1. Electronic Spectroscopy Ultraviolet and visible

2. Where in the spectrum are these transitions?

3. Why should we learn this stuff
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.
“hypsochromic” shift 
“bathochromic” 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 for a simple enone π n π*
-*; max=218
=11,000
n-*; max=320
=100
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):
h 170nm photon
LUMO g antibonding molecular orbital
HOMO u bonding molecular orbital

7. How Do UV spectrometers work?
Rotates, to achieve scan
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
Two photomultiplier inputs, differential voltage drives amplifier.

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 L•cm/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
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.
Absorbance
Wavelength, , generally in nanometers (nm)
0.0
400
800
1.0
200
maxwith certain extinction  UV
Visible

11. Solvents for UV (showing high energy cutoffs)
Water 205
CH3CN 210
C6H
Ether 210
EtOH 210
Hexane 210
MeOH 210
Dioxane 220
THF 220
CH2Cl2 235
CHCl3 245
CCl4 265
benzene 280
Acetone 300
Various buffers for HPLC, check before using.

12. Organic compounds (many of them) have UV spectra
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.
From Skoog and West et al. Ch 14

13. An Example--Pulegone Frequently plotted as log of molar extinction 
So at 240 nm, pulegone has a molar extinction of 7.24 x 103
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. Showing atoms whose MO’s contribute most to the bands
The orbitals involved
Showing atoms whose MO’s contribute most to the bands

16. The Quantitative Picture
(power in) P
(power out)
Transmittance:
T = P/P0
Absorbance:
A = -log10 T = log10 P0/P
B(path through sample)
The Beer-Lambert Law (a.k.a. Beer’s Law):
A = ebc
Where the absorbance A has no units, since A = log10 P0 / P
e 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.
Attached group increment, nm
Extend conjugation +30
Addn exocyclic DB +5
Alkyl +5
O-Acyl 0
S-alkyl +30
O-alkyl +6
NR2 +60
Cl, Br +5
Homoannular, base 253 nm
Acyclic, base 217 nm
heteroannular, base 214 nm

19. Similar for Enones O x  b  b g d,+ 202 227 239 215
Base Values, add these increments…  b g
d,+
X=H 207
X=R 215
X=OH 193
X=OR 193
Extnd C=C
+30
Add exocyclic C=C +5
Homoannular diene
+39
alkyl
+10
+12
+18 OH
+35
+50
OAcyl +6
O-alkyl
+17
+31
NR2
S-alkyl
Cl/Br
+15/+25
+12/+30
With solvent correction of…..
Water
EtOH
CHCl
Dioxane
Et2O
Hydrcrbn -11

20. Some Worked Examples
Base value x alkyl subst exo DB total 232 Obs. 237
Base value x alkyl subst exo DB total 234 Obs. 235
Base value ß alkyl subst total 239 Obs. 237

21. Distinguish Isomers!
Base value x alkyl subst exo DB total 239 Obs. 238
Base value x alkyl subst total 273 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, 2nd Ed. Ch. 14

24. UV of Benzene in heptane
Group
K band ()
B band()
R band
Alkyl
208(7800)
260(220) --
-OH
211(6200)
270(1450)
-O-
236(9400)
287(2600)
-OCH3
217(6400)
269(1500)
NH2
230(8600)
280(1400) -F
204(6200)
254(900)
-Cl
210(7500)
257(170)
-Br -I
207(7000)
258/285(610/180)
-NH3+
203(7500)
254(160)
-C=CH2
248(15000)
282(740)
-CCH
248(17000)
278(6500
-C6H6
250(14000)
-C(=O)H
242(14000)
328(55)
-C(=O)R
238(13000)
276(800)
320(40)
-CO2H
226(9800)
272(850)
-CO2-
224(8700)
268(800)
-CN
224(13000)
271(1000)
-NO2
252(10000)
280(1000)
330(140)
Benzenoid aromatics
UV of Benzene in heptane
From Crewes, Rodriguez, Jaspars, Organic Structure Analysis

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)
Fluorescence is a process that involves a singlet to singlet transition.
Phosphorescence is a process that involves a triplet to singlet transition