organic structure analysis professor marcel jaspars

Organic Structure Analysis

Professor Marcel Jaspars

Aim

• This course aims to extend student’s knowledge and experience with nuclear magnetic resonance (NMR) and mass spectroscopy (MS), by building on the material taught in the 3rd Year Organic Spectroscopy course, and also to develop problem solving skills in this area.

Learning Outcomes

By the end of this course you should be able to:• Assign 1H and 13C NMR spectra of organic

molecules.• Analyse complex first order multiplets.• Elucidate the structure of organic molecules

using NMR and MS data.• Use data from coupling constants and NOE

experiments to determine relative stereochemistry.

• Understand and use data from 2D NMR experiments.

Synopsis• A general strategy for solving structural problems using

spectroscopic methods, including dereplication methods.• Determination of molecular formulae using NMR & MS• Analysis of multiplet patterns to determine coupling constants.

Single irradiation experiments. Spectral simulation.• The Karplus equation and its use in the determination of relative

stereochemistry in conformationally rigid molecules. • Determination of relative stereochemistry using the nuclear

Overhauser effect (nOe). • Rules to determine whether a nucleus can be studied by NMR &

What other factors must be taken into consideration.• Multinuclear NMR-commonly studied heteronuclei. • Basic 2D NMR experiments and their uses in structure

determination.

Books

• Organic Structure Analysis, Crews, Rodriguez and Jaspars, OUP, 2009

• Spectroscopic Methods in Organic Chemistry, Williams and Fleming, McGraw-Hill, 2007

• Organic Structures from Spectra, Field, Sternhell and Kalman, Wiley, 2008

• Spectrometric Identification of Organic Compounds, Silverstein, Webster and Kiemle, Wiley, 2007

• Introduction to Spectroscopy, Pavia, Lampman, Kriz and Vyvyan, Brooks/Cole 2009

Four Types of Information from NMR

N

HN

HN

NH

NH

O

OO

OO

HN

O

HN

O OH

SMe

HO

1.62.43.24.04.85.66.47.2f1 (ppm)

6.9

8

0.9

21.

14

1.1

22.

03

1.1

3

3.8

33.

89

2.8

5

3.9

1

3.8

7

1.8

71.

08

0.9

9

0.9

3

0.8

90.

95

1.9

8

2.2

9

1.9

6

2.0

0

10.

00

102030405060708090100110120130140150160170

f1 (ppm)

10.

01

13.

73

18.

37

20.

22

28.

07

30.

01

37.

59

38.

70

47.

18

47.

60

52.

31

59.

30

59.

40

60.

11

99.

99

114.

95

125.

98

128.

07

128.

35

129.

02

136.

88

140.

86

155.

42

170.

56

172.

97

173.

40

173.

48

N

HN

HN

NH

NH

O

OO

OO

HN

O

HN

O OH

SMe

HO

1H NMR Chemical Shifts in Organic Compounds 12 11 10 9 8 7 6 5 4 3 2 01 -1

12 11 10 9 8 7 6 5 4 23 1 -1

RHS

R NH2

R OH

NH

H

RCONH2

CHC

ZH

H

H2C

O

HRC=CR2

H2C=CR2

R-CH2-R

M-CH3

CH3-O-

CH3-N

CH3-S-

CH3-Ph

-CO-CH2-

CH3-CO-

CH3-C-X

CH3-C=C-

Ph-CH2-

-CH2-C=C-

HO2C-

HCO-

OH

RCONHR

Z

Z

R

N H

Alcohols, Halogens

Amides

Phenol -OH

Alcohols-OHThiols-SHAmines-NH2

Methylenes (-CH2-)

Cyclopropyl

Metals (M-CH3)

ppm

ppm

Carboxylic acids-OH

AldehydesHeteroaromatics

AromaticsAlkenes

-HC-O- ó -HC-X

-CH2-O- ó -

CH2-X

Alkynes

Methyl groups (R-Me)

0

Z=O, N, S

Z=O, N, S

Amines R-NH2 -CH-N- ó -CH2-N

Relative to TMS (0 ppm)

13C NMR Chemical Shifts in Organic Compounds

220 210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0

KetoneAldehydeAcidEster, AmideThioketoneAzomethine

CN Nitr ile

Heteroaromatic

AlkeneAromatic

CC Alkyne

C Quat.

Halogen(C Tertiary)

Halogen(C Secondary)

Halogen(C Primary)

Halogen

230

220 210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0

*Relative to TMS (0 ppm)

ppm

ppm

Cyclopropanes

230

sp2

sp

sp3

4o

3o

2o

1o

Epoxides

4o Quaternary, 3o Tertiary, 2o Secondary, 1o Primary

C=OC=OC=OC=OC=SC=N-

C=N-

C=CC=CC=C

C-O-

CH-N-

C-N-

CH2-S-

C-S

CH2-X

C-XCH-CCH-O-

CH3-C

CH-S-

CH3-O-

CH-XCH2-CCH2-O-

CH3-N-

CH2-N-

CH3-S-CH3-X

Heteroaromatic

1.01.41.82.22.63.03.43.84.24.65.05.45.86.26.6

3.8

4

2.4

7

2.5

0

2.5

2

1.1

7

1.1

8

1.0

0

46

6.4

94

73.

91

48

1.2

9

69

1.1

86

98.

52

70

5.9

17

13.

43

72

1.0

38

10.

15

81

6.7

18

24.

70

83

1.6

18

38.

41

18

34.6

11

841.1

71

847.7

2

19

79.3

41

986.1

4

20

76.8

52

091.1

9

32

22.8

23

229.6

23

237.1

63

243.9

6

5 Minute Problem #1MF = C6H12OUnsaturated acyclic ether

Six Simple Steps for Successful Structure Solution

• Get molecular formula. Use combustion analysis, mass spectrum and/or 13C NMR spectrum. Calculate double bond equivalents (DBE).

• Determine functional groups from IR, 1H and 13C NMR• Compare 1H integrals to number of H’s in the MF.• Determine coupling constants (J’s) for all multiplets.• Use information from 3. and 4. to construct spin

systems (substructures)• Assemble substructures in all possible ways, taking

account of DBE and functional groups. Make sure the integrals and coupling patterns agree with the proposed structure.

Double Bond Equivalents

DBE = [(2a+2) – (b-d+e)]

2

C2H3O2Cl =

CaHbOcNdXe

Tabulate Data

Shift (ppm)

Int. Mult (J/Hz) Inference

6.48 1H dd, 14, 7

4.17 1H d, 14

3.97 1H d, 7

3.69 2H t, 7

1.65 2H quint, 7

1.42 2H sext, 7

0.95 3H t, 7

Solution

Shift Prediction

01234567PPM

Prediction

Organic Structure Analysis, Crews, Rodriguez and Jaspars

Molecularformula

Functionalgroups

Substructures

Very secure3D molecular

structure

MS, NMR

NMR, IR

UV

NMR

X-RAY

UnsaturationNumber(UN)

Working 2D

structures

List of working

2D structures

Reasonable3D molecular

structure

New 2Dmolecularstructure

NMR

ORD Totalsynthesis Molecular

modeling

Knownmolecularstructure

Dereplicate by MF

Dereplicate

Draw all isomers

NMR, MS, IR, UV

by structure

Purecompound

THE PROCESS OF STRUCTURE ELUCIDATION

DereplicationDediscovery

13C spectrum/DEPT-135

Low Resolution MS

m/z 335 [M+H]+

APT Formula C20H29

High Resolution MS

m/z 335.2222 [M+H]+

Molecular Formula C20H30O4

Crystal Unit Cell Dimensions

DATABASE

Substructures

1D NMR Spectra 2D NMR Spectra

Dereplication

1H NMR

Me

HOH

H

Me

200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10ppm

6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm

13C NMR

O

OHH

Dereplication

OH

O

H

H H

TestosteroneC19H28O2

m/z 288.2089

Accurate massm/ z 288.2080 ( 3 ppm)

Molecular formulaC19H28O2

DEPT-135 FormulaC19H27

Low Resolution MSm/z 288

Su

bst

ruct

ure

s

2D NMR

1H NMR

13C NMR

Me

HOH

H

O

Me

OHH

Other Search ParametersUVmax

Crystal unit cell dimensionsMS isotope cluster

Taxonomy


Determining the Molecular Formula Using NMR and MS Data

CCH2

CH, CH3

C

DEPT-135


140 130 120 110 100 90 80 70 60 50 40 30 20 10ppm

DEPT-135: CH, CH3, CH2

13C NMR

MS Errors

• Experimental accurate mass measurement (from MS) was 136.1256 suggesting C10H16 is the correct formula.

• The error between calculated and experimental mass is:

• 136.1256 - 136.1248 = 0.0008 = 0.8 mmu

Formula dbe Accurate mass

C10H16 3 136.1248

C9H12O 4 136.0885

C8H8O2 5 136.0522

C7H4O3 6 136.0159

C9H14N 3.5 136.1123

C8H12N2 4 136.0998

1361256 1361248

13612565 9 10 5 96. .

.. .

ppm

Molecular Formula Calculators

James Deline MFCalc

Isotope Ratio Patterns:C100H200

For 12Cm13Cn

1403.61404.6

Determining molecular formulae by HR-ESI/MS

ThermoFinnigan LTQ Orbitrap,

Xcalibur software

examples from MChem group practicals 2009(Rainer Ebel)

O

10

HN

N-(2-hydroxy-3-methylphenyl)acetamide

OH

Chemical Formula: C9H11NO2Exact Mass: 165.0790

[M+H]+

O

H2N

OH

Chemical Formula: C9H12NO2+

Exact Mass: 166.0863

O

10

HN

N-(2-hydroxy-3-methylphenyl)acetamide

OH

Chemical Formula: C9H11NO2Exact Mass: 165.0790

[M+Na]+

O

HN

OH

Na

Chemical Formula: C9H11NNaO22+

Exact Mass: 188.0677

9

HN

2-chloro-N-(4-hydroxyphenyl)acetamide

HOO

Cl

Chemical Formula: C8H8ClNO2Exact Mass: 185.0244

Analysis of isotope patterns

H2N

HOO

Cl

Chemical Formula: C8H9ClNO2+

Exact Mass: 186.0316m/z: 186.0322 (100.0%), 188.0292 (32.0%), 187.0355 (8.7%), 189.0326 (2.8%)

experimental

calculated

“monoisotopic peak”(mainly 12C7

13C1H935Cl14N16O2)

5 Minute Problem #2

NH

MeO

NH

MeO

A

B

• Al Kaloid, an Honours Chemistry Student at Slug State University has synthesised either A or B below. He is uncertain which one it is but he’s tabulated the 13C NMR shifts and their multiplicities. Can you help Al by determining which one it is?

Answer

Base Values:

ChemDraw

134.6

109.9

23.9

23.4

23.820.9

NH

133

110

23

22

21

20

134.6

109.9

134.6

109.9

23.9

23.4

23.820.9

23.9

23.4

23.820.9

55.8

55.8

NH

NH

O

O

The NMR effect

I = 1/2E

Bo off Bo on

I = -1/2(, antiparallel)

I = -1/2(, parallel)

Ha Hb

Ha

HbNo coupling

Coupling

Spin-spin coupling (splitting)

Spin-spin coupling (splitting)

Origin of spin-spin coupling

Ha

C C

Hb Ha

C C

Hb

Bo

parallel antiparallel

Coupling in ethanol

H

HO

H

H

H

H

Coupling is mutualHb

HO

Hb

Hb

Ha

Ha

Coupling in ethanol

• To see why the methyl peak is split into a triplet, let’s look at the methylene protons (CH2).

– There are two of them, and each can have one of two possible orientations- aligned with, or against the applied field

– This gives rise to a total of four possible states:

Hence the methyl peak is split into three, with the ratio of areas 1 : 2 : 1

Coupling in ethanol

• Similarly, the effect of the methyl protons on the methylene protons is such that there are 8 possible spin combinations for the three methyl protons:

The methylene peak is split into a quartet.The areas of the peaks have the ratio of 1:3:3:1.

Pascal’s triangle

n relative intensity multiplet

0 1 singlet

1 1 1 doublet

2 1 2 1 triplet

3 1 3 3 1 quartet

4 1 4 6 4 1 quintet

5 1 5 10 10 5 1 sextet

6 1 6 15 20 15 6 1 septet

Coupling patterns

First Order

• In CH3CH2OH we could explain coupling by the n + 1 rule, this is called 1st order coupling

Second Order

• Like CH3CH2OH expect 7 lines but get many more. /J < 6

Common Coupled Spin Systems

Complex 1st Order Spin Systems

Iterative application of the n + 1 rule

4.04.14.24.34.44.54.64.74.84.95.05.15.25.35.45.55.65.75.85.9

5 Minute Problem #3. Given the 1H NMR chemical shifts and coupling constants for allyl alcohol, explain

the observed spectrum (OH peak omitted)

H

H

H

OH

5.05

5.15 4.00

5.85H

H

H

OH

10.4 Hz

1.8 Hz

5.2 Hz1.8 Hz

1.8 Hz

17 Hz

5.1155.1205.1255.1305.1355.1405.1455.1505.1555.1605.1655.1705.1755.1805.185

20

48.2

1

20

49.9

3

20

51.6

4

20

53.3

3

20

65.4

3

20

67.1

5

20

68.8

6

20

70.5

5

A doubled quartet (dq)

What about this?

5.8105.8255.8405.8555.8705.8855.9005.915

1.0

0

A (ddt)5.86

23

25.0

7

23

30.2

3

23

35.4

2

23

40.6

6

23

42.2

7

23

45.8

0

23

47.4

3

23

52.6

4

23

57.8

5

23

62.9

9

ddt

5 5

5 510

17

5.8155.8255.8355.8455.8555.8655.8755.8855.8955.9055.915

4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm

4.0 3.5 3.0 2.5 2.0 1.5 ppm

Real Spectrum

Spin Simulation

Pro-R and Pro-S

Ph OH

H

H

1Ph OH

H

H

Ph

OH

HH

Homotopic, Enantiotopic, Diastereotopic

Methyl groups

Chemical Equivalence/Magnetic Non Equivalence

H

Cl

Cl

H

H

H

What is going on?

Cl

H

H

Cl

H

H

ResultH

Cl

Cl

H

H

H

8 Hz

2Hz

A

B

B

A

2Hz

8 HzA

B

J AB'J AB' J AB' J AB'

J AB JAB

Expect:

Using Coupling Constants

Glucose

4.04.24.44.64.85.05.25.45.65.86.06.26.4

O

H

HO

H

HO

H

OH

OHHH

OH

12

3

4 56

Glucose

-173.0o 9.50 Hz

O

H

HO

H

HO

H

OH

OHHH

OH 52.7o 3.51 Hz

169.4o 9.5 Hz

-170.5o 9.13 Hz

H

HO CH

CHO

12

3

45

6

C2-C3

C1-C2

H

HO COH

OH

5 Minute Problem #4• Work out which of 2.1 and 2.5 is equatorial and which is axial. Also work

out the 3 dihedral angles for 2.1, 2.5, 2.8, 6.8.• There are also peaks at: 6.80, 1H, d, J = 0.5 Hz; 1.95, 3H, s; 0.93, 9H, s.

O

ClCl

HH H

H2.10

2.50

0.93

2.806.80

1.95

30 Hz

Solution

1.52.02.53.03.54.04.55.05.5 ppm

Removing CouplingsChanging Solvents

OH

HaHb

Hc

CDCl3

C6D6

Ha, t Jab = Jac

Ha, dd Jab ≠ Jac

a = bCoupled to HcJab = Jac

a ≠ bEach coupled to HcJab ≠ Jac

1.52.02.53.03.54.04.55.05.5 ppm

Removing CouplingsSpin decoupling

OH

HaHb

Hc

CDCl3

CDCl3irradiated

Irradiate at 4.11 ppm

Signal due to Hadisappears

Coupling due to Hais removedSee Hb, Hc at b, cWith mutual Jbc

Spin Decoupling

OFF

A↑ X↑

A↑ X↓

A↓ X↑

A↓ X ↓

Spin Decoupling• Two spins, A (A), X (X) with JAX

• Irradiate X with RF power, A loses coupling due to X

ON

A↑ X↑ ↓

A↓ X ↑ ↓

Average of X ↑ and X ↓

Nuclear Overhauser Effect

Size of NOE

-1

-0.5

0

0.5

0.01 0.1 1 10 100w

0t

c

h

fast tumbling slow tumbling

Effect of NOE on 13C NMR(90o)x

a. Proton Decoupled

decouple

b. Distortionless Enhancement by Polarization T ransfer (DEPT, q=135o, t1= 3.5

ms)(90o)x

(90o)x (180o)x

(180o)x

decouplet1t1t1

qy

HOHO

12

3

4

5

678

9

10

pinanediol (2)

13C

1H

13C

1H

2426283032343638404244464850525456586062646668707274

1

2 6 4

7

3 10/5/9

8

13C – 1H NOE at equilibrium (small molecule)

C↑H ↑

C↓H↓

C↓H ↑

C↑H ↓

●●●●●

●●●●

●

13C – 1H NOE irradiation on H

C↑H ↑

C↓H↓

C↓H ↑

C↑H ↓

●●●

●●

●●●

C

C

Hsat

Hsat

●●

13C – 1H NOE irradiation on H left ON

C↑H ↑

C↓H↓

C↓H ↑

C↑H ↓

●●●

●●

●●●

C

C

Hsat

Hsat

●●

13C – 1H NOE result

C↑H ↑

C↓H↓

C↑H ↓

●●●●

●

●●●●

C

C

Hsat

Hsat

●

1H-1H NOE example

1.01.52.02.53.03.54.04.55.05.56.0

1

2

H3

H1H4

HO

ClZ-2-chloro-dodec-2,11-dien-1-ol

1 3 4

H H

HO

Cl

HH

H

E-2-chloro-dodec-2,11-dien-1-ol

1 3

4

nOe nOe

H2O

1H – 1H NOE at equilibrium (small molecule)

S↑I ↑

S↓I↓

S↓I ↑S↑I ↓

●●●●

●●●●

I S

1H – 1H NOE irradiation on S

S↑I ↑

S↓I↓

S↓I ↑S↑I ↓

●●●

●●●●

W1S (sat)

W1S (sat)

W1I

W1I

I S

●

1H – 1H NOE irradiation on S left ON

S↑I ↑

S↓I↓

S↑I ↓

●●●

●●●●

W1S (sat)

W1S (sat)

W1I

W1I

●

1H – 1H NOE result

S↑I ↑

S↓I↓

S↑I ↓

●●●½

●●●½ ½

W1S (sat)

W1S (sat)

W1I

W1I

½

I S

NOE 3D example

Me Me

Me

H

H

H

O

Me Me

Me

H

H

H

OH

H

H

Me Me

Me

H

H

H

OH

H

H


Random orientation of magnetic dipoles

(a) No Bo

Mo

x

yBo

Mxy = 0

(b) Bo on; prior to resonance

Net polarization Mz is due to

population excess in higher

energy state

The magnetic vectors

precess about Bo at

the Larmor frequency o

M z

y

x

(c) At resonance o = 1

The magnetic vectors

precess in phase with

frequency 1.

After resonance the return

to the equilibrium in (b)

occurs by the loss of Mxy via

dephasing of nuclear

dipoles by T2 and increase

in Mz by spin inversion

due to T1.

Events Accompanying Resonance


Mo

x

y

z

Bo

Excess of spinpopulation alongthe direction ofapplied magneticfield.

(90o)x

x

y

z

Bo

After 90o pulse

magnetization

is tipped into

the xy plane.

M

time t2

M=Magnetization which produces the FID. It decays as magnetization in xyplane diminishes after resonance

FT

frequency f2

preparation detection

ONE-PULSE SEQUENCE


ONE-PULSE SEQUENCE

1H

(90o)x

Preparation Detection

Fourier Transformation

0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7t1 (sec)

1.42.02.63.23.8f1 (ppm)

FT

Relaxation and Peak Shape

Rotational Correlation Time tc

wo = 2o

Nuclear spin

Example Atomic mass

Atomic number

Spin, I

13C, 1H, 17O, 15N, 3H

Odd Odd or Even

1/2, 3/2, 5/2 etc

12C, 16O Even Even 0

2H, 14N Even Odd 1, 2, 3 etc

6 1 8

6 8

1

1

7

7

Receptivity

Nucleus C Relative Receptivity Relative receptivity

29Si 4.7% -5.32

13C 1.1% 6.73

1H 100% 26.75

Multinuclear NMR

15N NMR Shifts

1000 900 800 700 600 500 400 300 100200 0

Amonium salt

Aliphatic amine PrimarySecondary

Tertiary

ppm

PrimarySecondary

TertiaryDimethyleneamine

Anilinium ion

Aniline

1000 900 800 700 600 500 400 300 100200 0ppm

Piperidine

Cyclic enamine

Aminophosphine

Guanidine

Urea, carbamate, lactame

Amide PrimarySecondary

TertiaryThiourea

Thioamide

NitramineIndole, pyrrole

ImideNitrile, isonitr ile

PyrrazoleDiazocompound

Diazonium salt

PyridineImine

Oxime

Nitrocompound

Nitrosocompound Ar-NO S-NON-NO

*Nitrometane as reference (380 ppm)

Pyrimidine

Pyridine

Pyrazine

Imidazole

InternalTerminal

Iso

N=N-N ImineAmineHidrazone, tr iazene

NO2 N

ImineAmine

31P NMR Shifts

700 600 500 400 300 200 100 0 -200-100 -300

R3P=CR2

Z3P

Z-P=P-Z

Z2N-P=NRArP=C=Z2

X =P(Y)=Z

Z6P-

ppm

ppm

Z-C P

R3-P-RR2P-PR2

Z3P=O

Z5P

700 600 500 400 300 200 100 0 -200-100 -300

ArP=C=Ar

ArP=C=O

Cl3P

(RO)3P (Ar)3P (Me)3P H3P

TMS-C P

R-C P

Cl(OR)2P=O

R2(OR)3P

R4HP

*Reference H3PO4 85%. Adapted from A Complete Introduction to Modern NMR Spectroscopy. R. S. Macomber. J. Wiley and Sons. 1998.

R2P-M+

ArP=C=Ar2

(TMS)3P

FC P

R3P=O

RO3P=O

R(OR)2P=O

Cl2RP=O

(OR)5P

Coupling

Effect of 31P on 1H NMR

0.01 0.02 0.03 0.03 0.03 0.51 0.54

4.00

6.18

8 7 6 5 4 3 2 ppm

P

O

OHO

Effect of 31P on 1H NMR

4.30 4.25 4.20 4.15 4.10 4.05 ppm

P

O

OHO

Effect of 31P on 13C NMR

60 50 40 30 20 ppm

O

P

O

O

O

1 2 34

5

4 3

1

5

The 2nd Dimension


BASIC LAYOUT OF A 2D NMR EXPERIMENT


How a 2D NMR experiment works

FT

FT

FT

FT

t2

t1

2t1

3t1

nt1

t1

f2

transformmatrix

t1

f2FT

FT

FT

FT f2

f1

n is the number of increments

Contour plot

TYPES OF 2D NMR EXPERIMENTS

• AUTOCORRELATED– Homonuclear J resolved– 1H-1H COSY– TOCSY– NOESY– ROESY– INADEQUATE

• CROSS-CORRELATED– Heteronuclear J resolved– 1H-13C COSY– HMQC– HSQC– HMBC– HSQC-TOCSY



STRATEGY BASED ON C-H CONNECTIVITY

H

C C

HHSQCHSQC &

H

C C

H

1H-1H COSY

C C

H

C C C

HSQC &H

C C C

HMBC

C C C


STRATEGY BASED ON C-H CONNECTIVITY

Get 1H, 13C NMR spectra

get multiplicities and integrals

Get one bond 13C-1H correlations

assign 1H resonances to 13C resonancesGet 1H-1H correlation data (eg COSY)

Check assignment of diastereotopic protonsusing COSY and HMQC

Assemble substructures using COSY data

Get long range 13C-1H correlation spectrum

(eg HMBC)Combine substructures into

all possible working structures

Check all working structures for consistencywith 2D NMR data

2D structure

Get 13C-1H correlation spectrum (eg HSQC)


STRATEGY BASED ON C-H CONNECTIVITY – DEPT DATA

CCH2

CH, CH3

CHCH CH

CH2

CH3 CH3

C


STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA

A B C D E F C

ab

c

d’d

ef



ATOM C (ppm) DEPT H (ppm)

A 131 CH 5.5

B 124 CH 5.2

C 68 CH 4.0

D 42 CH2 3.0

2.5

E 23 CH3 1.5

F 17 CH3 1.2



diastereotopic protons


STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA

ab

c

d

d'

ef

a b c d d' e f



ATOM C (ppm) DEPT H (ppm) COSY (HH)

A 131 CH 5.5 b, c, d/d’, f

B 124 CH 5.2 a, d/d’, f

C 68 CH 4.0 a, d/d’, e

D 42 CH2 3.0

2.5

a, b, c, d, e

E 23 CH3 1.5 c, d/d’

F 17 CH3 1.2 a, b


STRATEGY BASED ON C-H CONNECTIVITY – COSY DATAHSQC suggests diastereotopic protons:3.08/2.44 ppm1.86/2.07 ppm


STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA

A B C D E F

ab

c

d

d'

ef

C

H

And many more…



ATOM C (ppm) DEPT H (ppm) COSY (HH) HMBC (CH)

A 131 CH 5.5 b, c, d/d’, f b, c, d, f

B 124 CH 5.2 a, d/d’, f a, d, f

C 68 CH 4.0 a, d/d’, e a, d, e

D 42 CH2 3.0

2.5

a, b, c, d, e a, b, c, e

E 23 CH3 1.5 c, d/d’ c, d

F 17 CH3 1.2 a, b a, b


STRATEGY BASED ON C-H CONNECTIVITYRETROSPECTIVE CHECKING

Pieces:

A

H

B

H

C

OH

H

D

HH

E

H H

H

F

H H

H

C 131 C 124 C 68 C 42 C 23 C 17

H 5.5 H 5.2H 4.0

H 3.0H 2.5

H 1.5 H 1.2

Possibilities:

F

A

B

C

D

E

OH

F

B

A

C

D

E

OH

E

B

A

C

D

F

OH

E

A

B

C

D

F

OH

F

D

A

B

C

E

OH

E

D

B

A

C

E

OH

E

D

A

B

C

E

OH

F

D

B

A

C

E

OH

F

A

B

D

C

E

OH

E

B

A

D

C

F

OH

E

A

B

D

C

F

OH

F

B

A

D

C

E

OH

Combinatorialexplosion


STRATEGY BASED ON C-H CONNECTIVITYRETROSPECTIVE CHECKING

MeF

B

A

D

C

MeE

OH

Hb

Ha

Hd Hd'

Hc

And similarly for COSY data


PROSPECTIVE CHECKING

Pieces: A

H

B

H

C

OH

H

D

HH

E

H H

H

F

H H

H

C 131 C 124 C 68 C 42 C 23 C 17

H 5.5 H 5.2H 4.0

H 3.0H 2.5

H 1.5 H 1.2

A B

Reason: only 2 sp2 C's

A B

F

HMBC: F-a, F-b, A-f, B-f

A B

F

DHMBC: D-a, D-b, A-d, B-d

A B

F

DC

OH

HMBC: C-d, C-a, D-c, A-cA B

F

DC

E

OH

HMBC: E-c, E-d, C-e, C-d


2D EXERCISE 1. For a simple organic compound the mass spectrum shows a molecular ion at m/z 98. The following data has been obtained from various 1D and 2D NMR experiments. Using this information determine the structure of the molecule in question and rationalise the 2D NMR data given.

Atom dC (ppm) dH (ppm) 1H - 1H COSY

(3 bond only)

1H 13C Long range

(2 - 3 bonds)

A 218 s - - A-b, A-c, A-d, A-e

B 47 t 1.8 dd b-d B-c, B-d, B-e, B-f

C 38 t 2.3 m c-e C-b, C-d, C-e

D 32 d 1.5 m d-b, d-e, d-f D-b, D-c, D-e, D-f

E 31 t 2.2 m e-c, e-d E-b, E-c, E-d, E-f

F 20 q 1.1 d f-d F-b, F-d, F-e


2D EXERCISE 2. For a simple organic compound the mass spectrum shows a molecular ion at m/z 114. The following data has been obtained from various 1D and 2D NMR experiments. Using this information determine the structure of the molecule in question and rationalise the 2D NMR data given.

An additional peak is present in the 1H NMR at 11.6 ppm (bs).

Atom dC (ppm) dH (ppm) 1H - 1H COSY

(3 bond only)

1H 13C Long range

(2 - 3 bonds)

A 178 s - - A-d, A-b

B 136 d 5.7 m b-c, b-d B-d, B-c, B-e

C 118 d 5.5 m c-b, c-e C-b, C-d, C-e, C-f

D 38 t 3.0 d d-b D-b, D-c

E 25 t 2.1 m e-c, e-f E-b, E-c, E-f

F 13 q 1.0 t f-e F-c, F-e

organic structure analysis professor marcel jaspars

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