nmrsymetry spin system(1)

8/10/2019 NMRsymetry Spin System(1)

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NMR

NuclearMagnetic Resonance

ProtonNMR:

Symmetry

Index NMR-basics H-NMR
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Homotopic protons: A2spin System

Chemical shift equivalence : Isochronous nuclei

These nuclei are interchangeable by symmetry or rapid exchange

H

Cl

H

Cl

Protons are equivalent in chiral and achiralenvironment


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Homotopic protons examples

CH2Cl2

A2

CH2CF2

A2X2

H

H Cl

ClCl

Cl

A4


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Enantiotopic ProtonsPlane of symmetry

Enantiotopic protonsare

Equivalent in Achiral environment (like CDCl3)

Non-equivalent in Chiral environment (optically active solvent)

A2X A2X3

H

F

H

Cl

H

Me

H

COOH

6.0

d=6.15

J = 53.6 Hz

2.5 2.0 1.5 1.0


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A2X2spin system??

H1

H2

F3

F4

AAXX

H1& H2=> same shift : chemically equivalent (homotopic)

3

JH1-F4=>cis

3JH2-F4=>trans

The two protonsare coupled to the same nuclei with different coupling!

Magnetic Non-Equivalence


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Magnetic Equivalence

H-1

F

F

3-H

Cl

H-4

2-H

Cl

The 2 Hgeminal to Fluorineare enantiotopic

The 2 Hgeminal to Chlorine are enantiotopic

The 2 H(1,3) and2 H(2,4)are chemically equivalent

Is this an A2M2X2spin system?

These protons are chemically equivalentbut are magnetically non-equivalentbecause they have different couplings with neighbors

JH1-H2 JH1-H4

AAMMXX spin system


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No symmetry: Asymmetric center

C

O

OH CCH

CH3

OH

H H

ABX

Protons A and B have different shifts: they are Diastereotopic

Accidental overlap can occur producing deceptively simple spin system

*

A B

H

H

COOH

H

Me

OH

H

H

COOH

H

Me

OH H

H

COOHH

Me

OH


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H1-NMR

C

O

OH CH2

CH

CH3

OH

4.0 3.5 3.0 2.5 2.0 1.5

4.20

2.70 2.60

1.20

7.0 6.5

OH

CH3

CH
http://localhost/var/www/apps/Chem-805-2003/NMR-H.ppt


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Dissymmetric center

HOOC

C

C

C

OH

HB1

HA1

COOH

HB2

HA2

H

Plane of symmetry

Enantiotopic groups

Diastereotopic protons

Ha1= Ha2 Hb1= Hb2

A2B2X


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AB AB

AB

X

X


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Me

CH CH2

Me

Me

H

HH H

COOH

HOOC

Me

Me

H

HH H

COOH

COOH

HA HB

Symmetrical : C2axisEnantiotopic protons

Symmetrical :s

planeDiastereotopic protons

Mixture of 2 isomers

2 dissymmetric centers in

symmetrical molecule


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Equivalence, non-equivalence and symmetry

OH

Cl

H2

Me1

Me2

H1

dH1 = dH2

d

Me1 =d

Me2d

H1

d

H2

OH

Cl

H2

Cl

Me2

H1

*

d

Me1

d

Me2

OH

Cl

Me1

Cl

CH3

Me2

*

O

O CH3

CH3

H2H1

H3

H4

Br

H5

H6

dH1 dH2 AB

dH3 dH4 AB

dH3=dH4 A2


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4.5 4.0 3.5 3.0 2.5 2.0 1.5

Example of dissymmetric spin system

ABX3CH3

O

O CH3

CH3

HH

H H

dA = 3.40 dB = 3.55

2JAB= 9.4 Hz

3JA-Me=3JB-Me= 7.0 Hz

3.60 3.50 3.40

dq


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Chemical Shift Non-Equivalence over a distance

Diastereotopic protons

COOH

CH3

H H

H H

NH2

CH3

*

AB

AB

2 doublets


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Magnetic Equivalence

O O

HB1

HA2

HB2

HA1

PhH

Enantiotopic protons:

Diastereototopic protons:

Magnetic equivalence

JA1-B1JA2-B1

A1and A2are

Magnetically different

AABB

HA1= HA2

HB1= HB2

HA1HB1

HA2HB2


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AABB


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AABB

N

Cl

HA'

HB'

HB

HA

2 sets of homotopic protons: magnetically non-equivalent


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AABB: para


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AABB: Ortho


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Spin System: Pople Notation

Each Chemical Shift is designated by a letter Dn-> Difference in Shift in Hz

J-> Coupling in Hz

If the ratio Dn/Jis Small(8),Lettersused to designate the shift are far

AM, AX This case give rise to first order type spectra Is is also refer to as weakly

coupled case


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Pople Notation

CH2 CHOR

OR

Y

A2X(if the shift difference of CH2and CHis largecompare to coupling).

A2B(if the shift difference of CH2and CHis smallcompare to coupling).

3 SpinsAMX-> if the 3 spins have large chemical shift differenceABX-> if 2 spins are close and 1 is far away

ABC-> if 3 spins are close

When nuclei have identical shift but different magnetic coupling,

prime symbol is used. For example:

AABB or AAXX


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H

H

H

AABBC

A : dd

A

M

X

JAX= Jcis= 10 Hz

JAM= Jtrans= 17 Hz

JMX = Jgem= 2 Hz

Jtrans

M X

Jcis

Vi l C li


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Virtual Coupling

First order

CH3O

2N

Virtual coupling

CH2-OH

CH

b

CH3broad

OH CH3

Same shift


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Virtual

Coupling

COOH

COOH

HB

HA

H

C

Me

HB

HA

A2B2CX3

ecause of the close shifts of ABCprotons we observe virtual coupling

Me broad doublet


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Virtual Coupling : Symmetrical chains

1) CO2MeCH2 CH2 CO2Me

2) CO2MeCH2 CH2 CH2 CO2Me

3) CO2MeCH2 CH2CH2 CH2 CO2Me

1 2 3 4

1 2 3 4 5

1 2 3 4 5

A2 A2 A4 Singlet

A2 X2 A2 A4X2 Triplet, Quintet

A2 X2 X2 A2 A2 A2 X2X2

Complex spectra

Same shift, different J with A/A

Virtual coupling


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Virtual Coupling

3) CO2MeCH2 CH2CH2 CH2 CO2Me

S i l M l l i h 2 hi l


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Symmetrical Molecules with 2 chiral centers

Ph

Ph

H

H

H1 H2Br

Br

1r, 3r; erythro

H1= H2

Enantiotopic protons

Magnetically non-equivalent

AAXX

Ph

Ph

H

H

H1 H2Br

Br

1r, 3s; Meso

H1= H2

diastereotopic protons

ABX2

Due to fast rotation, J is average

A2X2


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Chiral Centers in Symmetrical Molecules

H1 H1

H

H2 H2

OH

HOH

H3 H3

COOH

COOH

Meso: plane of symmetry

H1H1 diastereotopic

H3H3 diastereotopic

H2H2 diastereotopic

Group1 = Group3

H1 H1

H

H2 H2

OH

H OH

H3 H3

COOH

COOH

Erythro: axis of symmetry

H1H1 diastereotopic

H3H3 diastereotopic

H2 = H2 enantiotopic

Group1 = Group3


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Chiral Centers in polymers

Isotactic polymerAB

X

R

HA

HB

X

R

HA

HB

X

R

HA

HB

X

R

HA

HB

X

R

HA

HB X

R HA

HB

X

R

HA

HB X

R HA

HB

Syndiotactic polymer

A = B A2

C l l ti Shift f i l li h ti d


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Calculating Shifts for simple aliphatic compounds

d= 0.23 + SSi(d)

CH3Cl d(calc)=2.76

d(exp.)=3.1

CH2Cl2 d(calc)=5.29d(exp.)=5.3

CHCl3 d(calc)=7.82

d

(exp.)=7.27


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Calculating Shifts for aliphatic compounds

d= 0.933 + SSi(d)

e.g. CH3-CO-CO-CH3

1 2 3

Subst. Effect value

- C2-C3 +0.244=O(at C2) +1.021

=O(at C3) +0.004

SSi(d) +1.231

d= 0.933 + 1.231 = 2.164

-CR3(at C3) -0.038

Experimental = 2.23


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Calculating

Shifts for

olefinic

compounds

Calculating Shifts for olefinic compounds


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Calculating Shifts for olefinic compounds

CC

H1 H2

OEtPhH1 d= 5.23 + Phgem+ OEttrans

+ 1.35 + (-1.28)

d= 5.32

H2 d= 5.23 + Phtrans+ OEtgem

+ (-0.10) + 1.18

d= 6.33

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

Calculating Shifts for olefinic compounds (deciding which isomer)


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Calculating Shifts for olefinic compounds (deciding which isomer)

CN

COOEt

H

Experimental: 8.22 ppm

* Double bond is further conjugated

COOEt

CN

H

Z isomer effect

Base 5.23

Ph (gem) 1.43

CN (cis) 0.78

COORconj *(trans) 0.33

Total 7.77

E isomer effect

Base 5.23

Ph (gem) 1.43

CN (trans) 0.58COORconj * (cis) 1.02

Total 8.26

Which one ??

Calculating Shifts


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Calculating Shifts

for aromatic

compounds

A ti b tit ti tt th


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Aromatic substitution pattern: ortho

8.00 7.90 7.80 7.70 7.60 7.50 7.40

CH3

CH3

O

O

AA XX

Typical spectra for ortho (symmetrical)

Aromatic substituent pattern: para


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Aromatic substituent pattern: para

Aromatic substituent pattern


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Aromatic substituent patternCH3O

NO2

CH3

O

NO 2

8.5 8.0

t

J=8.1

dt

J=7.7, 1.5

ddd

J=8.1, 2.2, 1.1

tJ=1.8



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8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4

CH3

O

NO 2

CH3

O

NO2

dd

J=7.7, 1.5

dd

J=8.1, 0.7

tdJ=7.4, 1.1 ~td

J=8.1, 1.5


C H NO


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5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0

C5H9NO4

1H

2H

3H

3HCH3

CH2

t

O

q

C

O

CH

CH3

NO2

d

q

C H O


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9 8 7 6 5 4 3 2 1 0

3.732.451.161. 00 0.93

9.55 9.50 9.45

7.0 6.9

6.15 6.10 6.05 2.45 2.40 2.35

1.20 1.15 1.10

C5H8O

CHO, d

J=8.1 Hz

dt,

J=15.8, 6.9

CHO

CH

ddt,

J=15.8, 8.1, 1.5

t,

J=7.4

CH

CH2 CH3

CH2

Trans J

C H O


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7 6 5 4 3 2 1 0

3.092.141.00

7.3 7.2

4.9 4.8 4.7 4.6

2Jgem= 1.5

C4H6O2 I = CH/2 + 1 = 2

CH3

C

s

CH=

dd

14.0, 6.6

O

O

CC

HH

H

dd

14.0, 1.5

Jtrans = 14.0

Jcis = 6.6

dd

6.6, 1.5


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N+

O-

O


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9.0 8.5 8.0 7.5

80 MHz

8.5 8.0 7.5

400 MHz

N

N

+O

-

O

S

N

3

3, d(2.4) 55, dd

9.2, 2.4

6

6, d(9.2)3

3, ddd

5.0, 1.5, 0.9

6

6, dt

7.9, 1.0

5

5, ddd

7.9, 7.4, 1.6

4

4, ddd

7.4, 5.0 , 1.0


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Proton and Heteronuclear NMR

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