Electronic SpectroscopyUltraviolet and visible
Where in the spectrum are these transitions?
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
Uses for UV, continuedKnowing 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
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 150-250 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 = 200-600 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
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
ππ
nππ
nπ*
ππ
nπ*π*
π*π*
π*π* π*
π*
-*; max=218 =11,000
n-*; max=320 =100
How Do UV spectrometers work?
Two photomultiplier inputs, differential voltage drives amplifier.
Matched quartz cuvettes
Sample in solution at ca. 10-5 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
Diode Array DetectorsDiode 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!
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
An Electronic SpectrumA
bsor
banc
e
Wavelength, , generally in nanometers (nm)
0.0400 800
1.0
200
UV Visiblemaxwith 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.
Solvents for UV (showing high energy cutoffs)
Water 205
CH3CN 210
C6H12 210
Ether 210
EtOH 210
Hexane 210
MeOH 210
Dioxane 220
THF 220
CH2Cl2235
CHCl3 245
CCl4 265
benzene 280
Acetone 300
Various buffers for HPLC, check before using.
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.
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
O
Can we calculate UVs?Electronic Spectra
Wavelength (nm)
Molar Absorptivity (l/mol-cm)
220 230 240 250 260 270 280 290 3000
10049
20097
30146
40194
50243
nacindolA
Electronic Spectra
Wavelength (nm)
Molar Absorptivity (l/mol-cm)
220 230 240 250 260 270 280 290 3000
10394
20789
31183
41578
51972
Nacetylindol
Semi-empirical (MOPAC) at AM1, then ZINDO for
config. interaction level 14
Bandwidth set to 3200 cm-1
The orbitals involved
Electronic Spectra
Wavelength (nm)
Molar Absorptivity (l/mol-cm)
200 210 220 230 240 250 260 270 280 290 3000
11097
22195
33292
44390
55487
Nacetylindol
Showing atoms whose
MO’s contribute most to the
bands
The Quantitative Picture• Transmittance:
T = P/P0
B(path through sample)
P0
(power in)P
(power out)• Absorbance:
A = -log10 T = log10 P0/P
• The Beer-Lambert Law (a.k.a. Beer’s Law):A = bc
Where the absorbance A has no units, since A = log10 P0 / P
is the molar absorbtivity with units of L mol-1 cm-1
b is the path length of the sample in cmc is the concentration of the compound in solution, expressed in mol L-1 (or M, molarity)
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.
Polyenes, and Unsaturated Carbonyl groups;
an Empirical triumphR.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 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
Similar for Enones
x
O O
X=H 207
X=R 215
X=OH 193
X=OR 193
215 202 227 239
Base Values, add these increments…
Extnd C=C +30Add exocyclic C=C +5
Homoannular diene +39
alkyl +10 +12 +18 +18
OH +35 +30 +50
OAcyl +6 +6 +6 +6
O-alkyl +35 +30 +17 +31
NR2
S-alkyl
Cl/Br +15/+25 +12/+30
With solvent correction of…..
Water +8
EtOH 0
CHCl3 -1
Dioxane -5
Et2O -7
Hydrcrbn -11
Some Worked Examples
O
Base value 217 2 x alkyl subst. 10 exo DB 5 total 232 Obs. 237
Base value 214 3 x alkyl subst. 30 exo DB 5 total 234 Obs. 235
Base value 215 2 ß alkyl subst. 24 total 239 Obs. 237
Distinguish Isomers!
HO2C
HO2C
Base value 214 4 x alkyl subst. 20 exo DB 5 total 239 Obs. 238
Base value 253 4 x alkyl subst. 20 total 273 Obs. 273
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
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
Benzenoid aromatics
From Crewes, Rodriguez, Jaspars, Organic Structure Analysis
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 210(7500) 257(170)
-I 207(7000) 258/285(610/180)
-NH3+ 203(7500) 254(160)
-C=CH2 248(15000) 282(740)
-CCH 248(17000) 278(6500
-C6H6 250(14000)
-C(=O)H 242(14000) 280(1400) 328(55)
-C(=O)R 238(13000) 276(800) 320(40)
-CO2H 226(9800) 272(850)
-CO2- 224(8700) 268(800)
-CN 224(13000) 271(1000)
-NO2 252(10000) 280(1000) 330(140)
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
Heterocycles
Nitrogen heterocycles are pretty similar to the benzenoid anaologs that are isoelectronic.
Can study protonation, complex formation (charge transfer bands)
Quantitative analysisGreat 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
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)