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Physical Chemistry Excited State Protonation: Equilibrium Objective: Determine the pK value of naphtol in its lowest excited electronic state Expected knowledge: Electronic states of aromatic molecules, vibronic structure of absorption and emission spectra, Förster cycle Literature: Atkins, de Paula, Physical Chemistry: Chapter 17 Background The electronic structure of a molecule determines its charge distribution, geometry, ionization potential, electron affinity and chemical reactivity. If the electronic structure of a molecule is changed (e.g. by the absorption or emission of a photon), we would accept such properties to be alter. For most organic molecules that contain an even number of electrons, the ground state is generally a singlet state (all electrons are paired), if it is not degenerate. The distributions of electrons in a given eigenstate of a molecule are generally described by molecular orbitals (MOs), ψ, which can be written as linear combination of atomic orbitals φ: (1) where the indices i,j indicate the numbering of the N atomic and molecular orbitals. Electron excitation due to the absorption of a photon brings about a change of the electron configuration. This change can generally be described as a transition from an occupied to an empty MO, but this terminology is misleading. Dipole allowed transitions do not change the spin state of the molecule, hence the only allowed transitions occur between different singlet states which are termed S0, S1, S2,...,Sn. Measuring properties of a molecule in its excited states can be challenging owing to the very short life time of such states. S1-states created by HOMO->LUMO transitions exhibit lifetimes in the micro- and nanosecond regime. Relaxation processes can involve photon emission or radiationless transitions which are caused by interactions between a molecule and its environment. If photon emission is involved the process is called luminescence. The respective transitions are illustrated by the Jablonski diagram in Figure 1. The process involves electronic as well as vibrational states of the molecule. Photon absorption transfers the molecule into vibrational excited states of an electronic singlet state, the transition properties are dictated by the displacement of the excited state along the vibrational coordinates of the molecule (FranckCondon principle). The relaxation process can involve three or even four steps. First the ψ j = cji φi i=1 N ∑ molecule relaxes into the vibrational ground state of the excited singlet state. Here the systems encounters two alternatives. The more likely one is the emission of a photon which transfers the system into excited vibrational states of the ground state, from where it relaxes into the vibrational ground state by transferring its vibrational energy to the surroundings . This process is called fluorescence. Alternatively, a process called intersystem crossing switches the system into an excited triplet state which exhibits a much longer lifetime than the excited singlet state. From there the system can return to the ground state due to photon emission. This process is called phosphorescence. Figure 1: Jablonski diagram illustrating the two different emission processes which together constitute luminescence (taken from http://www.photobiology.info/Visser-Rolinski.html). Figure 2: The absorption and emission spectrum of quinine illustrates the Stokes shift between absorption and emission (taken from absorption spectrum and fluorescence spectrum). Figure 2 shows the so called Stokes shift between the absorption and emission spectrum of another aromatic molecule, e.g. quinine. The energy difference is a measure of the combined vibrational relaxation energy produced in the excited state and the ground state. Figure 3: Structure of 2-naphtol (http:2-naphthol). The molecule to be investigated in the experiment is 2-naphtol (ArOH), which is shown in Figure 3. In water it is a weak acid, forming the hydronium ion and its conjugate base, the naphthoxy ion, ArO- . (2) In this experiment, the pK value of this reaction will be measured in the excited electronic state and compared with that in the ground state. This information indicates how the change in electronic structure alters the charge density at the oxygen atom. The deprotonation process increases the Helmholtz free energy of the ground as well as of the excited state. However, if the pK-values of the two states are different and if the entropy change associated with the deprotonation/protonation is state independent, the internal energies of the latter must be different. Energy conservation dictates a corresponding change of the zero-point transition energy of the absorption and a concomitant shift of fluorescence emission. The thermodynamic cycle is visualized in Figure 4. Figure 4: Illustration of how different internal energy ΔU and ΔU* contribution to the naphthoxy ion change the zero-point transition energy for the S0->S1 transition (taken from https://www.mtholyoke.edu/courses/magomez/ ChemicalThermodynamics/Labs/2napthol.pdf) ArOH + H2O← ⎯⎯ K ⎯⎯→ArO− + H3O− The relationship between deprotonation induced frequency shifts of the absorption and emission spectrum and the differences between the pK-values in ground and excited state can be derived as follows. Assuming constant pressure and no measureable change of the sample volume, the protonation processes are associated with the following change of the Helmholtz energy: (3) Thus (4) From the Förster cycle in Figure 4 we can deduce the following relationship: (5) where NA denotes the Avogardo number, h is Planck’s constant and νArOH, νArO- are the frequencies of the zero-point transitions for the indicated protonation states. By combining eqs. (4) and (5), one obtains: (6) where c is the speed of light. Note that the frequencies have been converted into wavenumbers. Eq. (6) can be employed in three different ways. The most secure method would be to measure the frequencies (wavenumbers) of the zero-point transitions, to which this equation is referring. Unfortunately, this can be done only somewhat indirectly. The simplest method is to obtain Δν from the shift of the maximum of the absorption spectrum. This method does not need any instrument correction, but it is based on the assumption that the deprotonation/protonation does not change the vibrational frequencies of the molecule. Alternatively, one can measure the corresponding shift of the fluorescence spectrum. Here, it might be necessary to correct the spectrum for the instrument response which can distort shape and maximum position of the fluorescence spectrum. The shift of both, the absorption and the fluorescence spectrum upon protonation/deprotonation will be utilized in this experiment. Procedure 1. Measure the absorption spectra of two solutions of ArOH (appr. 2·10-4M) at low and high pH, at which one of the protonation state should be predominantly populated. Use concentrated stock solutions of HCl and NaOH to produce hydronium and hydroxy ion concentrations of 0.02 M. Ensure that the analyte concentration is the same at both pH. Label and save these solutions. Record their absorption spectra. ΔΔU = ΔUS0 − ΔUS1 = −RT ln Ka,S1 Ka,S0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Δpk = pKS1 − pKS2 = NAhc 2.303RT ν! ArO− −ν!ArOH ⎡ ⎣ ⎤ ⎦ 2. Measure absorption spectra of ArOH (same concentration as used for 1) at intermediate pH. Adjust pH values by using ammonium chloride buffer solutions, e.g. NH4OH/NH4Cl = 0.1M/ 0.1 M, 0.1/0.2 M, 0.2/0.1 M. Prepare these solutions from 1.0 M stock solutions of NH4OH and NH4Cl. Obtain at least three spectra at different pH, which should be measured before and after the measurements. The temperature has to be kept constant. Overlapping all your spectra should yield an isosbestic point. If this doesn’t happen, something went wrong with the experiment. 3. Measure the fluorescence spectra for the two protonation states. Data Analysis Since the total concentration of the analyte is kept constant, we can write: (7) Assuming that both species absorb at λmax(ArOH), Beer-Lambert’s law implies: (8) where ε denotes the molar absorptivity of the indicated species. They can be obtained from the spectra recorded in step 1, assuming that the path length of the employed cuvette is 1 cm. Eq. (8) can then be used to obtain [ArOH] and, by means of eq. [7], [ArO- ]. Produce a Henderson-Hasselbach plot to obtain the pKs0 value of the molecule (you learnt how to do this in an earlier experiment). Assume that the activity coefficient is 1. Finally, in the last step use the λmax-values of absorption and fluorescence spectra to obtain the pK value in the excited state by means of eq. (6). You will find this to be non-trivial for the absorption spectrum, since the protonation is changing the overall band structure. You will have to use all the λmax values observed in the region between 310 and 330 nm and calculate the respective pK and ΔΔU. In the Discussion section, you will compare the pK values thus obtained from the absorption spectra with the corresponding value obtained from the fluorescence spectra as well as with literature values. Based on these comparisons you have to identify the correct λmax and thus also the correct ΔΔU. In your error analysis section, perform a detailed error analysis which fully accounts for all error bars and the statistical errors of ΔpK and the respective ΔΔU. In the discussion section, these errors should be discussed. Discuss the reason for discrepancies between Δλmax-values obtained from fluorescence and absorption spectra. Discuss possible reasons for the pK-shift. Why are the λmax values obtained from the absorption spectra not assignable to the zero-point energy of the S0->S1 transition? What justifies the assumption that the protonation does not affect the vibrational states of the molecule? What is a possible reason for the difference between the absorption band structures of the deprotonated and protonated state? Finally, calculate the heat in kJ/mol that the non-radiative part of the S1->S0 transition transfers to the solvent.

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