Electronic Spectroscopy Uv-visible

Electronic Spectroscopy Ultraviolet 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, 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 λ for a chromophore containing a phenol, carboxylic acid, etc.

“hypsochromic” shift

“bathochromic” shift

λ

The UV Absorption process •σ → σ * and σ → π * transitions: high-energy, accessible in vacuum UV (λ max <150 nm). Not usually observed in molecular UV-Vis. •n → σ * and π → σ * transitions: non-bonding electrons (lone pairs), wavelength (λ max ) in the 150-250 nm region. •n → π * and π → π * transitions: most common transitions observed in organic molecular UV-Vis, observed in compounds with lone pairs and multiple bonds with λ 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

What are the nature of these absorptions?

π* π* π* n

π* π* π* n

π* π* π* n π -π *; λ

ε

max

Example for a simple enone

n-π *; λ

=218

=11,000

max

=320

ε

=100

Example: π → π * transitions responsible for ethylene UV absorption at ~170 nm calculated with ZINDO semi-empirical excited-states methods (Gaussian 03W):

hν

HOMO π

u

bonding molecular orbital

170nm photon

LUMO π

g

antibonding molecular orbital

How Do UV spectrometers work? Rotates, to achieve scan

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

Two photomultiplier inputs, differential voltage drives amplifier.

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

Model from Agilent literature. Imagine replacing “cell” with a microflow cell for HPLC!

Disadvantage, resolution is 1 nm, vs 0.1 nm for normal UV

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 ε 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 Make solution of concentration low enough that A≤ 1 (Ensures Linear Beer’s law behavior)

1.0

λ max with certain UV extinction ε

Visible

Even though a dual beam goes through a solvent blank, choose solvents that are UV transparent.

Absorbance

Can extract the ε 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.

0.0 200

400

Wavelength, λ , generally in nanometers (nm)

800

Solvents for UV (showing high energy cutoffs) Water

205

THF

220

CH3C≡ N

210

CH2Cl2 235

C6H12

210

CHCl3

245

Ether

210

CCl4

265

EtOH

210

benzene

280

Hexane

210

Acetone

300

MeOH

210

Dioxane

220

Various buffers for HPLC, check before using.

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

An Example--Pulegone Frequently plotted as log of molar extinction

ε

So at 240 nm, pulegone has a molar extinction of 7.24 x 103 Antilogof 3.86

O

Can we calculate UVs? M olarAbsorptivity (l/m ol-cm )

ElectronicSpectra

50243

40194

30146

20097

10049 nacindolA 0 220

W avelength (nm ) 230

240

250

260

270

280

290

300

ElectronicSpectra

M olarAbsorptivity (l/m ol-cm ) 51972

41578

Semi-empirical (MOPAC) at AM1, then ZINDO for config. interaction level 14 Bandwidth set to 3200 cm

-1

31183

20789

10394

N acetylindol 0 220

W avelength (nm 230

240

250

260

270

280

290

300

The orbitals involved

Electronic Spectra

Molar Absorptivity (l/mol-cm) 55487

44390

33292

22195

11097

0 200

210

220

230

240

250

260

270

280

290

Showing atoms whose MO’s contribute Nacetylindolmost to the Wavelength (nm) bands 300

The Quantitative Picture • Transmittance: T = P/P0

• Absorbance: A = -log10 T = log10 P0/P

P0 (power in)

P (power out)

B(path through sample)

• The Beer-Lambert Law (a.k.a. Beer’s Law): A = ε bc Where the absorbance A has no units, since A = log10 P0 / P ε 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 triumph R.B. Woodward, L.F. Fieser and others Predict λ max for π⇒π * in extended conjugation systems to within ca. 2-3 nm.

Homoannular, base 253 nm

Acyclic, base 217 nm

heteroannular, base 214 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

β

β Ο

α X=H 207

x

X=R 215 X=OH 193 X=OR 193 With solvent correction of…..

227

202 215 Base Values, add these increments…

β

α

239

γ

δ ,+

Extnd C=C

+30

Add exocyclic C=C

+5

Homoannular diene

+39

alkyl

+10

+12

OH

+35

+30

OAcyl

+6

+6

+6

+6

+35

+30

+17

+31

+15/+25

+12/+30

Water

+8

EtOH

0

O-alkyl

CHCl3

-1

NR2

Dioxane

-5

S-alkyl

Et2O

-7

Cl/Br

Hydrcrbn -11

O

O

+18

+18 +50

Some Worked Examples

O

Base value 2 x alkyl subst. exo DB total Obs.

217 10 5 232 237

Base value 3 x alkyl subst. exo DB total Obs.

214 30 5 234 235

Base value 2 ß alkyl subst. total Obs.

215 24 239 237

Distinguish Isomers! Base value 4 x alkyl subst. exo DB total Obs.

214 20 5 239 238

Base value 4 x alkyl subst. total Obs.

253 20 273 273

HO2C

HO2C

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…

λ

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, 2

nd

Ed. Ch. 14

Benzenoid aromatics UV of Benzene in heptane

From Crewes, Rodriguez, Jaspars, Organic Structure Analysis

Group

K band (ε ) B band(ε )

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)

-C≡ CH

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)

-C≡ N

224(13000)

271(1000)

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 nonaqueous titrations Example here gives detn of endpoint for bromcresol green

Isosbestic points Single clear point, can exclude intermediate state, exclude light scattering and Beer’s law applies

Binding studies Form I to form II

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)