1. Applications of Spectrophotometry (Chapter 19)

3. Red shift of max with increasing conjugation CH 2 =CHCH 2 CH 2 CH=CH 2 max =185 nm CH 2 =CHCH=CH 2 max =217 nm vs. Red shift of max with # of rings Benzene max =204 nm Naphthalene max =286 nm

4. Polar solvents more likely to shift absorption maxima Shifts of max with solvent polarity n  * hypsochromic/blue shift  * bathochromic/red shift

5. heptane methanol Hypsochromic shift

6. Generally, extending conjugation leads to red shift “particle in a box” QM theory; bigger box Substituents attached to a chromophore that cause a red shift are called “auxochromes” Strain has an effect… max 253 239 256 248

7. Generally, extending conjugation leads to red shift “particle in a box” QM theory; bigger box

8. Determination of Concentration – Multicomponent System C B 1 2 Absorbance C+B Absorbance Wavelength (nm) 2 1

9. Determination of Concentration – Multicomponent System C B 1 2 Absorbance Wavelength (nm) (1) Measure  B and  C at 1 and 2 (pure substances)

10. Determination of Concentration – Multicomponent System C+B Absorbance Wavelength (nm) 2 1 (2) Measure A  at 1 and 2 (mixture)

11. Determination of Concentration – Multicomponent System C+B Absorbance Wavelength (nm) 2 1 A 1 = x  B 1 + y  C 1 A 2 = x  B 2 + y  C 2 x = molarity B y = molarity C

12. Isobestic Points Brocresol green HIn (acid form)In - (base form) When one absorbing species is converted to another, it is apparent in the absorption spectrum.

13. Isobestic Points [OH - ] pH [H + ] pH I see an isobestic point!!!

14. Isobestic Points The Total concentration of Bromocresol Green is constant throughout the reaction HIn (acid form)In - (base form) [Hin] + [In - ] = F bromocresol green A = b (  x C x +  y C y ) C x + C y = C But at the isosbestic point both molar absorptivities are the same!  x +  y = 

15. Isobestic Points HIn (acid form)In - (base form)  x +  y =  Therefore, the absorbance does not depend on the extent of reaction (i.e. on the particular concentrations of x and y) A = b (  x C x +  y C y ) = b  C x + C y ) = b  C An isobestic point is good evidence that only two principal species are present in a reaction.

16. Isobestic Points - Application Oximetry

17. HbHbO 2 O2O2 deoxyhemoglobinOxyhemoglobin 100 % O 2 0 % O 2

18. Measuring the equilibrium constant (The Scatchard Plot) Biochemistry example: The cellular action of a hormone begins when the hormone (L) Binds to it’s receptor protein (R) in a tight and specific way. The binding thus gives rise to conformational changes which change the biological activity of the receptor (an enzyme, an enzyme regulator, an ion channel, or a regulator of gene expression. R + LRL The binding depends of the concentration of the concentration of the components. K a is the association constant and K d is the dissociation constant

19. Measuring the equilibrium constant (The Scatchard Plot) When binding has reached equilibrium, the total number of binding sites, B max = [R] + [RL] whereas, the number of unbound sites would be [R] = B max – [RL]. The equilibrium constant would then be This expression can then be rearranged to find the ratio of bound to unbound (free) hormone

20. Measuring the equilibrium constant (The Scatchard Plot)

21. The method of continuous variation (for isomolar solution) (Job’s method) A + n BAB n Call this C Absorbance Mole fraction (  = n b / n a + n b ) 0 1 Allows for the determination of the stoichiometry of the Predominant product.

22. The method of continuous variation (for isomolar solution) (Job’s method) Requirements: The system must conform to Beer's law. Only single equilibrium Equimolar solutions M A = M B Constant volume 1 = V T = V A + V B K reasonably greater than 1 pH and ionic strength must be maintained constant

23. The method of continuous variation (for isomolar solution) (Job’s method) A + n BC Start Change Equil. M(1-x)Mx0 -C-nC+C M(1-x)-CMx-nCC M(1-x) is a concentration

24. Corrected Absorbance Mole fraction (  = n b / n a ) 01 A3XA3XAXAX 2 0.50.250.66