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Electronic Spectroscopy Ultraviolet and visible. Where in the spectrum are these transitions?

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Presentation on theme: "Electronic Spectroscopy Ultraviolet and visible. Where in the spectrum are these transitions?"— Presentation transcript:

1 Electronic Spectroscopy Ultraviolet and visible

2 Where in the spectrum are these transitions?

3 Why should we learn this stuff? After all, nobody solves structures with UV any longer! Many organic molecules have chromophores that absorb UV UV absorbance is about 1000 x easier to detect per mole than NMR Still used in following reactions where the chromophore changes. Useful because timescale is so fast, and sensitivity so high. Kinetics, esp. in biochemistry, enzymology. Most quantitative Analytical chemistry in organic chemistry is conducted using HPLC with UV detectors One wavelength may not be the best for all compound in a mixture. Affects quantitative interpretation of HPLC peak heights

4 Uses for UV, continued Knowing UV can help you know when to be skeptical of quant results. Need to calibrate response factors Assessing purity of a major peak in HPLC is improved by “diode array” data, taking UV spectra at time points across a peak. Any differences could suggest a unresolved component. “Peak Homogeneity” is key for purity analysis. Sensitivity makes HPLC sensitive e.g. validation of cleaning procedure for a production vessel But you would need to know what compounds could and could not be detected by UV detector! (Structure!!!) One of the best ways for identifying the presence of acidic or basic groups, due to big shifts in for a chromophore containing a phenol, carboxylic acid, etc. “bathochromic” shift “hypsochromic” shift

5 The UV Absorption process    * and    * transitions: high-energy, accessible in vacuum UV ( max <150 nm). Not usually observed in molecular UV-Vis. n   * and    * transitions: non-bonding electrons (lone pairs), wavelength ( max ) in the nm region. n   * and    * transitions: most common transitions observed in organic molecular UV-Vis, observed in compounds with lone pairs and multiple bonds with max = nm. Any of these require that incoming photons match in energy the gap corrresponding to a transition from ground to excited state. Energies correspond to a 1-photon of 300 nm light are ca. 95 kcal/mol

6 What are the nature of these absorptions? Example:    * transitions responsible for ethylene UV absorption at ~170 nm calculated with ZINDO semi-empirical excited-states methods (Gaussian 03W): HOMO  u bonding molecular orbital LUMO  g antibonding molecular orbital h 170nm photon Example for a simple enone π π n π π n π* π π n  -  *; max =218  =11,000 n-  *; max =320  =100

7 How Do UV spectrometers work? Two photomultiplier inputs, differential voltage drives amplifier. Matched quartz cuvettes Sample in solution at ca M. System protects PM tube from stray light D2 lamp-UV Tungsten lamp-Vis Double Beam makes it a difference technique Rotates, to achieve scan

8 Diode Array Detectors Diode array alternative puts grating, array of photosens. Semiconductors after the light goes through the sample. Advantage, speed, sensitivity, The Multiplex advantage Disadvantage, resolution is 1 nm, vs 0.1 nm for normal UV Model from Agilent literature. Imagine replacing “cell” with a microflow cell for HPLC!

9 Experimental details What compounds show UV spectra? Generally think of any unsaturated compounds as good candidates. Conjugated double bonds are strong absorbers Just heteroatoms are not enough but C=O are reliable Most compounds have “end absorbance” at lower frequency. Unfortunately solvent cutoffs preclude observation. You will find molar absorbtivities  in Lcm/mol, tabulated. Transition metal complexes, inorganics Solvent must be UV grade (great sensitivity to impurities with double bonds) The NIST databases have UV spectra for many compounds

10 An Electronic Spectrum Absorbance Wavelength,, generally in nanometers (nm) UV Visible max with certain extinction  Make solution of concentration low enough that A≤ 1 (Ensures Linear Beer’s law behavior) Even though a dual beam goes through a solvent blank, choose solvents that are UV transparent. Can extract the  value if conc. (M) and b (cm) are known UV bands are much broader than the photonic transition event. This is because vibration levels are superimposed on UV.

11 Solvents for UV (showing high energy cutoffs) Water205 CH 3 C  N210 C 6 H Ether210 EtOH210 Hexane210 MeOH210 Dioxane220 THF220 CH 2 Cl CHCl CCl benzene280 Acetone300 Various buffers for HPLC, check before using.

12 Organic compounds (many of them) have UV spectra From Skoog and West et al. Ch 14 One thing is clear Uvs can be very non-specific Its hard to interpret except at a cursory level, and to say that the spectrum is consistent with the structure Each band can be a superposition of many transitions Generally we don’t assign the particular transitions.

13 An Example--Pulegone Frequently plotted as log of molar extinction  So at 240 nm, pulegone has a molar extinction of 7.24 x 10 3 Antilog of 3.86

14 Can we calculate UVs? Semi-empirical (MOPAC) at AM1, then ZINDO for config. interaction level 14 Bandwidth set to 3200 cm -1

15 The orbitals involved Showing atoms whose MO’s contribute most to the bands

16 The Quantitative Picture Transmittance: T = P/P 0 B(path through sample) P 0 (power in) P (power out) Absorbance: A = -log 10 T = log 10 P 0 /P The Beer-Lambert Law (a.k.a. Beer’s Law): A =  bc Where the absorbance A has no units, since A = log 10 P 0 / P  is the molar absorbtivity with units of L mol -1 cm -1 b is the path length of the sample in cm c is the concentration of the compound in solution, expressed in mol L -1 (or M, molarity)

17 Beer-Lambert Law Linear absorbance with increased concentration--directly proportional Makes UV useful for quantitative analysis and in HPLC detectors Above a certain concentration the linearity curves down, loses direct proportionality--Due to molecular associations at higher concentrations. Must demonstrate linearity in validating response in an analytical procedure.

18 Polyenes, and Unsaturated Carbonyl groups; an Empirical triumph R.B. Woodward, L.F. Fieser and others Predict max for π  * in extended conjugation systems to within ca. 2-3 nm. Homoannular, base 253 nm heteroannular, base 214 nm Acyclic, base 217 nm Attached groupincrement, nm Extend conjugation+30 Addn exocyclic DB+5 Alkyl+5 O-Acyl 0 S-alkyl+30 O-alkyl+6 NR2+60 Cl, Br+5

19 Similar for Enones  x    X=H 207 X=R 215 X=OH 193 X=OR Base Values, add these increments… Extnd C=C+30 Add exocyclic C=C +5 Homoannular diene+39 alkyl OH OAcyl+6 O-alkyl NR 2 S-alkyl Cl/Br+15/+25+12/+30    With solvent correction of….. Water +8 EtOH 0 CHCl 3 -1 Dioxane -5 Et2O -7 Hydrcrbn -11

20 Some Worked Examples Base value217 2 x alkyl subst. 10 exo DB 5 total232 Obs.237 Base value214 3 x alkyl subst. 30 exo DB 5 total234 Obs.235 Base value215 2 ß alkyl subst. 24 total239 Obs.237

21 Distinguish Isomers! Base value214 4 x alkyl subst. 20 exo DB 5 total239 Obs.238 Base value253 4 x alkyl subst. 20 total273 Obs.273

22 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

23 Interpretation of UV-Visible Spectra Transition metal complexes; d, f electrons. Lanthanide complexes – sharp lines caused by “screening” of the f electrons by other orbitals One advantage of this is the use of holmium oxide filters (sharp lines) for wavelength calibration of UV spectrometers. See Shriver et al. Inorganic Chemistry, 2 nd Ed. Ch. 14

24 Benzenoid aromatics From Crewes, Rodriguez, Jaspars, Organic Structure Analysis UV of Benzene in heptane Group K band (  ) B band (  ) R band Alkyl208(7800)260(220)-- -OH211(6200)270(1450) -O - 236(9400)287(2600) -OCH 3 217(6400)269(1500) NH 2 230(8600)280(1400) -F204(6200)254(900) -Cl210(7500)257(170) -Br210(7500)257(170) -I207(7000)258/285(610/180) -NH (7500)254(160) -C=CH 2 248(15000)282(740) -C  CH 248(17000)278(6500 -C 6 H 6 250(14000) -C(=O)H242(14000)280(1400)328(55) -C(=O)R238(13000)276(800)320(40) -CO 2 H226(9800)272(850) -CO (8700)268(800) -C  N 224(13000)271(1000) -NO 2 252(10000)280(1000)330(140)

25 Substituent effects don’t really add up Can’t tell any thing about substitution geometry Exception to this is when adjacent substituents can interact, e.g hydrogen bonding. E.g the secondary benzene band at 254 shifts to 303 in salicylic acid In p-hydroxybenzoic acid, it is at the phenol or benzoic acid frequency

26 Heterocycles Nitrogen heterocycles are pretty similar to the benzenoid anaologs that are isoelectronic. Can study protonation, complex formation (charge transfer bands)

27 Quantitative analysis Great for non- aqueous titrations Example here gives detn of endpoint for bromcresol green Binding studies Form I to form II Isosbestic points Single clear point, can exclude intermediate state, exclude light scattering and Beer’s law applies Binding of a lanthanide complex to an oligonucleotide

28 More Complex Electronic Processes Fluorescence: absorption of radiation to an excited state, followed by emission of radiation to a lower state of the same multiplicity Phosphorescence: absorption of radiation to an excited state, followed by emission of radiation to a lower state of different multiplicity Singlet state: spins are paired, no net angular momentum (and no net magnetic field) Triplet state: spins are unpaired, net angular momentum (and net magnetic field)


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