CHE 232 MtWTF 8-8:50pm 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 CO2, H2O 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. • 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. • 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. • 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 Methane CH4 Though most C atoms have an atomic mass of 12, 1.1% have a mass of 13. Thus, 13CH4 is responsible for the peak at m/z = 17. This is called the M + 1 peak. • 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, 35Cl and 37Cl, which occur naturally in a 3:1 ratio. – The larger peak, M, which corresponds to the compound containing 35Cl. The smaller peak, the M+2 peak, corresponds to the compound containing 37Cl. – 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, 79Br and 81Br 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. Cl Cl -CH3 -Cl -CH3 Br -Br -CH3 Gas Chromatography/Mass Spectrometry (GC/MS) 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). 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. Forensic Mass Spectrometry •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 J. Yinon, Ed., Forensic Applications of Mass Spectrometry, CRC Press, 1995 pentane 1-pentene 1-pentyne What is the mass of the molecular ion of C3H6O? 3 C’s 3 x 12 = 36 6 H’s 6x1=6 1O 1 x 16 = 16 36 + 6 + 16 = 58 M = M+•(m/z) = 58 What molecular ions would you expect for C4H9F? 4 C’s 4 x 12 = 48 9 H’s 9x1=9 1F 1 x 19 = 19 48 + 9 + 19 = 76 A molecular ion peak at m/z = 76 What molecular ion peak would you expect for C5H11Cl? 5 C’s 5 x 12 = 60 5 C’s 11 H’s 11 x 1 = 11 11 H’s 11 x 1 = 11 1 Cl 1 x 35 = 35 1 Cl 1 x 37 = 37 60 + 11 + 35 = 106 C5H1135Cl (m/z) = 106 ? 5 x 12 = 60 60 + 11 + 37 = 108 C5H1137Cl (m/z) = 108 So the molecular ion peak of C5H11Cl consist of two peaks at 106 and 108 in a 3:1 ratio. CNH2N+2 CNH2N+2 CNH2N CNH2N-2 C3H4 (2(3)+2-5-1)/2=1 O Br (2(4)+2-7+1)/2=2 O NH2 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 C6 is not a reasonable formula Subtract 1 carbon and add 12 H’s C5H12 Step 2 – Calculate Degrees of Unsaturation (2n+2-#H’s)/2 (2(5) + 2 – 12)/2 = 0 Step 3 – Incorporate O into the formula (-CH4 when adding O) O C4H8O O (2(4) + 2 – 8)/2 = 1 Add another O atom O C3H4O2 O O O (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 amu 105 – 14 = 91 Step 2 – Determine max # C’s 91/12 = 7.5 C7NH? Step 3 – Add enough H’s to make up the rest of the mass. C7NH? 7 x 12 = 84 7 H’s gives C7NH7. (2(7.5) + 2 – 7)/2 = 5 HN 1 x 14 = 14 105 – (84 + 14) = 7 Step 4 – Add an O atom. C7NH7 C6NOH3 O (2(6.5) + 2 – 3)/2 = 6 N 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 = 65 146 – 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 C5BrH5 (2(5) + 2 – 6)/2 = 3 Br C6H6 C7H8 C8H10 m/z = 78 m/z = 92 m/z = 106 b.p. = 80.1C b.p. = 110.6C b.p. = 138.3C 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 Benzene Toluene p-xylene Electromagnetic spectrum Infrared region is 2.5 – 25 m Infrared Spectroscopy Absorption of infrared light (heat) by a compound. Different functional groups absorb at different wavelengths. Absorptions are recorded in wavenumber = 1/ 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. 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. 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 Bonds to H C-H, N-H, O-H 2500-2000 Triple Bonds 2000-1500 Double Bonds CC, CN C=C, C=O, C=N 1500-400 Single Bonds CC, CO, CN, CX O-H stretch appears at 3200-3600, C-H at 3000. -OH -CH N-H stretch appears at 3200-3500, C-H at 3000. -NH -CH CC stretch appears at 2250, C-H at 3000. - CC -CH CN stretch appears at 2250, C-H at 3000. -CH - CN C=O stretch appears at 1650-1800, C-H at 3000. -CH - C=O C=C stretch appears at 1650, C-H at 3000. - C=C -CH Which of these two isomers, cyclopentane or 1-pentene, is this the IR spectra of? 3000 cm-1 1650 cm-1 What IR peaks would you expect for the two molecules with the formula C2H6O? O OH Ethanol -OH 3200 - 3600 cm-1 -CH 3000 cm-1 Dimethyl ether -CH 3000 cm-1 IR spectra of ethanol. The spectra of dimethyl ether would look similar minus the large –OH stretch. 3000 cm-1 3200 - 3600 cm-1 Which of the three compounds below matches the IR spectra? O O 3000 cm-1 OH 1650 1800 cm-1 Using the mass and IR spectra, determine the identity of the compound containing only C, H and O? 58 100 Determine the # of C’s. 100/12 = 8 C’s So C8H4 8 x 12 = 96 100 – 96 = 4 DOUS=7 We know there is at least one oxygen so subtract a – CH4 group. O C7O DOUS=8 also not a likely formula So get rid of 1 C and add 12 H’s C6H12O DOUS=1 looks much better Now add one more O for more possibilities. C5H8O2 DOUS=2 O C6H12O O C5H8O2 O O O O O O 1650-1800 cm-1 1650 cm-1 OH OH 7 4 7 4 2 7 3 3 4 3 2 4 10 5