Introduction The identification of the structure of chemical substances is an activity fundamental to all varieties of research in organic chemistry. In some instances, the chemist may wish to establish that an unknown substance is identical to a previously reported compound. In other cases, the structure of an entirely new unknown compound must be deduced. This experiment is designed to simulate these common research situations. The aim of the experiment is to separate and identify the components of a binary organic mixture. Unknown samples will consist of approximately 2 g of a mixture of two organic compounds. One unknown in each sample is an organic acid (e.g. carboxylic acid, phenol or sulfonic acid). The other unknown in each sample is a neutral organic compound (e.g., ketone, ester, amide etc.). The unknowns will be separated by extraction (monitored by thin layer chromatography). The separated compounds will then be purified by recrystallization. The structure of the neutral unknown will be deduced by analysis of the 1D NMR ( 1 H NMR, 13 C NMR, DEPT 1 ), 2D NMR (gCOSY, 2 HETCOR 3 or HMQC 4 ), FT-IR and GC/MS spectral data together with analytical and data supplied by the teaching staff. UV-Vis analysis is optional. The acid unknown will be identified by infrared, 1 H NMR, 13 C NMR spectral data and comparison of its spectral characteristics with those of previously reported compounds, and by the preparation of a suitable derivative (amide). References Zubrick, J. W.; The Organic Lab Survival Manual, 3rd Ed.; Wiley: New York, 1992. (Z) Pasto, D. J.; Johnson, C.R.; Miller, M.; Experiments and Techniques in Organic Chemistry; Prentice-Hall: Englewood Cliffs, NJ, 1992. (PJM) Lambert, J. B.; Shurvell, H. F.; Lightner, D. A.; Cooks, R. G.; Organic Structural Spectroscopy; Prentice-Hall: Upper Saddle River, NJ, 1998. (LSLC) Silverstein, R. M.; Bassler, G. C.; T.C. Morrill, T. C.; Spectrometric Identification of Organic Compounds, 5th edition; Wiley: New York, 1991. McLafferty, F.; Tureen, F.; Interpretation of Mass Spectra; University Science Books: Mill Valley, CA, 1993. Lists of acid unknowns are contained in S. Rappaport, Handbook of Tables for Organic Compound Identification, 3rd Edition, Chemical Rubber Co. (1967). Procedure TLC Analysis of Mixture (read Z: chap.26; or PJM: section 2.6.4) Determine analytical TLC conditions that clearly resolve your unknown components. Be sure you are familiar with the effect of a compound's polarity on its mobility on a TLC plate. Your objective is to find a TLC solvent and stain such that a sample can be quickly and conveniently analyzed for the presence/absence of the two components of the mixture. It is desirable to find a solvent system that elutes one or both of the two components somewhere near the middle of the plate. TLC Plates - precoated plastic-backed silica gel plates containing a fluorescent indicator will be provided. Eluents – will find out what is the effective solvent to separate components. For example you can try a single solvent such as hexane, ether, dichloromethane, ethyl acetate, etc or mixtures of two or three components. Spot plates with a glass capillary using solution of approximately 5 mg of unknown in 0.5 mL of a suitable solvent (Et 2 O or Et 2 O:CH 2 Cl 2 , etc.). Always spot a sample of the original mixture on the same plate as a test sample for definitive comparison purposes. If you desire to spot a neutralized aqueous solution, you must remove the water in vacuo prior to eluting the plate. Visualization Nondestructive - When the plate has developed, remove it from the chamber and allow the solvent to evaporate. Place the plate under the short-wave UV lamp. Highly conjugated molecules, e.g. aromatic systems, will show up as dark spots on a green background. Beware that minor impurities can often show up rather strongly under UV light while an unknown with no significant UV chromophore will be completely invisible Destructive - Staining with ethanolic p -anisaldehyde-H 2 SO 4 /CH 3 COOH solution is very effective for visualizing a variety of organic molecules. The plate is simply dipped in the solution and then placed on a hot plate (100-150 °C). Phosphomolybdic acid solution is another very effective stain for most organic compounds. Ce IV (SO 4 ) 2 solution in sulfuric acid may also be useful. Organic molecules will appear as a brown spot on a white background. Testing of the Extraction Procedure Determine the acidic properties of your unknowns by conducting small-scale extractions of a solution of your mixture. In a small test tube, try to dissolve 5-10 mg of your mixture in 0.5 mL or so of ether. If the mixture is not soluble, some dichloromethane can be added. Add a portion of 5% NaHCO 3 or 5% NaOH and swirl the test tube gently to mix. Ether alone or mixtures of ether and hexane are the best solvents for the extraction of neutral organic compounds from ionized forms of organic acids because these ionic species have very little if any solubility in these solvents. The addition of significant portions of dichloromethane or the use of dichloromethane alone may extract both the neutral compound and the organic salt. Determine by TLC analysis whether your unknown components are still present in the organic phase (you can remove a portion of the upper organic layer with a pipette). Pay particular attention to whether the products of the extraction are identical to the components of the original mixture. After removal of the ether layer, the aqueous layer can be washed with a fresh portion of solvent and then analyzed by TLC for the other component of the mixture (you must allow the water to dry from the plate in order for this to work.) Check to make sure that your acid component is not affected upon neutralization of aqueous extracts - reactions of organic compounds with acid or base are not uncommon. With some skill, the small-scale extraction is extremely quick and easy to perform (however a larger scale test extraction may also be performed using a small separatory funnel). Separation of Mixture by Extraction (read Z: chap. 15 or PJM sections 2.1 2.5) Based on the results of the small-scale extractions, separate the components of your mixture by an appropriate extraction procedure. A 60 mL or 125 mL separatory funnel should be effective for this purpose. Use no more than 2/3 of your mixture (save the remainder in case of an accident!). Two or three extractions with small quantities of solvent are more efficient than a single large extraction. Whether or not further extractions are necessary can be ascertained by analyzing the most recent extraction by TLC relative to the combined extractions. The organic solutions should be dried by filtration through anhydrous magnesium sulfate before evaporation using a Buchner funnel or fritted glass funnel (available from the stockroom). The rate at which anhydrous magnesium sulfate will dry an organic solvent is extremely rapid - just be sure to carefully pour ethereal solutions from the top of the separatory funnel without allowing water to pour in. In fact, a good practice is to pass each sequential organic extract through a fritted glass funnel containing anhydrous magnesium sulfate (later extractions serve to wash through the compound from the first portions). In order to separate 1 gram of your mixture, you will likely find that a total of 25 mL aqueous base and 150 mL of organic solvent is adequate for the entire operation. When removing the solvent on the rotary evaporator, it is advisable to use a dish of lukewarm water under the flask rather than steam since some of the unknowns may not be very stable. Upon confirming that the neutral compound has been effectively removed, neutralize the aqueous layer and isolate the acid component of your mixture. Record the weights of original and separated material. Purification of Solid Unknowns (read: Z: chap. 13 or PJM: section 2.2) Purify both unknowns by recrystallization. Some of the unknowns will solidify upon concentration. For those that don't it is a good idea to save a small portion of the original solid mixture as seed crystals for recrystallization of one component. Good solvents for recrystallization include ether, hexane, dichloromethane, ethyl acetate and mixtures of these three solvents. An all-purpose procedure involves initially adding 1-2 mL of ether to the solid in a 50 mL flask. If the solid starts to dissolve rapidly, then add some hexane until it begins to crash out, then warm gently to dissolve. If the solid appears to dissolve slightly, then warm ether is probably the best choice. If the solid does not dissolve at all, then dichloromethane can be added and perhaps a more polar solvent like ethyl acetate or ethanol. Once the material has been dissolved in the optimum solvent, it can be allowed to cool slowly to room temperature on the bench at which time a very small seed crystal can be added if crystallization has not already taken place. Then the mixture may be placed in the refrigerator or freezer to complete crystallization. Be warned that for many compounds, a fluffy crystalline product may be a lot less material than expected. Crystals that have been isolated on the filter can usually be effectively freed from solvents such as ether, hexane and dichloromethane by sucking air through for 10-15 minutes. Oven drying is not recommended because most of the unknowns will either melt or decompose. Record the amount recovered, the melting point and melting range for each of your unknowns. A sharp melting range is the best indication of purity and a prerequisite to investing a lot of time in spectral analysis. NMR Spectral Analysis (read Z: chap. 33) Obtain 1 H NMR spectra (in CDCl 3 ) of both of your unknowns. If not soluble in CDCl 3 , try deuterated acetone or DMSO - ask your TA. Be sure that your sample is pure and is not wet with solvent and that your NMR tube is also free of any solvent that may have been used to wash it out. Such solvent peaks can obscure important peaks in the spectrum. Once you obtain a good spectrum, be sure to print out all of the information that you will need to analyze the spectrum including integrals, expansions and peak listings - having to redo spectra because of these two problems can lead to a great deal of congestion on the instruments. The TAs can give you a copy of the 1 H NMR spectrum of your neutral unknown if you find that your spectrum contains some minor inadequacy or ambiguity. However, if your acid unknown contains stray peaks, it would be quite unfortunate. As soon as possible, check each 1 H NMR spectrum carefully for purity and try to assign a number of protons to each peak or cluster of peaks based on the integration. Also, the sooner you begin analysis of the multiplet patterns, the better. First read the section on analysis of 1 H NMR coupling patterns, then make sure you understand how to do the practice problems on the First Report. As soon as possible, check each 1 H NMR spectrum carefully for purity and try to assign a number of protons to each peak or cluster of peaks based on the integration. Check for the presence of common impurities such as ether, acetone, etc. as well as the deuterated solvent peak(s) and water (which appears at various places in different solvents). Once you have determined which peaks belong to your compound, make a list of them in your worksheet. In the table, organize the chemical shift, multiplicities ( s inglet, d oublet, t riplet). Some of the peaks in the spectrum may appear as unmanageable multiplets ( m ), however the more carefully and patiently you analyze the peaks, the more information you will glean from the spectrum. Such analysis of complex multiplets ( dd, ddd, dt , etc.) and matching of the coupling constants allows one to trace out the spin system of the unknown. If the 1 H NMR of the neutral prove to be acceptable , then you can proceed to record the 13 C, DEPT, gCOSY and HETCOR NMRs. IR Analysis (read Z: chap. 32 or PJM: chapter 5) Obtain an IR spectrum of each component of your unknown mixture. Using the FTIR spectrophotometer, IR spectra can be measured on solutions of solid samples and thin-films of liquid samples. Identify in your notebook the principal absorptions in each spectrum and determine which functional groups are most likely present. UV Spectral Analysis (Optional) GC-Mass Spectral Analysis You will obtain the high-resolution mass spectral data for your neutral unknown. This will be accomplished using the Hewlett-Packard 5890 Series II GC/MS. Sign up in advance to use this instrument in 4-474 and ask your instrument TA or Ray Dove for help using this delicate equipment. As mass spectrometry is the most sensitive analytical tool available to the organic chemist, be careful to prepare a sample that is not too concentrated. Usually, a solution containing 1 mg of your neutral unknown dissolved in 0.5 mL of dichloromethane is sufficient to obtain a strong spectrum. Small disposable sample tubes are available for this purpose. After you have completed parts E, F and H, you should prepare to show your TAs that your spectral characterization of your two unknowns is acceptable to precede. You will then receive the elemental analysis of your neutral unknown. Organic Acid Unknown: Preparation of Derivatives Locate the m.p. of your organic acid unknown in the Tables for the Identification of Organic Compounds , keeping in mind that it could be a carboxylic acid, a phenol or a sulfonic acid. Examine the entries in the appropriate table within 6 °C of your measured melting point. Eliminate as many structures as possible on the basis of your NMR and IR data. Choose a derivative of your acid unknown which will allow you to confirm its identity or distinguish between two possibilities. Not all derivatives are listed for each entry in the Tables for the Identification of Organic Compounds . Procedures for preparation of common derivatives of almost any functional group are presented in PJM: chap. 9. Preparation of Acid Chloride and conversion to Amide or Anilide Generally, various acids are converted to amides or anilides via the acid chloride. Thionyl chloride is used to convert the unknown acid to the acid chloride. However, the range of possible unknowns covers a very wide spectrum of reactivity (aliphatic, aromatic, sterically hindered/unhindered, etc.). Some acids will react rapidly with SOCl 2 alone with gentle warming whereas sterically hindered or electron-deficient acids may only react after an hour or so at reflux or not at all. In either case, a couple of drops of N,N-dimethylformamide (DMF) serve as an efficient catalyst for the reaction between a carboxylic acid and thionyl chloride. This allows most acids to be converted to the corresponding acid chloride at room temperature over 1-3 hours (as observed by evolution of SO 2 and HCl gas). Questions you might consider: how does DMF catalyze the first reaction? how much volume of gas evolution will you be looking for this reaction? Caution Thionyl chloride is corrosive and its vapors are highly noxious. This procedure must be carried out in an efficient fume hood.) Procedure Weigh 50-150 mg of your unknown carboxylic acid into a 25 mL flask. Add 2-3 drops of N,N-dimethylformamide as a catalyst followed by 2.5-5 mL thionyl chloride. Attach a reflux condenser and by means of a piece of tygon tubing, attach the top of the reflux condenser to a mineral oil bubbler. Warm the mixture gently to reflux with a heating mantle and note any significant evolution of gases as an indication of the reaction. Based on your observations as well as the expected reactivity of your acid, decide how long to reflux the reaction. Cool the acid chloride reaction to room temperature. In the fume hood, cautiously pour or pipette the mixture into 25 mL of concentrated ammonia solution contained in a 250 mL beaker. The amide will precipitate from the aqueous solution and can be collected by filtration. Obtain a m.p. and an IR spectrum of your amide derivative. A Stepwise Guide to Structure Elucidation In order to determine the structure of your neutral unknown, you will draw upon the several powerful spectroscopic and analytical methods described above. Detailed below is a systematic approach to structure elucidation. While reading about each of the five stages of the structure elucidation process, follow the example to see how this is applied to a sample unknown. Determination of Molecular Formula (and Unsaturation number) Identification of Unsaturated Groups Analysis of 1 H NMR Coupling Patterns Assemblage of Molecular Subunits Confirmation of Structure Determination of Molecular Formula (and Unsaturation number) The following data will be important for determining the molecular formula: Elemental Analysis, Mass Spectrum Highest Ion Pattern, 1 H NMR (integration) and 13 C NMR (# of lines) . The goal is to find a molecular formula that is consistent with all of these. Elemental Analysis In research, samples are sent to outside laboratories for the purpose of obtaining % composition of C and H (N, S are more expensive). 2-5 milligrams of material is subject to complete combustion in an atmosphere of O2 and the resulting H2O and CO2 accurately weighed to obtain %C and %H within ± 0.4%. In practice, stray solvents and impurities can lead to incorrect results. Your TA's have calculated an elemental analysis of your unknown for the elements C, H, N, S which is within ± 0.4% for each element. Since you are guaranteed of the ± 0.4% range, you can apply a high level of confidence in this data. GC-MS Highest Ion Pattern A single large peak eluting from the gas chromatograph is good evidence that your neutral unknown was pure and passed through the GC unharmed. Each molecule passing into the Mass Spec is first converted by electron ionization (EI) to the molecular ion (M •+ ) in which the molecule has simply lost one electron. For nearly all of the unknowns, some of the M •+ ions will survive and show up in the spectrum. If this is the case, then the Elemental Analysis data and GC-MS may be used directly to determine the number of C, H, N and S atoms in the molecular formula (check out the example to see how easy this is). If the Highest Ion Pattern of your mass spectrum is not M •+ , then figuring out the molecular formula will be a fair amount more difficult. The following symptoms suggest that you don't have M •+ The calculations using the elemental analysis and the Highest Ion Pattern do not span an integer for each element. (see example) The mass of M •+ does not agree with the Nitrogen Rule. (The Nitrogen Rule states that a molecule or M •+ ion with an even number of nitrogen atoms must have an even mass. And similarly, a molecule or M •+ ion with an odd number of nitrogen atoms must have an odd mass.) The Highest Ion Pattern of the mass spectrum will generally display at least three peaks at consecutive m/e values if a sufficiently strong M •+ signal has been detected. The ratio of intensities of M •+ , (M+1) •+ and (M+2) •+ result from the statistical presence of (A+1) and (A+2) isotopes of the individual elements. The M+1 peak % results primarily from the number of carbons in the organic molecule. The M+2 peak % will generally be 1% or less that of the M peak % for small to medium sized organic molecules containing only C,H,N and O. A sizeable M+2 peak indicates the presence of (A+2) elements, most notably Cl and Br . Below are shown several possible isotopic patterns for the M •+ region of the spectrum. A isotope =100% A+1 (relative %) A+2 (relative %) 1H 2D (0.015%) - - - 12C 13C (1.1%) - - - 14N 15N (0.37%) - - - 16O 17O (0.04%) 18O (0.20%) 19F - - - - - - 28Si 29Si (5.1%) 30Si (3.4%) 31P - - - - - - 32S 33S (0.80%) 34S (4.4%) 35Cl - - - 37Cl (32.5%) 79Br - - - 81Br (98%) 127I - - - - - - M peak=100% M+1 peak %* M+2 peak % C1 1.10% 0.00% C2 2.20% 0.01% C3 3.30% 0.04% C4 4.40% 0.07% C5 5.50% 0.12% C6 6.60% 0.18% C7 7.70% 0.25% C8 8.80% 0.34% C9 9.90% 0.44% C10 11.00% 0.54% C11 12.10% 0.67% *add 0.37% for each N 1 H NMR (integration) The integration of the 1 H NMR spectrum is the most reliable method for determining the number of protons in the molecular formula. Only two problems can interfere with this: Stray peaks in the 1 H NMR spectrum may be mistakenly counted. Exchangeable protons (e.g., NH, OH) are broadened and may be exchanged with D 2 O present in some deuterated solvents (e.g. d 6 -acetone) thus giving a lower integral. Occasionally, these peaks (e.g., -COOH) may be overlooked altogether. 13 C NMR (# lines) The number carbons should be checked with the number of lines in the 13 C NMR. First, make sure that you know which peaks are from the deuterated solvent. Since 13 C NMR spectra are generally acquired with broadband 1 H decoupling, all resonances will appear as sharp singlets unless they are coupled to other nuclei such as 2D (I = 1), 19F (I = 1/2) or 31P (I = 1/2). So # Lines > #C means that your unknown has one or more of these atoms. # Lines < #C means that your unknown has groups for which pairs or sets of carbons are equivalent by virtue of symmetry or free rotation. Unsaturation Number For a molecule having molecular formula C x H y N z O n , the number of double bonds + number of rings = \(1 \cdot x - 0.5 \cdot y + 0.5 \cdot z + 0 \cdot n + 1\). This is a direct result of the known valences of these elements (C=4,H=1,N=3,O=2). For other elements, all that is needed is to know the valence of the element. Halogens can safely be said to have valence = 1 for the majority of organic compounds. However phosphorus may have valence = 3 or 5. And sulfur may have valence = 2,4 or 6 (e.g., CH 3 SCH 3 , CH 3 SOCH 3 , PhSO 3 H). Identification of Unsaturated Functional Groups Careful examination of the IR, and 13 C NMR spectra will reveal all of the unsaturated functional groups that are present in your unknown. The first stage of this search involves examination of the IR spectrum. Although several regions of the IR spectrum can reveal useful information about the unknown (see pages xx-xx), at this stage, we are only interested in examining the two regions of the IR spectrum that will help us identify special unsaturated functional groups. In the 2300 to 2000 cm -1 region, any peak that is observed, no matter how small, is likely to be diagnostic of one of these functional groups. By contrast, in the 1800 to 1650 cm -1 region, a carbonyl peak will always be strong, very likely the strongest peak in the spectrum. -C≡N, -C≡C-, >C=C=C< Region (2300 cm-1 to 2000 cm-1) -C≡N, -C≡C-, >C=C=C< Region (2300 cm-1 to 2000 cm-1).1 -C≡N, -C≡C-, >C=C=C< Region (2300 cm-1 to 2000 cm-1).2 -C≡N, -C≡C-, >C=C=C< Region (2300 cm-1 to 2000 cm-1).3 C=C=C< Region (2300 cm-1 to 2000 cm-1)" class="lt-chem-308529">cm-1 NaN comments 13C ppm C=C=C< Region (2300 cm-1 to 2000 cm-1)" class="lt-chem-308529">2250 ν(-C≡N) sharp, weak to med intens, almost always observed 117-123 C=C=C< Region (2300 cm-1 to 2000 cm-1)" class="lt-chem-308529"> NaN NaN ppm C=C=C< Region (2300 cm-1 to 2000 cm-1)" class="lt-chem-308529">2150 ν(RC≡CH) sharp, weak to med intens, check for ν(C-H) at 3300 ppm C=C=C< Region (2300 cm-1 to 2000 cm-1)" class="lt-chem-308529">2260-2190 ν(R-C≡C-R') sharp, weak to med intens, obsd only for R,R' different ppm C=C=C< Region (2300 cm-1 to 2000 cm-1)" class="lt-chem-308529">1950 ν(>C=C=C<) sharp, strong allene 200-210 >C=O Region (1800 cm-1 to 1650 cm-1) >C=O Region (1800 cm-1 to 1650 cm-1).1 >C=O Region (1800 cm-1 to 1650 cm-1).2 >C=O Region (1800 cm-1 to 1650 cm-1).3 C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">cm-1 NaN comments 13C ppm C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">1800 Acid Chloride CO Bond Order <2 170-180 ppm C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">1820,1760 Anhydride two peaks are observed (ns nas) 160-172 ppm C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">1735 Ester RCO2R' 165-175 ppm C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">1755 Carboate ROCO2R' 150-155 ppm C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">1735 Urethane ROCONR'2 154-160 ppm C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">1720 Aldehyde/Ketone aldehyde has n(CH) at 2750 cm-1 190-210 ppm C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">1650 Amide CO Bond Order <2 164-172 ppm C=O Region (1800 cm-1 to 1650 cm-1)" class="lt-chem-308529">1630 Urea R2NCONR'2 ppm For each of the peaks listed in the table, a range of actual positions can be observed. For carbonyl groups, phenyl substituents or double bonds attached to the C=O group donate electron density to the carbonyl group and lower the C=O stretching frequency. A carbonyl group in a small ring has a higher stretching frequency. Further evidence taken from the IR spectrum must be taken cautiously. If one peak is found in the carbonyl region, perhaps only one C=O group is present. However, since the C=O region spans only 170 cm-1 and even the sharpest peaks are several cm-1 wide, two C=O groups could show up in the same position by coincidence. Alternatively, the molecule could have two C=O groups related by symmetry. If two peaks are found in the carbonyl region, perhaps the molecule contains two C=O groups which are not related by symmetry. However, there are also exceptions to this assumption. Symmetrical anhydrides display two peaks in the carbonyl region because of the close proximity of the two C=O groups. One peak represents the symmetrical stretching of the two C=O groups and the other peak represents antisymmetric stretching. Each of the functional groups above may be confirmed by the presence of the appropriate peaks in the 13 C NMR spectrum, and the exact positions in the IR and 13 C NMR may be used later to help identify the exact nature of the functional group. The remaining downfield resonances in the 13 C NMR can then be used to identify C=C double bonds and aromatic rings present in the molecule. Below is shown a diagram of the expected chemical shift ranges of functional groups in the 13 C spectrum. Although nearly all peaks beyond 100 ppm in the 13 C spectrum are sp 2 , sometimes sp 3 carbons bearing two or three electronegative atoms may be found this far downfield (chemical shift effects are more or less additive). If the molecule contains no symmetry, then each line represents a single carbon in the 13 C NMR spectrum and the number of sp 2 carbons can simply be counted. An unknown, which displays symmetry in the 13 C NMR, requires determining which lines represent symmetrical carbons. Unfortunately, the number of carbons cannot be determined by integration of the 13 C spectrum (carbons bearing no hydrogen substituents will give smaller peaks). One good way to approach this problem is to use the integration of the 1 H NMR spectrum to identify protons that are related by symmetry and then extrapolate this knowledge to the 13 C NMR spectrum. The first thing you should do when you get your 13 C NMR spectrum is to make sure you know which peaks are from the deuterated solvent (CDCl 3 , CD 3 COCD 3 , etc.) and possible impurities. If your 13 C NMR spectrum displays exactly the number of lines equal to the number of carbons, then your unknown has no symmetry. If your unknown displays fewer lines than number of carbons, then some of the carbons in your unknown are related by symmetry or free rotation. Analysis of 1 H NMR Coupling Patterns Construction of Spin-Spin Coupling System The first stage of the analysis is to match up the coupling constants for all of the first order multiplets. For each multiplet in proton NMR spectrum, there must be some other resonance that shares the same coupling constant. For molecules that contain phosphorus or fluorine, many nearby protons may couple to these atoms. For molecules that do not contain these atoms, every coupling constant must be shared by another proton in the 1 H NMR spectrum. Usually, coupling constants will match to within approximately 0.2 Hz. Furthermore, note that if a methyl group is a doublet, then some other proton in the spectrum must appear as a 1:3:3:1 quartet or have this pattern combined with some other coupling constants. It is important that multiplets be interpreted correctly as described on pages 30-40. Once the coupling constants are matched between each other in the 1H NMR spectrum, the task is to construct a spin system showing their relationships within the molecule. In order to do this, some decisions must be made about whether a given H-H coupling constant is geminal, vicinal, or long-range . If a coupling constant is large (>3 Hz), it may safely be assumed to be geminal or vicinal and these should be considered first. A series of vicinal coupling constants can be used to construct a chain of the organic compound. The coupling information, if analyzed correctly, offers valuable and reliable information about the structure of a new compound. DEPT spectra provides you information for distinguishing CH 3 , CH 2 and CH groups. The signals of quaternary carbon nuclei and CDCl 3 are missing. gCOSY spectrum provides information regarding coupled pairs of protons separated by two or three bonds. In the chart that correlates signals of a normal 1 H NMR with each other, cross-peaks appear if spin coupling is present. This techniques is useful to find connectivity and very often the chemical structure. HETCOR is a two dimensional C, H correlation experiment that yields cross signals for all the protons and 13 C nuclei which are connected by a J CH over one bond. Assemblage of Molecular Subunits Methyl and ethyl substituents in an unknown structure can reliably be assigned connectivity to different groups based on their chemical shift values. Since the chemical shifts of peaks in the 1 H NMR spectrum are subject to subtle effects of solvent and structure chemical shifts of other protons can be difficult to interpret beyond the order of 0.5 to 1 ppm. Analysis of peak fragments in Mass Spectrum Peaks in the mass spectrum can be very helpful in piecing together the molecular subunits found above. For molecules that contain bromine or chlorine atoms, it becomes especially easy to follow the fragments that contain the halogen atom isotopic pattern. Confirmation of Structure The final stage of the analysis involves bringing together all of the data of the unknown compound and confirming that all of the data can be correctly rationalized for the structure proposed. Determination of Molecular Formula (and Unsaturation number) The GC-mass spectrum shows a highest ion pattern at m/e 209 which will be assumed to be M •+ . From the intensity of the (M+2) •+ peak, one Cl is clearly present ( 35 Cl 76%, 37 Cl 24% nat. abund.). Using the elemental analysis data, the number of C,H,N,S atoms can be obtained: \[\begin{array} {llr} 56.94\% \times 209 \div 12.01 = 9.91 & 57.74\% \times 209 \div 12.01 = 10.05 & \quad \text{? 10 C } \\ 3.84\% \times 209 \div 1.01 = 7.95 & 4.64\% \times 209 \div 1.01 = 9.60 & \quad \text{? 8 or 9 H } \\ 5.94\% \times 209 \div 14.01 = 0.886 & 6.74\% \times 209 \div 14.01 = 1.005 & \quad \text{? 1 N } \end{array}\] The ± 0.4% error does not allow us to figure out exactly how many H atoms are in the molecular formula. However, from the 1 H NMR integration, it is clear that there are 8 H. The 13 C NMR spectrum shows 10 lines, which is consistent for 10 C if the molecule has no symmetry. Adding up the groups so far, 10 x 12 + 8 x 1 +1 x 14 + 1 x 35 = 177. 209 - 177 = 32. (Note that 35 is used for the mass of the 35Cl isotope corresponding to the larger peak at m/e 209.) The molecular formula is therefore C 10 H 8 NO 2 Cl . Unsaturation # = 10 - 4.5 + 0.5 + 1 = 7. Molecular Formula & MW Elemental Anal. calc. obsd. GC-MS Highest Ion Pattern 1H NMR integration 13C NMR # of lines C10H8NO2Cl MW 209.64 C 57.34% H 4.24% N 6.34% S 0.0% each ±0.4% NaN 1:2:1:2:2 10 lines Unsaturation Number (# of double bonds or rings): 7 Identification of Unsaturated Functional Groups A cyano group (2254 cm -1 ) and a carbonyl group (1732 cm -1 ) are clearly present in the IR spectrum and are entered into the worksheet. The fact that 10 lines are found in the 13 C NMR spectrum for 10 C means that each line represents exactly one carbon of the structure. In this example, six peaks left in the 120-150 ppm region of the 13 C NMR represent six sp 2 carbons (formally three double bonds). Since four protons are found in the aromatic region of the 1 H NMR, a disubstituted benzene ring satisfies this data and accounts for the remaining 4 units of unsaturation. This aromatic ring must be either ortho or meta substituted, not para. (Free rotation would result in only four 13 C lines for a para-disubstituted aromatic ring.) Unsaturated Functional Group IR (cm-1) 13C NMR ppm # of lines, #C 1H NMR ppm #H Unsaturation No. of FG -C≡N 2254 118.0 ppm 1 line = 1 C - - - NaN 2 >C=O 1732 165.0 ppm 1 line = 1 C - - - NaN 1 n(CH) 3050 133.6, 133.5, 131.8, 131.4, 130.2, 127.6 6 lines = 6C 7.90, 7.59, 7.56, 7.47 4H 4 NaN Remaining Molecular Symmetry: None Remaining Molecular Symmetry: None Remaining Molecular Symmetry: None Remaining Molecular Symmetry: None Remaining Unsaturation: None. Remaining Unsaturation: None. Remaining Unsaturation: None. Analysis of 1 H NMR Coupling patterns In order to figure out whether the aromatic ring is ortho or meta disubstituted, we need to analyze the 1 H NMR spectrum carefully. The coupling constants between aromatic protons will tell the relationship of the two protons to each other (J ortho ˜ 6-8 Hz, J meta ˜ 2 Hz, J para < 1 Hz). Careful analysis of the three ddd patterns leads to the numerical coupling constants shown in the table. Next, as many of the coupling constants are matched (within a ± 0.2 Hz error allowance) as possible. Protons that share a common coupling constant are assigned as coupled to each other by that coupling constant. H b is coupled to H d with J = 7.0 Hz so Hb and Hd are ortho to each other. Since Hb also has a J = 8.0 Hz coupling constant, another proton is also ortho to H b . Same thing for H d . Therefore, the aromatic ring is ortho- disubstituted and the proton assignments are as shown below. 1H NMR ( 500 MHz, d6-acetone solvent) 1H NMR ( 500 MHz, d6-acetone solvent).1 1H NMR ( 500 MHz, d6-acetone solvent).2 1H NMR Spin System label \(\delta \) (ppm) (multiplicity, #H, J values) NaN a 7.90 (ddd, 1H, Jac=0.5, J =2.0, J =8.0 Hz) NaN b 7.59 (ddd, 1H, J =2.0, Jbd=7.0, J =8.0 Hz) NaN c 7.56 (ddd, 1H, Jac=0.5, J =2.0, J =8.0 Hz) NaN d 7.47 (ddd, 1H, J =2.0, Jbd=7.0, J =8.0 Hz) NaN e 4.57 (t, 2H, Jef=6.0 Hz) NaN f 3.04 (t, 2H, Jef=6.0 Hz) NaN Similarly, resonances e and f are each 6.0 Hz triplets and integrate to 2 protons each. This is consistent with the assignment for adjacent methylene groups. For each methylene group, the two protons are equivalent to each other in the 1 H NMR spectrum. Therefore, the methylene groups are not diastereotopic Assemblage of Molecular Subunits The four subunits identified in the previous two sections are entered into the worksheet at this stage, and the remaining subunits are a chlorine substituent and an oxygen substituent. This molecule is therefore a chain Cl-A-B-C-D-C? N, where A,B,C,D are the four disubstituted groups. Several approaches could be investigated to figure out the order of the groups. One could begin by suggesting that the methylene group resonating at 4.57 ppm in the 1 H NMR spectrum is so far downfield that it must be attached to oxygen. Similarly, the values of other peaks in the IR, 1 H NMR and 13 C NMR spectra may be considered. 1H NMR Subunits Remaining 13C NMR subunits Remaining subunits NaN NaN NaN One very effective method of determining the arrangement of groups in a molecule that is a chain of subunits is to analyze the fragmentation ions in the mass spectrum. Keep in mind that each peak in the spectrum represents the detection of a stable positive ion - either a cation (A + ) or a radical cation (A •+ ) - resulting from a single cleavage event. In this case, it is especially easy due to the presence of the Cl substituent. The very small m/e 174 peak that is (M-Cl) + doesn't reveal much about the structure. However, each of the fragment ions 156,139 and 111, each displaying the characteristic chlorine isotopic pattern are very helpful. m/e 111 : The 111 fragment is the easiest fragment to identify. The isotopic pattern reveals that it bears the Cl atom (m/e 35). The Cl atom is connected to the benzene ring (m/e 76). m/e 139 : The next highest fragment bears the Cl group too and is 28 mass units larger than m/e 111. So either the >C=O group or the -CH2CH2- group (m/e 28) is next in the chain. m/e 156 : The 156 peak bears the chlorine substituent and resulted from a cleavage of 53 mass units from M •+ . What mechanism results in the loss of 53 mass units from M •+ given the subunits listed above? Ans: A rearrangement occurred losing the cyano group (26 mass units) and the C 2 H 3 portion (27 mass units) of the -CH 2 -CH 2 - piece. (McLafferty Rearrangement) m/e (M+1)% (M+2)% Possible Assignments m/e.1 (M+1)%.1 (M+2)%.1 Possible Assignments.1 209 13.0% 36.1% M•+ 139 8.7% 33.5% 35+76+28=139 174 - - - - - - (M-35)+ 128 - - - - - - ? 156 7.8% 30.4% (M-27-CN)•+ 111 9.1% 34.5% 35+76=111 Confirmation of Structure From the preceding analysis, the molecular subunits are assembled as follows: Cl - ortho-disubstituted aromatic ring - carbonyl - oxygen - CH 2 CH 2 - and - C? N. The final step of the analysis involves drawing the structure carefully and bringing all remaining spectral data together for consideration. In the 1 H NMR spectrum, because a mirror plane of symmetry runs down the length of the molecule, the protons denoted e are equivalent to each other and so are observed as a single resonance. Same for the protons labelled f. Although e and f are rigorously an AA'XX' system, apparently free rotation of the bond between them allows each to be observed as triplets. In the 13 C NMR spectrum, the table on pages 41-42 can be used to assign many of the carbon resonances in the spectrum. However, because the aromatic carbon resonances are bunched together, they cannot be individually assigned. For the mass spectrum, it is good practice to draw the mechanism for two or more of the prominent peaks in the spectrum. Shown below is the McLafferty Rearrangement that leads to the peak at m/e 156 and the 37 Cl isotope peak at m/e 158. Footnotes D istortionless E nhancement by P olarization (population) T ransfer: display separate spectra for CH3, CH2 and CH carbon signals. Field g radient selection CO rrelation S pectroscop Y : homonuclear (for example H vs H) correlation spectroscopy HET eronuclear chemical shift COR relation by Polarization Transfer: carbon vs hydrogen chemical shift correlation. H eteronuclear M ultiple Q uantum C oherence, carbon vs hydrogen chemical shift correlation.