7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0Chemical Shift (ppm)
500 MHz, CDCl3
5H 1H 2H 2H 2H 2H 1H 3H 4H 9H 6H
1
23
4
56
8
9
1011
12
7
Me
ONO O
Ph
OSi
5H 1H 2H 3H 2H 2H 3H 4H 9H 6H
Problem 1. The addition of lithium dimethylcuprate to an alkyne produced a mixture of E and Z isomers which were separated by HPLC. Please assign the resonances in the spectra below and determine which isomer is which.
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0Chemical Shift (ppm)
500 MHz, CDCl3
5H 1H 2H 2H 2H 2H 1H 3H 4H 9H 6H
1
23
4
56
8
9
1011
12
7
Me
ONO O
Ph
OSi
5H 1H 2H 3H 2H 2H 3H 4H 9H 6H
Z isomer
E isomer
allylic proton: downfield due to A(1,3) strain
“normal” allylicproton shift
By far the simplest spectrum to understand, begin here.
COSY
axes: proton, protoncorrelations: off-diagonal peaks are 2-3 bond couplings between protonspurpose: assign protons to spin systems
spin system: a set of protons sharing through-bond (J) couplings
O123
4
5 6
1-2-3, 4-5, 6
HSQC
axes: proton, carboncorrelations: 1-bond C,H couplingspurpose: number proton spectrum, match each proton with a carbon, identifyCH2 pairs
1. HSQC: 1-bond C,H2. COSY: 2,3-bond H,H3. HMBC: 2,3-bond C,H
proton
carbon
down (CH2)up (CH, CH3)
proton 1 is directlyattached to carbon A
1
A
B
2 3
Numbering the SpectrumEven if the proton spectrum overlaps, HSQC will usually separate the peaksenough so they can be numbered (convention: left to right).
protoncarbon
1
A
C
2,3
B
4
1
2
3
4- although 2 and 3 overlap in the 1D spectrum, they have very different carbon shifts so the HSQC separates them
Next, use COSY to determine the composition of each spin system.
HMBC
proton
carbon
proton 1 is 2 or 3 bondsaway from carbon A
1
A
B
2
proton 2 is 2 or 3 bondsaway from carbon A
one-bond artifact:proton 2 is directlyattached to carbon B(look for doublets)
axes: proton, carboncorrelations: 2-3 bond couplings between protonspurpose: connect spin systems
protons 2 and 3 are on thesame carbon; i.e., 2/3 isa methylene (CH2) pair
- all experiments are proton-detected
proton
proton
A B C D
A
B
C
D
A-B-C, D
- only give a number for each unique chemical shift (e.g., a methyl group only gets one number)
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsSummary of 2D NMR Experiments
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsTypes of Structural Determination Problems(1) Spectral Assignment
- given a structure and a spectrum, verify that the spectrum matches the compound and assign all of the peaks to atoms in the molecule - real world scenario: reaction product analysis
(2) Unknown Structural Elucidation
- given the spectra of an unknown compound, determine its structure and then assign of its peaks - real world scenario: novel natural product, unexpected outcome of a reaction
WorkflowThese steps apply to both problems.
(0) Compound Purity
Has the compound been purified to the best of your ability? Is the baseline of the proton spectrum clean? Dirty spectra are hard to interpret and can make conclusions difficult to draw. Consider semi-preparative HPLC or preparative TLC prior to analysis. (Elemental analysis remains the gold standard of purity. A high resolution mass "hit" does not guarantee any level of purity.)
(1) Data Acquisition
- 1D spectra: proton, carbon (optional) - 2D spectra: COSY-45, HSQC, HMBC - of course, not all of these data will be necessary for simpler compounds
(2) Tabulate Data
- make it a habit to tabulate the data before analysis to avoid confirmation bias
1 2 3 4
1H 2H 1Hintegrals:
5
1H
Here is a hypothetical spectrum:
(a) Adopt the convention of labeling the peaks from left to right. This gives you a unique name for every resonance.
(b) Mass Balance: Are there the right number of protons and carbons?
(c) Start with the HSQC: Sometimes, the peaks will overlap. Use the HSQC to uniquely label every proton, as well as determine direct attachments between protons and carbons:
2
3
xx
xx
13C axis
1H axis Here, peaks 2 and 3 overlap in the proton spectrum, but the greater chemical shift range of the carbon axis disperses the peaks so that they can be distinguished.
(d) Multiplicity: HSQC phase gives DEPT information. Red/up= CH or CH3; blue/down=CH2. Write down all the methylene pairs for reference. These will help you distinguish geminal coupling from vicinal or long-range couplings in the COSY.
(e) Quaternary carbons: These do not appear in the HSQC, but should appear in the carbon or HMBC. Isolated, but clearly resolved peaks are often excellent "entry points" for analysis. Make a list of these carbons.
(f) Be methodical and double-check everything.
ID (1H) (13C) Hs type J (Hz) COSY HMBC
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...CH2 pairs: 4/5, ...
- peaks are listed by number from high to low chemical shift- HSQC: connect each proton to its directly attached carbons; find methylene pairs- COSY: 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 HSQC- quaternary carbons can be found from HMBC or the 1D spectrum carbon- HMBC: watch for one-bond peaks; more intense peaks like methyl groups are more likely to show long-range correlations; sp2 systems: 2J is small, but 3J is large (bigger for anti than syn)
xx xx4 5
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsWorkflow(2) Tabulate Data(d) Multiplicity
(g) Bookkeeping: Here is the format I use. A similar format is used in many natural products papers.
4/5 is a methylene pair.
(3) Generate Spin Systems- Use COSY to build up spin systems. Each "component" of the spin system is a methyl group, a methylene pair, or a methine (from the HSQC). Double-headed arrows represent (putative) vicinal couplings, with dashed lines for long-range couplings:
4/5 8 1 2
long-range coupling(You might not know which ones are long-range. Use yourchemical intuition and look at the peak intensities andasymmetry about the diagonal. You might have to changeyour diagram if you find it to be inconsistent later.)
(4) Connect Spin SystemsLook for HMBC correlations that connect a proton in onespin system to a carbon in another spin system:
Curved arrows indicate HMBC correlations. If you find sucha connection, that tells you that the spin systems must beadjacent in the head-to-tail sense shown. Note that thesedo not have to be mutual like COSY couplings:
145.23
4/5 8 1 6 72
6 7
33.37
3
3
If that doesn't work, you can look for a carbon, possibly quaternary, that protons in both spin systems have commonHMBC correlations to:
CH
C CH
CH
C CH
does notmean
finding thisHMBC
correlation
will bepresent
4/5 8 1 6 72 3
35.57
2.0 1.5 1.0Chemical Shift (ppm)
3.45 3.35Chemical Shift (p... Chemical Shift (ppm)
1 2 3 4 5 H2O 6 7 8 9-12 13 14
exchangeable
3.5 3.0 2.5 2.0 1.5 1.0 0.5F2 Chemical Shift (ppm)
16
24
32
40
48
56
64
F1 C
hem
ical
Shi
ft (p
pm)
2
(3)
(4)
(5)
6
(9)
(10)
12
(13)
14
CH or CH3
(CH2)
1
11
8
HSQC Spectrum
3.5 3.0 2.5 2.0 1.5 1.0 0.5F2 Chemical Shift (ppm)
16
24
32
40
48
56
64
72
F1 C
hem
ical
Shi
ft (p
pm)
1 2 3 4 5 6 7 8
HMBC
- arrows indicate one-bond peaks- these artifacts appear as doublets in proton dimension
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsWorkflow(5) Generate FragmentsIn many cases, complex, overlapping signals will prevent youfrom drawing out all the spin systems of the molecule. Inparticular, unknown structural elucidations will require you togenerate and connect fragments, which are tied together bywhat is visible in a spin system and HMBC. Here is how I puttogether menthol, which has some overlapping signals and onecontinuous spin system:
H3C CH3
X X
2, 25.82
8, 50.12
14, 16.0711, 21.00
Fragment Structures Vicinal COSYCorrelations
Key HMBCCorrelations
H
H
H3C CH3
X X
H
H
11-2 2-14
2-8H3C CH3
X X
- methyls 11, 14: COSY to 2- 2: COSY to 8
Key Evidence
11-14
11-2 14-8
- HSQC: 2, 8 = CH
- 11, 14: mutual HMBC- 11, 14: HMBC to 2, 8
- HMBC: 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, HMBC correlations are from protons to carbons- Xs represent non-protons
Write down your reasoning (even though it's boring). Bytheir nature, these problems are very complicated, and there's a99.9% chance you will not remember what peak is at 16.07 ppmseveral years from now. If you write down what you werethinking, you and others will be able to follow your process lateron without repeating the entire analysis. It's also likely you willmake some mistakes, which will be easier to track down this way.
(6) Entry Points
When you look at a COSY or HMBC 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. Here are some ideas:
- carbonyl region - aromatic rings and olefins - methyl groups - quaternary carbons
(7) Expand and Connect Fragments
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. Once you have exhausted all the easy data, work on what's left to expand your existing fragments, and if possible, connect them.
4, 13, 34.51
12, 22.10
6
XHO
CH2X
CH3
HO5-13, 9-4
- COSY, HMBC correlations connect termini
(8) List Full AssignmentsThis is self-explanatory:
1
2
3, 10 4, 13
5, 9
6
78
11
12
14
71.53
25.82
45.03 34.5123.10
31.62
50.12
21.00
22.10
16.07
HO HO
You should also file all the FIDs and your "good" notes of whatyou've done in the same place. If you feel the compound will gointo your thesis, take the time to write it out in journal format.
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsWorkflow(9) Check Answer
- Is your flat structure consistent with all of the NMR data?- Are there any other possibilities?
(10) Interpret Relative Stereochemistry
- H,H coupling constants - nOe/rOe correlations - (C,H) coupling constants - render structures in 3D and highlight diagnostic data - details: see Lecture 9TroubleshootingMy proton spectrum looks bad.
Do you have paramagnetic ions, particulates, or otherimpurities? Purify your compound by flash chromatography orHPLC, or at least filter it through some cotton. A very pure0.5 mg is preferable to a dirty 5 mg.
Is the sample volume correct? Samples that are too short aredifficult to shim.
Is this a S/N issue? If you have <0.5 mg, a detailed 2D NMRanalysis with standard equipment will be very difficult. You willsave time by going back to make more sample first.Remember, a lot of these experiments depend on theobservation of carbon-13 satellites, which are very small!
Are you in the intermediate exchange regime? Try changingthe temperature or field strength (higher or lower). Trychanging from CDCl3 to d6-DMSO.
Is this a shimming issue? Is your NMR tube bad? Try anothertube. Try recalling the default shims and manually shimming.Try using the gradient autoshim on deuterium-gradient-enabledspectrometers, followed by simplex autoshimming. Turn thespin off and adjust the XY shims as well.
My spectrum looks nice, but the peaks overlap.
Are any multiplets or even parts of multiplets exposed? Trychanging the solvent (benzene titration?) first. Then, try1D-TOCSY with a variety of mixing times to deconvolute thespectrum.
I don't have a lot of sample.
Did you set the experimental parameters correctly? Did youtune the probe and calibrate the pulse widths? These will makea substantial difference.
Try a Shigemi tube. Tubes with plugs of different magneticsusceptibilities are available. Samples in these expensive tubeswill be hard to shim, but have about double the S/N. These arenice for getting carbon-13 spectra. You can also try a smallerdiameter tube (but you have to have the right spinner and probe.)
Desperate? Try a cryo-probe at higher field strengths.
My sample has a lot of solvent in it.
Did you use a deuterated solvent? (Duh.) Co-evaporation ofyour sample with (regular) chloroform helps. If that's notenough, try heptane or toluene (may require higher vacuum).
I have nice data, but I don't understand it.
Did you check your data table? One mistake can make the dataunintelligible or give a ridiculous answer. Is everythingconsistent?
Try another angle. Have a coffee (or a beer) and start from anew entry point. Try having someone else look at the data.Just don't try the same strategy over and over. Rethink yourassumptions. Above all, think about the chemistry. Wheredid the compound come from? What is its reactivity?
courtesy Prof. Andy Phillips (Yale)
courtesy Prof. Andy Phillips (Yale)
4.0 3.5 3.0 2.5 2.0 1.5 1.0Chemical Shift (ppm)
2.81.12.01.01.11.11.01.01.02.0
192 184 176 168 160 152 144 136 128 120 112 104 96 88 80 72 64 56 48 40 32 24 16 8 0Chemical Shift (ppm)
13.9
1
25.2
727
.10
42.2
046
.14
48.3
5
59.8
4
173.
99
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 2Please assign the proton and carbon resonances of ethyl nipecotate. (500 MHz, CDCl3; spectra are courtesy ofDr. Jeffrey Simpson, MIT. See Simpson, Chapter 9 for his treatment of this problem.)
NH
O
O
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 2Please assign the proton and carbon resonances of ethyl nipecotate. (500 MHz, CDCl3; spectra are courtesy ofDr. Jeffrey Simpson, MIT. See Simpson, Chapter 9 for his treatment of this problem.)
NH
O
O
ID (1H) (13C) Hs Type J (Hz) COSY 1
2
3
4
5
6
7
8
9
10
11
Quaternary Carbons: Methylene Pairs:
4.0 3.5 3.0 2.5 2.0 1.5 1.0F2 Chemical Shift (ppm)
10
15
20
25
30
35
40
45
50
55
60
F1 C
hem
ical
Shi
ft (p
pm)
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 2
NH
O
O
This is the HMQC spectrum. Please number the protons. What are the methylene pairs in this molecule? Are there any quaternary carbons?
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 2
NH
O
O
This is the COSY spectrum. Try to label the off-diagonal peaks.
4.0 3.5 3.0 2.5 2.0 1.5 1.0F2 Chemical Shift (ppm)
1.0
1.5
2.0
2.5
3.0
3.5
4.0
F1 C
hem
ical
Shi
ft (p
pm)
4.0 3.5 3.0 2.5 2.0 1.5 1.0F2 Chemical Shift (ppm)
10
15
20
25
30
35
40
45
50
55
60
F1 C
hem
ical
Shi
ft (p
pm)
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 2
NH
O
O
The numbering I get is shown below.
1
2 4
3 5
6
7
9 10
11A multiplet analysis in ACD/NMR Processor gives:
1H NMR (500 MHz, CDCl3) ppm 1.05 (t, J=7.16 Hz, 3 H), 1.25 (ddd, J=10.49, 3.70, 2.59 Hz, 1 H), 1.40 - 1.53 (m, 2 H), 1.72 - 1.83 (m, 1 H), 2.17 - 2.28 (m, 1 H), 2.38 - 2.49 (m, 1 H), 2.60 (dd, J=12.34, 9.38 Hz, 1 H), 2.72 (dt, J=12.34, 3.70 Hz, 1 H), 2.95 (dd, J=12.34, 3.46 Hz, 1 H), 3.86 - 3.98 (m, 2 H).
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 HSQC-DEPT!
The carbon at 173.99 is quaternary (the ester).8
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 2
NH
O
O
4.0 3.5 3.0 2.5 2.0 1.5 1.0F2 Chemical Shift (ppm)
1.0
1.5
2.0
2.5
3.0
3.5
4.0
F1 C
hem
ical
Shi
ft (p
pm)
1 2 3 4 5 6 7 8/9 10 11
1
2
3
4
5
6
7
8/9
10
11
1-11
2-4
3-5
4-6
(2-6)
5-106-8/9(?)
7-8/9(?)
10-8/9(?)
Here’s how I annotate COSY spectra. I have bolded the correlations that do not correspond to geminalcouplings (guessed for aliphatic region) and are clear. Parentheses indicate weak couplings.
3-10
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 2
NH
O
O
Here’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 (1H) (13C) Hs Type J (Hz) COSY 1 3.92
59.84 2 m -- 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 --
10 1.25 25.27 1 ddd 10.5, 3.7, 2.6
3, 5
11 1.05 13.91 3 t 7.2 1
Quaternary Carbons: 173.99 Methylene Pairs: 1, 2/4, 3/5, 7/8, 9/10
(1) Entry Points: Start with the obvious. The ethyl group is 1 and 11, and there’s only one methine in this molecule:
NH
O
O
1
11
6
(2) Amine: Two sets of downfield, diastereotopicprotons must be adjacent to the amine. These must be 2/4 and 3/5. There is a clear COSY correlation between 2 and 6. Therefore, by process of elimination:
NH
O
O6
2/43/5
4.0 3.5 3.0 2.5 2.0 1.5 1.0F2 Chemical Shift (ppm)
1.0
1.5
2.0
2.5
3.0
3.5
4.0
F1 C
hem
ical
Shi
ft (p
pm)
1 2 3 4 5 6 7 8/9 10 11
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 2
NH
O
O
(4) Check: Do these assignments make sense based on chemical shift? ChemDrawpredictions certainly seem to be congruent with the assignments:
1-11
2-4
3-5
4-6
(2-6)
5-106-8/9(?)
7-8/9(?)
10-8/9(?)
3-10
(3) Aliphatics: We need to distinguish between 9/10 and 7/8. Note that 10 has COSY correlations to 3 and 5. (Why is one correlation stronger than the other?) Thus:
NH
O
O6
2/43/5
9/10
7/8 173.991 11
1.55;1.45
2.76;2.73
2.0
3.21;2.96
2.331.98;1.73 4.12
1.30
NH
O
O
(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).
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 3.2Chemical Shift (ppm)
1.11.02.01.05.81.11.01.0
104 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58Chemical Shift (ppm)
60.1
461
.35
62.4
3
69.2
6
71.1
372
.45
72.6
274
.02
76.3
9
81.4
4
92.2
4
103.
74
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3Please assign the proton and carbon resonances of sucrose (500 MHz, D2O).
OH
OHHO
O
HO
O
OHOH
HO
HOO
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 3.2F2 Chemical Shift (ppm)
58
60
62
64
66
68
70
72
74
76
78
80
82
84
86
88
90
92
F1 C
hem
ical
Shi
ft (p
pm)
3.65 3.60 3.55 3.50 3.45 3.40 3.35 3.30F2 Chemical Shift (ppm)
60
61
62
63
64
65
66
67
68
69
70
71
72F1
Che
mic
al S
hift
(ppm
)
inset
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3Please assign the proton and carbon resonances of sucrose (500 MHz, D2O).
HMQC
arrows: double intensity
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 3.3F2 Chemical Shift (ppm)
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
F1 C
hem
ical
Shi
ft (p
pm)
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3
COSY-90
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 3.2F2 Chemical Shift (ppm)
58
60
62
64
66
68
70
72
74
76
78
80
82
84
86
88
90
92
F1 C
hem
ical
Shi
ft (p
pm)
3.65 3.60 3.55 3.50 3.45 3.40 3.35 3.30F2 Chemical Shift (ppm)
60
61
62
63
64
65
66
67
68
69
70
71
72F1
Che
mic
al S
hift
(ppm
)
inset
HMQC
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3Numbering the protons requires some careful bookkeeping.
1
2
3
4
5 6
9
7
8
11
10
12
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 3.3F2 Chemical Shift (ppm)
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
F1 C
hem
ical
Shi
ft (p
pm)
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3
COSY-901 2 3 4 5-8 9 10 11 12
1
2
3
4 5-
8 9
10 1
1 12
2-3
3-4
4-5-8(?)
9-11
12-5-8(?)
9-12
1-11
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3Here’s some space for your data table:
ID (1H) (13C) Hs Type J (Hz) COSY 1
2
3
4
5
6
7
8
9
10
11
12
Quaternary Carbons: Methylene Pairs:
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3I get this data table:
ID (1H) (13C) Hs Type J (Hz) COSY 1 5.22
92.24 1 d 3.9 11
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
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3Spin Systems.
Based on chemical shift arguments, 1 is the anomeric proton. It may help if I re-draw sucrose:
OHO
HO
OH
HOH
O
O
OH
HO
OH
OH
Because this is a disaccharide, there are two major spin systems (10 is on its own). Here is how I draw them out:
1 = anomeric
11 dd
9 t
12 t
5/6, 7, or 8
2 d
3 t
4 m
5/6, 7, or 8
OHO
HO
OH
HOH
O
O
OH
HO
OH
OH
The left-hand spin system can be drawn out:
1
119
12
Which of 5/6, 7, or 8 must 12 be connected to?
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3Spin Systems.
1 = anomeric
11 dd
9 t
12 t
5/6, 7, or 8
2 d
3 t
4 m
5/6, 7, or 8
OHO
HO
OH
HOH
O
O
OH
HO
OH
OH
(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 Other Spin System
Note that proton 2 is a doublet. This can only occur one way in the other spin system
1
119
127
2 3
4
(3) What about 4?
We must decide if proton 4 is connected to 5/6 or 8. Looking closely at the COSYspectrum, it certainly looks like 4 is coupling to the right-hand part of the overlappingmess. 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?
3.70 3.65 3.60 3.55F2 Chemical Shift (ppm)
3.50
3.55
3.60
3.65
3.70
F1 C
hem
ical
Shi
ft (p
pm)
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 3Spin Systems.
1 = anomeric
11 dd
9 t
12 t
5/6, 7, or 8
2 d
3 t
4 m
5/6, 7, or 8
OHO
HO
OH
HOH
O
O
OH
HO
OH
OH
(4) Final Assignments
1
119
127
2 3
4
It remains only to assign the quaternary carbon (of which there is only one) and theremaining 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?
10
103.74
5/6
8
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 2.5Chemical Shift (ppm)
1.11.21.22.02.12.10.90.90.7
naringenin
O
O
OH
OH
OH
H2O
solvent: acetone-d6
Please assign this spectrum.
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 2.5Chemical Shift (ppm)
1.11.21.22.02.12.10.90.90.7
naringenin
O
O
OH
OH
OH
phenol OH
phenol OH
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 2.5Chemical Shift (ppm)
1.11.21.22.02.12.10.90.90.7
naringenin
O
O
OH
OH
OH
- number from left to right
- one number per unique chemical shift
1
23
4
56
7
8 9
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0F2 Chemical Shift (ppm)
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
135
F1 C
hem
ical
Shi
ft (p
pm)
naringenin
O
O
OH
OH
OH
gHSQC
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0F2 Chemical Shift (ppm)
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
135
F1 C
hem
ical
Shi
ft (p
pm)
naringenin
O
O
OH
OH
OH
blue = CH2
- two peaks, same horizontal line = same carbon = CH2
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0F2 Chemical Shift (ppm)
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
135
F1 C
hem
ical
Shi
ft (p
pm)
naringenin
O
O
OH
OH
OH
red = CH
integrates to 2H
symmetry makesthese protonsequivalent
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0F2 Chemical Shift (ppm)
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
135
F1 C
hem
ical
Shi
ft (p
pm)
naringenin
O
O
OH
OH
OH
red = CH
integrates to 1H
integrates to 2H
note: this is two peaks
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0F2 Chemical Shift (ppm)
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
135
F1 C
hem
ical
Shi
ft (p
pm)
naringenin
O
O
OH
OH
OH
red = CH
-re-numbering is necessary4
5
67
8
9 10x
208 200 192 184 176 168 160 152 144 136 128 120 112 104 96 88 80 72 64 56 48 40 32 24Chemical Shift (ppm)
43.2
079.6
5
95.5
496
.51
102.
94115.
8011
5.88
128.
7413
0.50
158.
40
164.
09165.
0116
7.00
196.
96
naringenin
O
O
OH
OH
OH
- combine accurate 1D 13C shift data with HSQC to generate table
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 2.5Chemical Shift (ppm)
1.11.21.22.02.12.10.90.90.7
naringenin
O
O
OH
OH
OH
1
23
4
5,67
8
9 10
Naringenin (500 MHz, acetone-d6) ID (1H) (13C) Hs Type J (Hz) COSY 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 CH2 pairs: 9/10 * exchangeable
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0F2 Chemical Shift (ppm)
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
F1 C
hem
ical
Shi
ft (p
pm)
4 5 6,7 8 9 10
4
5
6,7
8
9
10
gCOSY
- number peaks on both axes4-5
- ignore peaks onthe diagonal
8-9 8-10
9-10
Naringenin (500 MHz, acetone-d6) ID (1H) (13C) Hs Type J (Hz) COSY 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 CH2 pairs: 9/10 * exchangeable
updated data
spin systems:
4 m – 5 m; 8 dd – 9/10 dd/dd
naringenin
O
O
OH
OH
HO
spin systems:
4 m – 5 m; 8 dd – 9/10 dd/dd
- 9/10 can be assigned immediately
(there’s only one CH2 in the molecule)naringenin
O
O
OH
OH
HO
196.96
8
9,10
geminal = 17.1
vicinal = 13.0, 3.0
Naringenin (500 MHz, acetone-d6) ID (1H) (13C) Hs Type J (Hz) COSY 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 CH2 pairs: 9/10 * exchangeable
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
naringenin
O
O
OH
OH
HO
196.96
8
9,10
4, 5
6, 7
Naringenin (500 MHz, acetone-d6) ID (1H) (13C) Hs Type J (Hz) COSY 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 CH2 pairs: 9/10 * exchangeable
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 3.0F2 Chemical Shift (ppm)
40
48
56
64
72
80
88
96
104
112
120
128
136
144
152
160
168
176
184
192
F1 C
hem
ical
Shi
ft (p
pm)
gHMBC
1-bond peak
1-bond andn-bond peaksindicates symmetry
one-bond peak
O
O
13C
OH
OH
HO
1H
three-bond peak
O
O
13COH
OH
HO
1H
note: couplings to sp2 carbons are usually 3-bond because 2JCH is small
6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5F2 Chemical Shift (ppm)
189
190
191
192
193
194
195
196
197
198
199
200
201
F1 C
hem
ical
Shi
ft (p
pm)
O
O
OH
OH
OH
gHMBC:carbonyl region
6,7 8 9 10
196.96
long-range (>3 bond correlations)
O
13C
OH
OH
HO
1H
O
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5F2 Chemical Shift (ppm)
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
F1 C
hem
ical
Shi
ft (p
pm)
O
O
OH
OH
OH
8
4 5 6,7 8 9 10
79.65
HSQC
8 4-79.65
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5F2 Chemical Shift (ppm)
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
F1 C
hem
ical
Shi
ft (p
pm)
O
O
OH
OH
OH
4 5 6,7 8 9 10
79.65
O13C
OH
OH
HO
O
1H
O13C
OH
OH
HO
O
1H
three bonds four bonds
4/5 4/5
O
OH
OH
HO
O
proton 4 must be ortho tocarbon at 79.65 ppm
HH
4
5
12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0F2 Chemical Shift (ppm)
111.5
112.0
112.5
113.0
113.5
114.0
114.5
115.0
115.5
116.0
116.5
117.0
117.5
118.0
118.5
119.0
119.5
120.0
120.5
F1 C
hem
ical
Shi
ft (p
pm)
5
115.88
HSQC
1 2 3 4 5
HMBC
- HMBC can be used to assign OH protons
O
O
OH
OH
HO
5
12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0F2 Chemical Shift (ppm)
111.5
112.0
112.5
113.0
113.5
114.0
114.5
115.0
115.5
116.0
116.5
117.0
117.5
118.0
118.5
119.0
119.5
120.0
120.5
F1 C
hem
ical
Shi
ft (p
pm)
115.88
1 2 3 4 5
HMBC
3-115.88
O 13C
O
OH
HO
O
3
1H
12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0F2 Chemical Shift (ppm)
111.5
112.0
112.5
113.0
113.5
114.0
114.5
115.0
115.5
116.0
116.5
117.0
117.5
118.0
118.5
119.0
119.5
120.0
120.5
F1 C
hem
ical
Shi
ft (p
pm)
115.88
1 2 3 4 5
HMBC
O
O
OH
HO
O
HH
4
5
3H115.88
can you explain these correlations?
12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0F2 Chemical Shift (ppm)
111.5
112.0
112.5
113.0
113.5
114.0
114.5
115.0
115.5
116.0
116.5
117.0
117.5
118.0
118.5
119.0
119.5
120.0
120.5
F1 C
hem
ical
Shi
ft (p
pm)
115.88
1 2 3 4 5
HMBC
O
OH
OH
HO
O
H4
115.88
O
OH
OH
HO
O
HH 5
O
OH
OH
HO
O
HH 5
115.88
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 6.7F2 Chemical Shift (ppm)
80
85
90
95
100
105
110
115
120
125
130
135
140
145
150
155
160
F1 C
hem
ical
Shi
ft (p
pm)
how can the remaining ipsocarbons be assigned?
O
OH
OH
HO
O
HH
4
5
3?
?
what are the relevant correlations in this region?
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 6.7F2 Chemical Shift (ppm)
80
85
90
95
100
105
110
115
120
125
130
135
140
145
150
155
160
F1 C
hem
ical
Shi
ft (p
pm)
3 4 5
158.40
4
5
130.50
8
O
OH
OH
HO
O
HH
4
5
38
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 6.7F2 Chemical Shift (ppm)
80
85
90
95
100
105
110
115
120
125
130
135
140
145
150
155
160
F1 C
hem
ical
Shi
ft (p
pm)
3 4 5
158.40
4
5
130.50
8
O
O
OH
HO
O
3H
158.40
O
OH
OH
HO
O
130.50
H 5
3JCH is large
correlation missing
O
OH
OH
HO
O
130.50
H 4
2JCH is small
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 2.50F2 Chemical Shift (ppm)
40
48
56
64
72
80
88
96
104
112
120
128
136
144
152
160
168
176
184
192
F1 C
hem
ical
Shi
ft (p
pm)
9 10
196.96
130.50
8
102.94
O
OH
OH
HO
O
8
9/10
130.50
196.96
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 2.50F2 Chemical Shift (ppm)
40
48
56
64
72
80
88
96
104
112
120
128
136
144
152
160
168
176
184
192
F1 C
hem
ical
Shi
ft (p
pm)
9 10
196.96
130.50
8
102.94
O
OH
OH
HO
O
8
9/10
130.50
196.96
102.94
H
6.10 6.05 6.00 5.95 5.90 5.85 5.80F2 Chemical Shift (ppm)
95
100
105
110
115
120
125
130
135
140
145
150
155
160
165
F1 C
hem
ical
Shi
ft (p
pm)
6, 7
6, 7 – 102.94
O
OH
OH
HO
O
102.94
H
H
confirmation of assignment
12.0 11.5 11.0 10.5 10.0 9.5F2 Chemical Shift (ppm)
95
100
105
110
115
120
125
130
135
140
145
150
155
160
165
F1 C
hem
ical
Shi
ft (p
pm)
1 2
102.9467 7/96.51
1-102.94
2-95.54 2-96.51O
OH
O OH
1
HO
H
7
96.51
O
OH
O
HO
O
102.94H
1
O
OH
OH
O
O
H
H
6
2
95.54
O
OH
OH O
2
96.51
OH
H7
12.0 11.5 11.0 10.5 10.0 9.5F2 Chemical Shift (ppm)
95
100
105
110
115
120
125
130
135
140
145
150
155
160
165
F1 C
hem
ical
Shi
ft (p
pm)
1 2
102.9476
1-165.01
1-167.00 (long-range?)
O
OH
OH
HO
O
102.94
three phenolic ipsocarbons remain unassigned:
167.00, 165.01, 164.09
1
2 6
7
-these correlations are still difficult to interpret
6.01 6.00 5.99 5.98 5.97 5.96 5.95 5.94 5.93 5.92 5.91 5.90 5.89F2 Chemical Shift (ppm)
162.5
163.0
163.5
164.0
164.5
165.0
165.5
166.0
166.5
167.0
167.5
168.0
168.5
169.0
F1 C
hem
ical
Shi
ft (p
pm)
6-164.09
7-165.01
7-167.00
-raising contour level deconvolutes peaks
6.01 6.00 5.99 5.98 5.97 5.96 5.95 5.94 5.93 5.92 5.91 5.90 5.89F2 Chemical Shift (ppm)
162.5
163.0
163.5
164.0
164.5
165.0
165.5
166.0
166.5
167.0
167.5
168.0
168.5
169.0
F1 C
hem
ical
Shi
ft (p
pm)
6-164.09
7-165.01
7-167.00
1-165.01
1-167.00 (weak)
from previous slide:
O
OH
OH
HO
O
102.94
three phenolic ipsocarbons remain unassigned:
167.00, 165.01, 164.09
1
2 6
7
2-167.00 (weak)
raising contour level shows this peak
6.01 6.00 5.99 5.98 5.97 5.96 5.95 5.94 5.93 5.92 5.91 5.90 5.89F2 Chemical Shift (ppm)
162.5
163.0
163.5
164.0
164.5
165.0
165.5
166.0
166.5
167.0
167.5
168.0
168.5
169.0
F1 C
hem
ical
Shi
ft (p
pm)
6-164.09
7-165.01
7-167.00
1-165.01
1-167.00 (weak) 2-167.00 (weak)
O
OH
O
HO
O
102.941
2 6
7H
H
165.01
167.00 164.09
O
OH
O
HO
O
102.941
2 6
7H
H
165.01
167.00 164.09
Overall Assignments:
9/104
583
196.96
130.50
158.40
Observed vs. Predicted Chemical Shifts – ACD/Labs NMR Predictor (v6)
2 3 4 5 6 7 82
3
4
5
6
7
8
pr
edic
ted
chem
ical
shi
ft (p
pm)
observed chemical shift (ppm)
y = 1.0153x - 0.0278R2 = 0.9979
40 60 80 100 120 140 160 180 200
40
60
80
100
120
140
160
180
200
pred
icte
d ch
emic
al s
hift
(ppm
)
observed chemical shift (ppm)
y = 1.0006x - 0.3724R2=0.99996
-exchangeable peaks not predicted well-otherwise, excellent agreement-DFT methods can alsobe used, but theirimplementation canbe tricky
ID Obsvd. (1H, ppm)
Calcd. (1H, ppm)
Obsvd. (13C, ppm)
Calcd. (13C 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
Molecular ModellingB3LYP/6-31g(d)
- two possible ground-state conformers ID (1H) (13C) Hs Type J (Hz) 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
O
O
OH
OH
HO
9/10
8
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)
observed coupling (Hz)
rel G: +2.7 kcal/mol rel G: +0.0 kcal/mol
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 4
An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D2O addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key ROESY correlations). ID (1H) (13C) Hs Type J (Hz) COSY Key HMBC Key ROESY 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:
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 4
An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D2O addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key ROESY correlations). ID (1H) (13C) Hs Type J (Hz) COSY Key HMBC Key ROESY 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:
Hints
(1) Entry Point: Methyl GroupsEvery methyl group has an HMBC correlation to a quaternary carbon. What functional group does this represent? What do the chemical shifts of 9-12 signify?
(2) Spin SystemsWhat are the spin systems in this molecule?
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 4
An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D2O addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key ROESY correlations). ID (1H) (13C) Hs Type J (Hz) COSY Key HMBC Key ROESY 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
5 dd/6 dd 7 m 2 t 1 t 3 dd 4 d
(2) Spin SystemsWhat are the spin systems in this molecule?
Apart from 5/6, everything is a methine.
(3) Connect FragmentsFrom the HMBC, note that: 5 dd/6 dd 7 m 2 t 1 t 3 dd 4 d
X
O
Me9
170.8
HMBC
X
O
Me11
169.6X
O
Me12
170.4X
O
Me10
169.4
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 4
An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D2O addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key ROESY correlations). ID (1H) (13C) Hs Type J (Hz) COSY Key HMBC Key ROESY 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:
This looks a lot like a sugar!
AcO
OR
OAc
OAc
OAc
OR
X
5 dd/6 dd
7 m
2 t
1 t
3 dd
4 d170.8
169.6
170.4
169.4
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 4
An unknown compound of molecular weight 348 has the spectral data shown below. A singlet near 3.2 ppm disappears on D2O addition. Please deduce its molecular formula, skeletal connectivity, assignments, and relative stereochemistry (show key ROESY correlations). ID (1H) (13C) Hs Type J (Hz) COSY Key HMBC Key ROESY 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. Here are the assignments and key ROE correlations:
OAcO
OH
AcO
OAc
OAc
5/6
7
21
3
4
8
9
11
12
10
O
7
AcO
2
1
OAc3
OAcOH
4
AcO
OTBS
O O
O
Me
ON
KHMDSTHF, rt
?
LR-ESI-MS: 389.2O
O
Please deduce the flat structure of the byproduct. NMR Data (500 MHz, 45% CDCl3 in C6D6) ID (1H) (13C) Hs Type J (Hz) COSY-45 Key HMBC NOE 1 5.66 131.5 1 m -- 2, 3, 4 2 5.13 119.0 1 dd 17.1, 1.5 1, 3, 4 3 5.02 119.0 1 dd 10.5, 1.2 1, 2 4 4.39 66.1 2 m -- 1, 2 119.0, 131.5, 168.2 5 3.80 56.6 1 d 2.4 10 41.3, 43.9, 71.2, 168.2, 200.5 11 6 3.61 71.2 1 dd 9.0, 3.2 9, 11 56.6 10 7 3.10 63.2 1 t 2.9x2 9, 11 71.2, 200.5, 203.4 8 2.81 43.9 1 d 19.5 10 20.5, 41.3, 56.6, 203.4 9 1.94 34.2 1 ddd 14.6, 9.3, 2.9 6, 7, 11 63.2, 200.5
10 1.76 43.9 1 dd 19.5, 2.0 5, 8 20.5, 41.3, 56.6, 71.2, 203.4 6 11 1.52 34.2 1 dt 14.6, 2.9x2 6, 7, 9 71.2, 203.4 5 12 0.94 20.5 3 s -- -- 41.3, 43.9, 56.6, 71.2 13 0.81 25.9 9 s -- -- 25.7 14 -0.10 -4.5 6 d -- -- -- Quaternary Carbons: 200.5, 203.4, 168.2, 41.3, 25.7; IR: three carbonyls present
Chem 117E. Kwan Lecture 8: Solving Structural Determination ProblemsProblem 5
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