1. CHE 232 MtWTF 8-8:50pm

2. Chemical Identification Comparison of Physical Properties –Boiling Point –Melting Point –Density –Optical rotation –Appearance –Odor Chemical Test –Elemental Analysis Burn the compound and measure the amounts of CO 2, H 2 O and other components that are produced to determine the empirical formula. Used today as a test of purity of compounds that have already been identified.

3. Spectroscopy – measures the interaction of a compound with electromagnetic radiation of different wavelengths. –Nuclear Magnetic Resonance Spectroscopy (NMR) measures the absorption of radio waves by C and H in a magnetic field. Different kinds of C and H absorb energy of different wavelengths. –Infrared (IR) Spectroscopy measures the absorption of infrared (heat) radiation by organic compounds. Different functional groups (C=O, -OH) absorb energy of different wavelengths. –Ultraviolet/Visible Spectroscopy (UV/Vis) measures the absorption of ultraviolet and visible light by  bonds in an organic compound. Bonds of different types and with different extents of conjugation (C=C, C=O, C=C–C=C, aromatic) absorb energy of different wavelengths.

4. Mass Spectrometry –Doesn’t involve the absorption of any type of light. –Used in determining the molecular weight and formula of a compound. –A compound is vaporized and ionized by bombardment with a beam of high-energy electrons. –The electron beam ionizes the molecule by causing it to eject an electron. –When the electron bean ionizes the molecule, the species formed is called a radical cation, and symbolized as M +. –The radical cation M + is called the molecular ion or parent ion. –The mass of M + represents the molecular weight of M.

5. Because M is unstable it decomposes to form fragments of radicals and cations that have a lower molecular weight than M +. The mass spectrometer measures the mass of these cations. The mass spectrum is a plot of the amount of each cation (relative abundance) versus its mass to charge ratio (m/z, where m is mass and z is charge) Since z is almost always +1, m/z actually measures the mass (m) of the individual ions

7. Methane CH 4 Though most C atoms have an atomic mass of 12, 1.1% have a mass of 13. Thus, 13 CH 4 is responsible for the peak at m/z = 17. This is called the M + 1 peak.

8. Alkyl Halides and the M+2 peak –Most elements have one major isotope. –However some halogens have more than one. Iodine and Fluorine are isotopically pure but…. –Chlorine has two common isotopes, 35 Cl and 37 Cl, which occur naturally in a 3:1 ratio. –The larger peak, M, which corresponds to the compound containing 35 Cl. The smaller peak, the M+2 peak, corresponds to the compound containing 37 Cl. –Thus, when the molecular ion peak consists of two peaks (M, M+2) in a 3:1 ratio, a Cl atom is present. –Bromine also has 2 isotopes, 79 Br and 81 Br in a 1:1 ratio. Thus when the molecular ion consists of two peaks (M, M+2) in a 1:1 ratio, the compound contains a Br atom.

11. Sample introduced into GC inlet vaporized at 250 °C, swept onto the column by He carrier gas & separated on column. Sample components emerge from column, flowing into the capillary column interface connecting the GC col-umn and the MS (He removed). Gas Chromatography/Mass Spectrometry (GC/MS)

12. Mass Spectrometry Gas Chromatography-Mass Spectrometry (GC-MS) To analyze a urine sample for tetrahydrocannabinol, (THC) the principle psychoactive component of marijuana, the organic compounds are extracted from urine, purified, concentrated and injected into the GC-MS. THC appears as a GC peak, and gives a molecular ion at 314, its molecular weight.

13. Forensic Mass Spectrometry J. Yinon, Ed., Forensic Applications of Mass Spectrometry, CRC Press, 1995 Analysis of Body Fluids for Drugs of Abuse Analysis of Hair in Drug Testing Sports Testing Analysis of Accelerants in Fire Debris Analysis of Explosives Use of Isotope Ratios

14. pentane

15. 1-pentene

16. 1-pentyne

17. What is the mass of the molecular ion of C 3 H 6 O? 3 C’s 3 x 12 = 36 6 H’s 6 x 1 = 6 1 O 1 x 16 = 16 36 + 6 + 16 = 58 M = M + (m/z) = 58

18. What molecular ions would you expect for C 4 H 9 F? 4 C’s 4 x 12 = 48 9 H’s 9 x 1 = 9 1 F 1 x 19 = 19 48 + 9 + 19 = 76 A molecular ion peak at m/z = 76

19. What molecular ion peak would you expect for C 5 H 11 Cl? 5 C’s 5 x 12 = 60 11 H’s 11 x 1 = 11 1 Cl 1 x 35 = 35 60 + 11 + 35 = 106 C 5 H 11 35 Cl (m/z) = 106 ? 5 C’s 5 x 12 = 60 11 H’s 11 x 1 = 11 1 Cl 1 x 37 = 37 60 + 11 + 37 = 108 C 5 H 11 37 Cl (m/z) = 108 So the molecular ion peak of C 5 H 11 Cl consist of two peaks at 106 and 108 in a 3:1 ratio.

20. C N H 2N+2

21. C3H4C3H4 C N H 2N C N H 2N-2

25. (2(3)+2-5-1)/2=1 (2(4)+2-7+1)/2=2

26. Suggest possible formulas for a molecular ion (m/z) of 72. Step 1 – Determine the maximum number of C’s. 72/12 = 6 carbons maximum C 6 is not a reasonable formula Subtract 1 carbon and add 12 H’s C 5 H 12 Step 2 – Calculate Degrees of Unsaturation (2n+2-#H’s)/2 (2(5) + 2 – 12)/2 = 0 Step 3 – Incorporate O into the formula (-CH 4 when adding O) C4H8OC4H8O (2(4) + 2 – 8)/2 = 1

27. Add another O atom C3H4O2C3H4O2 (2(3) + 2 – 4)/2 = 2 Adding O adds one degree of unsaturation. Suggest possible formulas for a molecular ion (m/z) of 105. Step 1 – If the mass of the molecular ion is odd it contains at least one N. N = 14 amu105 – 14 = 91 Step 2 – Determine max # C’s 91/12 = 7.5C 7 NH?

28. Step 3 – Add enough H’s to make up the rest of the mass. C 7 NH? 7 x 12 = 84 1 x 14 = 14 105 – (84 + 14) = 7 7 H’s gives C 7 NH 7. (2(7.5) + 2 – 7)/2 = 5 Step 4 – Add an O atom. C 7 NH 7  C 6 NOH 3 (2(6.5) + 2 – 3)/2 = 6

29. Suggest a structure for a molecular ion peak that has 2 peaks 144 and 146 in a 1:1 ratio. Step 1 – Since we have an M and M + 2 peak as the molecular ion, we know that there is a halogen. Also since they occur in a 1: 1 ration we know it’s Br. 144 – 79 = 65146 – 81 = 65 Step 2 – Determine max # C’s 65/12 = 5 Carbons Step 3 – Add enough H’s to make up the rest of the mass. 5 x 12 = 60 144 – (60 + 79) = 5 H’s C 5 BrH 5 (2(5) + 2 – 6)/2 = 3

30. C 6 H 6 m/z = 78 b.p. = 80.1  C C 7 H 8 m/z = 92 b.p. = 110.6  C C 8 H 10 m/z = 106 b.p. = 138.3  C Since the sample consists of three components, the GC spectrum will have 3 peaks. Their order will be benzene, toluene and p-xylene in order of increasing boiling point. And the mass spectra of these compounds will have molecular ion peaks corresponding to their molecular weights

31. Benzene

32. Toluene

33. p-xylene

34. Electromagnetic spectrum Infrared region is 2.5 – 25  m

35. Absorption of infrared light (heat) by a compound. Different functional groups absorb at different wavelengths.. Absorptions are recorded in wavenumber = 1/ Infrared Spectroscopy Using this scale, the IR region is 4000-400 cm -1. Chemical bonds are not static, they have different vibrational modes, such as bending and stretching.

36. Different kinds of bonds vibrate at different frequencies, therefore they absorb different wavelengths of radiation. IR spectroscopy distinguishes between the different types of bonds, thus allowing the identification of the functional groups present. IR spectra are a plot of wavelength or wavenumber (x axis) versus transmittance (y axis). Transmittance is a measure of the light that isn’t absorbed by the sample.

37. There are two sections of the IR spectra, the functional group region (greater than 1500) and the fingerprint region (less than 1500). We will be concerned with the functional group region.. Bond strength and the wavelength of absorption are proportional. Thus, the stronger the bond the higher the wavelength of absorption. 4000-2500 2500-2000 2000-1500 1500-400 Bonds to H C-H, N-H, O-H Triple Bonds Double Bonds Single Bonds C  C, C  N C=C, C=O, C=N C  C, C  O, C  N, C  X

38. O-H stretch appears at 3200-3600, C-H at 3000. -OH   -CH

39. N-H stretch appears at 3200-3500, C-H at 3000. -NH   -CH

40. C  C stretch appears at 2250, C-H at 3000.  -CH  - C  C

41. C  N stretch appears at 2250, C-H at 3000. - C  N   -CH

42. C=O stretch appears at 1650-1800, C-H at 3000. - C=O   -CH

43. C=C stretch appears at 1650, C-H at 3000. - C=C   -CH

44. Which of these two isomers, cyclopentane or 1-pentene, is this the IR spectra of? 1650 cm -1 3000 cm -1

46. What IR peaks would you expect for the two molecules with the formula C 2 H 6 O? EthanolDimethyl ether -OH 3200 - 3600 cm -1 -CH 3000 cm -1

47. IR spectra of ethanol. The spectra of dimethyl ether would look similar minus the large –OH stretch. 3200 - 3600 cm -1 3000 cm -1

48. Which of the three compounds below matches the IR spectra? 1650 - 1800 cm -1 3000 cm -1

49. Using the mass and IR spectra, determine the identity of the compound containing only C, H and O? 100 58

50. Determine the # of C’s. 100/12 = 8 C’s 8 x 12 = 96 100 – 96 = 4 So C 8 H 4 We know there is at least one oxygen so subtract a – CH 4 group. C 7 O DOUS=8 also not a likely formula So get rid of 1 C and add 12 H’s C 6 H 12 O DOUS=1 looks much better Now add one more O for more possibilities. C 5 H 8 O 2 DOUS=2 DOUS=7

51. C 6 H 12 O C5H8O2C5H8O2

52. 1650-1800 cm -1 1650 cm -1

62. 7 7 4 4 7 2 3

63. 4 3 4 2 3105