Problem 1. Anthracene and a chiral derivative of anthracene

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Colin Stokes
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1 Molecular Photophysics 330 Physical rganic Chemistry 6C50 Thursday November 5 004, h This exam consists of four problems that have an equal weight in the final score Most problems are composed in such a way that it is not required know the answer of a previous (sub)problem to solve the next one. It is allowed to use the printouts of the course during the exam. Problem. Anthracene and a chiral derivative of anthracene Athracene (see below) has two allowed optical transitions in the UV region of the spectrum. The first transition ( 0 ) occurs around 400 nm (3 ev) and has a transition dipole moment along the x direction. The second transition ( 0 ) occurs near 50 nm (5 ev) and the transition dipole moment for this transition is along the y direction. x 5 ev y 3 ev 0 a. The molar absorption coefficient, which is related to the probability for absorption of a photon, for the 0 transition is more than ten times higher than for the 0 transition. From this experimental fact we can draw conclusions on the relative magnitude of the transition dipole moments µ for the two transitions if we assume that the Franck-Condon factors for both transitions are about the same. Explain whether µ ( 0 ) > µ ( 0 ) or that µ ( 0 ) < µ ( 0 ). Molecule I (see below) contains two anthracene units. In first approximation we may neglect the exchange of electrons between the two units and describe the interactions between the units in the excited state by transition dipole-transition dipole interactions.

2 I The rotation about the single bond joining the two anthracene units is hindered by the presence of the methyl groups and the molecule exists in two mirror image related forms (enantiomers) A and B. n the left side the absorption spectrum (notice the logarithmic scale) and circular dichroism (CD) spectrum of one of the enantiomeric forms is shown. A B Me Me Me Me b. Determine which of the forms has been measured (A or B) and explain your answer. c. In the CD spectrum shown above the CD of the 0 transition near 400 nm (3 ev) has not been measured. If it would have been measured, could this part of the CD spectrum also have been used to distinguish between enantiomer A and B? Explain your answer.

3 Problem. pins and states In this problem we consider the electronic configuration of the butadiene radical cation. There are four π orbitals. The ground state wave function is denoted by Ψ. π π π π π π π π Ψ Ψ The ground state wave function Ψ consists of the product of an orbital function and a spin function. The spin function of Ψ can be written as: σ = ( α() β () α(3) β () α() (3)) or abbreviated: σ ( αβα βαα ) α = For the radical cation an excited state Ψ exists which is illustrated above. The spin function of Ψ can be written as: σ = α() α() (3) or abbreviated: σ = ααα α We now consider optical excitation of the radical cation from its ground state Ψ to its excited state Ψ. The calculation for the transition dipole moments involves the overlap of the two spin functions: σ σ = ααα ( αβα βαα ) 3

4 a. Calculate the overlap of the spin functions and decide whether the transition is spin-allowed or spin-forbidden. b. To verify the result obtained under a. one can calculate the value of the total spin moment s (spin quantum number) for both states by calculating the eigenvalues of the operator. If the transition is allowed the spin quantum number s should be the same for both molecules. Calculate the value of s for σ and σ and confirm your result obtained under a. Use the relationships: = = = = z h z α = 0 α = h β β = h α β = 0 α α = β α = 0 α β = 0 β β = By applying the operator to σ we can generate a new spin state σ 3. By multiplying this spin state with an appropriate orbital function we have generated an additional excited state Ψ 3 c. Give an expression for this new spin state σ 3 and calculate whether the transition from the ground state Ψ to the excited state Ψ 3 is spin allowed. Problem 3. A cyclic diketone. We consider a cyclic diketone:, cyclopentanedione. This molecule has C v symmetry and is depicted in its standard orientation below: z σ v ' y For the spectroscopic properties the following orbitals play an essential role: 4

5 π* π* n n π π The character table of the C v point group is given below: a. Determine the symmetry labels of the π-orbitals π, π, π, π and the nonbonding orbitals n and n, which involve the free electron pair on the oxygen atom in a p y atomic orbital. These orbitals can be used to describe the lowest four singlet states of the diketone ( 0,,, 3 ) as illustrated below. how that the symmetry label Γ of each of the three states correspond to: Γ( 0 ) = A, Γ( ) = B, Γ( ) = A, and Γ( 3 ) = B. π n n π 0 3 5

6 b. Demonstrate that the transitions 0 (near 500 nm) and 0 3 (in the ultraviolet, ~50 nm) are dipole-allowed and determine their polarization directions. In the absorption spectrum of the cyclic diketone a very weak absorption band is observed near 350 nm, apparently due to a transition from the ground state to another state whose energy lies between that of the and 3 state. c. We now investigate whether this state could be identified as the state, which is mentioned above. Is the transition 0 allowed? If not, can molecular vibrations help to induce a small probability for the transition? If so, suggest a possible irreducible representation of a vibration, which may induce such probability. An alternative interpretation of the weak absorption band is that it arises from a transition between the singlet ground state and an unidentified triplet state at higher energy. d. Are such transitions allowed? If not, name a mechanism, which may induce some probability for such a transition. The three lowest triplet states of the diketone can be represented as follows: π n n π T T T 3 6

7 d. Based on their energy relative to the 0 ground state, one of the three triplet T, T and T 3 can be excluded in the alternative explanation of the weak transition at 350 nm involving a transitions between 0 and a higher lying triplet state. Which triplet state can be excluded? Explain your answer. Problem 4. Perylenebisimid and oligo(phenylene vinylene) in organic solar cells Perylenebisimid (PERY, see below) is a molecule that shows a very bright and intense fluorescence in CH Cl solution with a maximum in intensity near a wavelength of 530 nm. R R R R HN NH R R R R PERY PV4 A drop of the solution is put on a quartz plate and the solvent is allowed to evaporate. As dry material on the quartz plate the fluorescence of the perylenebisimid molecules is strongly reduced in intensity, and very weak in comparison with that from the solution. To explain this it may be assumed that perylenebisimid molecules form linear -dimensional aggregates upon removing the solvent. a. uggest how the perylenebisimid molecules should be arranged in the linear aggregate in order to explain the almost complete quenching of luminescence. When perylenebisimid is mixed with the oligo(phenylene vinylene) (PV4) interesting photophysical properties can be observed such as energy and electron transfer. Many of the observations can be explained by considering the energies of the Highest ccupied Molecular rbital (HM) and the Lowest Unoccupied Molecular rbital (LUM) of both PERY and PV4. 7

8 LUM.5 ev.3 ev HM PERY PV4 In the ground state (before photoexcitation) the HM of PERY and PV4 are both occupied by two electrons. b. Is electron transfer possible between ground state PERY and ground state PV4? Explain your answer drawing a HM-LUM diagram showing the occupation of orbitals. The excited state of PERY (denoted below as PERY*) may be described by promotion of one electron from the HM to the LUM. Likewise the state of the PV4 (denoted below as PV4*) may also be described by a HM-LUM excitation. c. Explain whether the following processes are likely to occur after excitation of PERY to its state, given the fact the PERY and PV4 are at very close distance:. energy transfer from PERY* to the PV4 yielding PV4- (PV4*)?. energy transfer from PERY* to PV4 yielding PV4-T? 3. electron transfer from PERY* to PV4 yielding PV4? 4. electron transfer from PV4 to PERY* yielding PV4? Explain your reasoning using a diagram. In formulating your answer also take into account spin selection rules. d. Using excitation light of 400 nm the PV4 molecule can be brought to its excited state (PV4*). Discuss whether the following processes are likely to occur:. energy transfer from PV4* to PERY yielding PERY- (PERY*)?. intersystem crossing from PV4* to PV4-T and energy transfer to, yielding singlet oxygen ( )? 8

9 3. electron transfer from PERY to PV4* yielding PV4? 4. electron transfer from PV4* to PERY yielding PV4? Explain your reasoning using a diagram. In formulating your answer also take into account spin selection rules. In the following you may assume that after excitation of PERY to its state, electron transfer from PV4 to PERY occurs. We now want to use the photoinduced electron transfer reaction between PERY and PV4 to construct a solar cell. To do so we use two different metal electrodes IT (indium-tin-oxide) and Al. For a metal the highest occupied electron level and the lowest unoccupied electron level have practically the same energy (also called the Fermi level). For the four materials the levels are summarized in the scheme below. LUM Al PERY PV4 IT Fermi level HM Fermi level Al PERY PV4 IT Constructing a solar, we start with the IT electrode at the bottom and deposit a layer of PV4 molecules on the IT. n top of the PV4 layer we deposit a layer of PERY molecules. Finally on top of the PERY layer, aluminum metal can be evaporated. e. Upon illumination of the device through the optically transparent IT electrode, a potential difference (voltage) between the two electrodes is formed, due to accumulation of charges. Explain at which electrode (Al or IT) the photogenerated electrons with negative charge and at which electrode the positively charged holes are collected. 9

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