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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
DEPARTMENT OF CHEMISTRY
5.32 Intermediate Chemical Experimentation
EXPERIMENT #2: QUALITATIVE ORGANIC ANALYSIS
I. INTRODUCTION
The identification of the structure of chemical substances is an activity fundamental to all varieties of research in organic chemistry. Chemists may wish to establish that an unknown substance is identical to a previously reported compound, or they may need to deduce the structure of an entirely new compound. 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), and the separated compounds will 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).
II. REFERENCES
- Zubrick, J. W.; The Organic Lab Survival Manual , 5th Ed.; Wiley: New York, 2001. (Z)
- Pasto, D. J.; Johnson, C.R.; Miller, M.; Experiments and Techniques in Organic Chemistry; Prentice-Hall: Englewood Cliffs, NJ, 1992. (PJM)
(^1) D istortionless E nhancement by P olarization (population) T ransfer: display separate spectra for CH 3 , CH 2
and CH carbon signals. (^2) Field g radient selection CO rrelation S pectroscop Y : homonuclear (for example H vs H) correlation
spectroscopy (^3) HET eronuclear chemical shift COR relation by Polarization Transfer: carbon vs hydrogen chemical shift
correlation. (^4) H eteronuclear M ultiple Q uantum C oherence, carbon vs hydrogen chemical shift correlation.
- 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).
III. PROCEDURE
Review the following laboratory techniques in the Digital Laboratory Techniques Manual , which can be found on the 5.32 website, before beginning this section.
DLTM:
**3. TLC – the Basics
- TLC – Advanced
- Reaction Workup I
- Reaction Workup II
- Filtration
- Recrystallization**
A. TLC ANALYSIS OF A MIXTURE (read Z: Ch 26 or PJM: section 2.6.4; DLTM: 3 & 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 that allow a quick and convenient analysis for the presence/absence of the two components of the mixture. The solvent system should elute one or both of the components near the middle of the plate.
- TLC Plates - precoated plastic-backed silica gel plates containing a fluorescent indicator will be provided.
- Eluents – 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 the TLC plates with a glass capillary, using a 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 a definitive comparison. To spot a neutralized aqueous solution, you must remove the water in vacuo prior to eluting the plate.
- Visualization a. 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.
pour ethereal solutions out the top of the separatory funnel without allowing water to pour in. A particularly good practice involves the passage of 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). 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, use a dish of lukewarm water under the flask rather than steam because 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.
D. PURIFICATION OF SOLID UNKNOWNS (read Z: Ch 13 or PJM: section 2.2; DLTM 9)
Purify both unknowns by recrystallization. Some of the unknowns will solidify upon concentration. For those that don't, 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 such as ethyl acetate or ethanol. After the material has been dissolved in the optimum solvent, it can be cooled slowly to room temperature on the bench. At this time, add a very small seed crystal if crystallization has not already taken place. After crystal formation occurs at room temperature, place the mixture in the refrigerator or freezer to complete crystallization. Be warned: 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 of material 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.
E. NMR SPECTRAL ANALYSIS
Obtain a 1 H NMR spectra (in CDCl 3 ) of both of your unknowns. If the compounds are not soluble in CDCl 3 , try deuterated acetone or DMSO - ask your TA. Be sure that your sample is pure and not wet with solvent. Your NMR tube must also be free of any solvent that may have been used to wash it out. Such solvent peaks can obscure important peaks in the spectrum. After you obtain a good spectrum, 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 (not acid) unknown if you find that your spectrum contains some minor inadequacy or ambiguity. 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). After 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. Read the section on analysis of 1 H NMR coupling patterns and make sure you understand how to do the practice problems on the First Report.
If the 1 H NMR of the neutral compound proves to be acceptable , then you can proceed to record the 13 C, DEPT, gCOSY and HETCOR NMRs.
F. IR ANALYSIS ( Optional ; 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 the principal absorptions in each spectrum in your notebook and determine which functional groups are most likely present.
G. UV SPECTRAL ANALYSIS (Optional)
H. GC-MASS SPECTRAL ANALYSIS
Obtain the high-resolution mass spectral data for your neutral unknown. This task will be accomplished by 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 with this delicate equipment. Because 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
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.
IV. A STEPWISE GUIDE TO STRUCTURE ELUCIDATION
To determine the structure of your neutral unknown, you will draw upon the several powerful spectroscopic and analytical methods described above. A systematic approach to structure elucidation is detailed below. 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.
A. Determination of Molecular Formula (and Unsaturation number) B. Identification of Unsaturated Groups C. Analysis of 1 H NMR Coupling Patterns D. Assemblage of Molecular Subunits E. Confirmation of Structure
A. 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 techniques.
1. 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). Two to five milligrams of material are subject to complete combustion in an atmosphere of O 2 , and the resulting H 2 O and CO 2 are 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. Because you are guaranteed the ± 0.4% range, you can apply a high level of confidence in this data.
2. GC-MS Highest Ion Pattern. A single large peak eluting from the gas chromatograph is 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 true for your compound, 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 •+:
3. 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 method:
- 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), giving a lower integral. Occasionally, these peaks (e.g., -COOH) may be overlooked altogether. 4. 13 C NMR (# lines)
The number carbon atoms 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. Because 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 (^2) D (I = 1), 19 F (I = 1/2) or 31 P (I = 1/2). If # Lines > #C, 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.
5. Unsaturation Number
For a molecule having a molecular formula Cx HyNzOn , the number of double bonds + number of rings = 1•x-0.5•y+0.5•z+0•n+1. This calucation 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 have a valence of 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).
B. 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, 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^ )
cm -1^ comments 13 C ppm
2250 (-C N) sharp, weak to med intens, almost always observed 117-123 ppm
2150 (RC CH) sharp, weak to med intens, check for (C-H) at 3300 ppm
2260-2190 (R-C C-R') sharp, weak to med intens, obsd only for R,R' different ppm
1950 (>C= C =C<) sharp, strong allene 200-210 ppm
>C=O Region (1800 cm -1^ to 1650 cm -1^ )
cm -1^ Comments 13 C ppm
1800 Acid Chloride
R Cl
O R C
O +
Cl
- (^) CO Bond Order >
170-180 ppm
1820,1760 Anhydride^ two peaks are observed ( s as ) 160-172 ppm
1735 Ester^ RCO 2 R' 165-175 ppm
1755 Carbonate^ ROCO 2 R' 150-155 ppm
1735 Urethane^ ROCONR' 2 154-160 ppm
1720 Aldehyde/Ketone^ aldehyde has (CH) at 2750 cm-1^ 190-210 ppm
1650 Amide
R C^ NR^2
O
R C^ NR^2
O -
CO Bond Order <
164-172 ppm
1630 Urea^ R 2 NCONR' 2 ppm
For each of the peaks listed in the table, a range of actual positions can be observed. 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.
R
O
R
Effect of Conjugation on ν (CO):
+ O -
CO Bond order < 2
Effect of Ring Size on ν (CO):
Me Me
O O
Me
O
Ph Ph
O
Me
OH
Me
O
O
O
O
O
Further evidence taken from the IR spectrum must used with care. If one peak is found in the carbonyl region, perhaps only one C=O group is present. However, because 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, exceptions to this assumption exist. Symmetrical anhydrides
NMR spectrum. Usually, coupling constants will match with each other to within approximately 0.2 Hz. Furthermore, 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. Multiplets be interpreted correctly as described in the Appendix. After the coupling constants are matched between each other in the 1 H NMR spectrum, their relationships must be shown using a spin system. To accomplish this task, 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 be assumed to be geminal or vicinal; 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.
2. DEPT spectra provide the information necessary for distinguishing CH 3 , CH 2 and CH groups. The signals of quaternary carbon nuclei and CDCl 3 are missing. 3. gCOSY spectra provide information regarding coupled pairs of protons separated by two or three bonds. In the chart that correlates signals of a normal 1H NMR with each other, cross-peaks appear if spin coupling is present. This technique is useful for finding connectivity and the chemical structure. 4. HETCOR is a two dimentional C, H correlation experiment that yields cross signals for all the protons and 13 C nuclei which are connected by a JCH over one bond.
D. 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. Because 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.
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.
E. 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.
2. 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. Because 10 lines are found in the (^13) C NMR spectrum for 10 C, 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). Because 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
>C=O
X Y
X or Y
ν(CH) 3050
118.0 ppm 1 line = 1 C 1 65.0 ppm 1 line = 1 C
133.6,133.5, 6 lines = 6C 131.8,131.4, 130 2 127 6
7.90,7.59, 4H
Remaining Molecular Symmetry: None. Remaining Unsaturation: None.
3. Analysis of 1H NMR Coupling patterns To determine whether the aromatic ring is ortho or meta disubstituted, the 1 H NMR spectrum must be analyzed 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, Jpara < 1 Hz). Careful analysis of the three ddd patterns leads to the numerical coupling constants shown in the table. Match as many of the coupling constants (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. Hb is coupled to Hd with
J = 7.0 Hz so Hb and Hd are ortho to each other. Because Hb also has a J = 8.0 Hz
coupling constant, another proton is also ortho to Hb , likewise for Hd. Therefore, the aromatic ring is ortho -disubstituted and the proton assignments are as shown below.
1H NMR ( 500 MHz, d6-acetone solvent) 1H NMR Spin System label (^) δ (ppm) (multiplicity, #H, J values)
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 -CH 2 CH 2 - 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)
Important Mass Spectrum Fragment Ions m/e (M+1) %
(M+2) %
Possible Assignments m/e (M+1) %
(M+2) %
Possible Assignments
209 13.0% 36.1% M•+^139 8.7% 33.5% 35+76+28=
174 - - - - - - (M-35)^ +^128 - - - - - -?
156 7.8% 30.4% (M-27-CN)^ •+^111 9.1% 34.5% 35+76=
5. 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, free rotation of the bond between them apparently allows each to be observed as triplets.
In the 13 C NMR spectrum, the table in the Appendix 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.
Cl
O
C
C
C
O
C N
H H
H (^) a H H
H (^) b
H (^) d
H (^) c
e
f Cl
O
C
CH 2
CH 2
O
C N
Aromatic C's (^) 60.
1 H NMR Assignments 13 C NMR Assignments
For the mass spectrum, drawing the mechanism for two or more of the prominent peaks in the spectrum is good practice. 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.
