lecture 3

Lectur e 3: 3: 2D NMR NMR Spectro Spectro sco py

E. Kwan

Chem 135

between sites, which is very useful if you want to know the structure of a molecule.

2D NMR NMR Spectro Spectro scopy sco py Eugene E. Kwan

Every 2D NMR experiment has the same general format: preparation xx xx 4 5

structural elucidation workflow

two frequency dimensions

absolute value vs. phase-sensitive modes

two-dimensional NMR spectroscopy HMBC

mixing

detection t2

t1

Scope of Lecture problem solving

evolution

COSY

HSQC

Two Frequency Frequency Dime Dimension nsion s 1D NMR spectra are called "1D" because they have one frequency dimension, but actually have an additional dimension, intensity:

(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° 90 ° pulse that generates transverse magnetization. Evolution: The resonances precess in the rotating frame (2) Evolution: The according to their offsets. This means that magnetization is "frequency-labeled" as a function of t 1. (3) Mixing: Magnetization 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 T hese 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). COSY-90 0 sequenc seq uenc e : This is best illustrated by the basic COSY-9

intensity H channel frequency 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

t2

t1 90x

90x

COSY stands for COrrelations COrrelations SPectroscopy 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.

Lecture Lectu re 3: 2D NMR NMR Spectr Spectr oscopy osc opy

E. Kwan t1

t2

experiment #1

t1

t2

experiment #2

t1

t2

Chem 135

(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:

experiment #3

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. However, 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:

Lecture Lectu re 3: 2D NMR NMR Spectr Spectr oscopy osc opy

E. Kwan t1

t2

experiment #1

t1

t2

experiment #2

t1

t2

Chem 135

(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:

experiment #3

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. However, 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:

Lectu re 3: 2D 2D NMR NMR Spectr osc opy op y

E. Kwan Flavors of COSY The basic COSY-90 sequence is:

H channel

t2

t1 90x

90x

Chem 135

(5) 1D 1D Curv es: 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 below are “projections” “projections” of the peaks peaks onto onto the axes. axes. Because increasing ni is very costly, costly, resolution resolution is poor. A better alternativ alternative e is to place the actual 1D spectra on these axes (acquired in separate experiments.)

In VNMR, this is requested with the "gCOSY" command. Here is the COSY-90 spectrum for ethyl acetate: O 0.5

O

1.0

(1) Cros Cross-peaks: s-peaks: In ethyl acetate, there’s an ethyl group, which means that a methyl methyl and a methylene methylene share a vicinal coupling.

1.5 2.0

(2) Isolated Peaks: Not everything has large enough couplings to give off-diagonal peaks (for example, this acetate). In general, COSY mostly shows 2J and 3J couplings.

2.5 3.0 3.5

(3) Diagon Diagon al: Of 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.

4.0

(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 F2 Chemical Shift (ppm)

1.5

4.5

1.0

0.5

) m p p ( t f i h S l a c i m e h C 1 F

Lectur e 3: 2D NMR Spectroscop y

E. Kwan

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:

Chem 135

(2) At sufficiently high resolution, crosspeaks are tilted. Typically, geminal couplings will appear with positive slope, while vicinal couplings will appear with negative slope. However, due to the variations in J, this is not always true. (3) COSY-45 is slightly less sensitive than COSY-90 (by about 15%), but since COSY-90 is already a very sensitive experiment, this is of no consequence. COSY-45 gives more simplification than COSY-60, and is therefore preferred. COSY-45 is the best absolu te value COSY experiment for rou tine use and sho uld b e used instead of COSY-90. The PFG version can be requested in VNMR with "gCOSY45."

COSY-90

COSY-45

(source: Claridge, pg 140) The long tails from the dispersive peaks can interfere with off-diagonal peaks. Therefore, COSY-90 data are presented in absolute-value mode as phase-twist lineshapes. tilting Just because something is popular doesn't necessarily mean it's the best. A popular absolute-value version of COSY is called COSY-. Here, the second 90° pulse is replaced by a smaller pulse of tip angle . Typically,  is 45° or 60°.

H channel

t2

t1 90x

x

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.

diagonal is less crowded

Lectur e 3: 2D NMR Spectro scopy

E. Kwan

Flavors o f COSY The difference between the two COSY spectra is particularly evident in this example from Reynolds and Enriquez:

Chem 135

Expansions clearly show the tilting effect:

COSY-90 geminal

COSY-45

vicinal

One 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 ). However, 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:

Lectur e 3: 2D NMR Spectroscop y

E. Kwan

Heteronuclear Correlation Spectrosc opy 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 o f X-nucleus (less sensitive) new strategy: " indirect" detection of H-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 (COSY, TOCSY), 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: H H H H H C C C O C C

one-bond (direct)

H H H

H H

C C C O C C

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: H H H

H H

C C C O C C

one-bond (direct)

H H H

H H

C C C O C C

multiple-bond (remote)

In this case, quaternary carbons can appear: In reality, various experimental considerations mean that the advantage is less than 32/4=8. In general, S N

3/ 2

  exc o bs

H H C O C C

However, directly-detected multiple-bond correlation experiments are insensitive and not in routine use.

Lecture 3: 2D NMR Spectroscopy

E. Kwan

Chem 135

HSQC - Heteronuc lear Sing le Quantum Cor relation Spectro sco py This is a proton-detected experiment that shows directly attached carbons (one bond). CH and CH 3 peaks are phased up (red) while CH2 peaks are phased down (blue). 1D curves are useful, but not required. This is the spectrum for menthol: Me

Me

(1) HO Me

(2)

(3)

Spreading Out: The 1D spectrum is quite crowded, but having the second dimension spreads out the peaks over a much larger area. 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. However, HSQC-APT sequences of this type can be acquired more quickly than edited DEPT spectra. Met hyl en es : These appear on the same horizontal line, since they’re two protons on the same carbon. This allows one to distinguish 2J from 3J (or >3J) in COSY spectra.

no t diastereotopic (double height)

diastereotopic

Lectu re 3: 2D NMR Spectr oscop y

E. Kwan

Chem 135

HMBC 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. HMBC 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=1JCH. Occasionally, this splitting is itself useful information. However, most of the time, one should be wary of these artifacts. (2) HMBC incorporates a delay which under ideal circumstances is 1/2JCH. Since long-range couplings occupy a relatively wide range of 5-25 Hz, this delay is often set at a compromise value of 60 ms (8 Hz). 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

32

40

48

56

arrows indicate one-bond peaks

64

72 3.5

3.0

2.5

2.0 F2 Chemical Shift (ppm)

1.5

1.0

0.5


Lectur e 3: 2D NMR Spectr osc opy

E. Kwan Summary o f 2D NMR Experiments

- all experiments are proton-detected

1. HSQC: 1-bond C,H 2. COSY: 2,3-bond H,H 3. HMBC: 2,3-bond C,H

Chem 135

COSY Next, use COSY 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

HSQC By far the simplest spectrum to understand, begin here.

spin system: a set of protons sharing through-bond (J) couplings

axes: proton, carbon correlations: 1-bond C,H couplings purpose: number proton spectrum, match each proton with a carbon, identify CH2 pairs 1 2 3

1 2 3

O 5

A

6

C

B

D

D 1-2-3, 4-5, 6

down (CH2) up (CH, CH3)

B

4

p r o t o n

C

c a r b o n

B A

A protons 2 and 3 are on the same carbon; i.e., 2/3 is a methylene (CH2) pair

proton proton 1 is directly attached to carbon A

A-B-C, D proton HMBC axes: proton, carbon correlations: 2-3 bond couplings between protons purpose: connect spin systems

Numbering the Spectrum Even if the proton spectrum overlaps, HSQC will usually separate the peaks enough so they can be n umbered (convention: left to right). 1

2,3

4

2

4

C B 3

A 1

proton

c a r b o n

- although 2 and 3 overlap in the 1D spectrum, they have very different carbon shifts so the HSQC separates them - only give a number for each unique chemical shift (e.g., a methyl group only gets one number)

1

2

c a r b o n

B

one-bond artifact: proton 2 is directly attached to carbon B (look for doublets)

A proton proton 1 is 2 or 3 bonds away from carbon A

proton 2 is 2 or 3 bonds away from carbon A


E. Kwan

Structural Elucidation Workflo w (1) Start with t he HSQC to label the pr oto n spect rum.

Here's a hypothetical proton spectrum, where peaks 2 and 3 are very close (peaks are labeled from left to right): 1

integrals: 1H

2

3

4

5



1H 1H

2H

The HSQC spectrum is very simple because it spreads out the peaks over two dimensions has little multiplet structure:

xx 2 xx 3 1

1 3

C a x i s

H axis

It will also identify methylene pairs, both by color (remember, CH and CH3 are opposite to CH 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 COSY crosspeak, which the HSQC will first identify as being geminal in nature.

xx xx 4 5

4/5 is a methylene pair.

Chem 135

(2) Tabu late th e Data. ID

(1H)

(13C)

Hs

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

--

1

br s

--

--

202.57

3 3.47 ... etc ...

type

J

(Hz)

COSY

HMBC

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; sp 2 systems: 2J is small, but 3J is large (bigger for anti than syn) (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

H H H

H H

C C C O C C

long-range coupling

Lectur e 3: 2D NMR Spectr oscopy

E. Kwan (7) Expand and Con nect 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. HO

HO

X 6

5-13, 9-4

CH2X 4, 13, 34.51

CH3

- COSY, HMBC correlations connect termini

12, 22.10

(8) List Full Assign ments

This is self-explanatory: 14 7 HO 1 3, 10

2

11

8

5, 9

6

4, 13

12

16.07

HO 50.12

21.00 25.82

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.

Chem 135


E. Kwan

Chem 135

Problem 1 Please assign the proton and carbon resonances of ethyl nipecotate. (500 MHz, CDCl 3; spectra are courtesy of Dr. Jeffrey Simpson, MIT. See Simpson, Chapter 9 for his treatment of this problem.) O O N H

2.0

1.0

4.0

3.5

1.0

3.0

1.0

1.1

1.1

1.0

2.5 Chemical Shift (ppm)

2.0

9 9 . 3 7 1

192

184

176

160

152

144

136

128

120

112

104 96 88 Chemical Shift (ppm)

80

72

64

1.1

2.8

1.5

4 8 . 9 5

168

2.0

56

5 4 0 3 . 1 . 2 . 8 6 2 4 4 4

48

40

1.0

0 7 1 . 2 . 7 5 2 2

32

24

1 9 . 3 1

16

8

0


E. Kwan

Chem 135

Problem 1 Please assign the proton and carbon resonances of ethyl nipecotate. (500 MHz, CDCl 3; spectra are courtesy of Dr. Jeffrey Simpson, MIT. See Simpson, Chapter 9 for his treatment of this problem.) O O N H

ID 1

(1H)

(13C)

Hs

2 3 4 5 6 7 8 9 10 11

Quaternary Carbons: Methylene Pairs:

Type

J

(Hz)

COSY


E. Kwan Problem 1

Chem 135

This is the HMQC spectrum. Please number the protons. What are the methylene pairs in this molecule? Are there any quaternary carbons?

O O N H

10

15

20

25

30 )

35

40

45

50

55

60

4.0

3.5

3.0


2.0

1.5

1.0

m p p ( t f i h S l a c i m e h C 1 F


E. Kwan Problem 1

Chem 135

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

O O N H 1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.0

3.5

3.0


2.0

1.5

1.0



E. Kwan Problem 1

Chem 135

The numbering I get is shown below.

O O N H

A multiplet analysis in ACD/NMR Processor gives:

10

11

1H NMR (500 MHz, CDCl 3)  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).

15

20

9

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!

7

10

25

8

30 )

The carbon at 173.99 is quaternary (the ester). 35

40

6 5

3

2

45

50

4

55

1 60

4.0

3.5

3.0


2.0

1.5

1.0



E. Kwan Problem 1

Chem 135

Here’s how I annotate COSY spectra. I have bolded the correlations that do not correspond to geminal couplings (guessed for aliphatic region) and are clear. Parentheses indicate weak couplings.

O O N H

1

2

3 4

5

6

7

8/9

10

11 1.0

1 1 0 1 9 / 8

108/9(?) 78/9(?)

7

68/9(?)

6 5

1.5

2.0

5-10

2.5

4-6

4

3-10

3

3-5

2

(2-6)

3.0

2-4 3.5

1-11

1

4.0

3.5

3.0


2.0

1.5

4.0

1.0



E. Kwan Problem 1

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.

O O

(1)

N H

Chem 135

En tr tr y Po in in ts ts : Start with the obvious. The ethyl group is 1 and 11, 11, and there’s only one methine in this this molecule molecule:: O

ID (1H) 1 3.92

(13C) 59.84

Hs 2

Ty p e m

(Hz)

--

COSY 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

11 1.05

13.91

3

t

10.5, 3.7, 2.6 7.2

J

1 6 O

11

N H

(2)

A m i n e: Two sets of downfield, diastereotopic protons must be adjacent to the amine. These must be 2/4 and 3/5. 3/5. There is is a clear COSY correlation between 2 and 6. Therefore, by process of elimination: O 6 O 2/4

3/5 N H

Quaternary Quaternary Carbons: 173.99 Methylene Pairs: 1, 2/4, 3/5, 7/8, 9/10

3, 5 1


E. Kwan

Chem 135

Problem 1 O O

(3)

N H

1 A l ip ip h at i cs cs : We need to distinguish between 9/10 and 7/8. 7/8. Note that 10 has COSY correlations to 3 and 5. (Why is one correlation stronger than the other?) Thus:

2

3 4

5

6

7

8/9

108/9(?)

173.99

6 9/10

11

1

O

78/9(?)

1.5

2.0

2/4

3/5 N H

(4)

11 1.0

O 7/8

10

68/9(?)

Ch ec k : Do these Do these assignments make sense based on chemical shift? ChemDraw predictions certainly seem to be congruent with the assignments:

1.55;1.45

3-10 3-5

(2-6)

3.0

2-4

4.12 O

2.5

4-6

O 1.98;1.73 2.33

5-10

1.30 3.5

2.76;2.73

(5)

N H 2.0

3.21;2.96

Ch ec k : Do the overlapping correlations make sense? Yes, they seem to actually be: 7-8 4.0 (geminal), 6/8 (vicinal), and (10-8 and 10-9).

1-11

3.5

3.0


2.0

1.5

4.0

1.0


Lectu re 3: 2D NMR Spectro sco py

E. Kwan

Chem 135

Problem 2 Please assign the proton and carbon resonances of sucrose (500 MHz, D2O). HO

HO OH O O

HO

O

HO

OH

OH OH

1.0

1.0

5.2

5.1

5.0

4.9

4.8

4 7 . 3 0 1

104

4.7

4.6

4.5

4.4

4 2 . 2 9

102

100

98

96

94

92

4.3 4.2 4.1 Chemical Shift (ppm)

4 4 . 1 8

90

88

86


78

1.1

5.8

4.0

3.9

9 3 . 6 7

2 2 5 0 . 6 . 4 . 4 2 2 7 7 7

76

74

3.8

72

3.7

3 1 . 1 7

1.0

2.0

3.6

1.0

3.5

3.4

6 2 . 9 6

70

1.1 3.3

3.2

3 5 4 4 . 3 . 1 . 2 1 0 6 6 6

68

66

64

62

60

58


E. Kwan

Chem 135

Problem 2

58

inset

60 62

60

arrows: double intensity

64

61

66

62 68 63 70 64 ) 65 66 67 68


72 74 76 78

69

80

70

82

71

84

72

3.65

3.60


3.35

86 88

3.30

90

HMQC 5.2

5.1

5.0

4.9

4.8

4.7

4.6

4.5


4.0

3.9

3.8

3.7

3.6

3.5

3.4

3.3

92 3.2



E. Kwan

Chem 135

Problem 2

COSY-90 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0 )

m p

p 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


4.0

3.9

3.8

3.7

3.6

3.5

3.4

3.3

t f i h S l a c i m e h C 1 F


E. Kwan

Chem 135

Problem 2 Numbering the protons requires some careful bookkeeping.

58

8

60 62

10

7

60

inset

64

61

66

62

12

63

11

64 ) 65 66 67 68


70

56 3

72

9

74

2

76 78

69

80

4

70

82

71

84

72

3.65

3.60


3.35

68

86 88

3.30

90

HMQC

1 5.2

5.1

5.0

4.9

4.8

4.7

4.6

4.5


4.0

3.9

3.8

3.7

3.6

3.5

3.4

3.3

92 3.2



E. Kwan

Chem 135

Problem 2

COSY-90 1

2

3

4 5-8 9

10

11 12

2 1

3.3

1 1

3.4

9-11 9-12

0 1

3.5

9

3.6

8 5

45-8(?)

4

3

125-8(?)

3.7 3.8 3.9

3-4

4.0 )

2

m p

2-3

p 4.1 (

4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0 5.1

1-11

5.2

1

5.2

5.1

5.0

4.9

4.8

4.7

4.6

4.5


4.0

3.9

3.8

3.7

3.6

3.5

3.4

3.3

t f i h S l a c i m e h C 1 F


E. Kwan Problem 2 Here’s some space for your data table:

ID 1

(1H)

(13C)

Hs

2 3 4 5 6 7 8 9 10 11 12

Quaternary Carbons : Methylene Pairs:

Type

J

(Hz)

COSY

Chem 135


E. Kwan Problem 2 I get this data table:

ID (1H) 1 5.22

(13C) 92.24

Hs 1

Type d

J

(Hz) 3.9

COSY 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

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 Methyl ene Pairs: 5/6, 8, 10

11, 12

Chem 135

Lecture 3: 2D NMR Spectro sco py

E. Kwan

Chem 135

Problem 2 Spin Systems.

Based on chemical shift arguments, 1 is the anomeric proton. It may help if I re-draw sucrose: HO HO HO

O

H

OH O

OH

OH

O OH HO

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

2d The left-hand spin system can be drawn out:

11 dd

3t

HO HO 12 HO

9t

4m

1 H

O

9

OH

11

O

OH

OH

O

12 t

OH

5/6, 7, or 8 HO

Which of 5/6, 7, or 8 must 12 be connected to? 5/6, 7, or 8

Lectur e 3: 2D NMR Spectro sco py

E. Kwan

Chem 135

Problem 2 Spin Systems.

1 = anomeric

2d

11 dd

(1) Number o f Proto ns

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.

3t

(2) The Other Spin System

9t

Note that proton 2 is a doublet. This can only occur one way in the other spin system

4m

HO

12 t

7

HO 12 HO

5/6, 7, or 8

9

1 H

O

OH

11

O

OH O

5/6, 7, or 8

2

OH 4 3

OH

HO

(3) What about 4?

3.50

3.55

3.60

3.65

3.70

3.70

3.65 3.60 F2 Chemical Shift (ppm)

3.55


We must decide if proton 4 is connected to 5/6 or 8. Looking closely at the COSY 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?


E. Kwan Problem 2 Spin Systems.

1 = anomeric

(4) Final Assignments

2d

HO

5/6

7

HO 12 HO

11 dd

Chem 135

3t

9

1 H

O

OH 10

11

O

OH

OH 4

O 103.74

9t

2

4m

3

8

OH

HO

12 t

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:

5/6, 7, or 8

- 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-d6

OH O

HO

O OH naringenin

Please assign this spectrum.

H2O

0.7

0.9 12.0

11.5

11.0

10.5

10.0

9.5

0.9 9.0

8.5

2.1 8.0

2.1

7.5 7.0 Chemical Shift (ppm)

2.0 6.5

6.0

1.2 5.5

1.2 5.0

4.5

4.0

3.5

1.1 3.0

2.5

OH O

HO

O OH naringenin

phenol OH

phenol OH

0.7

0.9 12.0

11.5

11.0

10.5

10.0

9.5

0.9 9.0

8.5

2.1 8.0

2.1


2.0 6.5

6.0

1.2 5.5

1.2 5.0

4.5

4.0

3.5

1.1 3.0

2.5

OH O

HO

- number from left to right - one number per unique chemical shift

O OH naringenin

5 6

1 4 3

2

8

9

1.2

1.1

7

0.7

0.9 12.0

11.5

11.0

10.5

10.0

9.5

0.9 9.0

8.5

2.1 8.0

2.1


2.0 6.5

6.0

1.2 5.5

5.0

4.5

4.0

3.5

3.0

2.5

gHSQC 35

OH O

40 45

HO

O

50

OH

55

naringenin

60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 135

7.5

7.0

6.5

6.0


4.5

4.0

3.5

3.0


35

OH O

HO

40

blue = CH2

O

45

- two peaks, same horizontal line = same carbon = CH 2

OH naringenin

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


4.5

4.0

3.5

3.0


35

integrates to 2H

OH O

40 45

HO

O

50

OH

55

symmetry makes these protons equivalent

naringenin

60 65 70 75 80 85 90 95 100 105 110

red = CH

115 120 125 130 135

7.5

7.0

6.5

6.0


4.5

4.0

3.5

3.0


35

OH O

40 45

HO

O

integrates to 1H

OH

50 55

naringenin

60 65

integrates to 2H

70 75

red = CH

80 85 90

note: this is two peaks

95 100 105 110 115 120 125 130 135

7.5

7.0

6.5

6.0


4.5

4.0

3.5

3.0


35

OH O

HO

40

x 9

O

10

45 50

OH

55

naringenin

60 65 70

8

75 80 85 90

6

95

7

100 105 110

red = CH

115 120

5

125 130

4 7.5

-re-numbering is necessary 7.0

6.5

6.0


4.5

4.0

3.5

3.0

135


OH O

HO

O OH naringenin

- combine accurate 1D 13C shift data with HSQC to generate table

4 7 . 8 2 1

8 8 . 5 1 1

5 6 . 9 7

6 9 . 6 9 1

208

200

1 0 . 5 9 6 0 . 1 4 0 6 1 0 . 7 6 1

192

184

176

168

0 5 . 0 3 1

0 4 . 8 5 1

160

152

144

136

128

0 8 . 5 1 1

4 9 . 2 0 1


0 2 . 3 4

1 5 . 6 4 9 5 . 5 9

96

88

80

72

64

56

48

40

32

24

Naringenin (500 MHz, acetone- d6) ID COSY J (Hz) (1H) (13C) Hs Type 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

OH O

HO

O OH naringenin

5,6 7

1 4 3

2

9

10

1.2

1.1

8

0.7

0.9 12.0

11.5

11.0

10.5

10.0

9.5

0.9 9.0

8.5

2.1 8.0

2.1


2.0 6.5

6.0

1.2 5.5

5.0

4.5

4.0

3.5

3.0

2.5

gCOSY 4

5

0 1

6,7

8

9

10

2.5

3.0

9

9-10

3.5

4.0

- ignore peaks on the diagonal

4.5 )

5.0

8

5.5

8-9 8-10

7 , 6

6.0

6.5

5

4-5 - number peaks on both axes

4 7.5

7.0

7.5 7.0

6.5

6.0


4.5

4.0

3.5

3.0


updated data Naringenin (500 MHz, acetone-d6) ID COSY (1H) (13C) Hs Type J (Hz) 1 s* --1 12.18 --1 s* --2 9.56 -1 s* --3 8.51 128.74 2 m -5 4 7.40 115.88 2 m -4 5 6.91 95.54 1 m --6 5.97 96.51 1 m --7 5.96 79.65 1 dd 13.0, 3.0 9, 10 8 5.46 43.20 1 dd 13.0, 17.1 8, 10 9 3.19 43.20 1 dd 17.1, 3.0 8, 9 10 2.74 Quaternary carbons: 196.96, 167.00, 165.01, 164.09, 158.40, 130.50, 102.94 CH2 pairs: 9/10 * exchangeable

OH

O

spin systems: 4 m – 5 m; 8 dd – 9/10 dd/dd

HO

O OH naringenin

Naringenin (500 MHz, acetone- d6) ID COSY J (Hz) (1H) (13C) Hs Type 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 geminal = 17.1 * exchangeable

OH

O

196.96

vicinal = 13.0, 3.0

9,10


HO

- 9/10 can be assigned immediately (there’s only one CH2 in the molecule)

O

8 naringenin

OH

Naringenin (500 MHz, acetone- d6) ID COSY J (Hz) (1H) (13C) Hs Type 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

OH

HO

- by elimination, 4 and 5 are on the mono-phenol - however, order is unclear - 6/7 must be on the bis-phenol

196.96 9,10

6, 7


O

O

8 naringenin

4, 5 OH

gHMBC OH

OH

O 1

O

40

1

H

H

48

13

HO

O

C

HO

O

1-bond peak

13

C

OH one-bond peak

OH

three-bond peak

56 64 72 80 88

1-bond and n-bond peaks indicates symmetry

96 104 ) 112 120 128 136 144 152 160 168 176

sp 2

note: couplings to carbons are usually 2 3-bond because JCH is small 12.0

11.5

11.0

10.5

10.0

9.5

9.0

8.5

8.0 7.5 7.0 F2 Chemical Shift (ppm)

6.5

6.0

5.5

184 192

5.0

4.5

4.0

3.5

3.0


gHMBC: carbonyl region 6,7

8

9

10

189

OH O

HO

190

191

O OH

192

193

194

195

196

196.96

197

198

long-range (>3 bond correlations)

OH

O 13

199

1

C

H 200

HO

O 201

OH 6.0

5.5

5.0


4.0

3.5

3.0

2.5


4

5

6,7

8

9

10 71

OH O 72

HO

O

73

OH

74

8

75

76

77

) m p p ( t f i h

S 78 l

79

79.65

80

HSQC

81

82

8

4-79.65

83

84

85

86 7.5

7.0

6.5

6.0


4.5

4.0

3.5

3.0

2.5

a c i m e h C 1 F

4

5

6,7

8

9

10 71

OH O 72

HO

O

73

OH

OH

74

OH

O 1

O

76

H 13

13

HO

O

75

C

HO

4/5

O

1

C

77

4/5

three bonds

79

four bonds

80

OH

O 4

81

H HO

O

H

5

82

83

OH

84

proton 4 must be ortho to carbon at 79.65 ppm

85

86 7.5

7.0

6.5

6.0


4.5

4.0

3.5

) m p p ( t f i h

S 78 l

OH

OH

79.65

H

3.0

2.5

a c i m e h C 1 F

HMBC 1

2

3

4

5

111.5 112.0 112.5 113.0

OH

O 113.5 114.0

HO

O

114.5

OH

115.0 )

m p

p 115.5 ( t f

i h S l

116.0 a c

5

115.88

i m e h

116.5 C 1 F

117.0

HSQC

117.5 118.0

5

118.5 119.0 119.5 120.0

- HMBC can be used to assign OH protons 12.5

12.0

11.5

11.0

10.5


9.0

8.5

120.5

8.0

7.5

7.0

HMBC 1

2

3

4

5

111.5 112.0 112.5 113.0 113.5

OH

O

114.0 114.5 115.0 )

m p

HO

13

O

115.88

p 115.5 ( t f

C

i h S l

116.0 a c

i m e h

1

O

H

116.5 C

3

1 F

3-115.88

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


9.0

8.5

8.0

7.5

7.0

HMBC 1

2

3

4

5

111.5 112.0 112.5

OH

O

113.0

4

113.5

H HO

H

O

O

114.0

5

114.5

115.88

115.0 )

H 3

m p

p 115.5 ( t f

i h S l

116.0 a c

115.88

i m e h

116.5 C 1 F

117.0 117.5 118.0

can you explain these correlations?

118.5 119.0 119.5 120.0 120.5

12.5

12.0

11.5

11.0

10.5


9.0

8.5

8.0

7.5

7.0

HMBC 1

2

3 OH

4 O

HO

O H

111.5 112.0

H OH

5

112.5

H

O

4

5 113.0 113.5

OH

114.0

HO

115.88

O

114.5

OH

115.0 )

m p

p 115.5 ( t f

i h S l

116.0 a c

115.88

i m e h

116.5 C 1 F

117.0

OH

O

117.5 118.0

H HO

H

O 115.88

5 118.5 119.0

OH

119.5 120.0 120.5

12.5

12.0

11.5

11.0

10.5


9.0

8.5

8.0

7.5

7.0

80

how can the remaining ipso carbons be assigned?

85

what are the relevant correlations in this region?

90

95

100

105

110

115

OH

S l a

O

c 120 i

4

125

H HO

H

O

5

130

135

OH

?

3

140

145

150

?

155

160 8.6

8.5

8.4

8.3

8.2

) m p p ( t f i h

8.1

8.0

7.9


7.4

7.3

7.2

7.1

7.0

6.9

6.8

6.7

m e h C 1 F

3

4

5

8

80

OH

O

85

4

90

H HO

H

O 8

5

OH

95

100

105

3

110

5

115

) m p p ( t f i h S l a

c 120 i

125

4 130.50

130

135

140

145

150

155

158.40

160 8.6

8.5

8.4

8.3

8.2

8.1

8.0

7.9


7.4

7.3

7.2

7.1

7.0

6.9

6.8

6.7

m e h C 1 F

3

4

5

8

80

OH

OH

HO

O

85

H

O

3

O

5

90

95

130.50

H4 HO

O

OH

100

JCH is large 105

130.50 2

OH 110

JCH is small

5

115

) m p p ( t f i h S l a

c 120 i

125

4 130.50

OH

HO

130

O

O O

3 H

correlation missing

135

140

145

158.40 150

155

158.40

160 8.6

8.5

8.4

8.3

8.2

8.1

8.0

7.9


7.4

7.3

7.2

7.1

7.0

6.9

6.8

6.7

m e h C 1 F

9

10

40 48 56 64 72

8

80 88 96

102.94

104 )

m p p (

t 112 f i

120

OH

130.50

196.96

O

128 C 1 F

136

9/10

144 152

HO

160

O 8

168

130.50

176

OH

184 192

196.96 3.45

3.40

3.35

3.30

3.25

3.20

3.15

3.10


2.85

2.80

2.75

2.70

2.65

2.60

h S l a c i m e h

2.55

2.50

9

10

40 48 56 64 72

8

80 88 96

102.94

104 )

m p p (

t 112 f i

102.94 OH

130.50

120

128 C

196.96

O

1 F

136

H

144

9/10

152

HO

O 8

160 168

130.50

OH

176 184 192

196.96 3.45

3.40

3.35

3.30

3.25

3.20

3.15

3.10


2.85

2.80

2.75

2.70

2.65

h S l a c i m e h

2.60

2.55

2.50

6, 7 95

100

105

110

102.94 OH

6, 7 – 102.94

115

120

O

125

H

130

135

HO

O

140

H

145

OH

150

confirmation of assignment

155

160

165

6.10

6.05

6.00


5.90

5.85

5.80


1

2

7 6

7/96.51

95

1 H

102.94

100

O

O

7H

105

2-95.54

96.51 HO

2-96.51

O OH

110

115

120

OH

96.51

O

OH

O

125

7 H

1-102.94

2H O

102.94 1

H

95.54 O

2H O

O H6

130

O

OH

OH

O

135

140

145

HO

O 150

OH

155

160

165

12.0

11.5


10.5

10.0

9.5


1

2

6 7

95

102.94 1 OH

102.94

100

105

O

110

7

115

HO 2

6

120

O

125

OH

1-165.01

130

three phenolic ipso carbons remain unassigned: 167.00, 165.01, 164.09

135

140

145

-these correlations are still difficult to interpret

150

155

160

1-167.00 (long-range?)

165

12.0

11.5


10.5

10.0

9.5


162.5

163.0

163.5

164.0

6-164.09

164.5

165.0

7-165.01

165.5

166.0

166.5

7-167.00

167.0

167.5

168.0

168.5

-raising contour level deconvolutes peaks 6.01

6.00

5.99

5.98

5.97

5.96


5.93

5.92

5.91

5.90

169.0

5.89


102.94 1 OH

from previous slide:

162.5

O

163.0

7 HO 2

6-164.09

163.5

6

164.0

O

164.5

OH 165.0

1-165.01

three phenolic ipso carbons remain unassigned: 167.00, 165.01, 164.09

7-165.01

165.5

166.0

166.5

1-167.00 (weak)

2-167.00 (weak)

7-167.00

167.0

167.5

168.0

raising contour level shows this peak 6.01

6.00

5.99

5.98

5.97

5.96


5.93

5.92

5.91

168.5

169.0

5.90

5.89


162.5

1 H

165.01 7

102.94 163.0

O

H

163.5

HO 2

6-164.09

O

6

167.00

164.0

O OH

164.09

164.5

165.0

1-165.01

7-165.01

165.5

166.0

166.5

1-167.00 (weak)

7-167.00

2-167.00 (weak)

167.0

167.5

168.0

168.5

169.0

6.01

6.00

5.99

5.98

5.97

5.96


5.93

5.92

5.91

5.90

5.89


Observed vs . Predicted Chemical Shifts – ACD/Labs NMR Predictor (v6) ID

8

1 2 3 4 5 6 7 8 9 10

7 ) m p p ( 6 t f i h s l a c i 5 m e h c d 4 e t c i d e r p 3

Obsvd. (1H, ppm) 12.18 9.56 8.51 7.40 6.91 5.97 5.96 5.46 3.19 2.74

Calcd. (1H, ppm) 9.87 9.87 9.87 7.38 6.90 6.20 6.19 5.44 3.18 2.72

Obsvd. (13C, ppm) ---128.74 115.88 95.54 96.51 79.65 43.20 43.20 196.96 167.00 165.01 164.09 158.40 130.50 102.94

y = 1.0153x - 0.0278 2

R = 0.9979 2 2

3

4

5

6

7

8

observed chemical shift (ppm) 200 180 ) m p 160 p ( t f i 140 h s l a c 120 i m e h 100 c d e t c 80 i d e r p 60

Overall Assignments: 1 H

165.01 7

102.94 O

196.96

O

H

9/10 4

y = 1.0006x - 0.3724

40

60

80

100

120

140

160

observed chemical shift (ppm)

180

6

O 8

167.00 164.09

2

R =0.99996

40

HO 2

200

130.50 158.40

5 3 OH

Calcd. (13C ppm) ---128.58 115.68 95.63 96.31 79.26 42.81 42.81 196.76 167.20 164.32 163.67 158.18 129.85 102.39

-exchangeable peaks not predicted well -otherwise, excellent agreement -DFT methods can also be used, but their implementation can be tricky

Molecular Modelling B3LYP/6-31g(d)

- two possible ground-state conformers OH

O

9/10 HO

O 8 OH

ID 1 2 3 4 5 6 7 8 9 10

(1H) 12.18 9.56 8.51 7.40 6.91 5.97 5.96 5.46 3.19 2.74

(13C) ---128.74 115.88 96.51 95.54 79.65 43.20 43.20

Hs 1 1 1 2 2 1 1 1 1 1

Type s* s* s* m m m m dd dd dd

J

(Hz)

-------13.0, 3.0 13.0, 17.1 17.1, 3.0

rel G: +0.0 kcal/mol

rel G: +2.7 kcal/mol

observed coupling (Hz)

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)

Lectur e 3: 2D NMR Spectros cop y

E. Kwan

Problem 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) 1 5.20 2 5.08 3 4.96 4 4.59 5 4.26 6 4.11 7 3.68 8 ~3.2 9 2.08 10 2.05 11 2.02 12 2.00 Quaternary

COSY (13C) Hs Type J (Hz) 73.1 1 t 9.5x2 3, 2 68.7 1 t 9.5x2 1, 7 71.6 1 dd 9.5, 8.0 4, 1 99.8 1 d 8.0 3 62.3 1 dd 12.3, 4.9 6, 7 62.3 1 dd 12.3, 2.5 5, 7 71.8 1 m -2, 5, 6 -1 br s --20.95 3 s --20.90 3 s --20.79 3 s --20.83 3 s --Carbons: 170.8, 170.4, 169.6, 169.4

Molecular formula:

Key HMBC 170.4 169.6 169.4 170.8 170.8

170.8 169.4 169.6 170.4

Key ROESY 4, 7 3 2 1, 7 7 7 1, 4, 5, 6

Chem 135


E. Kwan

Chem 135


ID (1H) 1 5.20 2 5.08 3 4.96 4 4.59 5 4.26 6 4.11 7 3.68 8 ~3.2 9 2.08 10 2.05 11 2.02 12 2.00 Quaternary


Key HMBC 170.4 169.6 169.4 170.8 170.8

Key ROESY 4, 7 3 2 1, 7 7 7 1, 4, 5, 6

170.8 169.4 169.6 170.4

Molecular formula:

Hints •

Entry Point: Methyl Groups Every 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 Systems What are the spin systems in this molecule?


E. Kwan

Chem 135


ID (1H) 1 5.20 2 5.08 3 4.96 4 4.59 5 4.26 6 4.11 7 3.68 8 ~3.2 9 2.08 10 2.05 11 2.02 12 2.00 Quaternary


Key HMBC 170.4 169.6 169.4

Key ROESY 4, 7 3 2 1, 7 7 7 1, 4, 5, 6

170.8 170.8

170.8 169.4 169.6 170.4

(2) Spin Systems What are the spin systems in this molecule? 5 dd/6 dd

7 m

1 t

3 dd

4d

5 dd/6 dd

7 m

2 t

2 t

Apart from 5/6, everything is a methine. (3)

Connect Fragments From the HMBC, note that:

1 t

3 dd

4d

HMBC O

O 170.8 X

Me 9

169.6 X

O 170.4 Me X 11

O 169.4 Me X 12

Me 10


E. Kwan

Chem 135


ID (1H) 1 5.20 2 5.08 3 4.96 4 4.59 5 4.26 6 4.11 7 3.68 8 ~3.2 9 2.08 10 2.05 11 2.02 12 2.00 Quaternary (4)


Key HMBC 170.4 169.6 169.4 170.8 170.8

Key ROESY 4, 7 3 2 1, 7 7 7 1, 4, 5, 6

170.8 169.4 169.6 170.4

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: 169.4 OAc

169.6 OAc

5 dd/6 dd AcO 170.8

2 t

OAc 170.4

This looks a lot like a sugar!

OR

4d

1 t

7 m OR

3 dd

X

lecture 3

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