Using NMR and IR Spectroscopy to Determine Structures Dr. Carl Hoeger, UCSD The following guidelines should be helpful in assigning a structure from NMR (both PMR and CMR) and IR data. At the end of this guide is a Spectroscopy Worksheet, it is a good idea to get in the habit of using this format when doing problems. It will help you keep everything straight. 1. If you can, determine the molecular formula, (normally you are given this or enough information to get at it); once you have that, calculate the index of hydrogen deficiency (IHD * or "degree of unsaturation"). This will tell you how many unsaturations total there are in the molecule (this is the total number of π-bonds + rings); HOWEVER it does NOT distinguish between rings and multiple bonds! Values of 4 and higher may indicate the presence of an aromatic ring (1 ring + 3 double bonds = 4): the NMR will be very helpful here! * IHD = (#C) # H 2 # Halogens 2 + # N 2 (oxygen and sulfur are ignored) + 1 2. Turn next to the IR spectrum (if you have one; otherwise jump to step #3). Look for distinctive stretches (see below) but don't get too caught up in trying to analyze the IR in detail at this point; you can return to it later if necessary. REMEMBER: the absence of an absorption is just as important as its presence! Distinctive IR stretches, in approximate order of importance, are: a) the carbonyl (C=O) absorption between 1690-1760 ; this strong band indicates either an aldehyde, ketone, carboxylic acid, ester, amide, anhydride or acyl halide. The aldehyde may be confirmed with C-H absorption from 2840 to 2720. b) the O-H or N-H absorption between 3200 and 3600. This indicates either an alcohol, N- H containing amine or amide, or carboxylic acid. For -NH 2 a doublet will be observed. c) the triple bond absorptions at 2100-2260 ; these can be small but since this region of the IR has virtually nothing else in it, they will be exposed. Remember that internal may be weak or nonexistent. d) the C-H absorption(s) between 3100 and 2850. Use 3000 as your break point. An absorption above 3000 indicates H s on C=C s, either alkene or aromatic. Confirm the aromatic ring by finding peaks at 1600 and 1500. Confirm alkenes with an absorption at 1640-1680. C-H absorption between 3000 and 2850 (below 3000 ) is due to aliphatic hydrogens (H s on sp 3 carbons). e) the C-O absorption between 1080 and 1300. These peaks are normally rounded like O-H and N-H peaks but not as broad and are prominent, but BE CAREFUL: this is the fingerprint region! Use this region to confirm your suspicions, rather than assign absolute structure. Carboxylic acids, esters, ethers, alcohols and anhydrides all contain this peak. 3. Move to the CMR if it is available (otherwise move to the next step). Since you will almost always have only the proton-decoupled 13 C spectrum, you will simply be looking at the number of resonances. Classify the resonances as aliphatic (sp 3 ), alkenyl (sp 2 ) or alkynyl (sp), aromatic, or carbonyl. Use this to begin to get an idea of what you might be looking at from a gross structural
standpoint. Remember that a simple CMR could indicate a highly symmetric molecule. Don t obsess at this point about putting together a structure; this just provides more pieces to the puzzle. 4. Now move to the 1 H NMR spectrum (PMR). Begin by using the integration to determine the number of H's contributing to each resonance. Measure the integration step heights (if necessary); remember that you need to measure the total integral for an absorption that is split (so, all three peaks of a 1:2:1 triplet are measured together, since they are all part of one resonance). For a complex splitting, like a non-first order pattern, measure the integral height from where it comes in to where it goes out. There are two ways to proceed at this point depending on what information you may have: i) Conversion factor method: if you know how many H s are in the molecule, you can take the sum of ALL the integral heights and divide that into the number of H s total. Now, if you multiply that number by the individual integral heights you will get the # of H s responsible for each integral: Total # H's = H's per mm; Total height of all integrals (mm) Integral ht in mm H's per mm = H's for that resonance ii) Ratio method: take the individual integral heights for each resonance and determine the ratios of the integrals. Compare these with the formula to get the numbers of H's (if you have the chemical formula). Thus, if two resonances have a 1:3 ratio, and the compound has 24 H's, then these represent 6H and18h respectively; similarly, if the compound had 8 H s this ratio would be 2:6. Always remember to be flexible, especially when the ratios are large (for example, it is much more difficult to measure a 1:10 ratio accurately than a 1:1 or a 2:3 ratio). For example, if you measure a ratio of 1:10 and the compound has only10 H's, then the ratio is probably really 1H:9H. Some structural clues are usually evident at this point: an upfield signal that integrates for 3 equivalent H's will be due to a CH 3 group (especially if it is a singlet). A peak due to 2 equivalent H's will usually be from a CH 2 group (remember to be flexible!) Similarly, a signal integrating for 6 H's probably indicates 2 equivalent CH 3 groups (or, less commonly, 3 CH 2 's), etc. Unfortunately, sometimes signals overlap and it's not possible to determine individual integrals for each do what you can and move on. 5. Take an overall look at the NMR and see if you can pick out signature resonance regions: Are there peaks downfield in the vinylic (δ 4.5-6) or aromatic (δ 6.5-8) regions? Is there a slightly broadened (D 2 O exchangeable) singlet in the δ 1-5 range characteristic of an alcohol OH? Is there a slightly broadened (D 2 O exchangeable) singlet in the δ 10-13 range for a CO 2 H? Is there a peak in the 9-11 range? Could be an aldehyde, especially if split. Are there peaks in the δ 3-4 range for H's on a C that is bonded to an electronegative atom? 6. Begin at this point to put together a table of what you see. Start by recording the chemical shifts, number of H s, and splitting of the signals (see #7). For simple splitting, it is customary to use abbreviations: s = singlet, d = doublet, t = triplet, q = quartet. Higher order splitting patterns will use combinations of these (see #7). For more complicated splitting, especially if it is not first-order, it is customary to use m for multiplet, a simple way of indicating it is too complicated for simple
analysis, or simply I can t figure it out. At this point it is also a good idea to write down all the bits and pieces you've identified, along with lines to represent bonds to other things (use symbols such as X to help you keep track of where you are in your detective work): Chemical Shift #H Splitting Possible Assign. δ 0.9 3 t CH 3, next to a CH 2 δ 3.6 2 q X CH 2, next to a CH 3 ; X is electronegative δ 6.9 7.3 5 m Add up the numbers of atoms accounted for by your collection of pieces and compare it to your chemical formula and see if anything is missing. For example, there might be an O or an S and a Br unaccounted for. (Note that the included lines indicate that S and O are divalent and Br is monovalent; N would have three lines since it is trivalent, etc. All of this will help you remember when you start to put pieces together). With this collection of molecule parts you can start to put them together like the pieces of a jigsaw puzzle. 7. Use the splitting information to start to refine your structure. Splitting patterns are generally very simple in flexible, non-cyclic (acyclic) molecules and the coupling constants will almost always be close to 7 Hz. The number of peaks indicates how many H's are next door to the H's that cause the signal. That is, for H a C C H b, the signal for H a will be split by H b and vice versa. If an absorption is split into N separate peaks, the H's causing the peak must be next to N-1 H's. Thus, a triplet indicates splitting by 2H's, a quartet indicates splitting by 3H's, etc. You should learn to recognize some characteristic patterns of isolated alkyl groups: ethyl: CH 2 -CH 3 appears as a 2 H quartet (CH 2, next to 3 H s) and a 3 H triplet (CH 3, next to 2 H s); t-butyl: C(CH 3 ) 3 appears as a 9 H singlet (3 CH 3 s, next to 0 H s); isopropyl: CH(CH 3 ) 2 is a 1 H septet (CH, next to 6 H s) and a 6 H doublet (2 CH 3 s, next to 1 H). Remember to count only the number of protons next to each group that contributes to the absorption. For example, in diethyl ether, CH CH 3 O CH 2 CH 2 3, the CH 3 on the left is next to 2 H s left left (on the adjoining CH 2 labeled left), so the CH 3 signal is a triplet, not a quintet (that is to say that the left CH 3 does not feel the spin states of the right CH 2 protons, so its absorption is not split by them). Also remember that equivalent H s do not split each other, so Cl-CH 2 CH 2 -Cl gives a singlet, not a triplet. (Note: hydrogen-bonded OH protons are normally are NOT split by (and do not split) neighboring H s (but they can, usually when run under dilute conditions, so be careful). THE FOLLOWING WILL BECOME IMPORTANT FOR ALKENES, AROMATIC COMPOUNDS, AND MORE COMPLEX SYSTEMS. Coupling constants (J) are dependent on dihedral angles, so in compounds with restricted rotation the splitting will often be more complex. For example, in cyclic compounds, the ring protons can only adopt certain dihedral angles relative to their neighbors, and the cis and trans neighbors will therefore have different coupling constants. To make things even more complex, the observed couplings often result from averaging over two or more right
conformations. The splitting patterns that result are sometimes uninterpretable (especially at low magnet field strength: i.e. under 300 MHz). Vinylic H s often show complex splitting too, but these patterns are usually interpretable with a little effort. Two cis vinylic H s will split each other by about 9 Hz, while trans H s will split each other by about 15 Hz. Often, the neighboring H s will split a vinylic proton by different amounts and produce striking multiplets or clusters of peaks, for example, a doublet of doublets (dd), doublet of triplets (dt), or a doublet of doublets of quartets (ddq); in these patterns the first splitting has a coupling constant larger than the second splitting. Thus, in a doublet of triplets J doublet > J triplet, so here a signal will be split first into a doublet, and then each of the resulting peaks split into a triplet. Such higher order splitting patterns can be analyzed by the use of "tree diagrams" (a good article on this can be found in J. Chem. Edu. (1995) 72, pp. 614-615). Such analyses are most useful once the structure has been narrowed down to a few possibilities and should only really be pursued during the final stages of structure analysis. 8. Put the pieces together in a way that's consistent with the splitting info and the chemical shifts. Occasionally you may find more than one way to do this, so you'll have to consider all the possibilities and perhaps use more subtle spectroscopic information to reach a decision. Sometimes you just have to live with the fact that you simply can not tell the difference. 9. After you've put the pieces together, check the structure against the spectra and any other data that are available. Do the expected splittings correspond to those actually observed? Are the expected chemical shifts in reasonable agreement with those observed? Is the structure consistent with the IR spectrum? If not, you may need to back up and try connecting things differently.
Spectroscopy Worksheet Dr. C. Hoeger I. Unknown ID: Molecular Formula = Mp/Bp = IHD = Physical Appearance: II. IR Data: Absorption (cm -1 ) intensity Possible structure Notes: III. NMR Data: Chemical Shift # H s splitting Possible structure IV. CMR Data: Possible Structure (use back for scratch work):