Spectroscopy_Workbook_1

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Unformatted text preview: Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ Experiment 8 Spectroscopy Workbook I This workbook consists of instruction and problems related to the elucidation of structural information based on one or more forms of spectroscopic data. It is often the case that one form of spectroscopy is not enough to determine the structure, which is why multiple techniques used in tandem are usually the standard practice of an organic chemist to characterize new materials or verify existing ones. Typically, mass spectrometry is used to ascertain the molecular weight (although it can tell much more if studied in detail) infrared spectroscopy is used to determine the functional groups present, and NMR is used to deduce the connectivity of the atoms. Together, these results give us a relatively undisputable structure. Part 1. Determining the Molecular Formula from Mass Spectral Analysis A. The Rule of Thirteen This method is used to determine the molecular formula from the observed molecular ion peak of the mass spectrum for hydrocarbons. Since the relative mass to charge ratio is determined in this technique, often the peak at the highest mass pertains to the compound of interest which we term the molecular ion. This is simply the compound minus 1 electron. Because the mass of an electron is fairly negligible compared to the mass of protons and neutrons, the chemical formula can be determined. Practice Question 1: The molecular ion is observed to be at m/z 72. What is the molecular formula? We know that carbon has a mass of 12 amu and hydrogen 1 amu and the sum 13. Dividing the mass of 72 by 13 gives the number of carbon atoms: 5. The remainder left over, 7, is added to the number of carbons, 5, to give the number of hydrogens: 12. Therefore, the formula is C5H12. Question 1: The molecular ion is observed to be at m/z 108. What is the molecular formula? Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ Question 2: The molecular ion is observed to be at m/z 54. What is the molecular formula? B. Applying the Rule of Thirteen to Hydrocarbons that Contain Oxygen You must be told the number of oxygen atoms in order to solve this type of question. When solving the formula for hydrocarbons containing oxygen, you must consider that the mass of oxygen, 16 amu, will replace a single carbon and four hydrogens, also 16 amu. Practice Question 2: Consider a molecular ion at m/z 92. What is the molecular formula if it includes 1 oxygen atom? First, apply the rule of thirteen to the molecular ion: 92/13 = 7 r1 which would give the hydrocarbon C7H8. Since the mass of one oxygen atom equal to CH4, we must subtract this from the hydrocarbon formula and add oxygen: C7H8  ­ CH4 + O = C6H4O Question 3: What is the molecular formula of a compound that exhibits a molecular ion at m/z 100 and is known by infrared spectroscopy to have an ester functional group? C. The Fundamental Nitrogen Rule The rule is that compounds that include the atoms C, H, O, and N will have an even mass to charge ratio if there are an even number (or zero) of nitrogen atoms. This is due to the fact that carbon has four bonds and will always give even numbers (1 carbon and 4 hydrogens = 16 amu). Oxygen maintains this even order because when it is doubly bonded to carbon it is in the place of 2 hydrogens and when it is singly boned, it carries either a single hydrogen to even it out (as in an alcohol) or another carbon chain (ether, ester). Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ Practice Question 3: Consider a molecular ion at m/z 87. What is the molecular formula if it includes nitrogen atom? Follow the normal rule of thirteen to arrive at C6H15. Because one nitrogen atom is equal to 14 amu, subtract CH2 and add N. C6H15  ­ CH2 + N = C5H13N Question 4 Consider a molecular ion at m/z 135. What is the molecular formula if it the infrared spectrum shows evidence of an amine? Question 5 Consider a molecular ion at m/z 99. What is the molecular formula if it the infrared spectrum shows evidence of an amide? Part 2. Determining the Degrees of Unsaturation from the Molecular Formula A. Saturation refers to the hydrogens of a hydrocarbon. A saturated fat, for instance, has no double bonds so we say it is saturated with hydrogens. Conversely, an unsaturated fat does have one or many double bonds. The degree to which something is saturated can give important structural information. The equation to calculate the degree of unsaturation (U) is given below: U= (2C+2) ­H 2 Where C is the number of carbons and H the number of hydrogens or halogens. Each nitrogen counts as ½ C. Other atoms are necessary to be factored into the equation (i.e. oxygen, sulfer). The unsaturation number implies certain functionality about a compound: Chem 6AL Russak Fall 2010 Unsaturation Number 0 1 2 3 4 Name_________________________________ TA/Section___________________________ Implication No double bonds or rings One ring or double bond One triple bond or 2 double bonds or 2 rings or some combination Many combos of the previous Usually means it is aromatic Practice Question 4 Consider the formula C4H10. What is the unsaturation number and what are some possible structures for this formula? U = [(2x4+2) ­10]/2 = 0 sites of unsaturation, which means no double bonds or rings. Possible structures that can be drawn are: Question 6 Use the derived molecular formula from question 4 to determine the degree of unsaturation. There may be many possible, but draw one possible structure based upon the formula and the degree of unsaturation. Question 7 Use the formula from practice question 2 to determine the degree of unsaturation. There may be many possible, but draw one possible structure based upon the formula and the degree of unsaturation. Chem 6AL Russak Fall 2010 Question 8 Name_________________________________ TA/Section___________________________ Part 3. Elucidating the Structure given 1H NMR spectra A. In order to effectively solve for a structure given certain molecular information and its proton NMR you should follow these steps: 1) Determine the degree of unsaturation 2) Consider any integration given 3) Conduct a spectral analysis that involves the number of signals and their position. 4) Investigate the splitting patterns and try to correlate them to functionality (i.e. a triplet and a quartet may mean an ethyl group) 5) Deduce information about the molecule based on absence of certain information (i.e. an aromatic structure is not possible because there are no signals between 7 ­8) 6) Idenitify the unknown by drawing the structure, labeling the non ­ equivalent protons with numbers or letters, and annotating which signals correspond to which proton Practice Question 5 Consider the following NMR spectra where the signals and integration are st(1H), s(3H), d(6H) from left to right. Given that a molecular ion peak shows up at m/z 86 and contains 1 oxygen, elucidate the structure. Be sure the annotate the structure and label the peaks. (see next page) Use the formula from question 5 to determine the degree of unsaturation. There may be many possible, but draw one possible structure based upon the formula and the degree of unsaturation. Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________  ­First, solve for the molecular formula which you should find to be C5H10O. Based on this, the degree of unsaturation is 1 which means a double bond or ring is possible in the structure. Solve for U in all further questions in this workbook.  ­Based on the integration, you have 1 hydrogen that sees 7, 3 hydrogens that see none, and 6 hydrogens that see 1. The prevalence of a septet (1H) and a doublet (6H) means there is an isopropyl group. 3 hydrogens that see none correlates to a methyl group.  ­Next, notice that there are 3 signals in the NMR which means 3 non ­ equivalent hydrogens. Based on the number of carbons, there must be either some symmetry or some carbons do not bear any hydrogens at all or possibly a combination of both.  ­Deduce information: There is most likely not an alcohol due to the lack of a singlet proton. The absence of any alkene protons tell us that there is no unsaturation, which means the unsaturation must be to the oxygen atom. This can be in the form of an aldehyde or a ketone. Based on the lack of a proton between 9 ­10 ppm, it can’t be an aldehyde.  ­Based on this information, the structure would be 3 ­methyl ­2 ­ butanone. Labeling the signals as such on the structure, the signals on the NMR spectrum above would be labeled from right to left: a, b, c O (c)H3C CH3(b) H ( a) CH3(c) Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ Question 9 The mass spectrum showed an M+ (m/z =92) to M+2 peak at roughly 3:1 intensity. Based on this and the 1H NMR below, determine the molecular formula, degrees of unsaturation, elucidate the structure, label and annotate the structure and the spectra. You will answer these same things for the subsequent questions. δ0.9 (3H), δ1.3 (2H), δ1.8 (2H), δ3.6 (2H) Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ Question 10 This compound has the same M+ and M+2 as question 9. δ1.0 (6H), δ1.9 (1H), δ3.5 (2H) Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ Question 11 A molecular ion at was found at 100. δ0.7 (2H), δ0.8 (12H), δ1.5 (2H) Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ δ1.0 (6H), δ2.4 (4H) Question 12 A molecular ion was found at 86 and has 1 oxygen. Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ Question 13 A molecular ion was found at 120 and has 1 oxygen. δ2.4 (3H), δ7.15 (2H), δ7.9 (2H), δ10.0 (1H) Zoom in view on peaks between 7 and 8 ppm Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ Question 14 A molecular ion was found at 102 and contains 1 oxygen. δ0.9 (6H), δ1.65 (4H), δ3.35 (4H) Chem 6AL Russak Fall 2010 Name_________________________________ TA/Section___________________________ For questions 15 ­19, determine the structure of each of the following unknown compounds by the same methods as previous (don’t forget to calculate the unsaturation number) Question 15: C3H7Br δ1.71 d(6H), δ4.29 heptet(1H) Question 16: C4H9Cl δ1.62 singlet Question 17: C4H8O δ1.1 t(3H), δ2.1 s(3H), δ2.5 quartet(2H) Question 18: C4H10O2 δ3.4 s(6H), δ3.6 t(4H) Question 19: C8H10O δ1.5 t(3H), δ4.1 quartet(2H), δ7.0 ­8.0 m(5H) ...
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