Introduction. The analysis of the outcome of a reaction requires that we know the full structure of the products as well as the reactants

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2 Introduction The analysis of the outcome of a reaction requires that we know the full structure of the products as well as the reactants

3 Spectroscopy and the Electromagnetic Spectrum Unlike mass spectrometry, infrared (IR), ultraviolet (UV) and nuclear magnetic resonance (NMR) spectroscopies: are nondestructive involve interaction of molecules with electromagnetic energy rather than with high-energy electron beam

4 The Electromagnetic Spectrum The electromagnetic spectrum is the range of electromagnetic energy, including IR, UV and visible radiation

5 The electromagnetic spectrum covers a continuous range of wavelengths and frequencies, radio waves to g rays High n Low n Low l High l

6 Wavelength, Frequency and Amplitude

7 Wavelength (l) is the distance from one wave maximum to the next Frequency (n) - is the # of waves that pass by a fixed point per unit time (s -1 or Hz) Amplitude - is the height of a wave, measured from midpoint to peak Wavelength x Frequency = Speed l (m) x n (s -1 ) = c l = c n n = c l Speed of light: C vacuum = 3.00 x 10 8 m/s

8 The Planck equation gives: e = hn = hc l where e = Energy of 1 photon (1 quantum) h = Planck s constant (6.62 x10-34 J.s) n = Frequency (s -1 ) l = Wavelength (m) c = Speed of light (3.00 x 10 8 m/s) Radiant energy is proportional to its frequency and inversely proportional to its wavelength

9 The Planck equation can be rewritten: N A hc E = N A e = = l 1.20 x 10-4 kj/mol l where E = Energy of Avogadro s number of photons N A = Avogadro s number e = Energy of 1 photon (1 quantum) h = Planck s constant (6.62 x10-34 J.s) c = Speed of light (3.00 x 10 8 m/s) l = Wavelength (m)

10 An absorption spectrum shows the wavelength on the x-axis and the intensity of the various energy absorptions expressed in % transmittance on the y-axis. Ethyl alcohol CH 3 CH 2 OH

11 Infrared Spectroscopy of Organic Molecules The infrared (IR) region is lower in photon energy than visible light Only m to m region is used by organic chemists for structural analysis

12 Absorption Spectrum IR energy in a spectrum is usually measured as wavenumber ~ Wavenumber (n) is the inverse of wavelength is proportional to frequency is expressed in cm -1 Wavenumber (cm -1 ) = 1 l (cm) Specific IR absorbed by organic molecule is related to its structure

13 Infrared Energy Modes Molecules are in constant motion (i.e bond stretching, contracting, bending ) Their energy is quantized

14 Infrared Energy Modes Combinations of atomic movements, such as bending and stretching of bonds between groups of atoms, are called normal modes IR energy absorption corresponds to specific modes

15 Infrared Energy Modes When a molecule is irradiated with electromagnetic radiation, energy is absorbed if the frequency of the radiation matches the frequency of the vibration. Energy absorption increases amplitude for the vibration

16 Infrared Energy Modes IR energy - is characteristic of the atoms in the group and their bonding - corresponds to the amount of energy needed to increase the amplitude of specific molecular vibrations

17 Interpreting Infrared Spectra Most functional groups absorb at about the same energy and intensity independent of the molecule in which they are. Characteristic IR absorptions can be used to confirm the presence of a functional group in a molecule are listed in Table 12.1

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19 Fingerprint Region of Infrared Absorption Spectrum IR spectrum has a lower energy region characteristic of molecule as a whole known as fingerprint region. Its range goes from 1500 cm -1 to 400 cm -1

20 Hexane 1-hexene 1-hexyne

21 Regions of Infrared Absorption Spectrum cm -1 N-H, C-H, O-H (stretching) N-H, O-H 3000 C-H cm -1 C C and C N (stretching)

22 Regions of Infrared Absorption Spectrum cm -1 double bonds C=O, C=C C=N (stretching) C=O C=C cm -1 Below 1500 cm -1 fingerprint region

23 Differences in Infrared Absorptions Molecules vibrate and rotate in normal modes, which are combinations of motions These are related to force constants Bond stretching dominates higher energy (frequency) modes

24 Differences in Infrared Absorptions Light objects connected to heavy objects vibrate fastest (at higher frequencies): C-H, N-H, O-H > C-O, C-N For two heavy atoms, stronger bond requires more energy (higher frequency): C C, C N > C=C, C=O, C=N > C-C, C-O, C-N, C-X

25 8. Infrared Spectra of Hydrocarbons C-H, C-C, C=C, C C have characteristic peaks

26 Example: Hexane

27 Alkenes

28 Example: 1-hexene

29 Alkynes (Terminal alkyne)

30 Example: 1-hexyne

31 Practice Problem: The IR spectrum of phenylacetylene is shown below. What absorption bands can you identify?

32 Infrared Spectra of Some Common Functional Groups Spectroscopic behavior of functional groups is discussed in later chapters Brief summaries are presented here

33 Alcohols Example: Cyclohexanol

34 Amines Example: Cyclohexylamine

35 Aromatic Compounds

36 Example: phenylacetylene Ring bonds cm -1

37 Carbonyl Compounds Strong, sharp C=O peak at 1670 to 1780 cm -1 The exact position of absorption within the range is characteristic of each type of carbonyl compound. It can often be used to identify aldehydes, ketones, and esters.

38 Aldehydes 1730 cm -1 in saturated aldehydes 1705 cm -1 in aldehydes next to double bond or aromatic ring

39 Example: Phenylacetaldehyde C=O 1725 cm -1

40 Ketones 1715 cm -1 in six-membered ring or acyclic ketones 1750 cm -1 in five-membered ring ketones 1690 cm -1 in ketones next to a double bond or an aromatic ring

41 Example: cyclohexanone 1715 cm -1 Ring bonds cm -1

42 Esters 1735 cm -1 in saturated esters 1715 cm -1 in esters next to aromatic ring or a double bond

43 Problem 1: Cyclohexane or Cyclohexene?

44 Problem 2: Propose structure(s) for unknown hydrocarbon

45 Problem 3: Propose structure(s) for unknown hydrocarbon

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