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 orbitalLUMO 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 Spectrum

Abs

orba

nce

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--PulegoneFrequently 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)

220 230 240 250 260 270 280 290 300

0

10049

20097

30146

40194

50243

nacindolA

Electronic Spectra

Wavelength (nm)

Molar Absorptivity (l/mol-cm)

220 230 240 250 260 270 280 290 300

0

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)

200 210 220 230 240 250 260 270 280 290 300

0

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 = bcWhere 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 cm

c 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 +30

Add 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 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

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)