CHEMISTRY 318 IR, MS, UV, NMR SPECTROSCOPY PROBLEM SET All Sections
CHEMISTRY 318 IR, MS, UV, NMR SPECTROSCOPY PROBLEM SET General Instructions for the 318 Spectroscopy Problem Set Consult the Lab Manual, the textbooks by Solomons and by Pavia, et al., and the following discussion to help you with the analyses. In the Lab Manual section, Spectroscopy I, there is a section titled "Using On-line Databases to Help Solve Organic Chemistry Spectroscopy Problems". You should review it so you can use on-line sources to help you analyze some of the problem spectra and as a source for literature spectra. The NMR example below is a continuation of the example begun in the section referred to above. The GMU Organic Chemistry web page (http://mason.gmu.edu/~sslayden/lab/index.html) has links to the databases. Short example: Suppose you also have the 1 H-NMR for this compound. If you give it even the briefest of examinations, you have confirmation of the presence of an aromatic ring (how?) and the absence of saturated alkyl groups (where would they appear in the NMR?). The Spectroscopic Tools Wizard (http://www.science-andfun.de/tools/) would be very useful in helping you determine the molecular environment of the hydrogens between 5-8 ppm. Going back to the substructure search form at NIST (http://webbook.nist.gov/chemistry/), you could require the presence of a benzene ring and the exclusion of simple alkyl groups. The return hit list is very short for the molecular weight of this unknown. Instructions Part I: Answer the questions in the space provided on the page. Part II: Each Combined Spectra Problem contains some combination of the following information: IR spectrum, NMR ( 1 H- and 13 C-) spectrum, mass spectrum or molecular ion peak, major UV peak, element % mass. On each spectrum, neatly label the major absorption peaks with the structural fragment that causes the absorption. On the answer sheet that follows each problem, fill in the requested information so that you assemble all the structural information that can be obtained from the spectra. Then, using this information, propose the structure and identity of the compound. You can confirm the identity from literature or on-line sources. A completed example is given. 2
I. 1.) Match the C 13 -NMR chemical shift values (ppm) on the right with the lettered methylene carbon atoms on the left. CH 3 O CH 2 C O CH 2 CH 2 CH 3 a b c 68.9 22.7 26.4 2.) Draw a picture showing the H 1 -NMR splitting patterns (singlet, doublet, etc.) for each methylene group in the structure above. Explain how you determined each splitting pattern. 3.) Draw an expanded structural formula for each compound below. How many different resonance peaks would you expect to find in the H 1 -NMR spectrum of each of the compounds (ignore spin-spin splitting)? Justify your answer. [If you have a problem with the drawing the structure or to confirm that your structure is correct look up the compound name in an on-line source.) a.) propene b.) ethyl methyl ether c.) propane 4.) Draw an expanded structural formula for each compound. How many resonance peaks should be observed in the hydrogen-decoupled C 13 -NMR spectrum of each of the following compounds? Justify your answer. [If you have a problem with the drawing the structure or to confirm that your structure is correct look up the compound name in an on-line source.) a.) toluene b.) p-chlorobenzoic acid c.) 2-propanol 3
5.) In the spectrum below, there are two sets of hydrogens (TMS is the internal standard). To measure the peak integrals, draw horizontal lines that extend the base line for each peak's integration, as shown with the dotted lines. Then, using a millimeter ruler, measure the length of each "stairstep". For the example, the lengths below are about 8mm and 25mm. The ratio of hydrogens giving rise to each peak is thus 1:3. The molecular structure has a methyl group with 3 equivalent hydrogens and a tert-butyl group with 9 equivalent hydrogens in a 1:3 ratio. The integrals give the relative number of hydrogens, not the absolute number of hydrogens. When the integral is measured for a multiplet in an H 1 -NMR spectrum, the individual stairstep for each part of the multiplet is ignored and the integral is measured over the entire multiplet. Thus the total peak area is proportional to the number of hydrogens for that chemical shift. Measure the integrals and state the relative number of hydrogens for the sets of hydrogens in the spectrum shown below. 4
II. Combined Spectra Problems For each set of spectra on the following pages, you should be able to deduce a unique molecular structure. Once you have deduced a structure, it is advisable to confirm the identification by finding authentic spectra in the literature or on the web. On each spectrum, neatly label the major absorption peaks with the structural fragment that causes the absorption. On the answer sheet that follows each problem, fill in the requested information so that you assemble all the structural information that can be obtained from the spectra. Then, using this information, propose the structure and identity of the compound. For each problem, show the steps you followed in deducing the unique molecular structure that is your answer. For some problems, you may be given the molecular formula or the mass spectral parent ion, and for others you are given the elemental analysis from which you may calculate the molecular formula. If you are given the elemental analysis, it is in the form: %C, % H, and where applicable, %N. In many cases, other elements (oxygen, halogen) are also present. The number of atoms, n, of a particular element X (C, H, N, etc.) in a compound can be obtained from the analysis data using the equation nx (%X) (RMMof compound) 100 (RAM of X) The relative molecular mass (RMM) of each compound in the set is given by the m/z value of the molecular ion, M +, in the mass spectrum. RAM is the relative atomic mass of each element (C=12.00, H=1.00, etc). Once the number of each of the elements C, H, N has been calculated, the other elements present can be deduced from a comparison of the RMM of the compound with the partial mass based on C, H, and N alone. Once the molecular formula of a compound is known, the following equation can be used to calculate the number of "units of unsaturation" (or "units of hydrogen deficiency" or "double bond equivalents") UU, in the molecule UU = 1/2(2n 4 + n 3 - n 1 + 2) Where n 4 = the number of tetravalent atoms present n 3 = the number of trivalent atoms present n 1 = the number of monovalent ions present An example problem and solution appear on the next two pages. 5
Example Spectra Problem The spectra here have been marked to indicate major absorptions. Other answers are entered on the next page.
Answer to Example Problem Mass Spectrum Analysis: The integral molecular weight of the unknown is 88. Because this is an even number, the compound has either no nitrogen atoms or an even number of N atoms. The highest m/z peaks do not show evidence of the presence of atoms such as Cl or Br (which would have two peaks in 75:25 and 50:50 ratios, respectively). Elemental Analysis: None was given IR Analysis: The absorption at 3000 cm 1 (and not to the left) shows only aliphatic (saturated) C H groups. The carbonyl, C=O, absorbs strongly at 1740 cm 1. The two intense peaks between 1000-1300 cm 1 indicate the carbonyl belongs to the ester group. 13 C-NMR analysis: The NMR shows 4 different kinds of carbon atoms. The absorption at 172 ppm is typical of the carbonyl carbon. The upfield absorptions at 14 and 21 ppm are likely CH 3 and/or CH 2 groups (although no C H splitting patterns are given with this spectrum). The absorption at 61 ppm could be a C next to an electronegative atom. 1 H-NMR analysis: The NMR shows three different kinds of CH groups. The integration of 2:3:3 suggests a methylene ( CH 2 ) and two methyls ( CH 3 ). The CH group absorbing as a singlet cannot be next to a neighboring CH group (n+1=1 and so n=0). The upfield triplet must be next to a C bonded to 2H s (n+1=3) and the downfield quartet must be next to a C bonded to 3 H s (n+1=4). The triplet integrates for 3H s and so must be a CH 3 group next to a methylene; the quartet integrates for 2 H s and so must be a CH 2 next to a methyl. Thus, there is an ethyl group present, CH 2 CH 3. Combined Analysis: The IR shows the only functional group present is an ester. There is no aromaticity or other unsaturation in the molecule (except for the C=O). This means that the absorption in the 13 C-NMR at 61 ppm cannot be due to a C=C group and the carbon causing the absorption is next to the oxygen atom of the ester. The presence of a C=O is confirmed in the 13 C-NMR. The ester group accounts for 44 amu of the 88 integral mass. The ethyl group is on one side of the ester functional group and accounts for 29 mass units. The remaining mass (88-44-29 = 15 and belongs to the singlet methyl group. The methyl is on the other side of the ester functional group. Because the quartet is so far downfield in the range for methylene 1 H-NMR absorption, the methylene must be bonded to the O of the ester. Unknown Identity: Thus, the unknown is identified as ethyl acetate. Structure: O CH 3 C O CH 2 CH 3 7
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Mass Spectrum Analysis: Answer to Problem 1 Elemental Analysis: IR Analysis: 13 C-NMR analysis: 1 H-NMR analysis: Combined Analysis: Unknown Identity: Structure: 9
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Mass Spectrum Analysis: Answer to Problem 2 Elemental Analysis: IR Analysis: 13 C-NMR analysis: 1 H-NMR analysis: Combined Analysis: Unknown Identity: Structure: 11
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Mass Spectrum Analysis: Answer to Problem 3 Elemental Analysis: IR Analysis: 13 C-NMR analysis: 1 H-NMR analysis: Combined Analysis: Unknown Identity: Structure: 13
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Mass Spectrum Analysis: Answer to Problem 4 Elemental Analysis: IR Analysis: 13 C-NMR analysis: 1 H-NMR analysis: Combined Analysis: Unknown Identity: Structure: 15
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Mass Spectrum Analysis: Answer to Problem 5 Elemental Analysis: IR Analysis: 13 C-NMR analysis: 1 H-NMR analysis: Combined Analysis: Unknown Identity: Structure: 17