organic structure analysis professor marcel jaspars

Organic Structure Analysis

Professor Marcel Jaspars

• 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.

• 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

1.62.43.24.04.85.66.47.2f1 (ppm)

102030405060708090100110120130140150160170

f1 (ppm)

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

RCONH2

HRC=CR2

H2C=CR2

R-CH2-R

CH3-O-

CH3-S-

CH3-Ph

-CO-CH2-

CH3-CO-

CH3-C-X

CH3-C=C-

Ph-CH2-

-CH2-C=C-

RCONHR

Alcohols, Halogens

Amides

Phenol -OH

Alcohols-OHThiols-SHAmines-NH2

Methylenes (-CH2-)

Cyclopropyl

Metals (M-CH3)

Carboxylic acids-OH

AldehydesHeteroaromatics

AromaticsAlkenes

-HC-O- ó -HC-X

-CH2-O- ó -

Alkynes

Methyl groups (R-Me)

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

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)

Cyclopropanes

Epoxides

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

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

C=CC=CC=C

CH2-S-

C-XCH-CCH-O-

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

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)]

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

UnsaturationNumber(UN)

Working 2D

structures

List of working

2D structures

Reasonable3D molecular

structure

New 2Dmolecularstructure

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

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

Dereplication

TestosteroneC19H28O2

m/z 288.2089

Accurate massm/ z 288.2080 ( 3 ppm)

Molecular formulaC19H28O2

DEPT-135 FormulaC19H27

Low Resolution MSm/z 288

2D NMR

1H NMR

13C NMR

Other Search ParametersUVmax

Crystal unit cell dimensionsMS isotope cluster

Taxonomy

Determining the Molecular Formula Using NMR and MS Data

CH, CH3

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. .

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)

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

Chemical Formula: C9H11NO2Exact Mass: 165.0790

[M+H]+

Chemical Formula: C9H12NO2+

Exact Mass: 166.0863

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

Chemical Formula: C9H11NO2Exact Mass: 165.0790

[M+Na]+

Chemical Formula: C9H11NNaO22+

Exact Mass: 188.0677

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

Chemical Formula: C8H8ClNO2Exact Mass: 185.0244

Analysis of isotope patterns

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

• 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

23.820.9

The NMR effect

I = 1/2E

Bo off Bo on

I = -1/2(, antiparallel)

I = -1/2(, parallel)

HbNo coupling

Coupling

Spin-spin coupling (splitting)

Origin of spin-spin coupling

parallel antiparallel

Coupling in ethanol

Coupling is mutualHb

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)

5.15 4.00

10.4 Hz

1.8 Hz

5.2 Hz1.8 Hz

1.8 Hz

5.1155.1205.1255.1305.1355.1405.1455.1505.1555.1605.1655.1705.1755.1805.185

A doubled quartet (dq)

What about this?

5.8105.8255.8405.8555.8705.8855.9005.915

A (ddt)5.86

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

1Ph OH

Homotopic, Enantiotopic, Diastereotopic

Methyl groups

Chemical Equivalence/Magnetic Non Equivalence

What is going on?

ResultH

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

Glucose

-173.0o 9.50 Hz

OH 52.7o 3.51 Hz

169.4o 9.5 Hz

-170.5o 9.13 Hz

HO COH

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.

2.806.80

Solution

1.52.02.53.03.54.04.55.05.5 ppm

Removing CouplingsChanging Solvents

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

CDCl3irradiated

Irradiate at 4.11 ppm

Signal due to Hadisappears

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

Spin Decoupling

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

A↑ X↑ ↓

A↓ X ↑ ↓

Average of X ↑ and X ↓

Nuclear Overhauser Effect

Size of NOE

0.01 0.1 1 10 100w

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

pinanediol (2)

2426283032343638404244464850525456586062646668707274

3 10/5/9

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 ↓

●●●

●●

●●●

●●

13C – 1H NOE irradiation on H left ON

C↑H ↑

C↓H↓

C↓H ↑

C↑H ↓

●●●

●●

●●●

●●

13C – 1H NOE result

C↑H ↑

C↓H↓

C↑H ↓

●●●●

1H-1H NOE example

1.01.52.02.53.03.54.04.55.05.56.0

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

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

nOe nOe

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

S↑I ↑

S↓I↓

S↓I ↑S↑I ↓

●●●●

1H – 1H NOE irradiation on S

S↑I ↑

S↓I↓

S↓I ↑S↑I ↓

●●●

●●●●

W1S (sat)

1H – 1H NOE irradiation on S left ON

S↑I ↑

S↓I↓

S↑I ↓

●●●

●●●●

W1S (sat)

1H – 1H NOE result

S↑I ↑

S↓I↓

S↑I ↓

●●●½

●●●½ ½

W1S (sat)

NOE 3D example

Random orientation of magnetic dipoles

(a) No Bo

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

(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

Excess of spinpopulation alongthe direction ofapplied magneticfield.

(90o)x

After 90o pulse

magnetization

is tipped into

the xy plane.

time t2

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

frequency f2

preparation detection

ONE-PULSE SEQUENCE

(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)

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

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

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

Nitrocompound

Nitrosocompound Ar-NO S-NON-NO

*Nitrometane as reference (380 ppm)

Pyrimidine

Pyridine

Pyrazine

Imidazole

InternalTerminal

N=N-N ImineAmineHidrazone, tr iazene

ImineAmine

31P NMR Shifts

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

R3P=CR2

Z-P=P-Z

Z2N-P=NRArP=C=Z2

X =P(Y)=Z

R3-P-RR2P-PR2

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

ArP=C=Ar

ArP=C=O

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

TMS-C P

Cl(OR)2P=O

R2(OR)3P

*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

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

8 7 6 5 4 3 2 ppm

Effect of 31P on 1H NMR

4.30 4.25 4.20 4.15 4.10 4.05 ppm

Effect of 31P on 13C NMR

60 50 40 30 20 ppm

1 2 34

The 2nd Dimension

BASIC LAYOUT OF A 2D NMR EXPERIMENT

How a 2D NMR experiment works

transformmatrix

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

HHSQCHSQC &

1H-1H COSY

HSQC &H

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

CH, CH3

CHCH CH

CH3 CH3

STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA

A B C D E F C

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

E 23 CH3 1.5

F 17 CH3 1.2

diastereotopic protons

STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA

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

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

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

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:

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:

Combinatorialexplosion

STRATEGY BASED ON C-H CONNECTIVITYRETROSPECTIVE CHECKING

Hd Hd'

And similarly for COSY data

PROSPECTIVE CHECKING

Pieces: A

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

Reason: only 2 sp2 C's

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

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

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

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

nmr slide

dereplication slide

solution slide

cahbocndxecahbocndxe

structure of organic

c nmr spectrum

multinuclear nmr

d nmr experiments

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