1. Unit One Part 10:infrared spectroscopy andmass spectrometry my last lecture...yippee

2. 10Unit OnePartinfrared spectroscopy?What isFunctional groups & IRMass spectrometry 3. determining structureUV uv-vis nmr irconjugation CH functionalgroups 4. determining structureUVthree common forms ofspectroscopy, all used to uv-visnmrir look at molecules (and determine their structure)...conjugationCHfunctionalgroups 5. determining structure UV three common forms of spectroscopy, all used touv-visnmr irlook at molecules (anddetermine theirstructure)...conjugationlast lecture...not did nmr CHfunctionalgoing to look at uv so that leaves ir and...groups 6. determining structureOCH3H3CNN H ONNCH3Mass:caffeine194.08 C8H10N4O2 well also look at massspectrometry (not a formof spectroscopy, but stilluseful!) 7. look at infraredspectroscopy rst what is infrared spectroscopy? 8. infrared spectroscopy molecule in excitedenergy state E2energy energybasically we shineenergy infrared radiation on a molecule and some of it is... energy molecule inenergy state E1 9. infrared spectroscopy molecule in excitedenergy state E2energy energy is absorbed, exciting themolecule...we measure what comesenergy through and note what has beenabsorbed (unlike nmr when we measured what was emitted) energymolecule in energy state E1 10. what energy is absorbedby the molecule? 11. atom 1 atom 2two atoms connected bybonda bond can be thought tobe the same as... 12. spring...two masses / blobs joined by a spring...mass 1 mass 2 13. bondsbonds vibrate (stretching andcontracting) in thevibrate same way... 14. likesprings...a spring would... 15. ...this means we can use Hookes law on extension and elasticity...Hookeslaw 16. ...of course, I hate mathsso Im not actually going to show you Hookes lawbut just tell you what it reveals...Hookeslaw 17. energy to vibrate bonddepends on...springthe force needed to vibrate a bond / spring depends on the strength of the bond / spring and...mass 1 mass 2 strength 18. energy to vibrate bonddepends on...spring...the difference in mass of the two ends(or atoms)...thismeans... mass 1 mass 2difference in mass 19. strong (short) bondrequires......so it is easy to seemultiple bonds in IR...more energy 20. light (hydrogen) atomvibrates faster...and bonds tohydrogen (CH, OHetc)more energy 21. bonds only vibratewith fixed frequencies and... bonds vibrate with certain frequencies 22. ...they will onlyabsorb energy of the same frequency orwavelength... bond will only absorbenergy of same frequency 23. energy / lightlong wavelength short wavelengthlow frequencyhigh frequency low energy high energy 24. energy / light ...so, how does all this mumbo-jumbo (Imean physics) effect IRspectroscopy?long wavelength short wavelengthlow frequencyhigh frequency low energy high energy 25. infrared spectroscopy 1 1 1 2 3O 2EtO CH3 3 3 the cartoon version 26. infrared spectroscopy 1 1 12 3O 2EtO CH3 3 3...shine IR on molecule...certain bonds will absorb certain wavelengths of energy and thus byobserving which wavelengths are missing we have a clue as to whatbonds are in the molecule... the cartoon version 27. infrared spectroscopy high lowenergy energy1 11 2 3 O2 EtO CH33 3 highlow the real version wave wavenumber number 28. infrared spectroscopy highlowenergyenergy1wavenumber is the1 inverse (1/) of the wavelength in cm and measures1 2 3O energy...2 EtO CH333 high low the real version wavewavenumbernumber 29. more on the theory ofIR can be found at:www.massey.ac.nz/~gjrowlan in Introduction to organicand bioorganic molecules and reactions 30. interpreting IR spectra energy to cause vibration 4000 3000 2000 15001000 cm-1O HC CC CCOchange in scaleN HC NC OCFC HC O CCl bonds totriple double single hydrogenbondsbondsbonds regions of four the spectrum are important... 31. interpreting IR spectra energy to cause vibration 4000 3000 2000 1500 1000 cm-1O HC CC CC Ochange in scaleN HC NC OC FC HC O C Cl bonds totriple double single hydrogenbondsbondsbondslight atoms (H) 32. interpreting IR spectra energy to cause vibration 4000 3000 2000 1500 1000 cm-1O HC CC CC Ochange in scaleN HC NC O CFC HC O C Cl bonds totriple double single hydrogenbondsbondsbondsstrongstrongbonds bonds 33. interpreting IR spectraenergy to cause vibration 40003000 2000 1500 1000 cm-1OH CC C C C Ochange in scaleNH CN C O C FCH CO C Cl bonds totriple doublesingle hydrogen bonds region isnt that actually fingerprint bonds bondsuseful...it is unique to each molecule but almost impossible to interpret...good ifyou have a computer database I guess...1500400cm1 ngerprint region 34. interpreting IR spectrathese two molecules have same functional group (ketone) andare almost identical in threeregions... OO1500400cm1 ngerprint region 35. interpreting IR spectraonly really differ in fingerprint...but I couldnt tell you what bondstretching caused this peak! OO1500400cm1 ngerprint region 36. bad slide alert! 37. functional group absorptions 38. functional group absorptions can predict roughly wheremost functional groups will come...youll be given this in an exam ifyou need it... 39. examples of IR spectra H H NNH2benzeneamine3480aniline3395two NH stretches visible...butnot for the reason you think (Im not going to tell you whyin case it confuses you!) 40. examples of IR spectra H3C H N NH 3443N-methylbenzenamine N-methylaniline 41. examples of IR spectra PhOH3224 O O H H HH OOPhPh phenol - H-bonding 42. examples of IR spectraPhOH3224O O hydrogen bonding causesH H the peak to be very broad as the HHstrength of H-bonds variesOO depending on factors likedistance...PhPhphenol - H-bonding 43. examples of IR spectra OH 3627H O O H 2,6-di-tert-butyl-4-methylphenol 44. examples of IR spectra OH 3627 Hno H-bonding so OH hasspecific strength bond and sharpO O peak (there is no H-bonding as the large tert-butyl groupsH prevent the two moleculesgetting close to each other...)2,6-di-tert-butyl-4-methylphenol 45. examples of IR spectra OC=C1642 C=Ohex-5-en-2-one 1718 46. examples of IR spectraC=C C=O1634O1674 note how putting the two groups in conjugation makes the bonds weaker (and hence have a lower wavenumber)(E)-hex-4-en-3-one 47. examples of IR spectraC=C C=O1634O1674 why does conjugation make the bonds weaker? Think about the resonance forms...(E)-hex-4-en-3-one 48. examples of IR spectra OH 3010 C=O 1712 OHObutanoic acid 49. examples of IR spectra NHC=O 3356 1662O 3184 1634HN H butanamide 50. examples of IR spectraweaker C=O stretch inamide due to a resonance form involving the nitrogenlone pair...see IR can tell usa lot of useful information... NH C=O 33561662O 31841634 H NHbutanamide 51. mass spectrometry 52. a mass spectroscopist issomeone who gures out whatsomething is by smashing it with ahammer & looking at the piecesJEOL (manufacturer) website 53. a mass spectrometerM+ M+ + e 2e 54. a mass spectrometer basically, what you need to know is that a mass spectrometer fires electrons at your compound knocking one electron off the compound to form a radical cation that is then detected...M+ M++ e2e 55. molecular massmolecular ion NH3 + +NH + e3 2ebecause the mass of an electron isvery very small...17.031 5.5 x 10417.030mass-to-charge ratio (m/z) 56. molecular massmolecular ion NH3 + +NH + ethe radical cation3 2e (or molecular ion)effectively has the samemass as the originalcompound...17.031 5.5 x 10417.030mass-to-charge ratio (m/z) 57. molecular massmolecular ion ...as a result massspectrometry gives us themolecular mass...it also gives a lot more info but thats for another day (or course) NH3 + +NH+ e 3 2e17.031 5.5 x 10417.030mass-to-charge ratio (m/z) 58. cyclohexane+C6H12 + e[C6H12 ]+2e m/z 84C = 6x12H = 12x1a simple examplem/z 85showing the massdue to 13Cof cyclohexane...isotope 59. cyclohexane +C6H12 +e[C6H12 ]+2em/z 84 C = 6x12 H = 12x1...all these otherpeaks are useful...but lets ignore m/z 85 due to 13Cisotope 60. 31isotopes35Cl37Clchlorine exists as twoisotopes (same elementdifferent mass due tonumber of neutrons) 61. 31isotopes35Cl37Clthere is 3 times as much Clmass 35 than Cl mass 37 (hence average is 35.5 as shown on most periodic tables) 62. isotopes C6H5Cl +e[C6H5Cl]+ + 2e the mass spectrum has twopeaks...one for each isotope and the relative size of these peaks will m/z 112 be 3 : 1... due to35Cl Cl m/z 114due to 37ClC6H535Cl Mr = M+ = 112C6H537Cl Mr = M+2 = 114 63. 11isotopes79Br81Br chlorine not the only element withisotopes...bromine exists as twoisotopes in equal proportion so thespectrum of a bromide... 64. isotopes C6H5Br+ e [C6H5Br]++ 2e...will have two peaks of equal intensity with a mass 2 apartm/z 156due to 79Brm/z 158due toBr 81BrC6H579Br Mr = M+ = 156C6H581Br Mr = M+2 = 158 65. what have....we learnt? to determine molecularstructure thebasics of infrared spectroscopy the basics of mass spectrometry 66. good luck!