Interlude on Spectroscopy

Transcription

1 Interlude on Spectroscopy interaction of radiation with matter type of interaction depends on the energy of the radiation use of the full electromagnetic spectrum is possible interaction happens through absorption spectrum is a plot of the absorption as a function of the energy used A λ, wavelength UV/Vis ν, wavenumber IR ν, frequency NMR

2 high energy Electromagnetic spectrum low energy Radio waves ionization M + hν M + + e molecular rotations microwave spectroscopy electronic transitions UV/Vis molecular vibrations IR nuclear spin transitions NMR

3 Chapter 2. Ultraviolet (and visible) spectroscopy electronic transitions which transitions? wavelength and intensity, λ max and ε what happens if we change the molecule? substituent effects on λ max and ε laws for the prediction of λ max in conjugated alkenes and ketones Reading: Pavia Chapter 7 Don t need 7.13, 7.14D and E, 7.15

4 1. General range of radiation: nm absorption leads to excitation of electrons E electronically excited state ΔE: absorbed energy, quantized electronic ground state of a molecule Question: How large is the energy gap needed (let s say at 200 nm) and what does that imply for the population of the two states? Planck s law E = hν = c h = λ Js 8 m / s nm 9 nm / m = J Boltzmann distribution N N u l ΔE kt J J / K 298K = e = e = 0 read: 1 upper over all lower : all molecules are in their electronic ground state

5 2. Nature of electronic excitations A. Electronic states electronic levels for the lowest-energy transition are HM and LUM E LUM ΔE HM ground state excited state E is quantized: only a specific ΔE leads to this promotion we should observe a line spectrum true for an atom, but not for a molecule a molecule vibrates and rotates

6 2. Nature of electronic excitations continued A. Electronic states every electronic state possesses vibrational and rotational states λ: nm μm cm E Rotational levels ΔE Rotational levels

7 2. Nature of electronic excitations continued A. Electronic states leads to line broadening: all ΔE are similar but slightly different A A λ max λ λ max λ (wavelength for maximum absorption) observed band due to poor resolution

8 2. Nature of electronic excitations continued B. Molecular orbitals occupied Ms n unoccupied Ms σ σ* π π* general orbital diagram E σ* π* n π σ possible transitions: n π* n σ* π π* E π σ* σ π* σ σ* molecules?

9 3. Selection rules govern which transitions are allowed, which forbidden allowed transition is observed forbidden transition is either not observed or very weak the promoted electron does not change spin hν only one electron is excited singlet transitions between orbitals of different symmetry are forbidden singlet : different symmetry π π* allowed n π* forbidden, but observed

10 4. Chromophores group of atoms that makes an absorption possible (at least two) important chromophores less important chromophores (first λ max too small/short) R 2 C=, R 2 C=S, R 2 C=N-R R-H, R--R, R-NH 2 R-N=N-R R 2 C=CR 2, R-C C-R, R-C N R-N 2 R, R, R H R NH 2 (aromatics in general) see Table 7.3

11 4. Chromophores continued probably the best-studied chromophore is the carbonyl group nm but forbidden, low intensity nm allowed, high intensity, but too short to be observed!

12 4. Chromophores continued typical UV-Vis spectrum of a carbonyl compound π π* What is this? C= plus C=C? n π* 235 nm 315 nm π π* becomes observable 189 nm 280 nm

13 4. Chromophores continued the combination of chromophores leads to shifts in band intensity: increased: hyperchromic effect decreased: hypochromic effect wavelength: increased: longer λ max : - bathochromic effect - red shifted π π* band - less energetic light - through conjugation C C C λ 1,max < λ 2,max decreased: shorter λ max : - hypsochromic effect - blue shifted n π* band - more energetic light - through heteroatoms C C C λ 1,max > λ 2,max A A H H λ [nm] λ [nm]

14 4. Chromophores continued reason for a red shifted π π* band through conjugation: HM LUM gap becomes smaller upon conjugation acyclic conjugated: M (interaction) diagram for the π-orbitals of ethene and butadiene LCA principle: linear combinations of atomic orbitals (p only) π * 4 3 nodes π * 1 node π * π * π * 3 2 nodes p C ΔE p 1 C ΔE 2 π 2 1 node π no node π π C C C C π 1 no node C C C C C C C C

15 4. Chromophores continued the red shift through conjugation can be dramatic: A B C β-carotene 2 C 40 H 56 : U = ½(56 0) = = 13, 11 conjugated C=C λ max 465 nm orange!

16 4. Chromophores continued reason for a red shifted π π* band through conjugation reason for a blue shifted n π* band through heteroatoms: inductive effect: energy of n is lowered π π* still shows a bathochromic shift: π-system is still extended through the heteroatom n π* π π* C C C λ 1,max > λ 2,max λ 1,max < λ 2,max

17 follows the Lambert-Beer law A I log I = 0 = εcl 5. Absorption ε is determined by the size of the absorbing system probability of the transition ε < 1000 log ε 2-3: low intensity absorption ε >> 1000 log ε 4-5: high intensity absorption plotted is A or ε or log ε sample solvent I I 0

18 6. Materials Central question: Where does the material absorb? solvent: - should be transparent in the region of interest - ideal UV-Vis spectrum: A λ [nm] short-wavelength cutoff: H 2, CH 3 CN 190 nm CHCl nm see Table 7.1 cuvette: quartz, transparent to 200 nm polymer, transparent to 220 nm (PMMA) optical glass, transparent to only 350 nm

19 Beam in a UV-Vis spectrometer 7. Spectrometer grating Vis UV movable parts Diode-array spectrophotometer no movable parts: faster

20 8. Spectrum UV-Vis spectra are often not published x-y data of the extrema are reported instead Info from this spectrum reported: λ max 230 nm log ε Info from this spectrum gained: - no long-wavelength absorption: not highly conjugated - longer-wavelength low intensity: probably forbidden, possibly n π* - shorter-wavelength high intensity: probably allowed, possibly π π*

21 9. Prediction of λ max of π π* transitions I. Dienes and higher conjugated C=C Why do we write and not? Conformation matters! acyclic dienes butadiene s-trans π 4 π nm 271 nm π 2 s-cis π 1 Interesting, but s-cis is not usually important in acyclic systems! cyclic dienes transoid cisoid use Woodward-Fieser rules to predict λ max

22 9. Prediction of λ max of π π* transitions continued I. Dienes and higher conjugated C=C Woodward-Fieser rules R.B. Woodward Nobel Prize 1965 calculate λ max from a base value acyclic (s-trans) transoid cisoid base value [nm] plus increments for structural features another conjugated C=C alkyl group or ring residue exocyclic C=C increment [nm] plus increments for other substituents see Table 7.5

23 9. Prediction of λ max of π π* transitions continued I. Dienes and higher conjugated C=C Examples = 219 nm = 244 nm x5 + 5 = 229 nm Colour code: base system increments not important for absorption x5 + 5 = 234 nm CH = 240 nm CH 3

24 9. Prediction of λ max of π π* transitions continued I. Dienes and higher conjugated C=C Examples x5 = 263 nm x5 + 2x5 = 283 nm Colour code: base system increments not important for absorption x30 + 5x5 + 3x5 = 353 nm

25 9. Prediction of λ max of π π* transitions continued II. Enones and higher conjugated C= (ketones only) Woodward rules calculate λ max from a base value acyclic (s-trans) 6-membered ring 5-membered ring base value [nm] plus increments for structural features another conjugated C=C alkyl group or ring residue increment [nm] 30 in α 10 in β 12 in γ 18 plus increments for other substituents δ exocyclic C=C γ β homocyclic (cisoid) diene 5 39 α see Table 7.7

26 9. Prediction of λ max of π π* transitions continued II. Enones and higher conjugated C= Examples = 237 nm Colour code: = 255 nm = 227 nm base system increments not important for absorption = 302 nm = 214 nm = 267 nm

27 9. Prediction of λ max of π π* transitions continued III. Aromatic compounds good chromophore: 3 conjugated C=C, cis, in one ring second primary band, allowed transition, usually not observed primary band, forbidden transition secondary band, forbidden transition often vibrational fine structure conjugation effects as discussed earlier Read 7.14 by yourself, focus on conjugation issues no quantitative treatment: λ max not easy to predict for substitution

28 Example An alkene C 12 H 16 and an unsaturated ketone C 11 H 14, both almost identical in structure ( ), show UV absorptions at 269 and 286 nm, respectively. Give their structures.