1. Unit 11:Data processing and analysis. A.Infrared spectroscopy B.Mass spectrometry C.X-ray diffraction/crystallography D.H NMR

2. A. Infrared spectroscopy What is infrared? An electromagnetic wave

3. Energy of electromagnetic radiation is carried in discrete packets of energy called photons or quanta. E = hν E = energy of a single photon of radiation h = 6.63 x 10 -34 J  s (Plank’s constant) ν = frequency of the radiation

4. Example: Calculate the energy of a photon of visible light with a frequency of 3.0 x 10 14 s -1. Express in kJ mol -1. E = hν E = ( 6.63 x 10 -34 J  s)(3.0 x 10 14 s -1 ) E = 1.989 x 10 -19 J

5. Note:. ν = c = 3.0 x 10 8 m  s -1 Type of em radiationTypical f (s -1 ) Typical (m) Radio waves (low energy)3 x 10 6 10 2 Microwaves3 x 10 10 10 -2 Infrared (IR)3 x 10 12 10 -4 Visible (ROYGBIV)3 x 10 15 10 -7 Ultraviolet (UV)3 x 10 16 10 -8 X-rays3 x 10 18 10 -10 Gamma rays (high energy)> 3 x 10 22 < 10 -14 Thus, ν= c/

6. Wavenumber c=λν  1/λ = ν/c In IR spectroscopy, the frequency of radiation is often measured as number of waves per centimeter (cm -1 ), also called wavenumber. Example: Calculate the wavenumber in cm -1 for an IR wave with a frequency of 3 x 10 13 s -1.

7. Infrared spectroscopy

8. normal vibrationvibration having absorbed energy By measuring the IR spectrum of a molecule we can determine what kind of motion the molecule has and therefore what kind of bonds are present in the molecule. A bond will absorb radiation of a frequency similar to its vibration Light atoms vibrate at higher freq. than heavier atoms; they absorb IR radiation of shorter wavelength ( more energy) Multiple bonds vibrate at higher freq. than single bonds; they absorb IR radiation of shorter wavelength ( more energy)

9. Symmetrical stretching Antisymmetrical stretching Scissoring RockingWaggingTwisting Infrared has the right energy to be absorbed by the polar bonds of a molecule. The IR radiations ( specific energy or frequency ) make the bonds stretch, bend, or vibrate.

10. Using IR to excite molecules Not all vibrations absorb IR. For absorption, there must be a change in bond polarity (dipole moment) as the vibration occurs. Thus, diatomic gas molecules such as H 2, Cl 2 and O 2 do not absorb IR.

11. Vibrations of H 2 O, SO 2 & CO 2 MoleculeAsymmetrical stretching Symmetrical stretching Symmetrical bending H2OH2O SO 2 CO 2 O O S -- -- ++ IR active O O S -- -- ++ O O S -- -- ++ H H O ++ ++ -- H H O ++ ++ -- H H O ++ ++ -- O O C -- -- ++ O O C -- -- ++ IR inactive O O C -- -- ++ IR active

12. Matching wavenumbers with bonds “fingerprint region” lots of overlap, so not very useful broad and strong very strong broad and strong Usually sharper than OH Data Booklet -table 26

13. https://www.youtube.com/watch?v=DDTIJgIh86E

14. IR spectrum of ethanol, CH 3 CH 2 OH

15. IR spectrum of ethyl ethanoate, CH 3 COOCH 2 CH 3 “fingerprint region” C=O C-H

16. FINGERPRINT REGION The 1400 cm -1 to 800 cm -1 range is the “fingerprint” region : its the pattern is characteristic of a particular compound

17. IR SPECTRUM OF A CARBONYL COMPOUND carbonyl compounds show a sharp, strong absorption between 1700 and 1760 cm -1 this is due to the presence of the C=O bond

18. IR SPECTRUM OF AN ALCOHOL alcohols show a broad absorption between 3200 and 3600 cm -1 this is due to the presence of the O-H bond

19. IR SPECTRUM OF A CARBOXYLIC ACID carboxylic acids show a broad absorption between 3200 and 3600 cm -1 this is due to the presence of the O-H bond they also show a strong absorption around 1700 cm -1 this is due to the presence of the C=O bond

20. http://undergrad-ed.chemistry.ohio- state.edu/anim_spectra/index.html Practice some more spectra

21. IR spetra of butanal and butanone

22. B.Mass spectrometry

23. The Mass Spectrometer http://www.youtube.com/watch?v=J-wao0O0_qM&feature=related

24. How it works: The sample is bombarded with a stream of high energy electrons. The collision is so energetic that it causes the molecule to break up into different fragments (ions). The fragments ( + ions) of a particular mass are detected and a signal is sent to a recorder. The strength of the signal is a measure of the number of ions with that charge/mass ratio that are detected.

25. Fragmentation Patterns : evidence for the structure of the compound The largest mass peak corresponds to a parent ion passing through the instrument unscathed, but other ions produced as a result of this break up are also detected. For each fragmentation, one of the products keeps the + charge and will be detected. Generally the most stable + ion is formed CH 3 -CH 2 -OH  CH 3 -CH 2 + + OH peak 29 no peak 17

26. Example: ethanol mass/charge relative abundance 06030 100 0 15 29 31 45 46

27. Example: ethanol

28. Note: This fragmentation will yield either CH 3 + and CH 2 OH or CH 3 and CH 2 OH +, yielding peaks at both 15 and 31

29. Data table 28-Mass spectral of fragment lost. MrMr loss of… 15CH 3 17OH 29C 2 H 5 or CHO 31CH 3 O 45COOH

30. C. X-ray diffraction / X-ray crystallography http://www.theguardian.com/science/video/2013/oct/09/100-years-x-ray-crystallography-video-animation

31. When X-rays shine on a crystal (orderly structure), they are reflected and produce an ordered diffraction pattern. The diffraction pattern produced by X-rays helps determine the electron density of the crystal. As the electron densities are related to the element’s electron configuration, we can also determine the identity of the atoms. Note: H atoms have a very low electron density ( 1e) and are not visible on the X-ray diffraction pattern. http://www.rsc.org/learn-chemistry/resource/res00000020/computational-chemistry#!cmpid=CMP00001685

32. Nuclear magnetic resonance (NMR) spectroscopy D.Nuclear magnetic resonance. H NMR spectroscopy.

33. Nuclear magnetic resonance (NMR) spectroscopy

34. When an external magnetic field is applied to a molecule the energy of its spinning protons splits in 2 separate levels. The spin aligned with the magnetic field will be at a lower energy state. The spin aligned against the magnetic field will be at a higher energy state. We can then get a proton change its spin from with the magnetic field to against the magnetic field by providing the right amount of energy. The moment the proton returns to its lower energy level and reverses its spin, it gives out the energy that shows as a peak - resonance- in the NMR spectrum.

35. The HNMR spectrum: its interpretation https://www.youtube.com/watch?v=gaGUpACXijE Chemical shifts vs TMS standard (tetramethylsilane) Different H groups ( types) IntegrationNumber of H in the group = area under the peak Multiplicity of peak (splitting) Number of adjacent H See Data Table 27: H NMR data

36. The TMS standard -All 12 H’s are in identical chemical environments, so one signal is recorded = zero point on the scale -The H are very shielded by the electrons of C that is more electronegative that silicon. Therefore they experience the strength of the magnetic field the least. -The signal doesn’t interfere with the signal given by H bonded to Carbon. The less the H will be shielded in a group, the greater the shift will be.

37. 5 4 3 2 1 0  CHEMICAL SHIFTS 3 environments = 3 signals Triplet d = 3.4 Sextet d = 1.9 Triplet d = 1.0 Signal for H’s on carbon 3 is shifted furthest downfield from TMS due to proximity of the electronegative halogen 123 TMS 1 H-NMR spectroscopy: 1-bromopropane

38. 5 4 3 2 1 0  INTEGRATION Area ratio from relative heights of integration lines = 2 : 2 : 3 Carbon 13 Carbon 22 Carbon 32 123 2 2 3 TMS 1 H-NMR spectroscopy: 1-bromopropane

39. 5 4 3 2 1 0  SPLITTING SPLITTING PATTERN Carbon 1 Chemically different hydrogen atoms on adjacent atoms = 2 2 + 1 = 3 The signal will be a TRIPLET 123 1 TMS 1 H-NMR spectroscopy: 1-bromopropane

40. 5 4 3 2 1 0  SPLITTING SPLITTING PATTERN Carbon 2 Chemically different hydrogen atoms on adjacent atoms = 5 5 + 1 = 6 The signal will be a SEXTET 123 2 TMS 1 H-NMR spectroscopy: 1-bromopropane

41. Peaks Three different signals as there are three chemically different protons. Shift Signals are shifted away from TMS signal, are nearer to the halogen ( H is deshielded more by Br ). Integration The integration lines show that the ratio of protons is 2:2:3 Splitting Signals include a triplet (d = 1.0) sextet (d = 1.8) triplet (d = 3.4) The signals due to the protons attached to carbon... C1 triplet (d = 1.0) coupled to the two protons on carbon C2 ( 2+1 = 3 ) C2 sextet (d = 1.8) coupled to five protons on carbons C1 and C3 ( 5+1 = 6 ) C3 triplet (d = 3.4) coupled to the two protons on carbon C2 ( 2+1 = 3 ) 4 3 2 1 0  SUMMARY TMS 123 1 H-NMR spectroscopy: 1-bromopropane