Qualitative Analysis of Unknown Compounds 1. Infrared Spectroscopy Identification of functional groups in the unknown All functional groups are fair game (but no anhydride or acid halides, no alkenes or alkynes) 2. Elemental Analysis Determination of the Empirical Formula of the Unknown Smallest whole number ratio of elements in the formula (mole ratio of elements) 3. Mass Spectral Analysis Determination of the Molecular Formula of the Unknown Actual number of atoms in the formula Molecular Weight of the Unknown Halide identification 4. Proton NMR Symmetry: the number of chemically different protons Chemical Shift: chemical environment of protons (e- rich or e- poor) Integration: ratio of protons Splitting Patterns: arrangement of neighboring protons (how many next door) 5. Carbon-13 NMR Symmetry: number of chemically different Carbons Chemical Shift: chemical environment of Carbons (e- rich or e- poor) Also: Splitting Patterns Number of attached protons (not required for unknowns, only for lab exam) Elemental Analysis the empirical formula Ex. Calculate the empirical formula for a compound whose elemental analysis is the following: %C = 47.35 %H = 10.60 %O = 42.05 How to do? Convert percentages to grams Assume the 100% from the formula is 100 grams total, and thus all the percents can become grams. Then, convert grams to moles by dividing by the atomic weights: %C = 47.35/12.011 = 3.94 %H = 10.60/1.008 = 10.516 %O = 42.05/ 15.999 = 2.62 To obtain the smallest whole number ratio (i.e. mole ratio of elements, the empirical formula), divide each by the smallest value: %C = 47.35/12.011 = 3.94/2.62 = 1.5 %H = 10.60/1.008 = 10.516/2.62 = 4 %O = 42.05/ 15.999 = 2.62/2.62 = 1 1
How do you deal with a half factor or a third factor? Multiply all values by 2 or 3 Empirical Formula = C 3 H 8 O 2 Mass Spectroscopy v Used to measure molecular mass v Typically used to determine the molecular weight of a compound v The Process: Ionization of a compound caused by loss of an electron, results in the formation of a species that is positively charged the cation radical called the Molecular Ion. No longer neutral, the molecular ion fragments after formation, into smaller pieces, some of which are radicals and some that are cations. The cations are detected in the instrument and sorted according to mass. The x-axis mass-to-charge ratio (m/z) = the mass of the fragments, because the charge is always +1. The y-axis of a mass spectrum is Relative Abundance = the relative number of times a cation fragment appears in the detector. The most abundant fragment is arbitrarily assigned the value of 100% (999/1000 on your tables) and the amounts of the other fragments is measured relative to this fragment, referred to as the Base Peak. The initial cation radical contains all the atoms of the original compound and is only missing one electron, of negligible weight. Thus the Molecular Ion has the same mass as the original compound. Using the molecular ion, and the empirical formula, one can determine the molecular formula for a compound. 2
Consider a compound whose empirical formula is C 2 H 5. This empirical formula weighs 29 a.m.u. The molecular ion appears at 58 a.m.u. What s the numerical relationship? Since the molecular ion weighs twice that of the weight of the empirical formula, you know you need to double the empirical formula to find the true molecular formula or (2 x C 2 H 5 ) or C 4 H 10. Determine the molecular formula for a compound whose molecular ion is 73 and contains carbon, hydrogen and nitrogen. Begin with the non-carbon/non-hydrogen atoms - Nitrogen, 14 amu. 73-14 = 59 How many Carbons fit into 59? 4 Carbons weigh 48. 59-48 = 11 Empirical Formula = C 4 H 11 N. Draw a structure that corresponds to this molecular formula, C 4 H 11 N. Determine how many unsaturations exist: UN = #C ½ (H + X) + ½ (N) +1 Possible structures? Identification of Halides: Review: Isotopes are variations of the same elements, with differing numbers of neutrons in the nucleus. 1. Chlorides: Cl 35, Cl 37 Isotopes occur in a 3:1 abundance ratio in nature. The molecular ion, by definition, must contain the Cl 35 isotope, and a second major peak appears two units higher for those molecules that contain the Cl 37 isotope. Required Peaks to identify: M + (total weight, including isotope Cl 35 ) = 78 (see below) M+2 (total weight, including isotope Cl 37 ) = 80 Below is the MS for 1-chloropropane: 3
2. Bromides: Br 79, Br 81 Isotopes occur in about a 1:1 abundance ratio in nature. The molecular ion, by definition, must contain the Br 79 isotope, and a second major peak appears two units higher for those molecules that contain the Br 81 isotope. Required Peaks to identify: M + (total weight, including isotope Br 79 ) = 108 (see below) M+2 (total weight, including isotope Br 81 ) = 110 Below is the MS for bromoethane: 4
How does one determine which peak is the molecular ion? Since it encompasses the molecular weight of the entire compound (minus only one electron), it will be one of the highest mass peaks, found on the far right of the spectrum but is NOT necessarily the furthest peak on the right, due to the existence of isotopes. Isotopes of atoms have different weights. While the average mass of hydrogen may be 1.0079 amu, the MS detector only sees the specific isotopes, hydrogen-1, deuterium (hydrogen-2) and/or tritium (hydrogen-3) involved in a fragment. You cannot use AVERAGES when dealing with your mass spectra. Stay away from the values shown on the periodic table. By definition, the molecular ion always contains the following isotopes only: Carbon-12, Hydrogen-1, Oxygen-16 and Nitrogen-14. The halides are always the lower of the two isotope pairs, Chlorine-35 and Bromine-79. Now: Tie Elemental Analysis and Mass Spectroscopy Together 1. Calculate the Empirical Formula. Consider Unknown Compound X has an empirical formula of C 6 H 15 N (101 a.m.u.): The Mass Spectrum for Compound X shows a series of peaks on far right of spectrum at 100, 101 and 102. Which peak is the molecular ion? The molecular ion MUST either MATCH or be a multiple of the weight of the empirical formula. If C 6 H 15 N weighs 101 amu, then the molecular ion must equal 101 or 202 or 303. When the weight of the molecular ion matches the weight of the empirical formula, you know that the empirical formula IS the molecular formula: C 6 H 15 N. 5
2. Consider Unknown Compound Y has an empirical formula of C 4 H 9 O (73 a.m.u.): As you can see, the Mass Spectrum for Compound Y shows multiple peaks above 73. AS such, C 4 H 9 O cannot be the molecular formula. Try doubling or tripling the weight of the empirical formula it to see where it matches the molecular ion (146, 219, etc.) On the spectrum, you ll see a molecular ion peak at 146. What is the molecular formula for Compound Y? The molecular ion must be 146 and the molecular formula is C 8 H 18 O 2. 6
Qual Lab: You will have TWO separate, different unknown compounds. For each you will: 1. Determine your empirical formulas 2. Determine your molecular ions and 3. Determine your molecular formulas for each unknown. 4. Analyze your IR s and proton NMR s to determine your functional groups and build the structure (carbon skeleton) of your molecule. 7