1. NMR Nuclear Magnetic Resonance
Proton NMR:
Symmetry
Index
NMR-basics
H-NMR

3. Homotopic protons: A2 spin System
Chemical shift equivalence : Isochronous nuclei
These nuclei are interchangeable by symmetry or rapid exchange
Protons are equivalent in chiral and achiral environment

4. Homotopic protons examples
CH2Cl2
CH2CF2
A2X2 A2 A4

5. Enantiotopic Protons Plane of symmetry A2X3 A2X d =6.15 J = 53.6 Hz
Enantiotopic protons are
Equivalent in Achiral environment (like CDCl3)
Non-equivalent in Chiral environment (optically active solvent)

6. A2X2 spin system??
AA’XX’
H1 & H2 => same shift : chemically equivalent (homotopic)
3JH1-F4 => cis
3JH2-F4 => trans
The two protons are coupled to the same nuclei with different coupling!
Magnetic Non-Equivalence

7. Is this an A2M2X2 spin system?
Magnetic Equivalence
The 2 H geminal to Fluorine are enantiotopic
The 2 H geminal to Chlorine are enantiotopic
The 2 H(1,3) and 2 H(2,4) are chemically equivalent
Is this an A2M2X2 spin system?
These protons are chemically equivalent but are magnetically non-equivalent because they have different couplings with neighbors
JH1-H2
JH1-H4 
AA’MM’XX spin system

8. No symmetry: Asymmetric center*
ABX
A B
Protons A and B have different shifts: they are Diastereotopic
Accidental overlap can occur producing deceptively simple spin system

9. H1-NMR OH
CH3 CH

10. Dissymmetric center  Plane of symmetry Enantiotopic groups Ha1 = Ha2
Hb1 = Hb2
Diastereotopic protons
A2B2X

11. X AB AB AB X

12. 2 dissymmetric centers in symmetrical moleculeH
COOH
HOOC Me H
COOH Me
Symmetrical : C2 axis
Enantiotopic protons
Symmetrical : s plane
Diastereotopic protons CH
CH2 HA HB
Mixture of 2 isomers

13. Equivalence, non-equivalence and symmetry
d H1 = d H2
d Me1 = d Me2 * *
d H1  d H2
d Me1  d Me2
d H1  d H2 AB
d H3  d H4 AB
d H3 = d H4 A2

14. Example of dissymmetric spin system
d A = 3.40
d B = 3.55
2JAB = 9.4 Hz
ABX3
3JA-Me = 3JB-Me = 7.0 Hz dq

15. Chemical Shift Non-Equivalence over a distance
Diastereotopic protons * AB AB
2 doublets

16. Magnetically different
Magnetic Equivalence
Magnetic equivalence
HA1 = HA2
HB1 = HB2
Enantiotopic protons:
JA1-B1  JA2-B1
A1 and A2 are
Magnetically different
Diastereototopic protons:
HA1  HB1
HA2  HB2
AA’BB’

17. AA’BB’

18. AA’BB’
2 sets of homotopic protons : magnetically non-equivalent

19. AA’BB’: para

20. AA’BB’: Ortho

21. 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/J is Small (<8), Letters used to designate the shift are close AB, ABC … This represent case of second order spectra: These spectra must be simulated with the help of quantum mechanic equations. Such programs are available on Nuts or Mestrec or Spectrometer software. This case is also called strongly coupled
If the ratio Dn/J is large (>8), Letters used 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

22. Pople Notation
A2X (if the shift difference of CH2 and CH is large compare to coupling).
A2B (if the shift difference of CH2 and CH is small compare to coupling).
3 Spins
AMX -> if the 3 spins have large chemical shift difference
ABX -> 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:
AA’BB’ or AA’XX’

23. A M X
Jtrans M X
Jcis
AA’BB’C
JAX = Jcis = 10 Hz
JAM = Jtrans = 17 Hz
JMX = Jgem = 2 Hz
A : dd

24. Virtual Coupling Virtual coupling First order Same shift CH2-OH
CH3 broad
CH2b

25. Virtual Coupling Me broad doublet A2B2CX3
Because of the close shifts of ABC protons we observe “virtual coupling”

26. Virtual Coupling : Symmetrical chains
1) CO2Me – CH2 – CH2 – CO2Me
A A A Singlet
2) CO2Me – CH2 – CH2 – CH2 – CO2Me
A X A A4X Triplet, Quintet
3) CO2Me – CH2 – CH2 – CH2 – CH2 – CO2Me
A X X A A2 A2’ X2 X2’
Complex spectra
Same shift, different J with A/A’
Virtual coupling

27. Virtual Coupling
3) CO2Me – CH2 – CH2 – CH2 – CH2 – CO2Me

28. Symmetrical Molecules with 2 chiral centersPh H H1 H2 Br Ph H H1 H2 Br
1r, 3r; erythro
1r, 3s; Meso
H1 = H2
H1 = H2
diastereotopic protons
ABX2
Enantiotopic protons
Magnetically non-equivalent
AA’XX’
Due to fast rotation, J is average
A2X2

29. Chiral Centers in Symmetrical MoleculesOH H3
H3’
COOH H1 H H2
H2’ OH H3
H3’
COOH
Meso: plane of symmetry
Erythro: axis of symmetry
H1  H1’ diastereotopic
H1  H1’ diastereotopic
H3  H3’ diastereotopic
H3  H3’ diastereotopic
Group1 = Group3
Group1 = Group3
H2  H2’ diastereotopic
H2 = H2’ enantiotopic

30. Chiral Centers in polymersX R HA HB
Isotactic polymer AB X R HA HB
Syndiotactic polymer
A = B  A2

31. Calculating Shifts for simple aliphatic compounds
d = SSi(d)
CH3Cl d(calc)=2.76
d(exp.)=3.1
CH2Cl2 d(calc)=5.29
d(exp.)=5.3
CHCl3 d(calc)=7.82
d(exp.)=7.27

32. Calculating Shifts for aliphatic compounds
d = SSi(d)
e.g. CH3-CO-CO-CH3
Subst. Effect value
- C2-C
=O (at C2)
=O (at C3)
-CR3 (at C3)
SSi(d)
d = = 2.164
Experimental = 2.23

33. Calculating Shifts for olefinic compounds

34. Calculating Shifts for olefinic compounds
d = Phgem + OEttrans C H1 H2
OEt Ph
(-1.28)
d = 5.32 H2
d = Phtrans + OEtgem
+ (-0.10)
d = 6.33

35. Calculating Shifts for olefinic compounds (deciding which isomer)
Experimental: 8.22 ppm
Which one ??
Z isomer effect
Base
Ph (gem) 1.43
CN (cis) 0.78
COORconj *(trans) 0.33
Total
E isomer effect
Base
Ph (gem) 1.43
CN (trans) 0.58
COORconj * (cis) 1.02
Total
* Double bond is further conjugated

36. Calculating Shifts for aromatic compounds

37. Aromatic substitution pattern: ortho
AA’ XX’
Typical spectra for ortho (symmetrical)

38. Aromatic substituent pattern: para

39. Aromatic substituent pattern
J=8.1 t
J=1.8 dt
J=7.7, 1.5
ddd
J=8.1, 2.2, 1.1

40. Aromatic substituent patterntd
J=7.4, 1.1 dd
J=8.1, 0.7
~td
J=8.1, 1.5 dd
J=7.7, 1.5

41. C5H9NO4
NO2 t 1H 2H 3H d
CH3 CH O C
CH3
CH2 O q q

42. C5H8O CH3 CH2 CH CH CHO Trans J CHO, d J=8.1 Hz CH2 t, J=7.4 ddt,

43. C4H6O2 I = C – H/2 + 1 = 2 H O CH3 s C C C H H O dd 6.6, 1.5 dd
Jtrans = 14.0
Jcis = 6.6 dd
14.0, 1.5
2Jgem = 1.5 dd
14.0, 6.6
CH=

45. 6’ 3
400 MHz 5’ 6 4’
6, d
(9.2)
3, d(2.4) 5 3’
5, dd
9.2, 2.4
6’, dt
7.9, 1.0
3’, ddd
5.0, 1.5, 0.9
5’, ddd
7.9, 7.4, 1.6
4’, ddd
7.4, 5.0 , 1.0
80 MHz

46. Proton and Heteronuclear NMR
NEXT
Proton and Heteronuclear NMR
Index
NMR-basics
H-NMR
NMR-Symmetry
Heteronuclear-NMR