NMR Nuclear Magnetic Resonance M Proton NMR Symmetry

NMR Nuclear Magnetic Resonance M Proton NMR: Symmetry Index NMR-basics H-NMR

Protons and other nuclei in NMR spectra can be classified as heterotopic, diastereotopic, enantiotopic and homotopic. Heterotopic and diastereotopic protons will have different chemical shifts and couplings to neighboring magnetic nuclei, enantiotopic and homotopic protons will have identical chemical shifts. They may or may not have identical couplings to other nuclei. Distinction can be made by the substitution test.

Chemical shift equivalence If a set of nuclei exist in identical environments, they are expected to have the same chemical shift. Such nuclei are called chemical shift equivalent or chemically equivalent. A pair of nuclei in a molecule are chemically equivalent if they are interchangeable through any symmetry operation of the molecule OR if they interchange by a rapid process (rapid with respect to the NMR timescale). If a pair of nuclei can be interchanged by rotation about an axis of symmetry of the molecule then they are chemically equivalent and are called homotopic. e. g. the pair of protons in dichloromethane are chemically equivalent.

Magnetic Equivalence (spin coupling equivalence) In a spin system, when two equivalent nuclei have identical relations with the same identical partners, they are “magnetically equivalent”. Only in this case it's possible to define them as a group and not individually. Two nuclei are magnetically equivalent when they have: 1 - The same chemical shift 2 - The same coupling constants. . . 3 - With the same partners !

At left, H 4 and H 6 are magnetically equivalent, because they are both coupled with H 5 and with no other. At right, H 2 and H 6 are related by symmetry, therefore they have the same chemical shift and the same coupling constants. Their partners, however, are different: H 2 is coupled with H 3, while H 6 is not (or not with the same intensity). In conclusion. H 2 and H 6 are NOT magnetically equivalent and must be declared separately.

On the other hand, the two pairs of equivalent protons in trans-dichlorocyclopropane are NOT magnetically equivalent, because each of the A protons is coupled differently to the two X protons (one is a trans coupling, the other a cis). In the Pople nomenclature, such magnetically inequivalent nuclei are given an AA' designation. Thus the dichlorocyclopropane is referred to as an AA'XX' system, where A and A' refer to protons that are symmetry equivalent but not magnetically equivalent. The spectrum will be much more complicated than two triplets.

In general any system which contains chemical shift equivalent but magnetically inequivalent nuclei of the AA' type will not give first order splitting patterns, although sometimes the spectra may appear to be first order ("deceptively simple" spectra). For example, X-CH 2 -Y systems are of the AA'XX' type, but the coupling constants JAX and JAX' are often close enough in size that apparent triplets are seen for each CH 2 group.

Two important generalizations: Coupling between symmetry equivalent but magnetically inequivalent nuclei typically will affect the appearance of the NMR spectrum. In fact, it is the coupling between the equivalent nuclei that is responsible for the complexity of spectra of the AA'BB'X. . type. Coupling between magnetically equivalent nuclei does not affect NMR spectra, cannot be detected, and thus can be ignored.

The Substitution Test for Equivalence of Protons For a pair of protons to be tested, replace one and then the other with another group (one not present in the molecule). Compare the two structures formed. If they are identical, the protons are homotopic, if they are enantiomers, the protons are enantiotopic, if they are diastereomers then the protons are diastereotopic, if they are structural isomers, the protons are heterotopic Homotopic Protons:

Enantiotopic Protons: Enantiotopic protons normally have identical chemical shifts. However, when the molecule is placed in a chiral environment (say with an optically active solvent, cosolvent or Lewis acid) then the protons can become diastereotopic. This is in contrast to homotopic protons, which are always identical.

Diastereotopic Protons: The concept of diastereotopicity was first introduced during the early days of NMR spectroscopy, when certain kinds of molecules gave unexpectedly complex NMR spectra, leading to some confusion about the origins of this hitherto undetected phenomenon. A typical situation where diastereotopic protons are seen is a CH 2 group in a chiral molecule (one with an asymmetric center, or other types of asymmetry).

A more subtle form of diastereotopism is demonstrated in the classical example of diethyl acetal below. Even though diethyl acetal has no asymmetric centers, the CH 2 group is diastereotopic. This can be shown by applying the substitution test, which creates a pair of diastereomers G and H. Thus the ethyl group forms an ABX 3 pattern. The key to understanding this type of diastereotopicity is that the molecule has a plane of symmetry (hence is achiral). However, there is no plane of symmetry that bisects the CH 2 protons, so they are nonequivalent.

The dibromocyclopropane spectrum illustrates this effect in a different context - the protons of the CH 2 Cl group are diastereotopic. However, the protons of the cyclopropane CH 2 group are not, since they are related by a plane of symmetry.

Not all CH 2 groups in chiral molecules are diastereotopic - in the following chiral molecules the CH 2 is on a C 2 axis of symmetry, and the protons are homotopic. In general, CH 2 groups (or other similar groups like CHMe 2, CHF 2, etc) will be diastereotopic when part of chiral molecules unless the CH 2 group is on a C 2 rotation axis (as in the molecules below).

Heterotopic Protons: The NMR Time Scale It is important to recognize that diastereotopic and magnetic equivalence effects are subject to the time scale of the NMR experiment, which is on the order of tenths of a second. Flexible molecules will often have several conformations, some of which may have lower symmetry than others. However, since these conformations are typically interconverting rapidly, the observed symmetry in the NMR spectrum will be that of the most symmetric conformation reachable. Thus cyclohexane is a sharp singlet at room temperature, whereas at -100 °C the ring inversion is slow on the NMR time scale, and a much more complex spectrum results.

Pople Nomenclature for Coupled Spin Systems The analysis of complex NMR patterns is assisted by a general labeling method for spin systems introduced by Pople. Each set of chemically equivalent protons (or other nuclei) is designated by a letter of the alphabet. Nuclei are labeled AX or AMX if their chemical shift differences are large compared to the coupling between them (Δδ >> 5 J). Nuclei are labeled with adjacent letters of the alphabet (AB, ABC, MN or XYZ) if they are close in chemical shift compared to the coupling between them (i. e. if they are strongly coupled).

If groups of nuclei are magnetically equivalent, they are labeled An. Bn, etc. Thus CH 3 groups are A 3, or X 3. A group of magnetically equivalent nuclei must have identical chemical shifts, and all members of the group must be coupled equally to nuclei outside the group. If nuclei are chemical shift equivalent but not magnetically equivalent, then they are labeled AA', BB'B'' or XX'. Thus in an A 2 X 2 system the A nucleus must have identical couplings to the two X nuclei. In an AA'XX' system, on the other hand, JAX ≠ JAX'. There are usually profound differences in the appearance of A 2 X 2 compared to AA'XX' patterns.

Homotopic protons: protons A 2 spin System Chemical shift equivalence : Isochronous nuclei These nuclei are interchangeable by symmetry or rapid exchange Protons are equivalent in chiral and achiral environment

Homotopic protons examples CH 2 Cl 2 A 2 CH 2 F 2 A 2 X 2 A 4

Enantiotopic Protons Plane of symmetry A 2 X 3 =6. 15 J = 53. 6 Hz Enantiotopic protons are protons Equivalent in Achiral environment (like CDCl 3) Non-equivalent in Chiral environment (optically active solvent)

A 2 X 2 spin system? ? AA’XX’ H 1 & H 2 => same shift : chemically equivalent (homotopic) 3 J H 1 -F 4 => cis 3 J H 2 -F 4 => trans The two protons are coupled to the same nuclei with different coupling! two protons different coupling Magnetic Non-Equivalence

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 H(2, 4) Is this an A 2 M 2 X 2 spin system? These protons are chemically equivalent but are magnetically nonequivalent because they have different couplings with neighbors JH 1 -H 2 JH 1 -H 4 AA’MM’XX spin system

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

H 1 -NMR OH CH CH 3

Dissymmetric center Plane of symmetry Enantiotopic groups Ha 1 = Ha 2 Hb 1 = Hb 2 Diastereotopic protons A 2 B 2 X

X AB AB X AB

2 dissymmetric centers in symmetrical molecule Me H H HOOC H H Me H COOH Me H H H COOH Me Symmetrical : s plane Diastereotopic protons Symmetrical : C 2 axis Enantiotopic protons CH CH 2 HA Mixture of 2 isomers HB

Equivalence, non-equivalence and symmetry * * d H 1 = d H 2 d Me 1 = d Me 2 d H 1 d H 2 d Me 1 d Me 2 d H 1 d H 2 AB d H 3 d H 4 AB d H 5 = d H 6 A 2

Example of dissymmetric spin system d A = 3. 40 2 J AB AB X 3 3 J A-Me d B = 3. 55 = 9. 4 Hz = 3 JB-Me = 7. 0 Hz dq

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

Magnetic Equivalence Magnetic equivalence Enantiotopic protons: HA 1 = HA 2 HB 1 = HB 2 Diastereototopic protons: HA 1 HB 1 HA 2 HB 2 JA 1 -B 1 JA 2 -B 1 A 1 and A 2 are Magnetically different AA’BB’

AA’BB’ Spin. Works => load AA’BB’

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

AA’BB’: para Spin. Works => load AA’BB’-para

AA’BB’: Ortho Spin. Works => load AA’BB’

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 Spin. Works 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 also refer to as weakly coupled case • •

Pople Notation A 2 X (if the shift difference of CH 2 and CH is large CH compare to coupling). A 2 B (if the shift difference of CH 2 and CH is small CH compare to coupling). 3 Spins AMX -> AMX if the 3 spins have large chemical shift difference ABX -> ABX if 2 spins are close and 1 is far away ABC -> 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’

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

Virtual Coupling Virtual coupling First order Same shift CH 2 -OH CH 3 broad CH 2 b

Virtual Coupling Me broad doublet A 2 B 2 CX 3 Because of the close shifts of ABC protons we observe “virtual coupling” coupling

Virtual Coupling : Symmetrical chains 1) 2) 3) 1 2 3 4 CO 2 Me – CH 2 – CO 2 Me A 2 A 4 1 2 3 4 5 CO 2 Me – CH 2 – CO 2 Me A 2 X 2 A 4 X 2 1 2 3 4 Singlet Triplet, Quintet 5 CO 2 Me – CH 2 – CO 2 Me A 2 X 2 A 2 A 2’ X 2’ Complex spectra Same shift, different J with A/A’ Virtual coupling

Virtual Coupling 3) CO 2 Me – CH 2 – CO 2 Me

Symmetrical Molecules with 2 chiral centers Ph Ph Br H H 2 H 1 H Br Ph 1 r, 3 r; erythro H 1 = H 2 Enantiotopic protons Magnetically non-equivalent AA’XX’ Due to fast rotation, J is average A 2 X 2 Br H H 1 H H 2 Br Ph 1 r, 3 s; Meso H 1 = H 2 diastereotopic protons ABX 2

Chiral Centers in Symmetrical Molecules COOH H 1 OH H H 2 ’ OH H H 3 ’ H 3 AB H 1 OH H AB H 2 ’ AB H H 3 OH H 3 ’ COOH Meso: plane of symmetry AB AA AB COOH Erythro: axis of symmetry H 1’ diastereotopic H 3’ diastereotopic Group 1 = Group 3 H 2’ diastereotopic H 2 = H 2’ enantiotopic

Chiral Centers in polymers X HA R HB X HA R HB X HB Isotactic polymer AB Syndiotactic polymer A = B A 2

Calculating Shifts for simple aliphatic compounds d = 0. 23 + SSi(d) CH 3 Cl d(calc)=2. 76 d(exp. )=3. 1 CH 2 Cl 2 d(calc)=5. 29 d(exp. )=5. 3 CHCl 3 d(calc)=7. 82 d(exp. )=7. 27

Calculating Shifts for aliphatic compounds d = 0. 933 + SSi(d) 1 2 3 e. g. CH 3 -CO-CO-CH 3 Subst. Effect - C 2 -C 3 =O (at C 2) =O (at C 3) -CR 3 (at C 3) value +0. 244 +1. 021 +0. 004 -0. 038 S S i (d ) +1. 231 d = 0. 933 + 1. 231 = 2. 164 Experimental = 2. 23

Calculating Shifts for olefinic compounds

Calculating Shifts for olefinic compounds Ph OEt C H 1 C d = 5. 23 + Phgem + OEttrans d = 5. 32 H 2 + 1. 35 + (-1. 28) d = 5. 23 + Phtrans + OEtgem d = 6. 33 + (-0. 10) + 1. 18

Calculating Shifts for olefinic compounds (deciding which isomer) Experimental: 8. 22 ppm Which one ? ? Z isomer Base Ph (gem) CN (cis) COORconj *(trans) effect 5. 23 1. 43 0. 78 0. 33 E isomer Base Ph (gem) CN (trans) COORconj * (cis) effect 5. 23 1. 43 0. 58 1. 02 Total 7. 77 Total 8. 26 * Double bond is further conjugated

Calculating Shifts for aromatic compounds

meta bromo nitro benzene Calculated shifts HA=8. 44 HB=7. 82 HC=7. 31 HD=8. 19 HA HD HB HC

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

Aromatic substituent pattern: para

Aromatic substituent pattern 4 1 3 t J=1. 8 2 dt J=7. 7, 1. 5 ddd J=8. 1, 2. 2, 1. 1 Calculated shifts 1 = 8. 86 2 = 8. 52 3 = 7. 53 4 = 8. 24 t J=8. 1

Aromatic substituent pattern Calculated shifts 1 = 8. 31 2 = 7. 74 1 = 7. 69 1 = 8. 08 dd J=8. 1, 0. 7 td J=7. 4, 1. 1 1 ~td J=8. 1, 1. 5 dd J=7. 7, 1. 5

NO 2 C 5 H 9 NO 4 CH 3 C O q 2 H q 1 H CH 3 O CH CH 2 d 3 H t 3 H

C 5 H 8 O I = 5 – 8/2 + 1 = 2 CH CHO, d J=8. 1 Hz ddt, J=15. 8, 8. 1, 1. 5 dt, J=15. 8, 6. 9 CH Trans J CH 2 CH 3 t, J=7. 4

C 4 H 6 O 2 H CH 3 O C C H Jtrans = 14. 0 Jcis = 6. 6 dd 14. 0, 6. 6 CH= C H O dd 6. 6, 1. 5 dd 14. 0, 1. 5 2 J gem = 1. 5 I = C – H/2 + 1 = 2 s

6’ 3 400 MHz 5’ 4’ 6 3, d(2. 4) 3’, ddd 5. 0, 1. 5, 0. 9 80 MHz 5, dd 9. 2, 2. 4 5 5’, ddd 7. 9, 7. 4, 1. 6 3’ 6’, dt 7. 9, 1. 0 6, d (9. 2) 4’, ddd 7. 4, 5. 0 , 1. 0

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