The Excited State of β-naphtol

Transcription

1 The Excited State of β-naphtol laboratory experiment in chemical physics

2 The Excited State of β-naphtol ntroduction n this laboratory we will use the absorption and fluorescence spectroscopy to study the proteolysis of β naphthol (a wee aromatic acid) in its ground and excited state. n the theoretical part of the lab we will derive an expression for the fluorescence intensity as a function of ph for the acid and base form of the b-naphthol using the reaction inetics of the proteolysis reaction. The ph dependence of the fluorescence will then be used to determine the pka value for the excited state. The theoretical and experimental principles give in this lab This lab instruction does not give comprehensive instruction on how to perform the lab; instead it gives the bacground for self study for both the theoretical and experimental part of the course. Theory central question in this lab is how the molecular properties changes after photoexcitation. Photoexcitation means that an electron in the molecule has moved into a higher orbital which is often an antibonding orbital. n electron in an antibonding orbital will change the bonding properties in the molecule such as bonds distances and angles, a proton for example might become wealy bounding in a molecule and it is therefore reasonable to assume that the acidbase equilibrium will be different in the excited state ac compared to the ground state. molecule in the excited state will also have a different electron distribution than the ground state and therefore we can assume the dipole moment will be different in both magnitude and direction in the excited state. This will influence both the absorption and fluorescence spectra. molecule in a solvent will be in a cage of solvent molecules, the solvent molecules will orient around the molecules to achieve the lowest possible energy for the system. What will happen to the molecule is excited? How fast can the solvent molecules reorient to around the new dipole of the excited state. Franc Condon principle tells us that it is reasonable to assume that the excitation of an electron is so fast, that nuclear changes have no time to happen. So will the solvent cage loo the same around the excited molecule immediately after absorption of light? Will the cage that surrounded the ground state to be optimal also for the excited state? f not, what can you expect to happen, and how fast will the change occur compared to the lifetime of the excited state? These factors will be important to consider when we loo at different methods for studying the acid-base equilibrium β-naphthol in both the ground and excited states.

3 n order to study the excited state properties of b-naphtol the sample will be excited by appropriate?+ wavelength. Under steady state condition, the following reaction scheme can be written. * [H + ] B * +H + B 2 B+H + K a,b, *, B *, 2 3, 4 K a, B re β-naphtol acid and base forms in the ground and excited state. re :a reaction rate for deactivation of * and B* by fluorescence, internal conversion and quenching.(/ τ fluorescence lifetime) are reaction rates for the protolysis equilibrium for the excited state s the acid constant for β-naphtol are excitation rate; x ωε X [X], where ω is a instrumental constant, ε X and [X] are the absorption coefficient and concentration of X.

4 ()-Determination of pka of the absorption spectra pka can be determined if we can measure the ratio [B] / [] at different ph. One way is to set a high and a low ph, determine the absorption coefficient for the base and acid form at two wavelengths and then measure the absorbance at intermediate ph values. Here we use a simpler method using a single wavelength. Under the assumption that the absorbance of the pure acid and base form ( 0 and 0 B) at this particular wavelength can be measured separately by setting the low and high ph, we can we write the ratio: ( ) B ( ) [ B] [ ] ε B ε [ ] [ ] B ε [ B] ε B ε B ε [] B + ε o ε B [] B o B tot o o B tot ( [ ]) B ε C tot [ ] ε B ( C tot [ B] ) () nd since tot + B and C tot [ ] [B ], and summing that the total content of beta-naphthol is a constant. [ ] and [B ] are the concentration of pure acid and pure base form. Observe that tot will change with ph, but and B are constants at a given wavelength. Question: Thin of an experiment to determine the pka and maes an estimate based on the absorption spectra (see attached figure). t what wavelength would you measure? How do you change the ph while eeping the β-naphthol concentration constant? () - Determination of pka* from the absorption spectra Under chemical equilibrium, the change in the Gibbs free energy under standard condition can be written as: ΔG - T ln where is the Boltzmann constant. This means that both K and K* can be expressed using the difference in (free-) energy between protolysed and non-protolysed form. nformation on the energy differences can be obtained from absorption spectra. The energy level scheme below can be used:

5 E hν and E B hν B, where ν is the absorption frequency corresponding to the transition between the lowest vibration levels in the ground and excited state (00- transition). ΔH > (ΔH )* is explain the red shift in the absorption spectra in the base form. Question: t is expected that ΔH and (ΔH ) * is positive? Motivate! You should now be able to write an expression of difference pka * - pka as a function of ν and ν B Hint: pply Hesss law on the energy diagram. Then use: ΔG ΔH -TΔS, ΔG - T lnk, lgkink/ln0 and the approximation ΔS (ΔS )*. Question: is this a reasonable approximation? What will be the largest contribution to the entropy of the reaction? One problem is that we do not now which wavelength in the absorption spectrum that correspond to the 0-0-transition. The spectral resolution is not sufficient to enable us to distinguish individual peas. Note that the maxima only indicate the most liely transition. good approximation is that λ 00 (λ abs,max + λ em,max ) / 2, i.e. the mean value for the of absorption and emissions maxima. Question: Explain with a simple figure why this approximation is reasonable. What factors affect the red shift between the emission and absorption spectra. Write an approximate expression of pka*, mae the change from frequency to wavelength. Use the given absorption spectra to provide an estimate of pka *. () ß-naphthol fluorescence We begin by formulating general expressions for how the fluorescence intensity from the excited acid and base form changes with changing ph. f we introduce expressions for how the concentration of * and B * changes as a function of time

6 as shown in the reaction scheme and assuming a steady sate condition, [ ] d * dt [ ] db* dt The concentration of excited state can be expressed as: 3 [ * ]+ [ * ] 4 [ B* ] 0 (2) 4 [ B* ]+ 2 [ B* ] 3 [ * ] B 0 (3) [ ] B + (4) Factor out from the numerator and from the denominator we obtain [ ] B (5) Multiply both numerator and denominator by 4 / 2 gives [ ] B + (6) f we use the fact that ω ε [], B ω ε B [B], ph, [B]/[]0 ph-pk a, and that []C tot -[B], where C tot is the total concentration of ß-naphthol, we obtain a very useful expression for the fluorescence of the acid form : D F * +0 ph pk a ph 2 + ε B 0 ph pk a ( ε ) ph 2 (7) D (P * ω ε C tot ) / is a constant if C tot is a constant. We have also used the fact that F * P * [Α ], where P * is the probability if fluorescence from the exited acid

7 form of the molecule. n expression for the base form fluorescence (F B * ) can be derived ia an approximately similar method F B* D B 0pH pk a +0 ph pk a ( ) ε 0 pk a ph εb ph 2 (8) where D B (P B * ω ε B C tot ) / 2. Question: perform this derivation. Hint: Multiply numerator and denominator by 3 and factor out from the numerator and / 2 the denominator) By examining the properties of expressions for F * and F B* as a functions of ph ranges (ph experimentally, a reasonable of ph is in the range 0 to 3), and at various excitation wavelengths, we can arrive at a method for determining pka and pka * respectively. (V) - Determination of pka of fluorescence spectra n this section we shall study in detail the functions F * f(ph) and F B * f(ph). First, we see that one can determine the pka also from the fluorescence measurements, i.e. the excited molecules can provide information on the characteristics of the ground state. Can we in the absorption spectrum find a wavelength where only the base absorbs? That is ε 0, ε B > 0?What happened to the expression for F B *? Tas: Determine the value of F B * when ph pka. Note that we already now how D B can be determined. t is necessary to estimate the value of 4 0 -ph / 2 : M - s -. f we assume that all collisions lead to reaction, i.e a diffusion controlled process The diffusion constant in water is of size 0 0 M - s s -, / 2 τ fluorescence lifetime. This can be easily determined for example with single photon counting. For a simple aromatic compounds τ is typically a few nanoseconds. We therefore have 4 0 -ph / pH << för ph>4. For a wea acid such as β-naphthol if is certainly that pka> 4. Suggest an experiment to determine the pka! Suggest an appropriate excitation wavelength and at which you would measure the emission. Outlining how F B *

8 should loo lie as a function of ph? (V) - Determination of pka * from fluorescence spectra t is a bit tricier to determine pka *. The easiest way is to use the expression for F * and select the appropriate values for ε, ε B. The method is based on the fact that we can show that at ph values higher than pka * but lower than the pka; the fluorescence intensity from both the acid base form are independent of ph, i.e it will form a plateau. Consider the expression for F *. f 5 <ph <pka -2, we have 0 ph-pk a << and 4 0 -ph / 2 <<. good approximation in this interval is therefore: F * D + 3 F * D + 3 R p (9) Rp is defined as the functions plateau value. We see that in principle the ratio 3 / can be determined. Tas: Show that the F * has a plateau even at very low ph values. 3 / can be estimated using the approximate value you wored out for pka*. Note: 2. Setch the form of the function, i.e. F * f(ph) for ph 0-3. f ph<<pk a we now that 0 ph-pk a <<.Then we can for reasonable values ε B /ε mae the simplification: ph F * D ph 2 (0) mmediately we see that for the ph value when 3 / 4 0 -ph / 2, the expression become nd we find the relationship + 3 F * D +2 R a () 3

9 R a 2 R p (2) From an experimental determination of the plateau value Rp, Ra can be calculated. From the experiential curve when the F * / D R a, we can obtain the ph value that satisfies the equality 3 / 4 0 -ph / 2 which is equivalent to 3 / 4 0 -ph / 2. Vi have already shown that K a * 3 / 4,, therefore K a * 0 -ph / 2. f the resulting ph is inserted into the expression pka* ph - lg ( / 2 ) can pk a * be determined if we now and 2. The fluorescence lifetimes are not so difficult to measure experimentally using for example time resolved single photon counting, for this lab however we can use the literature values: s - and s -. Tas: We can similarly use F B * f(ph) to determine the pka*. Show that in this case. R a R p / (+ R p ) if ε /ε B (is there such a wavelength?), Setch F B*! Experiment: t remains now to design experiments in which the pka and pka * can be determined. We have access to the absorption and fluorescens spectrophotometers and a ph meter! Mae a summary of the experiments to be performed, the wavelengths of excitation and emission to be used in the various experiments. Miscellaneous: - Beta-naphthol (2-hydroxy) is a potential carcinogen. Gloves are recommended. - Keep in mind that beta-naphthol is a light-sensitive substance. References: () - Laws W.R. ; Brand L. Journal of Physical Chemistry, Vol. 83, page 759th (2) - van Stam J.; Löfroth J.-E. Journal of Chemical Education, Vol. 63, page 8 (3) - Laowicz, Principles of Fluorescence Spectroscopy,

10 bsorption spectra of β-naphtol at different ph values: Curve (a), ph3.3; (b), ph0.0; (c), ph9.6; (d), ph9.0; (e), ph0.0

11 Fluorescence spectra of β-naphtol at different ph values Curve (a), ph3.6; (b), ph9.2; (c), ph8.3; (d), ph7.8; (e), ph2.8; (f), ph.0