nmrsymetry spin system(1)

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    NMR

    NuclearMagnetic Resonance

    ProtonNMR:

    Symmetry

    Index NMR-basics H-NMR

    http://localhost/var/www/apps/Chem-805-2003/chem805-index.ppthttp://localhost/var/www/apps/Chem-805-2003/NMR-intro.ppthttp://localhost/var/www/apps/Chem-805-2003/NMR-H.ppthttp://localhost/var/www/apps/Chem-805-2003/NMR-H.ppthttp://localhost/var/www/apps/Chem-805-2003/NMR-intro.ppthttp://localhost/var/www/apps/Chem-805-2003/chem805-index.ppt
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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

    Aromatic substituent pattern

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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

    Aromatic substituent pattern

    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

    NEXT

    http://localhost/var/www/apps/Chem-805-2003/NMR-het.ppthttp://localhost/var/www/apps/Chem-805-2003/NMR-het.ppthttp://localhost/var/www/apps/conversion/NMR-het.ppthttp://localhost/var/www/apps/Chem-805-2003/NMRsymetry.ppthttp://localhost/var/www/apps/Chem-805-2003/NMR-H.ppthttp://localhost/var/www/apps/Chem-805-2003/NMR-intro.ppthttp://localhost/var/www/apps/Chem-805-2003/chem805-index.ppt