IntroductionTheoryAbsorption and fluorescence spectroscopyExperimentalData analysisError analysisWritten laboratory reportReferencesReferencesExperiment 6a: Excited-state properties of 2-naphthol (the acidity constants)(Dated: October 29, 2009)I. INTRODUCTIONThe electronic structure of a molecule determines such physical and chemical properties as its charge distribution,geometry (and therefore the dipole moment), ionization potential, electron affinity, and of course, chemical reactivity.If the electronic structure of a molecule were to be changed, one would expect its physical and chemical propertiesto be altered. Such a rearrangement can occur when a molecule is raised to an electronically excited state via theabsorption of a quantum of light (e.g. a photon) whose energy matches the energy gap between the ground andexcited states.For most organic molecules that contain an even number of electrons, the ground state is characterized by having allelectron spins paired; the net spin angular momentum is zero, and such an arrangement is called a singlet state. Whenconsidered in terms of molecular orbitals (MOs), electronic excitation involves the promotion of an electron from afilled MO to a higher, unoccupied MO. This new electronic configuration, which characterizes the electronically excitedstate, may be one in which the two electrons reside in the singly occupied MO’s with opposite spins. Accordingly,this electronically excited state is also a singlet. The ground, and lowest electronically excited, singlet states are oftendenoted as S0and S1, respectively. Higher excited singlet states are referred to as S2, S3, ..., Sn.Although measurements of the physical and chemical properties of a molecule in its ground state can be carriedout, more or less, at leisure (assuming that the molecule is thermally stable), the examination of these propertiesin its excited states is severely hampered by the fact that these states are very short-lived. For most molecules, S1states have lifetimes ranging from 10−6– 10−11s (e.g. µs to some ps). Excited states are metastable; they undergodecay processes that dissipate the energy they possess relative to more stable products. For example, the excited stateof a molecule may, in general: spontaneously return (spontaneous emission; lifetime 10−6to 10−9s) to the groundstate via photon emission (fluorescence), convert electronic excitation into vibrational energy (eventually generatingheat), undergo bond dissociation or rearrangement (“fast”) or a change in electron spin multiplicity (“slow”). Becausespontaneous emission from an excited state (i.e., fluorescence) often takes place very rapidly, fluorescence can be usedas a probe, or measurement, of excited state concentration (e.g., fluorescence assay). In addition, fluorescence studiescan provide information about the physical and chemical properties of these short-lived singlet states. This field ofexperimentation is called photochemistry or photophysics depending on what is being studied. In this experiment, theground and excited state acidity constants of 2-naphthol (ArOH; see Fig. 1) will be determined. For more informationon optical spectroscopy, see Refs. [1–3].FIG. 1: Structure of 2-naphthol (ArOH). Also called β-naphthol.II. THEORYIn aqueous solution, ArOH behaves as a weak acid, forming the hydronium ion and its conjugate base, the naphthoxyanion, ArO−:ArOH + H2O ⇋ H3O++ ArO−(1)It is instructive to measure the acidity constant of ArOH in its lowest excited electronic state, denoted as K∗a, and tocompare this value with that of the ground state Ka. This information indicates how the change in electron structurealters the charge density at the oxygen atom. The experimental method is best introduced in terms of the energy-leveldiagram shown in Fig. 2.Typeset by REVTEXFIG. 2: A schematic diagram of the ground (S0) and first excited singlet state (S1) energies of free naphthol and its conjugatebase, the naphthoxy anion, in aqueous solution. For transitions shown, the up arrow indicates absorption and down arrowfluorescence.The relative energies of the free acid and its conjugate base (e.g. the naphthoxy anion) are indicated for both theelectronic ground (S0) and lowest excited (S1) states in aqueous solution (see Fig. 2). Each anion is elevated withrespect to its free acid by an energy, ∆H and ∆H∗, respectively. These are the enthalpies of deprotonation. Boththe ground state acid and its conjugate base can be transformed to their respective excited states via the absorptionof photons of energy hvArOHand hvArO−. For simplicity, these absorptive transitions in Fig. 2 are shown to be equalto the fluorescence from the excited to the ground states of the acid and conjugate base. Note that the ground andexcited state vibrational levels that are involved in the transitions are not indicated.The free energy of deprotonation of ArOH can be expressed in terms of the enthalpy and entropy of deprotonation,and the equilibrium (ionization) constant:∆G = ∆H − T∆S = −RT ln(Ka) (for the ground state) (2)∆G∗= ∆H∗− T ∆S∗= −RT ln(K∗a) (for the excited state)for the S0and S1states, respectively. If we make the assumption that the entropies of dissociation of ArOH andArOH∗are equal, it follows that∆G − ∆G∗≈ ∆H − ∆H∗= −RT lnµKaK∗a¶(3)and thus from Fig. 2, using a Hess’ law approach (i.e., by considering a closed loop consisting of the absorption,deprotonation in excited state, emission, and protonation in the ground state), it can be deduced thatNAhνArOH+ ∆H∗− NAhνArO−− ∆H = 0 (4)⇒ ∆H − ∆H∗= NA(hνArOH− hνArO−)where h is Planck’s constant and ν’s are the transition frequencies (Hz). Avogadro’s number, NA, has been includedto put each energy term on a molar basis. Combining Eqs. (3) and (4) and rearranging giveslnµKaK∗a¶=hNA(νArO−− νArOH)RT(5)Thus knowledge of the energy gap between the ground and first excited states for both the free acid and its conjugatebase leads to an estimate for K∗a, if Kais known. The analysis presented above, which accounts for the observedthermodynamic and spectroscopic energy differences, was first developed by F¨orster [4]. This approach is thus oftenreferred to as a F¨orster cycle. Eq. (5) can be recast into the following form:2lnµKaK∗a¶=log³KaK∗a´log(e)≈ 2.303 logµKaK∗a¶= 2.303 (log(Ka) − log(K∗a)) (6)Recall that pKa= − log(Ka)