Chem 6AL Name____________________ Winter 2016 TA/Section_____________________ Spectroscopy Workbook I 1 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 sufficient to determine the structure of an organic compound, which is why multiple techniques are often used in tandem. Typically mass spectroscopy 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 for hydrocarbons (compound containing only carbon and hydrogen). Depending on the type of ionization method used (hard or soft) the molecular ion can be either the peak with the highest mass to charge ratio (m/z) or the peak present in the highest abundance. Most mass spectra that we will see use soft ionization methods, in which the molecular ion will be the ion with the highest m/z. We can use the molecular ion to determine the molecular formula. Practice Question 1: The molecular ion is observed to be at m/z 72. What is the molecular formula? We know that carbon atom has a mass of 12 amu and hydrogen has a mass of 1 amu, for a sum of 13 amu. Dividing the mass by 72 give 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. Thus the molecular formula is C5H12. Question 1: The molecular ion is observed to be at m/z 108. What is the molecular formula? Question 2: The molecular formula is observed to be at m/z 54. What is the molecular formula? Not for Credit, Practice OnlySpectroscopy Workbook I 2 B. Applying the Rule of Thirteen to Hydrocarbons that Contain Oxygen We can use the Rule of Thirteen to determine the molecular formula of compound that contain more than just carbon and hydrogen. When solving the formula for compound containing carbon, hydrogen, and oxygen you must consider that the mass of oxygen (16 amu) will replace a carbon and four hydrogens, which together also have a mass of 16 amu. Practice Question 2: Consider a molecular ion at m/z 92. What is the molecular formula if there is one oxygen present in the molecule? First, apply the rule of thirteen to the molecular ion to get a formula of C7H8. Since the mass of oxygen atom is equal to the mass of a CH4 group, 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 with 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 fundamental nitrogen rule states that for compounds whose molecular ion contains no atoms other than C, H, N, or O atoms and that has an even m/z must contain either no nitrogen atoms or an even number of nitrogen atoms. A compound whose molecular ion has an odd m/z value must contain an odd number of nitrogen atoms. Practice Question 3: Consider a molecular ion at m/z 87. What is the molecular formula if it includes a nitrogen atom? Follow the rule of thirteen to arrive at the molecular formula C6H15. Because one nitrogen atom weighs 14 amu (equal to CH2), 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 the infrared spectrum shows evidence of an amine? Not for Credit, Practice OnlySpectroscopy Workbook I 3 Part 2: Determining the Degree of Unsaturation from the Molecular Formula Saturation refers to the number of hydrogens on 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 more double bonds. The degree to which something is saturated can give important structural information. The equation to calculate the degree of unsaturation (U) is: U = C + 1 −12(H + X − N) Where C is the number of carbon atoms, H is the number of hydrogens, X is the number of halogens, and N is the number of nitrogens. The degree of unsaturation is a powerful number that can tell us a lot about the potential structure of a compound. As degrees of unsaturation increases, the number of potential π-bonds and rings increase. Degrees of Unsaturation Implication 0 No π-bonds or rings 1 One π-bond or ring 2 Two π-bonds, two rings, or one of each 3 Three π-bonds, three rings, or some combination that adds up to 3. 4 Four π-bonds, three rings, or some combination that add up to 4. For most compounds we will see, this will imply the compound contains an aromatic ring Practice Question 4: Consider the formula C4H10. What is the unsaturation number and what are some possible structure for a compound with this molecular formula? U = 4 + 1 −12(10)= 0 The degree of unsaturation of this compound is zero, therefore it cannot contain any π-bonds or rings. Possible structures are: Question 5: Look at your molecular formula from Question 4. What is the degree of unsaturation of this compound? Draw two potential structures of this compound. Not for Credit, Practice OnlySpectroscopy Workbook I 4 Question 6: Look at your molecular formula from Question 2. What is the degree of unsaturation of this compound? Draw two potential structures of this compound. Note: You should solve for U in all further questions in this workbook. Part 3: Elucidating the Structure When Given 1H-NMR Spectra In order to effectively solve for a structure given certain molecular information and its 1H-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 structure (e.g. a triplet and a quartet may mean an ethyl group) 5. Deduce the information about the molecule based on absence of certain information (e.g. an aromatic structure is not possible because there are no signals between 7-8 ppm) 6. Identify the unknown by