreases the π!π* energy gap, the structure on the left is most likely B, the molecule with the longer wavelength absorption.

6. You’re studying the photochemistry of cyclohexanone. a) What would be the first singlet excited state of cyclohexanone? Give its abbreviation and draw an energy diagram for the relevant orbitals involved. Since this is a carbonyl compound, S1 would correspond to 1(n!π*):

π* n π

b) When you irradiate a dilute solution of cyclohexanone (solvent = cyclohexane) by LFP, you observe the formation of the following radical products. What is the possible origin of these products?

The 1(n! *) state has an unpaired electron on the oxygen atom, and thus behaves like an alkoxyl radical. It can therefore participate in typical radical reactions, such as atom abstractions. For example, abstracting a hydrogen atom from the solvent results in the observed products: 1

O

O

* behaves like:

H

O

H

O

H

H +

+ solvent

7. The UV-vis absorption spectrum of the molecule at right is shown below. What electronic transition in the molecule is responsible for the absorption at right in the spectrum? Give its abbreviation and draw an energy diagram for the relevant orbitals involved.

O

N H

H N

O

A piece of white cloth is stained with a solution of this molecule, a known industrial dye. What colour would the dyed cloth appear to be? The cloth is then treated with a solution of bleach and the apparent colour disappears. What chemical change to the molecule is responsible for this loss of colour?

V"""I"""""""""B""""""""""G""""""""Y""""O""""""""""R"

This is a carbonyl–containing compound, and thus there are two possible electronic transitions: n→π* and π→π*. The absorption spectrum shows these two major bands – the one in the visible region (centred around λ = 620 nm) is the lower energy transition, and is therefore n→π*: π∗

π∗

hν n

n

π

π S0

S1

This molecule is known as indigo, the industrial dye used for blue jeans. As we can see from the absorption spectrum, indigo absorbs strongly the longer visible wavelengths – red, orange, yellow, and green (see labels above) – and as a result, the apparent colour to our eyes is the complementary colour, a deep blue. We now realize what is responsible for “colour”: absorption in the visible region. Of course, these longer wavelengths mean that the HOMO-LUMO gap (i.e. the S0→S1 transition) must be small (the orbitals need to be close to one another). We saw in class that there is one effect principally responsible for decreasing the HOMO-LUMO gap: conjugation. In general, increasing the degree of conjugation in a molecule decreases the gap, leading to further and further red-shifted absorption spectra. Here are some more examples:

N N

Mg N N

O

β−CAROTENE (appears orange)

AcO O OR CHLOROPHYLL-a (appears green)

So, if colour arises from conjugation, then the lack of colour must be caused by the lack of conjugation. This is the effect of adding bleach. “Bleach” is an aqueous solution of NaOCl. The hyperchlorite ion, OCl–, as well as its conjugate acid, HOCl, are extremely good oxidizing agents (the chlorine atom is in an undesirable oxidation state, +1, and wants to be reduced). This is why we use these products to remove stains and in disinfection: they efficiently oxidize organic material. So, if we oxidize our starting material, indigo, we will correspondingly break the conjugated system and increase the HOMO-LUMO gap (notice that it is not necessary to specify exactly where the oxidation occurs, because any change in the π system of indigo will disrupt the conjugation). As a result of this widening, the wavelength associated with that absorption must get shorter – i.e. the S0→S1 transition undergoes a significant blue-shift, out of the visible, and into the UV. So, to our eyes, the bleached cloth will appear white again, as there is no absorption in the visible region (all wavelengths are reflected, and we perceive this as being “white”). Note the association between the three concepts here: molecular structure, absorption, and colour. Now, you can think about why certain objects appear colourful to our eyes, and what must be happening to molecular structures if the colour changes!

8. A biochemist is studying thionine (T), a ligand that forms a complex with trypsine (S). The figure below is the Job plot for the interaction of the two molecules. Calculate the value of the stoichiometric factor « n ». 0.7

S + nT → (STn)

∆Absorption

0.6 0.5

max = 0.75 =

0.4

n n +1

∴n = 3

0.3 0.2 0

0.2 0.4 0.6 0.8 Mole fraction of thionine

1

Suggested Problems from Pavia, 4th edition, Chapter 7: #1, #4, #5, and #6. Suggested Problems from Pavia, 5th edition, Chapter 10: #1, #4, #5, and #6.

Practice Problem Set #3 – Fluorescence 1. Briefly describe the differences and similarities between a spectrophotometer and a fluorimeter. 2. What is a quantum yield? Describe an experimental procedure that can be used to determine a fluorescence quantum yield. 3. The absorption and fluorescence spectra of pyrene are shown below. a) Which curve is the absorption and which is fluorescence? b) The fluorescence spectrum of pyrene was recorded using 355 nm as the excitation wavelength. What would be the overall qualitative effect on the spectrum if λex was decreased to 315 nm?

4. Derivatives of the molecule coumarin are commonly used as fluorophores.

a) Explain why coumarin would be a good fluorophore, using photochemical concepts (e.g. kF, ΦF etc.). b) We wish to determine the quantum yield of fluorescence of a derivative of coumarin, called C428, by comparing the intensity of its fluorescence with that of rhodamine (ΦF = 1.00). Solutions of each fluorophore are prepared and their fluorescence spectra are measured. In the fluorimeter, a ratio of 0.88 is calculated for the relative fluorescence intensities of C428 and rhodamine, respectively. What is the value of ΦF of C428? c) If the observed singlet excited state lifetime of C428 is 43 ns, what is the actual value of the fluorescence lifetime, τF? d) In the determination of ΦF, it is necessary to “optically match” the two solutions of C428 and rhodamine – meaning, the absorption of C428 at λex must be exactly equal to the absorption of rhodamine at λex. Can you explain why this step is necessary?

5. DAPI, shown below, is a popular DNA stain. H2N

N H

NH

NH2 NH

a) The main absorption band of DAPI is centred at 350 nm. Would it be possible to excite DAPI to S1 via absorption of two photons of λ = 700 nm? Why or why not? b) Does DAPI fluoresce when free in solution? Explain why or why not, using photochemical concepts (e.g. kF, ΦF etc.). c) How does the fluorescence change when DNA is added? Why? Explain, using photochemical concepts (e.g. kF, ΦF etc.). 6. An experimental fluorescence spectrum for anthracene was surprisingly missing the (0,0) band...