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Qualitative Analysis of Unknown Compounds 1. Infrared

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Qualitative Analysis of Unknown Compounds 1. Infrared Qualitative Analysis of Unknown Compounds 1. Infrared Spectroscopy • Identification of functional groups in the unknown • Will take in lab today 2. Elemental Analysis • Determination of the Empirical Formula of the Unknown • Smallest whole number ratio of elements in the formula (mole ratio of elements) • Given on form with Assignment ID number 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 • Hand-out with both graph and table 4. Proton NMR • Symmetry: Number of chemically different protons • Chemical Shift: Chemical environment of protons • Integration: Ratio of protons • Splitting Patterns: arrangement of neighboring protons • Hand-out with spectrum (x-axis only) Elemental Analysis –the empirical formula Remember, in order to calculate the empirical formula, you need to take the percentages from the elemental analysis and assume these are gram amounts which need to be converted to moles (because formulas are mole ratios). Ex. Calculate the empirical formula for a compound whose elemental analysis is the following: %C = 68.31 %H = 11.47 %O = 20.22 What next? %C = 68.31/12.011 = 5.687 %H = 11.47/1.008 = 11.380 %O = 20.22/16.00 = 1.264 What next? %C = 68.31/12.011 = 5.687/1.264 = 4.499 ≈ 4.5 %H = 11.47/1.008 = 11.380/1.264 = 9.003 ≈ 9 %O = 20.22/16.00 = 1.264/1.264 = 1 What do you do when the value comes up at a “half”, as in 4.5 or 2.5? Multiply all of your values by by 2… Empirical Formula: C9H18O2 (MW of 158 g/mol) What do you do if you have a third (as in 3.3333) or two thirds (4.66667)? Multiply all by 3! There are only three options when you do the math: (a) it should result in perfect or almost perfect whole numbers, (b) possible half values (multiply by 2) or (c) or possibly thirds (one-third or two thirds values – multiply by 3). Anything else is a sure sign of a mathematical error! Formulas are helpful when determining structures. Not so much the empirical formula - being only the smallest whole number ratio, it doesn’t contain all of the atoms in the molecule you are studying so it is better to know the molecular formula before determining a structure. When you know the molecular formula, you can have some added guidance by being able to determine a possible structure. The Unsaturation Number (or Degrees of Unsaturation) can be calculated from the molecular formula. Remember that every degree of unsaturation is equivalent to a loss of 2 hydrogen atoms from the maximum allowed in an alkane with the same number of carbon atoms. This occurs every time there is a ring structure or a pi bond (double bond has one pi bond, triple bond has two pi bonds). How many unsaturations in: O NH 2 Two rings (2), 7 pi bonds = 9 unsaturations. Each of the following has two degrees of unsaturations, and fit the formula C6H10O: O O O H O To calculate the Unsaturation Number (degree of unsaturations) from a formula: #C – ½(#H + #X) + ½(#N) + 1 2 Calculate the number of unsaturations in: C12H16INO2 UN = 12 – ½(17) + ½(1) + 1 = 12 – 8.5 + 0.5 + 1 = 5 Always helpful to keep in mind: if your molecular formula results in an Unsaturation Number of 4 or greater, there’s a very strong chance that you have an aromatic ring in your structure (ring structure plus three pi bonds). It is also always helpful to calculate this information as soon as you have a formula as it may prevent you from making incorrect assumptions. Consider C8H18O2 – with two oxygen atoms, you may automatically jump to the conclusion that you have a carboxylic acid or ester functional group but when you calculate the UN (= 0), you realize that neither of these functional groups are possible, as they both have one unsaturation in their structure. O O OH OR 1 unsaturation Perhaps you have two alcohols or an alcohol and an ether: OH O no unsaturations To determine the molecular formula, we will need to discuss mass spectroscopy. Mass Spectroscopy: Purpose: Determination of a compound’s molecular weight (as well as structural information) Here’s the basic process: A small sample to be studied is vaporized then ionized by being bombarded by high energy electrons (like the molecule was zapped with a lightning bolt). The energy of the electrons in that lightning bolt correspond to >1000 kcal/mol of energy. Under these conditions, the organic molecule is destroyed, breaking into a number of fragments and we can use the fragmentation to help us determine structural information. When the high-energy electron (e-) impacts an organic molecule (M), one of the molecule’s valence electrons (e-) is ejected, resulting in the formation of an ion call the Molecular Ion (M+•): 3 M + e- → M+• + e- + e- This new ion is a “radical cation”. A radical is a species that contains an unpaired electron (but is neutral in charge) and a cation is a species missing an electron, resulting in an overall positive charge. One this happens, the molecular ion can then begin to fragment to give a number of other cations and neutral, radical species. These fragments are then passed through a magnetic field where they are separated and detected according to mass. Only the charged species (cations) can be detected by mass spectroscopy since neutral species (radicals) cannot be affected by the magnetic field. Technically, we are actually measuring the “mass to charge ratio” (m/z, where m = mass and z = charge) for each ion but since the charge is virtually almost always +1, this really is simply the mass of the fragment. The Molecular Ion: The molecular ion is still the entire molecule, minus only one electron, the weight of which is negligible, thus the weight of the molecular ion equates to the molecular weight of the entire molecule. Shown below is the mass spectrum for Butane. Knowing that carbon weighs 12 and hydrogen atoms weigh 1 each, butane (C4H10) has a molecular weight of 58. Find the molecular ion (M+•) peak that corresponds to the MW. The mass spectrum graph is a plot of Mass-to-Charge Ratio, m/z, (X axis) versus Abundance (Y axis). 4 • Mass-to-Charge Ratio can be simplified to “mass of fragment” since the charge on each is +1. • Abundance is the relative number of times that each fragment passed through the detector. The more often the fragment formed and passed by, the larger the abundance for the fragment (the larger the “peak”). What can you get from knowing the molecular weight? You can figure out a molecular formula. 1. Let’s assume you know you have a compound that only has carbons and hydrogens and whose molecular weight (a.k.a. molecular ion) is 58 a.m.u. What can you determine? Start by determining how many carbon atoms can fit into a weight of 58. • How many carbon atoms fit into 58? § 4 carbons weigh 48 § 58 – 48 = 10. • That leaves you 10 left over for hydrogens. • Molecular formula: C4H10. 2. What if the compound also has oxygen in its structure (with a molecular ion of 58)? What do you have to save room for in this case? • Start by subtracting 16 (from the oxygen) from the total of 58. § 58-16 = 42. • How many carbons fit into 42? § 3 carbons weigh 36. § 42 – 36 = 6 • That leaves 6 left over for hydrogens • Molecular Formula: C3H6O. Fragmentation occurs when the molecular ion is produced with sufficient internal energy to cause spontaneous dissociation. This is similar to the spontaneous dissociation of water in the dehydration of an alcohol to form a stable carbocation. Fragmentation occurs in a “logical” fashion to form stable intermediates, both cations and radicals. Tertiary systems are more stable than secondary and these are more stable than primary. The fragments that we see on the spectrum are called “daughter peaks”. The largest peak on the spectrum is called the Base Peak, whose relative intensity is set to 100 (“999” on the table on your unknown’s spectrum). The base peak MAY be the same as the molecular ion but more often than not, they are different peaks. Notice 5 the large number of peaks found to the left of the molecular ion – those are from the fragment ions that formed. Note the tiny peak at m/z = 59? This is called an M+1 peak (one amu higher than the molecular ion). How did THAT get in there? What weighs more than the entire molecule? To understand this, we need a little reminder on isotopes… 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. Here is a table summarizing mass and natural abundance for several isotopes relevant to organic molecules and mass spectroscopy: 6 Isotope Atomic Mass Natural Abundance (%) 1H 1.0079 99.985 2H (D) 2.010 0.015 12C 12 98.9 13C 13.003 1.1 14N 14.003 99.63 15N 15 0.37 16O 15.995 99.76 17O 16.999 0.038 18O 17.999 0.2 19F 19.998 100 35Cl 34.969 75.77 37Cl 36.966 24.23 79Br 78.918 50.69 81Br 80.916 49.31 127 I 126.905 100 The most important atoms in organic molecules include carbon (12C), hydrogen (1H), oxygen (16O) and nitrogen (14N).
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