E. Kwan Lecture 3: 2D NMR Spectroscopy Chem 135 2D NMR Spectroscopy Eugene E. Kwan between sites, which is very useful if you want to know the structure of a molecule. Every 2D NMR experiment has the same general format: preparation evolution mixing detection Scope of Lecture problem solving structural elucidation workflow two frequency dimensions two-dimensional NMR spectroscopy MBC xx xx 4 5 SQC absolute value vs. phase-sensitive modes Two Frequency Dimensions 1D NMR spectra are called "1D" because they have one frequency dimension, but actually have an additional dimension, intensity: intensity frequency CSY 2D NMR means that there are two frequency dimensions. 1D peaks tell you something about a particular chemical site: what it's chemical environment is like, how many nuclei are present, how many nuclei are near the site, etc. But there's no mechanism for telling you anything about the the connections (1) Preparation: Some sequence of pulses is used to generate states that are poised to interact in a useful way. This is typically a 90 pulse that generates transverse magnetization. (2) Evolution: The resonances precess in the rotating frame according to their offsets. This means that magnetization is "frequency-labeled" as a function of t 1. (3) Mixing: Magnetization is transfered through bond (or (through space or chemical exchange). (4) Detection: The magnetization that did not get transferred during the mixing period will appear at the same frequency during the detection period. These are diagonal peaks of frequency ( A, A ). Magnetization that was at frequency A but moved to frequency B during the mixing period will precess at an off-diagonal frequency ( A, B ). This is best illustrated by the basic CSY-90 sequence: channel t 1 t 2 90 x t 1 CSY stands for Crrelations SPectroscopy and is a method for finding homonuclear, through-bond correlations. (There are other methods for finding heteronuclear, through-bond correlations.) Implicitly, the above diagram means that we run a series of experiments, with a fixed values of t 1 every time. 90 x t 2
E. Kwan Lecture 3: 2D NMR Spectroscopy Chem 135 experiment #1 experiment #2 experiment #3 t 1 t 2 t 1 t 2 t 1 t 2 (1) As with 1D spectra, this may also involve intermediate zero-filling, apodization, or linear prediction. (2) Quadrature detection, i.e., the discrimination of positive vs. negative frequencies is possible and necessary, but is complicated and will not be considered in this lecture. (3) Recall that 1D spectra have a real and imaginary part and that phase correction ensures that the real part has a purely absorption lineshape. In 2D spectra, there are two real parts and two imaginary parts (one for each dimension). In an ideal world, the real part of both would be absorptive as well: The entire experiment generates a 2D data matrix. Fourier transformation of the rows, followed by the columns gives the final 2D spectrum: (Red = negative contour; black = positive countour) If this is the case, we can present the data in a phase-sensitive format. owever, in some experiments, this is impossible, and it is necessary to mix the real and imaginary parts to give an absolute value format: absolute value = Sqrt[real 2 + imaginary 2 ] These partially absorptive/dispersive peaks do not have a standard Lorentzian shape and instead appear as broader phase-twisted shapes:
E. Kwan Lecture 3: 2D NMR Spectroscopy Flavors of CSY The basic CSY-90 sequence is: channel 90 x t 1 90 x In VNMR, this is requested with the "gcsy" command. ere is the CSY-90 spectrum for ethyl acetate: t 2 Chem 135 (5) 1D Curves: These are not part of the 2D experiment. They do not have to be drawn at all, but it is customary to add them to for reference. The curves shown below are projections of the peaks onto the axes. Because increasing ni is very costly, resolution is poor. A better alternative is to place the actual 1D spectra on these axes (acquired in separate experiments.) 0.5 (1) Cross-peaks: In ethyl acetate, there s an ethyl group, which means that a methyl and a methylene share a vicinal coupling. (2) Isolated Peaks: Not everything has large enough couplings to give off-diagonal peaks (for example, this acetate). In general, CSY mostly shows 2 J and 3 J couplings. (3) Diagonal: f course, every proton is coupled to itself. There s nothing interesting to see here. (4) Shape: The spectrum is a square, since we are correlating a spectrum to itself. (6) Symmetry: If A is coupled to B, then B is coupled to A, so there is reflection symmetry about the diagonal. 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5
E. Kwan Lecture 3: 2D NMR Spectroscopy Chem 135 A detailed quantum-mechanical treatment, which is outside the scope of this course, shows that for an AX system, the crosspeaks have doubly absorptive lineshapes, but the diagonal peaks have doubly dispersive lineshapes: (2) At sufficiently high resolution, crosspeaks are tilted. Typically, geminal couplings will appear with positive slope, while vicinal couplings will appear with negative slope. owever, due to the variations in J, this is not always true. (3) CSY-45 is slightly less sensitive than CSY-90 (by about 15%), but since CSY-90 is already a very sensitive experiment, this is of no consequence. CSY-45 gives more simplification than CSY-60, and is therefore preferred. CSY-45 is the best absolute value CSY experiment for routine use and should be used instead of CSY-90. The PFG version can be requested in VNMR with "gcsy45." CSY-45 CSY-90 (source: Claridge, pg 140) The long tails from the dispersive peaks can interfere with off-diagonal peaks. Therefore, CSY-90 data are presented in absolute-value mode as phase-twist lineshapes. Just because something is popular doesn't necessarily mean it's the best. A popular absolute-value version of CSY is called CSY-. ere, the second 90 pulse is replaced by a smaller pulse of tip angle. Typically, is 45 or 60. channel 90 x t 1 x t 2 tilting diagonal is less crowded This has several important consequences: (1) The diagonal is compressed. Since we don't care about the diagonal, and it can interfere with nearby cross-peaks, this is good.
E. Kwan Lecture 3: 2D NMR Spectroscopy Flavors of CSY The difference between the two CSY spectra is particularly evident in this example from Reynolds and Enriquez: CSY-90 Expansions clearly show the tilting effect: geminal vicinal Chem 135 CSY-45 ne post-acquisition strategy that is sometimes used to enhance S/N is called "symmetrization" and is based on the idea that the spectrum should have reflection symmetry about f 1 =f 2 (VNMR: foldt). owever, because f 1 is usually digitized better than f 2, crosspeaks are usually more resolved on one side of the diagonal. Therefore, peaks that appear more strongly on one side of the diagonal may be long-range peaks. In addition to losing this information, false crosspeaks may appear from the coincidental symmetry of t 1 noise:
E. Kwan Lecture 3: 2D NMR Spectroscopy eteronuclear Correlation Spectroscopy For heteronuclear experiments, one has the option to prepare, evolve, or mix the magnetization on either proton or carbon (the X-nucleus). old strategy: "direct" detection of X-nucleus (less sensitive) new strategy: "indirect" detection of -nucleus (better) From Claridge, page 191: Chem 135 where exc is the initially excited spin and obs is the observed nucleus. In dual-band probes, there is always a coil on the inside (more sensitive) and a coil on the outside (less sensitive). For inverse-detection probes, the proton coil is on the inside; direct-detection probes have the carbon coil on the inside. Regardless of the detection scheme, the goal of all of these experiments is to connect protons with carbons. Unlike proton-proton correlation experiments (CSY, TCSY), we have the possibility of protons being directly (one-bond) or remotely (multiple-bond) connected. For an inverse-detected experiment, we see carbons that are directly or remotely attached to protons: C C C C C C C C C C one-bond (direct) multiple-bond (remote) (This means that quaternary carbons, which do not have any attached protons, do not appear in inverse-detection experiments.) Remote correlations can be transmitted through heteroatoms, but this is by no means required. Conversely, for a direct-detection experiment, we see protons that are directly or remotely attached to carbons: C C C C C C C C C C In reality, various experimental considerations mean that the advantage is less than 32/4=8. In general, S N exc 3/2 obs one-bond (direct) multiple-bond (remote) In this case, quaternary carbons can appear: C C C owever, directly-detected multiple-bond correlation experiments are insensitive and not in routine use.
E. Kwan Lecture 3: 2D NMR Spectroscopy Chem 135 SQC - eteronuclear Single Quantum Correlation Spectroscopy This is a proton-detected experiment that shows directly attached carbons (one bond). C and C 3 peaks are phased up (red) while C 2 peaks are phased down (blue). 1D curves are useful, but not required. This is the spectrum for menthol: Me Me Me (1) Spreading ut: The 1D spectrum is quite crowded, but having the second dimension spreads out the peaks over a much larger area. (2) DEPT: The additional APT pulses are a kind of extra fanciness that require more time during which signal can decay, so there is some sensitivity loss. owever, SQC-APT sequences of this type can be acquired more quickly than edited DEPT spectra. not diastereotopic (double height) diastereotopic (3) Methylenes: These appear on the same horizontal line, since they re two protons on the same carbon. This allows one to distinguish 2 J from 3 J (or >3 J) in CSY spectra.
E. Kwan Lecture 3: 2D NMR Spectroscopy Chem 135 MBC Spectra This is a proton-detected experiment that shows carbons that are 2-3 bonds away from protons. It is absolute value. (1) A common artifact is the appearance of one-bond correlations. MBC is "tuned" to detect the small couplings arising from long-range interactions, but this is not perfect. These one-bond artifacts appear as doublets in f 2, with J= 1 J C. ccasionally, this splitting is itself useful information. owever, most of the time, one should be wary of these artifacts. (2) MBC incorporates a delay which under ideal circumstances is 1/2J C. Since long-range couplings occupy a relatively wide range of 5-25 z, this delay is often set at a compromise value of 60 ms (8 z). This means that not all of the correlations will appear, and certainly not with equal intensity. Additionally, note that three-bond couplings are often larger than two-bond ones. 1 2 3 4 5 6 7 8 16 24 arrows indicate one-bond peaks 32 40 48 56 64 72 3.5 3.0 2.5 2.0 1.5 1.0 0.5
E. Kwan Lecture 3: 2D NMR Spectroscopy Summary of 2D NMR Experiments 1. SQC: 1-bond C, - all experiments are proton-detected 2. CSY: 2,3-bond, 3. MBC: 2,3-bond C, SQC By far the simplest spectrum to understand, begin here. axes: proton, carbon correlations: 1-bond C, couplings purpose: number proton spectrum, match each proton with a carbon, identify C 2 pairs 1 2 3 B A proton proton 1 is directly attached to carbon A carbon down (C 2 ) up (C, C 3 ) Numbering the Spectrum Even if the proton spectrum overlaps, SQC will usually separate the peaks enough so they can be numbered (convention: left to right). C B A 1 1 2,3 2 3 proton 4 4 carbon protons 2 and 3 are on the same carbon; i.e., 2/3 is a methylene (C 2 ) pair - although 2 and 3 overlap in the 1D spectrum, they have very different carbon shifts so the SQC separates them - only give a number for each unique chemical shift (e.g., a methyl group only gets one number) CSY Next, use CSY to determine the composition of each spin system. axes: proton, proton correlations: off-diagonal peaks are 2-3 bond couplings between protons purpose: assign protons to spin systems spin system: a set of protons sharing through-bond (J) couplings 2 3 1 4 5 6 1-2-3, 4-5, 6 MBC axes: proton, carbon correlations: 2-3 bond couplings between protons purpose: connect spin systems B A 1 proton D C B A 2 proton 1 is 2 or 3 bonds away from carbon A A B C D carbon proton A-B-C, D one-bond artifact: proton 2 is directly attached to carbon B (look for doublets) proton 2 is 2 or 3 bonds away from carbon A Chem 135 proton
E. Kwan Lecture 3: 2D NMR Spectroscopy Structural Elucidation Workflow (1) Start with the SQC to label the proton spectrum. (2) Tabulate the Data. ID ( 1 ) ( 13 C) s type J (z) CSY MBC Chem 135 ere's a hypothetical proton spectrum, where peaks 2 and 3 are very close (peaks are labeled from left to right): integrals: 1 2 3 4 1 2 1 The SQC spectrum is very simple because it spreads out the peaks over two dimensions has little multiplet structure: xx xx 4 5 xx 2 xx 3 1 axis It will also identify methylene pairs, both by color (remember, C and C 3 are opposite to C 2 ) and by the fact that they occur on the same line. In this case, 4/5 is apparently a methylene pair. Methylene pairs will always share a geminal coupling. With diastereotopic protons, these will give rise to a CSY crosspeak, which the SQC will first identify as being geminal in nature. 1 5 13C axis 4/5 is a methylene pair. 1 5.76 145.23 1 d 5.1 2 152.12 2 3.76 72.45 1 d 5.1 1,3 32.47 3 3.47 -- 1 br s -- -- 202.57... etc... Quaternary Carbons: 35.57, 54.32, 202.57... C 2 pairs: 4/5,... - peaks are listed by number from high to low chemical shift - SQC: connect each proton to its directly attached carbons; find methylene pairs - CSY: if 1 is coupled to 2, then check that 2 is coupled to 1; however, both partners of a methylene pair may not show couplings to a common partner (peak 3) - exchangeable protons do not appear in the SQC - quaternary carbons can be found from MBC or the 1D spectrum carbon - MBC: watch for one-bond peaks; more intense peaks like methyl groups are more likely to show long-range correlations; sp 2 systems: 2 J is small, but 3 J is large (bigger for anti than syn) (3) Generate Spin Systems - Use CSY to build up spin systems. Each "component" of the spin system is a methyl group, a methylene pair, or a methine (from the SQC). Double-headed arrows represent (putative) vicinal couplings, with dashed lines for long-range couplings: 4/5 8 1 2 long-range coupling C C C C C
E. Kwan Lecture 3: 2D NMR Spectroscopy (You might not know which ones are long-range. Use your chemical intuition and look at the peak intensities and asymmetry about the diagonal. You might have to change your diagram if you find it to be inconsistent later.) (4) Connect Spin Systems Look for MBC correlations that connect a proton in one spin system to a carbon in another spin system: 145.23 33.37 4/5 8 1 2 6 7 Curved arrows indicate MBC correlations. If you find such a connection, that tells you that the spin systems must be adjacent in the head-to-tail sense shown. Note that these do not have to be mutual like CSY couplings: finding this MBC correlation C C C does not mean If that doesn't work, you can look for a carbon, possibly quaternary, that protons in both spin systems have common MBC correlations to: 35.57 C C C 3 will be present 4/5 8 1 2 6 7 3 (5) Generate Fragments In many cases, complex, overlapping signals will prevent you from drawing out all the spin systems of the molecule. In particular, unknown structural elucidations will require you to generate and connect fragments, which are tied together by what is visible in a spin system and MBC. ere is how I put together menthol, which has some overlapping signals and one continuous spin system: Fragment Structures 2, 25.82 3 C C 3 11, 21.00 14, 16.07 X Key Evidence X 8, 50.12 - methyls 11, 14: CSY to 2-2: CSY to 8 Vicinal CSY Correlations 11-2 2-14 3 C C 3 2-8 X X 11-2 - SQC: 2, 8 = C Chem 135 Key MBC Correlations 3 C C 3 X - 11, 14: mutual MBC - 11, 14: MBC to 2, 8 X 11-14 14-8 - MBC: double-headed arrows to indicate mutual correlations; single-headed arrows to indicate a one-way correlation - if you feel comfortable, you can write "11-2" instead of "11-25.82" to speed things up, but remember, MBC correlations are from protons to carbons - Xs represent non-protons Write down your reasoning (even though it's boring). By their nature, these problems are very complicated, and there's a 99.9% chance you will not remember what peak is at 16.07 ppm several years from now. If you write down what you were thinking, you and others will be able to follow your process later on without repeating the entire analysis. It's also likely you will make some mistakes, which will be easier to track down this way. (6) Entry Points When you look at a CSY or MBC spectrum, you will see a lot of peaks. Where should you start generating fragments from? In general: clearly resolved peaks with unambiguous or characteristic chemical shifts. ere are some ideas: - carbonyl region - aromatic rings and olefins - methyl groups - quaternary carbons
E. Kwan Lecture 3: 2D NMR Spectroscopy (7) Expand and Connect Fragments Chem 135 Don't bite off too much at a time. Make a small fragment with correlations you feel confident in assigning. Then, move on to another entry point and generate another fragment. Fragments with uncertain correlations are hard to use. nce you have exhausted all the easy data, work on what's left to expand your existing fragments, and if possible, connect them. X C 2 X 6 4, 13, 34.51 C 3 12, 22.10 (8) List Full Assignments This is self-explanatory: 14 7 1 2 8 11 5, 9 3, 10 6 4, 13 12 5-13, 9-4 - CSY, MBC correlations connect termini 16.07 21.00 25.82 50.12 71.53 23.10 45.03 34.51 31.62 22.10 You should also file all the FIDs and your "good" notes of what you've done in the same place. If you feel the compound will go into your thesis, take the time to write it out in journal format.
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 1 Please assign the proton and carbon resonances of ethyl nipecotate. (500 Mz, CDCl 3 ; spectra are courtesy of Dr. Jeffrey Simpson, MIT. See Simpson, Chapter 9 for his treatment of this problem.) Chem 135 N 2.0 1.0 1.0 1.0 1.1 1.1 1.0 2.0 1.1 2.8 4.0 3.5 3.0 2.5 2.0 1.5 1.0 Chemical Shift (ppm) 173.99 59.84 48.35 46.14 42.20 27.10 25.27 13.91 192 184 176 168 160 152 144 136 128 120 112 104 96 88 80 72 64 56 48 40 32 24 16 8 0 Chemical Shift (ppm)
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 1 Please assign the proton and carbon resonances of ethyl nipecotate. (500 Mz, CDCl 3 ; spectra are courtesy of Dr. Jeffrey Simpson, MIT. See Simpson, Chapter 9 for his treatment of this problem.) ID ( 1 ) ( 13 C) s Type J (z) CSY 1 Chem 135 N 2 3 4 5 6 7 8 9 10 11 Quaternary Carbons: Methylene Pairs:
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 1 This is the MQC spectrum. Please number the protons. What are the methylene pairs in this molecule? Are there any quaternary carbons? Chem 135 N 10 15 20 25 30 35 40 45 50 55 60 4.0 3.5 3.0 2.5 2.0 1.5 1.0
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 1 This is the CSY spectrum. Try to label the off-diagonal peaks. Chem 135 N 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.0 3.5 3.0 2.5 2.0 1.5 1.0
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 1 The numbering I get is shown below. Chem 135 N A multiplet analysis in ACD/NMR Processor gives: 1 NMR (500 Mz, CDCl 3 ) ppm 1.05 (t, J=7.16 z, 3 ), 1.25 (ddd, J=10.49, 3.70, 2.59 z, 1 ), 1.40-1.53 (m, 2 ), 1.72-1.83 (m, 1 ), 2.17-2.28 (m, 1 ), 2.38-2.49 (m, 1 ), 2.60 (dd, J=12.34, 9.38 z, 1 ), 2.72 (dt, J=12.34, 3.70 z, 1 ), 2.95 (dd, J=12.34, 3.46 z, 1 ), 3.86-3.98 (m, 2 ). The methylene pairs are 1, 2/4, 3/5, 7/8, 9/10, and 11. This would have been a lot easier with a better-resolved SQC-DEPT! The carbon at 173.99 is quaternary (the ester). 3 5 6 7 8 9 10 11 10 15 20 25 30 35 40 45 2 4 50 55 1 60 4.0 3.5 3.0 2.5 2.0 1.5 1.0
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 1 ere s how I annotate CSY spectra. I have bolded the correlations that do not correspond to geminal couplings (guessed for aliphatic region) and are clear. Parentheses indicate weak couplings. Chem 135 N 1 2 3 4 5 6 7 8/9 10 11 1.0 1 2 3 4 5 6 7 8/9 10 11 2-4 3-5 4-6 (2-6) 7-8/9(?) 6-8/9(?) 1-11 10-8/9(?) 5-10 3-10 1.5 2.0 2.5 3.0 3.5 4.0 4.0 3.5 3.0 2.5 2.0 1.5 1.0
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 1 ere s my data table below. At this point, overlap makes the spin systems hard to get at. But we can use some common sense. ID ( 1 ) ( 13 C) s Type J (z) CSY 1 3.92 59.84 2 m -- 11 Chem 135 N (1) Entry Points: Start with the obvious. The ethyl group is 1 and 11, and there s only one methine in this molecule: N 6 (2) Amine: Two sets of downfield, diastereotopic protons must be adjacent to the amine. These must be 2/4 and 3/5. There is a clear CSY correlation between 2 and 6. Therefore, by process of elimination: 1 11 2 2.95 48.35 1 dd 12.3, 3.5 4, 6 (w) 3 2.72 46.14 1 dt 12.3, 3.7 5, 10 4 2.60 48.35 1 dd 12.3, 9.4 2, 6 5 2.43 46.14 1 m -- 3, 10 6 2.22 42.20 1 m -- 4, 2 (w) 7 1.78 27.10 1 m -- 8 1.46 27.10 1 m -- 9 1.46 25.27 1 m -- 6 10 1.25 25.27 1 ddd 10.5, 3.7, 3, 5 2.6 11 1.05 13.91 3 t 7.2 1 3/5 N 2/4 Quaternary Carbons: 173.99 Methylene Pairs: 1, 2/4, 3/5, 7/8, 9/10
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 1 Chem 135 N 1 2 3 4 5 6 7 8/9 10 11 (3) Aliphatics: We need to distinguish between 9/10 and 7/8. Note that 10 has CSY correlations to 3 and 5. (Why is one correlation stronger than the other?) Thus: 10-7/8 173.99 8/9(?) 6 1 11 9/10 7-8/9(?) 3/5 2/4 N (4) Check: Do these assignments make sense based on chemical shift? ChemDraw predictions certainly seem to be congruent with the assignments: 1.55;1.45 1.98;1.73 4.12 2.33 1.30 2.76;2.73 3.21;2.96 N 2.0 (5) Check: Do the overlapping correlations make sense? Yes, they seem to actually be: 7-8 (geminal), 6/8 (vicinal), and (10-8 and 10-9). 2-4 3-5 1-11 4.0 3.5 3.0 2.5 2.0 1.5 1.0 4-6 (2-6) 6-8/9(?) 5-10 3-10 1.0 1.5 2.0 2.5 3.0 3.5 4.0
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 Please assign the proton and carbon resonances of sucrose (500 Mz, D 2 ). Chem 135 1.0 1.0 1.1 5.8 1.0 2.0 1.0 1.1 5.2 5.1 5.0 4.9 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2 Chemical Shift (ppm) 103.74 92.24 81.44 76.39 74.02 72.62 72.45 71.13 69.26 62.43 61.35 60.14 104 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 Chemical Shift (ppm)
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 Chem 135 58 inset 60 62 arrows: double intensity 60 61 62 63 64 65 66 67 68 69 70 71 72 64 66 68 70 72 74 76 78 80 82 84 86 3.65 3.60 3.55 3.50 3.45 3.40 3.35 3.30 MQC 88 90 92 5.2 5.1 5.0 4.9 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 Chem 135 CSY-90 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0 5.1 5.2 5.2 5.1 5.0 4.9 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 Numbering the protons requires some careful bookkeeping. Chem 135 58 8 inset 60 60 7 10 62 64 61 62 63 64 65 66 67 68 2 3 5 6 9 11 12 66 68 70 72 74 76 78 69 70 4 80 82 71 84 72 86 1 3.65 3.60 3.55 3.50 3.45 3.40 3.35 3.30 MQC 88 90 92 5.2 5.1 5.0 4.9 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 Chem 135 CSY-90 1 2 3 4 5-8 9 10 11 12 1 2 3 4 5-8 9 10 11 12 2-3 3-4 4-5-8(?) 9-11 1-11 9-12 12-5-8(?) 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0 5.1 5.2 5.2 5.1 5.0 4.9 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3
E. Kwan Lecture 3: 2D NMR Spectroscopy Chem 135 Problem 2 ere s some space for your data table: ID ( 1 ) ( 13 C) s Type J (z) CSY 1 2 3 4 5 6 7 8 9 10 11 12 Quaternary Carbons: Methylene Pairs:
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 I get this data table: ID ( 1 ) ( 13 C) s Type J (z) CSY 1 5.22 92.24 1 d 3.9 11 Chem 135 2 4.02 76.39 1 d 8.8 3 3 3.86 74.02 1 t 8.6x2 2, 4 4 3.70 81.44 1 m -- 3 5 3.66 72.45 1 m -- 6 3.64 72.45 1 m -- 7 3.63 62.43 1 m -- 8 3.62 60.14 2 m -- 9 3.57 72.62 1 t 9.7x2 11, 12 10 3.48 61.35 2 s S 11 3.36 71.13 1 dd 10.0, 3.9 1, 9 12 3.28 69.26 1 t 9.5x2 9 Quaternary Carbons: 103.74 Methylene Pairs: 5/6, 8, 10
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 Spin Systems. Chem 135 Based on chemical shift arguments, 1 is the anomeric proton. It may help if I re-draw sucrose: Because this is a disaccharide, there are two major spin systems (10 is on its own). ere is how I draw them out: 1 = anomeric 2 d The left-hand spin system can be drawn out: 11 dd 9 t 3 t 4 m 12 9 11 1 12 t 5/6, 7, or 8 5/6, 7, or 8 Which of 5/6, 7, or 8 must 12 be connected to?
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 Spin Systems. Chem 135 1 = anomeric 11 dd 9 t 12 t 5/6, 7, or 8 2 d 3 t 4 m 5/6, 7, or 8 3.50 3.55 3.60 3.65 (1) Number of Protons Note that 12 is connected to a methine. Therefore, it cannot be connected to 5/6 or 8, and by process of elimination, must be connected to 7. (2) The ther Spin System Note that proton 2 is a doublet. This can only occur one way in the other spin system (3) What about 4? 12 9 7 11 1 4 We must decide if proton 4 is connected to 5/6 or 8. Looking closely at the CSY spectrum, it certainly looks like 4 is coupling to the right-hand part of the overlapping mess. That means we can make a reasonable guess (but not a firm conclusion) that it's 8. What steps could you take to confirm this? 2 3 3.70 3.70 3.65 3.60 3.55
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 2 Spin Systems. (4) Final Assignments Chem 135 1 = anomeric 11 dd 9 t 2 d 3 t 4 m 5/6 12 9 7 11 1 103.74 10 2 4 3 8 12 t 5/6, 7, or 8 5/6, 7, or 8 It remains only to assign the quaternary carbon (of which there is only one) and the remaining methylene (also only one left). Questions you should know the answers to: - Why do all the protons have large couplings except 1? - Why aren't the hydroxyl protons visible? - Why was this problem difficult, and what could you have done to make life easier for yourself?
solvent: acetone-d 6 naringenin Please assign this spectrum. 2 0.7 0.9 0.9 2.1 2.1 2.0 1.2 1.2 1.1 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 Chemical Shift (ppm)
naringenin phenol phenol 0.7 0.9 0.9 2.1 2.1 2.0 1.2 1.2 1.1 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 Chemical Shift (ppm)
- number from left to right naringenin - one number per unique chemical shift 1 5 6 2 3 4 7 8 9 0.7 0.9 0.9 2.1 2.1 2.0 1.2 1.2 1.1 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 Chemical Shift (ppm)
gsqc 35 40 45 50 naringenin 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 135 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0
35 blue = C 2 40 45 naringenin - two peaks, same horizontal line = same carbon = C 2 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 135 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0
35 integrates to 2 40 45 50 naringenin symmetry makes these protons equivalent 55 60 65 70 75 80 85 90 95 100 105 red = C 110 115 120 125 130 135 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0
35 40 45 naringenin integrates to 1 50 55 60 integrates to 2 65 70 red = C note: this is two peaks 75 80 85 90 95 100 105 110 115 120 125 130 135 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0
35 x 9 10 40 45 50 naringenin 55 60 65 6 7 8 70 75 80 85 90 95 100 105 5 red = C 110 115 120 125 4 -re-numbering is necessary 130 135 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0
naringenin 115.80 102.94 96.51 95.54 43.20 79.65 115.88 128.74 - combine accurate 1D 13 C shift data with SQC to generate table 196.96 167.00 165.01 164.09 158.40 130.50 208 200 192 184 176 168 160 152 144 136 128 120 112 104 96 88 80 72 64 56 48 40 32 24 Chemical Shift (ppm)
naringenin Naringenin (500 Mz, acetone-d 6 ) ID ( 1 ) ( 13 C) s Type J (z) CSY 1 12.18 -- 1 s* -- 2 9.56 -- 1 s* -- 3 8.51 -- 1 s* -- 4 7.40 128.74 2 m -- 5 6.91 115.88 2 m -- 6 5.97 95.54 1 m -- 7 5.96 96.51 1 m -- 8 5.46 79.65 1 dd 13.0, 3.0 9 3.19 43.20 1 dd 13.0, 17.1 10 2.74 43.20 1 dd 17.1, 3.0 Quaternary carbons: 196.96, 167.00, 165.01, 164.09, 158.40, 130.50, 102.94 C 2 pairs: 9/10 * exchangeable 1 5,6 7 2 3 4 8 9 10 0.7 0.9 0.9 2.1 2.1 2.0 1.2 1.2 1.1 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 Chemical Shift (ppm)
gcsy 4 5 6,7 8 9 10 4 5 6,7 8 9 10 4-5 - ignore peaks on the diagonal - number peaks on both axes 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 9-10 8-9 8-10 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5
updated data Naringenin (500 Mz, acetone-d 6 ) ID ( 1 ) ( 13 C) s Type J (z) CSY 1 12.18 -- 1 s* -- -- 2 9.56 -- 1 s* -- -- 3 8.51 -- 1 s* -- -- 4 7.40 128.74 2 m -- 5 5 6.91 115.88 2 m -- 4 6 5.97 95.54 1 m -- -- 7 5.96 96.51 1 m -- -- 8 5.46 79.65 1 dd 13.0, 3.0 9, 10 9 3.19 43.20 1 dd 13.0, 17.1 8, 10 10 2.74 43.20 1 dd 17.1, 3.0 8, 9 Quaternary carbons: 196.96, 167.00, 165.01, 164.09, 158.40, 130.50, 102.94 C 2 pairs: 9/10 * exchangeable spin systems: 4 m 5 m; 8 dd 9/10 dd/dd naringenin
Naringenin (500 Mz, acetone-d 6 ) ID ( 1 ) ( 13 C) s Type J (z) CSY 1 12.18 -- 1 s* -- 2 9.56 -- 1 s* -- 3 8.51 -- 1 s* -- 4 7.40 128.74 2 m -- 5 6.91 115.88 2 m -- 6 5.97 95.54 1 m -- 7 5.96 96.51 1 m -- 8 5.46 79.65 1 dd 13.0, 3.0 9 3.19 43.20 1 dd 13.0, 17.1 10 2.74 43.20 1 dd 17.1, 3.0 Quaternary carbons: 196.96, 167.00, 165.01, 164.09, 158.40, 130.50, 102.94 C 2 pairs: 9/10 * exchangeable geminal = 17.1 spin systems: 4 m 5 m; 8 dd 9/10 dd/dd vicinal = 13.0, 3.0-9/10 can be assigned immediately (there s only one C 2 in the molecule) 8 naringenin 196.96 9,10
Naringenin (500 Mz, acetone-d 6 ) ID ( 1 ) ( 13 C) s Type J (z) CSY 1 12.18 -- 1 s* -- 2 9.56 -- 1 s* -- 3 8.51 -- 1 s* -- 4 7.40 128.74 2 m -- 5 6.91 115.88 2 m -- 6 5.97 95.54 1 m -- 7 5.96 96.51 1 m -- 8 5.46 79.65 1 dd 13.0, 3.0 9 3.19 43.20 1 dd 13.0, 17.1 10 2.74 43.20 1 dd 17.1, 3.0 Quaternary carbons: 196.96, 167.00, 165.01, 164.09, 158.40, 130.50, 102.94 C 2 pairs: 9/10 * exchangeable 196.96 spin systems: 4 m 5 m; 8 dd 9/10 dd/dd - by elimination, 4 and 5 are on the mono-phenol - however, order is unclear - 6/7 must be on the bis-phenol 6, 7 8 naringenin 4, 5 9,10
gmbc 1 1 40 48 13 C one-bond peak 13 C three-bond peak 1-bond peak 56 64 72 80 1-bond and n-bond peaks indicates symmetry 88 96 104 112 120 128 136 144 152 160 168 note: couplings to sp 2 carbons are usually 3-bond because 2 J C is small 176 184 192 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0
gmbc: carbonyl region 6,7 8 9 10 189 190 191 192 193 196.96 194 195 196 197 long-range (>3 bond correlations) 13 C 1 198 199 200 201 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5
4 5 6,7 8 9 10 71 72 73 74 8 75 76 79.65 SQC 77 78 79 80 81 82 8 4-79.65 83 84 85 86 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5
4 5 6,7 8 9 10 71 72 73 74 75 1 76 79.65 13 C 13 C 4/5 4/5 three bonds four bonds 4 5 1 77 78 79 80 81 82 83 proton 4 must be ortho to carbon at 79.65 ppm 84 85 86 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5
MBC 1 2 3 4 5 111.5 112.0 112.5 113.0 113.5 114.0 114.5 115.88 SQC 5 115.0 115.5 116.0 116.5 117.0 117.5 5 118.0 118.5 119.0 119.5 - MBC can be used to assign protons 120.0 120.5 12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0
MBC 1 2 3 4 5 111.5 112.0 112.5 113.0 113.5 114.0 114.5 115.88 13 C 1 3 3-115.88 115.0 115.5 116.0 116.5 117.0 117.5 118.0 118.5 119.0 119.5 120.0 120.5 12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0
MBC 1 2 3 4 5 111.5 112.0 112.5 115.88 4 5 3 115.88 113.0 113.5 114.0 114.5 115.0 115.5 116.0 116.5 117.0 117.5 can you explain these correlations? 118.0 118.5 119.0 119.5 120.0 120.5 12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0
MBC 1 2 3 4 5 111.5 112.0 115.88 4 115.88 5 112.5 113.0 113.5 114.0 114.5 115.0 115.5 116.0 116.5 117.0 117.5 5 118.0 118.5 115.88 119.0 119.5 120.0 120.5 12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0
how can the remaining ipso carbons be assigned? what are the relevant correlations in this region? 80 85 90 95 100 105 110 4 5 115 120 125 130 135? 3 140 145 150? 155 8.6 8.5 8.4 8.3 8.2 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 160
3 4 5 8 8 4 5 80 85 90 95 100 3 105 110 5 4 130.50 115 120 125 130 135 140 145 150 158.40 8.6 8.5 8.4 8.3 8.2 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 155 160
3 4 5 8 80 85 5 4 130.50 4 130.50 2 J C is small 158.40 3 correlation missing 130.50 3 J C is large 5 90 95 100 105 110 115 120 125 130 135 140 145 150 158.40 8.6 8.5 8.4 8.3 8.2 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 155 160
9 10 40 48 56 64 8 72 80 88 102.94 130.50 196.96 96 104 112 120 128 136 9/10 144 152 8 130.50 160 168 176 184 196.96 192 3.45 3.40 3.35 3.30 3.25 3.20 3.15 3.10 3.05 3.00 2.95 2.90 2.85 2.80 2.75 2.70 2.65 2.60 2.55 2.50
9 10 40 48 56 64 8 72 80 88 102.94 130.50 102.94 196.96 9/10 96 104 112 120 128 136 144 152 8 130.50 160 168 176 184 196.96 192 3.45 3.40 3.35 3.30 3.25 3.20 3.15 3.10 3.05 3.00 2.95 2.90 2.85 2.80 2.75 2.70 2.65 2.60 2.55 2.50
6, 7 95 100 105 110 102.94 6, 7 102.94 115 120 confirmation of assignment 125 130 135 140 145 150 155 160 165 6.10 6.05 6.00 5.95 5.90 5.85 5.80
1 2 102.94 7 7/96.51 6 1 7 96.51 2-95.54 2-96.51 95 100 105 110 115 120 1-102.94 102.94 1 2 95.54 6 96.51 7 2 125 130 135 140 145 150 155 160 165 12.0 11.5 11.0 10.5 10.0 9.5
1 2 6 7 102.94 1 102.94 95 100 105 7 110 115 1-165.01 2 6 1-167.00 (long-range?) three phenolic ipso carbons remain unassigned: 167.00, 165.01, 164.09 -these correlations are still difficult to interpret 120 125 130 135 140 145 150 155 160 165 12.0 11.5 11.0 10.5 10.0 9.5
162.5 163.0 163.5 6-164.09 164.0 164.5 7-165.01 7-167.00 165.0 165.5 166.0 166.5 167.0 167.5 168.0 -raising contour level deconvolutes peaks 168.5 169.0 6.01 6.00 5.99 5.98 5.97 5.96 5.95 5.94 5.93 5.92 5.91 5.90 5.89
from previous slide: 7 1 102.94 162.5 163.0 163.5 1-165.01 6-164.09 7-165.01 2 6 three phenolic ipso carbons remain unassigned: 167.00, 165.01, 164.09 164.0 164.5 165.0 165.5 166.0 166.5 1-167.00 (weak) 7-167.00 2-167.00 (weak) 167.0 167.5 168.0 raising contour level shows this peak 168.5 169.0 6.01 6.00 5.99 5.98 5.97 5.96 5.95 5.94 5.93 5.92 5.91 5.90 5.89
162.5 165.01 7 1 102.94 163.0 163.5 6-164.09 2 6 167.00 164.09 164.0 164.5 1-165.01 7-165.01 1-167.00 (weak) 7-167.00 2-167.00 (weak) 165.0 165.5 166.0 166.5 167.0 167.5 168.0 168.5 169.0 6.01 6.00 5.99 5.98 5.97 5.96 5.95 5.94 5.93 5.92 5.91 5.90 5.89
bserved vs. Predicted Chemical Shifts ACD/Labs NMR Predictor (v6) predicted chemical shift (ppm) 8 7 6 5 4 200 180 3 y = 1.0153x - 0.0278 R 2 = 0.9979 2 2 3 4 5 6 7 8 observed chemical shift (ppm) ID bsvd. ( 1, ppm) Calcd. ( 1, ppm) bsvd. ( 13 C, ppm) Calcd. ( 13 C ppm) 1 12.18 9.87 -- -- 2 9.56 9.87 -- -- 3 8.51 9.87 -- -- 4 7.40 7.38 128.74 128.58 5 6.91 6.90 115.88 115.68 6 5.97 6.20 95.54 95.63 7 5.96 6.19 96.51 96.31 8 5.46 5.44 79.65 79.26 9 3.19 3.18 43.20 42.81 10 2.74 2.72 43.20 42.81 196.96 196.76 167.00 167.20 165.01 164.32 164.09 163.67 158.40 158.18 130.50 129.85 102.94 102.39 predicted chemical shift (ppm) 160 140 120 100 80 60 40 y = 1.0006x - 0.3724 R 2 =0.99996 verall Assignments: 1 102.94 165.01 196.96 7 9/10 4 5 8 2 6 3 167.00 164.09 -exchangeable peaks not predicted well -otherwise, excellent agreement -DFT methods can also be used, but their implementation can be tricky 40 60 80 100 120 140 160 180 200 observed chemical shift (ppm) 130.50 158.40
Molecular Modelling B3LYP/6-31g(d) - two possible ground-state conformers ID ( 1 ) ( 13 C) s Type J (z) 8 9/10 1 12.18 -- 1 s* -- 2 9.56 -- 1 s* -- 3 8.51 -- 1 s* -- 4 7.40 128.74 2 m -- 5 6.91 115.88 2 m -- 6 5.97 96.51 1 m -- 7 5.96 95.54 1 m -- 8 5.46 79.65 1 dd 13.0, 3.0 9 3.19 43.20 1 dd 13.0, 17.1 10 2.74 43.20 1 dd 17.1, 3.0 rel G: +2.7 kcal/mol rel G: +0.0 kcal/mol observed coupling (z) dihedral (8-C-C-9): 47.2 (13.0) dihedral (8-C-C-10): 71.0 (3.0) dihedral (8-C-C-9): 177.8 (13.0) dihedral (8-C-C-10): 59.5 (3.0)
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 4 An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D 2 addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key RESY correlations). Chem 135 ID ( 1 ) ( 13 C) s Type J (z) CSY Key MBC Key RESY 1 5.20 73.1 1 t 9.5x2 3, 2 170.4 4, 7 2 5.08 68.7 1 t 9.5x2 1, 7 169.6 3 3 4.96 71.6 1 dd 9.5, 8.0 4, 1 169.4 2 4 4.59 99.8 1 d 8.0 3 1, 7 5 4.26 62.3 1 dd 12.3, 4.9 6, 7 170.8 7 6 4.11 62.3 1 dd 12.3, 2.5 5, 7 170.8 7 7 3.68 71.8 1 m -- 2, 5, 6 1, 4, 5, 6 8 ~3.2 -- 1 br s -- -- 9 2.08 20.95 3 s -- -- 170.8 10 2.05 20.90 3 s -- -- 169.4 11 2.02 20.79 3 s -- -- 169.6 12 2.00 20.83 3 s -- -- 170.4 Quaternary Carbons: 170.8, 170.4, 169.6, 169.4 Molecular formula:
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 4 An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D 2 addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key RESY correlations). Chem 135 ID ( 1 ) ( 13 C) s Type J (z) CSY Key MBC Key RESY 1 5.20 73.1 1 t 9.5x2 3, 2 170.4 4, 7 2 5.08 68.7 1 t 9.5x2 1, 7 169.6 3 3 4.96 71.6 1 dd 9.5, 8.0 4, 1 169.4 2 4 4.59 99.8 1 d 8.0 3 1, 7 5 4.26 62.3 1 dd 12.3, 4.9 6, 7 170.8 7 6 4.11 62.3 1 dd 12.3, 2.5 5, 7 170.8 7 7 3.68 71.8 1 m -- 2, 5, 6 1, 4, 5, 6 8 ~3.2 -- 1 br s -- -- 9 2.08 20.95 3 s -- -- 170.8 10 2.05 20.90 3 s -- -- 169.4 11 2.02 20.79 3 s -- -- 169.6 12 2.00 20.83 3 s -- -- 170.4 Quaternary Carbons: 170.8, 170.4, 169.6, 169.4 Molecular formula: ints Entry Point: Methyl Groups Every methyl group has an MBC correlation to a quaternary carbon. What functional group does this represent? What do the chemical shifts of 9-12 signify? (2) Spin Systems What are the spin systems in this molecule?
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 4 An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D 2 addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key RESY correlations). Chem 135 ID ( 1 ) ( 13 C) s Type J (z) CSY Key MBC Key RESY 1 5.20 73.1 1 t 9.5x2 3, 2 170.4 4, 7 2 5.08 68.7 1 t 9.5x2 1, 7 169.6 3 3 4.96 71.6 1 dd 9.5, 8.0 4, 1 169.4 2 4 4.59 99.8 1 d 8.0 3 1, 7 5 4.26 62.3 1 dd 12.3, 4.9 6, 7 170.8 7 6 4.11 62.3 1 dd 12.3, 2.5 5, 7 170.8 7 7 3.68 71.8 1 m -- 2, 5, 6 1, 4, 5, 6 8 ~3.2 -- 1 br s -- -- 9 2.08 20.95 3 s -- -- 170.8 10 2.05 20.90 3 s -- -- 169.4 11 2.02 20.79 3 s -- -- 169.6 12 2.00 20.83 3 s -- -- 170.4 Quaternary Carbons: 170.8, 170.4, 169.6, 169.4 (2) Spin Systems What are the spin systems in this molecule? 5 dd/6 dd 7 m 2 t 1 t 3 dd 4 d Apart from 5/6, everything is a methine. (3) Connect Fragments From the MBC, note that: 5 dd/6 dd 7 m 2 t 1 t 3 dd 4 d MBC 170.8 X Me 9 169.6 X Me 11 170.4 X Me 12 169.4 X Me 10
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 4 An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D 2 addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key RESY correlations). Chem 135 ID ( 1 ) ( 13 C) s Type J (z) CSY Key MBC Key RESY 1 5.20 73.1 1 t 9.5x2 3, 2 170.4 4, 7 2 5.08 68.7 1 t 9.5x2 1, 7 169.6 3 3 4.96 71.6 1 dd 9.5, 8.0 4, 1 169.4 2 4 4.59 99.8 1 d 8.0 3 1, 7 5 4.26 62.3 1 dd 12.3, 4.9 6, 7 170.8 7 6 4.11 62.3 1 dd 12.3, 2.5 5, 7 170.8 7 7 3.68 71.8 1 m -- 2, 5, 6 1, 4, 5, 6 8 ~3.2 -- 1 br s -- -- 9 2.08 20.95 3 s -- -- 170.8 10 2.05 20.90 3 s -- -- 169.4 11 2.02 20.79 3 s -- -- 169.6 12 2.00 20.83 3 s -- -- 170.4 Quaternary Carbons: 170.8, 170.4, 169.6, 169.4 (4) Esters: From chemical shift arguments, these are acetates, no methyl esters or methyl ketones. Also, every proton in the major spin system is on an oxygen. Thus, we have: Ac 170.8 5 dd/6 dd R 169.6 Ac 7 m 2 t This looks a lot like a sugar! 1 t Ac 170.4 169.4 Ac 3 dd R 4 d X
E. Kwan Lecture 3: 2D NMR Spectroscopy Problem 4 An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D 2 addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key RESY correlations). Chem 135 ID ( 1 ) ( 13 C) s Type J (z) CSY Key MBC Key RESY 1 5.20 73.1 1 t 9.5x2 3, 2 170.4 4, 7 2 5.08 68.7 1 t 9.5x2 1, 7 169.6 3 3 4.96 71.6 1 dd 9.5, 8.0 4, 1 169.4 2 4 4.59 99.8 1 d 8.0 3 1, 7 5 4.26 62.3 1 dd 12.3, 4.9 6, 7 170.8 7 6 4.11 62.3 1 dd 12.3, 2.5 5, 7 170.8 7 7 3.68 71.8 1 m -- 2, 5, 6 1, 4, 5, 6 8 ~3.2 -- 1 br s -- -- 9 2.08 20.95 3 s -- -- 170.8 10 2.05 20.90 3 s -- -- 169.4 11 2.02 20.79 3 s -- -- 169.6 12 2.00 20.83 3 s -- -- 170.4 Quaternary Carbons: 170.8, 170.4, 169.6, 169.4 In fact, it is acetylated glucose. Note the large trans di-axial couplings. ere are the assignments and key RE correlations: 5/6 Ac 9 Ac 11 7 2 1 Ac 12 3 4 8 Ac 10 Ac Ac 2 7 Ac 1 3 4 Ac