5. Carbon-13 NMR Symmetry: number of chemically different Carbons Chemical Shift: chemical environment of Carbons (e- rich or e- poor)

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Leonard Barber
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1 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 alide 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) Elemental Analysis the empirical formula Ex. Calculate the empirical formula for a compound whose elemental analysis is the following: %C = % = %N = ow to do? Convert percentages to grams Assume the 100% from the formula is 100 grams total, and thus all the percentages can become grams. Then, convert grams to moles by dividing by the atomic weights: %C = 71.22/ = % = 14.94/ = %N = 13.84/ = To obtain the smallest whole number ratio (i.e. mole ratio of elements, the empirical formula), divide each by the smallest value: %C = 71.22/ = 5.930/0.988 = 6.00 % = 14.94/ = 14.82/0.988 =

2 %N = 13.84/ = 0.988/0.988 = 1.00 Empirical Formula: C 6 15 N (with a weight of 101 a.m.u.) One more example: Calculate the empirical formula: %C = % = %O = %C = 47.35/ = 3.94/2.62 = 1.5 % = 10.60/1.008 = /2.62 = 4 %O = 42.05/ = 2.62/2.62 = 1 ow do you deal with a half factor or a third or two-thirds factor? Multiply all values by 2 or 3 Empirical Formula = C 3 8 O 2 Mass Spectroscopy v Used to measure molecular mass v Typically used to determine the molecular weight of a compound v Can also determine structural information 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. 2

3 The x-axis mass-to-charge ratio, or more simply, the mass of the fragments, because the charge is always +1. The y-axis of a mass spectrum is intensity, or the relative number of times a cation fragment appears in the detector. The most abundant fragment is referred to as the Base Peak and is assigned the a value of 100% (999/1000 on your tables) and the amounts of the other fragments are measured relative to this most abundant fragment. The initial cation radical, the Molecular Ion, 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. Consider a compound whose empirical formula is C 3 8. This empirical formula has a mass of 44 a.m.u. The molecular ion appears at 44 a.m.u. Since the molecular ion matches the empirical formula, the empirical formula must be the same as the molecular formula. Now, consider a compound whose empirical formula is C 2 5. This empirical formula has a mass of 29 a.m.u. The molecular ion appears at 58 a.m.u. Since the molecular ion weighs 3

4 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 5 ) or C Consider a hydrocarbon whose molecular ion is 84. Determine a possible molecular formula. 1. What s the maximum number of Carbons in 84? 7 Carbons would weigh 84 but leaves us with no possible hydrogens. 6 Carbons would weigh 72, with 12 left over. This means that C 6 12 is a possible molecular formula. Don t go crazy though someone might want to get creative and say ow about C 5 24? While this mathematically weighs 84, it isn t even remotely physically possible! C 5 can only bond to 12! What if you know the compound has C, and O in its structure and the molecular ion is 84? Oxygen weighs 16. With this in mind, the C and must be formulated from Carbons weigh 60. This leaves 8 for the number of hydrogen atoms and an empirical formula of C 5 8 O. Random Question: Draw a structure that corresponds to this molecular formula, C 5 8 O. This is easier to do if you have some idea of how many rings or multiple bonds you have (i.e. how many unsaturations?): Number of Unsaturations = #C ½ ( + X) + ½ (N) +1 ow many unsaturations in C 5 8 O? Possible structures? Consider a compound whose molecular ion is 73 and contains nitrogen. What might be the molecular formula? Note Odd Value for Molecular Ion = presence of an odd number of nitrogen atoms!! 4 Carbons weigh 48 and 1 nitrogen weighs 14. Subtotal would be 62. That would leave 11 hydrogen atoms possible and thus would lead to C 4 11 N. The most important atoms in organic molecules include carbon ( 12 C), hydrogen ( 1 ), oxygen ( 16 O) and nitrogen ( 14 N). These each have one major isotope.** These isotopes are the only isotopes found in the molecular ion. owever, carbon-13 is abundant enough (1.1%) that it can be observed on a mass spectrum. Thus the M+1 peak in the mass spectrum of butane is due to the relatively small number of butane molecules where one of the four carbon atoms is a carbon-13 isotope, instead of carbon-12. 4

5 C 3 C 2 C 2 C 3 MW = 58 C 3 C 2 C 2 C 3 MW = 59 (1 13 C isotope indicated) **Isotopes are atoms of a particular element that differ in their number of neutrons. The atomic masses we see on the period table are actually the weighted average of all isotopes based on their percentages (natural abundance) and masses (variation caused by differing numbers of neutrons in the nucleus), for any given element. The chance of most molecules having two 13 C isotopes is extremely small so you ll never see an M+2 caused by two 13 C isotopes. The instrument could never detect that microscopic percentage. Thus you won t see a peak for M+2 caused by carbon-13 isotopes. OWEVER, there is something called an M+2 that occurs and its caused by chlorine and bromine atoms. These show significant M+2 peaks in their mass spectra as they have alternative isotopes that are relatively abundant. As a result, we can use mass spectroscopy to detect and determine the presence and identity of chlorine and bromine in organic molecules. Identification of alides: 1. Bromides: Br 79, Br 81 Isotopes occur in about a 1:1 abundance ratio (50.7 : 49.3) in nature. The molecular ion, by definition, must contain the Br 79 isotope, and a second 5

6 major peak appears two units higher for those molecules that contain the Br 81 isotope. This means you should have two peaks (M +. and M+2) of almost equal intensity. M + (total weight, including isotope Br 79 ) M+2 (total weight, including isotope Br 81 ) Below is the MS for bromomethane: Roughly 1:1 intensity M + C 79 Br (M+2) + C 81 Br + C3 Notice that the molecular weight was 94 or 96 (whole numbers). If you use the values from the periodic table, you probably would have calculated: (12.011) + 3(1.0079) = and perhaps rounded that to 95. If you look at the mass spectrum, you will see that the fragment indicated by the peak at 95 is much smaller than the peak at 94. Remember that the atomic masses given on the period table are a weighted average of the masses of each isotope based on their natural abundance. Since 79 Br and 81 Br are present in about a 1:1 ratio, this averages out to an atomic mass of ~80, but there is no 80 Br. You must always calculate molecular masses for mass spectroscopy using an individual isotope and do not use the average masses from the periodic table. 2. 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 for Cl: M + (total weight, including isotope Cl 35 ) 6

7 M+2 (total weight, including isotope Cl 37 ) The molecular ion peaks for the two chlorine isotopes are observed at m/z = 50 (M +, 100%) and m/z = 52 (M+2, 33%). M + + C3 C 35 Cl (M+2) + C 37 Cl Roughly 3:1 intensity ow 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 weight peaks, found on the far right of the spectrum but is NOT always the actual furthest peak on the right, due to the existence of isotopes. The mass spectrometer measures the size of each positively charged fragment that travels through the detector. This includes the weight of the specific atoms involved in each fragment. Isotopes of atoms have different weights. While the average weight of hydrogen may be a.m.u., 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. 7

8 By definition, the molecular ion always contains the following isotopes only: Carbon-12, ydrogen-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. Unknown Compound X has an empirical formula of C 6 15 N (101 a.m.u.). The mass spectrum is shown below: 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? What is the molecular formula for Compound X? 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 15 N. 8

2. Unknown Compound Y has an empirical formula of C 4 9 O (73 a.m.u.): As you can see, the mass spectrum for Compound Y shows multiple peaks above 73. Can C 4 9 O be the molecular formula?

9 2. Unknown Compound Y has an empirical formula of C 4 9 O (73 a.m.u.): As you can see, the mass spectrum for Compound Y shows multiple peaks above 73. Can C 4 9 O be the molecular formula? Can 73 be the molecular ion? No If its not a direct match, it must be a multiple of the weight of the empirical formula. Try doubling or tripling the weight of the empirical formula, to see where it matches a peak on the far right hand side. 73 x 2 = 146. On the spectrum, you ll see a molecular ion peak at 146. What? You can t see it?? It s plain as day Even the little ones count 9

What is the molecular formula for Compound Y? The molecular ion is found at 146 and the molecular formula is C 8 18 O 2. Qual Lab: You have been assigned TWO separate, different unknown compounds.

10 What is the molecular formula for Compound Y? The molecular ion is found at 146 and the molecular formula is C 8 18 O 2. Qual Lab: You have been assigned TWO separate, different unknown compounds. For each you should now: 1. Determine your empirical formulas 2. Determine your molecular ions and 3. Determine your molecular formulas for your two unknowns. 4. Analyze your IR s and proton NMR s to determine your functional groups and build the structure (carbon skeleton) of your molecule. Proton NMR is next Next week Carbon-13 NMR the final confirmation of your molecular structure. 10

Qualitative Analysis of Unknown Compounds

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