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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. 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 Aromatics appears at 1500 and 1600, C-H at 3000. Monosubstitution on an aromatic ring shows at around 700. -CH - Aromatic - Monosubstituted