4Me 11 5 8 10 Si O 12 9 N O Ph 1 7 O NMR Solutions.pdf · draw. Consider semi-preparative HPLC or...

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


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)

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:


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?


NH

O

O

This is the COSY spectrum. Try to label the off-diagonal peaks.


1.0

1.5

2.0

2.5

3.0

3.5

4.0

F1 C

hem

ical

Shi

ft (p

pm)


10

15

20

25

30

35

40

45

50

55

60

F1 C

hem

ical

Shi

ft (p

pm)


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


NH

O

O


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


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


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


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)


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)


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


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)


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?


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)


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.


1.11.21.22.02.12.10.90.90.7

naringenin

O

O

OH

OH

OH

phenol OH

phenol OH


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


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


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


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


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


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


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



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



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


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


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


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?


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


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?


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


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


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


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


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


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


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


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:



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?



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




(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




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


4Me 11 5 8 10 Si O 12 9 N O Ph 1 7 O NMR Solutions.pdf · draw. Consider semi-preparative HPLC or...

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Transcript of 4Me 11 5 8 10 Si O 12 9 N O Ph 1 7 O NMR Solutions.pdf · draw. Consider semi-preparative HPLC or...