Course Notes NMR 1 (first NMR lecture)shimizu-uofsc.net/Ken_D._Shimizu_Group_Website/Group... · 2018-07-28 · Purity. Evidence for documenting compound purity should include one

Day 1

NMR Spectroscopy from Organic Chemists’ POV

i.e. Why do we care?

1) Identify compound (or verify its structure)

2) Verify purity

3) Characterize Physical Propertiesa) polarityb) conformationc) recognition propertiesd) electronic properties (conductivity, color, fluorescence etc)

To Publish Paper(s)!!

Purity. Evidence for documenting compound purity should include one or more of the following:• A standard 1D proton NMR spectrum or proton-decoupled carbon NMR spectrum showing at most trace peaks not attributable to the assigned structure. A copy of a spectrum with a signal-to-noise ratio sufficient to permit seeing peaks with 5% of the intensity of the strongest peak should be included in the supporting information. The normal full range of chemical shifts should be displayed (usually 0–10 ppm for proton; 0–220 ppm for carbon). For new compounds, copies of both proton and carbon spectra are required (see ‘Identity’ above).• Combustion elemental analytical values for carbon and hydrogen (and nitrogen, if present) agreeing with calculated values within 0.4%.• Quantitative gas chromatographic analytical data for distilled or vacuum-transferred samples, or quantitative HPLC analytical data for materials isolated by column chromatography or separation from a solid support. The stationary phase, solvent (HPLC), detector type, and percentage of total chromatogram integration represented by the product peak should be reported. Alternatively, a copy of the chromatogram may be included in the supporting information.• Electrophoretic analytical data obtained under conditions that permit observing impurities present at the 5% level.• For known solid compounds, a narrow melting point range that is in close agreement with a cited literature value.The type of evidence appropriate for demonstrating a compound’s purity will necessarily depend on the method of preparation, the compound’s air and thermal stability, the complexity of its structure, the nature of reasonably likely impurities, and the amount of sample available. For example, combustion analysis would not be an appropriate choice for the product of an isomerization reaction; a proton NMR spectrum would need to be supplemented with other evidence when reasonably likely impurities (including unreacted starting materials) do not have unique resonances or are NMR-silent (for example, an inorganic salt). A narrow melting point range is not sufficient by itself to document the purity of a new compound. MS accurate mass (HRMS) data may be used to support a molecular formula assignment but cannot serve to document compound purity.

The Journal upholds a high standard for compound characterization to ensure that compounds being added to the chemical literature have been correctly identified and can be synthesized in known yield and purity by the reported preparation, isolation, and purification methods. For all new compounds, evidence adequate to establish both identity and degree of purity (homogeneity) must be provided. Purity documentation must be provided for known compounds whose preparation by a new or improved method is reported. For combinatorial libraries containing more than 20 compounds, complete characterization data must be provided for at least 20 diverse members. Authors may be asked to provide copies of original spectra or analytical reports if an editor or reviewer raises a question about any of the reported results.

Identity. Evidence for documenting the identity of new compounds should include both proton and carbon NMR data and either MS accurate mass (HRMS) or elemental analysis data.

J Org Chem (Author Guidelines)

PURITY COMPUTATIONAL DATA in SI*IDENTITY

New Known

Weight a

nd percen

tage y

ield

Physica

l stat

e / m

p range i

f crys

t. solid

IR UV-Vis

1H N

MR

13C N

MR

MS MS Accurat

e mas

s (HRMS)

Optical

rotat

ion/ORD/C

D

Enantio

meric/D

iaster

eomeri

c rati

o

X-ray

[ORTEP an

d CIF in

SI*]

<> Copy of 1

H/13C N

MR spec

trum in

SI*

Copy of c

hromato

gram in

SI*

Quant. G

C, HPLC, e

lectro

phoresis

Elemen

tal an

alysis

<> Cartes

ian co

ordinate

s or Z

-matr

ix

# of im

aginary

freq

uencie

s

Total

energ

y

Identify versus Purity

2

Information in 1D 1H NMR

Observable Feature in Spectra

1. number of peaks2. chemical shift (ppm)3. integration of peaks4. splitting pattern

(singlet, doublet etc.)5. coupling constants6. peak shape (sharp or broad)7. presence of other peaks

Structural or purity characteristics

1. number of different types of protons2. chemical environment of proton3. ratio of different types of protons4. number of adjacent protons of

different types5. relative orientation of adjacent protons6. exchangeable protons7. purity

2








CCHOH

HH

H

H

2








CCHOH

HH

H

H

2








CCHOH

HH

H

H

2








2

2

2

CCHOH

HH

H

H

triplettriplet doublet of quartet

2








2

2

2

CCHOH

HH

H

H

7.5 Hz5.0 Hz 5.0 and 7.5 Hz

2








2

2

2

CCHOH

HH

H

H

2

ALL structral information must match the proposed structure







2

2

2

CCHOH

HH

H

H

2








2

2

2

CCHOH

HH

H

H

PURITY COMPUTATIONAL DATA in SI*IDENTITY

New Known

Weight a

nd percen

tage y

ield

Physica

l stat

e / m

p range i

f crys

t. solid

IR UV-Vis

1H N

MR

13C N

MR

MS MS Accurat

e mas

s (HRMS)

Optical

rotat

ion/ORD/C

D

Enantio

meric/D

iaster

eomeri

c rati

o

X-ray

[ORTEP an

d CIF in

SI*]

<> Copy of 1

H/13C N

MR spec

trum in

SI*

Copy of c

hromato

gram in

SI*

Quant. G

C, HPLC, e

lectro

phoresis

Elemen

tal an

alysis

<> Cartes

ian co

ordinate

s or Z

-matr

ix

# of im

aginary

freq

uencie

s

Total

energ

y

Identify versus Purity

1. number of peaks = number of chemically inequivalent protons

Symmetry Conformational Equilibrium?

•look for plane, pointC and S symmetry

•is it in slow orfast exchange atthe temperature

Molecular andlocal chirality

OH

racemic

CH3

OH

OH

O

Br

Cl

H

H CC

C

Cl Br

CH3H3C

CH3H NN

Br

CH3

H

H

H

H

HCH3

H3C

H H

H

O

H NCH3

CH3

CH3

1. number of chemically equivalent protons

ab c

# of different types of protons

3

CommentsThere is a high degree of symmetry in this molecule so there are only 3 different types of protons. There is the concern that the axial and equatorial protons and methyl groups might be different. But due the arrangement of the methyl groups, one will always be axial and the other will be equatorial. This means that the molecule will flip between two equal chair conformations.

H

HH

H Ha a

bc

c'

4

The methylene (CH2) protons (c and c’) are different and are diastereotopic. They are often referred to as endo and exo. The reason is that the methylene bridge makes the top and bottom face of the cyclohexane ring different. This is common in bicyclic systems. Note: that the methylene (CH2) bridge protons (a) are chemically equivalent due to symmetry in this system. However, if there are substituents on the cyclohexane ring, the two Ha protons could also be diastereotopic.

CH3H

H

H

H

H

HCH3a

b

ab

c

cc'

c' 4 This molecule also has a high degree of symmetry. The two CH3 groups are the same due to C2 symmetry. The adjacent methyne protons (b) are also the same. Both methylene groups are diastereotopic (c and c’) but the methylene groups are the same due to the C2 symmetry.

NN

a

ba

b

c

c

3 This molecule also has C2 symmetry. So, the protons on each ring are the same.


3

Comments

All four methyl groups are chemically equivalent. The reason is that there is free rotation around the b-c single bond allowing the exchange of the up and down methyl groups.

5 All the aromatic protons in this molecule (a-d) are different. The three methyl group protons (e) are all the same due to free rotation of the methyl group single bond.

3

The two methyl groups appear to be chemically equivalent because it looks like they can exchange by rotation around the O=C-N single bond. However, the O=C-N of amides and formamides rotates slowly on the NMR timescale due to its partial double bond character (which is evident from the minor zwitterionic resonance structure.) This makes the two groups attached to amide and formamide nitrogens chemically inequivallent at room temperature. If the NMR sample is heated, then they become chemically equivalent due to faster rotation around the O=C-N bond.

1 or 2At room temperature, all of the protons on cyclohexane are the same. However, if cyclohexane is cooled then the ring flip between the two chair forms becomes slow on the NMR timescale. Under these low temperature conditions, the axial and equatorial protons become chemically inequivalent.

a a

aab b

c

Br

CH3

b

c

a

d e

O

H NCH3

CH3

O

H NCH3

CH3a

b

b'

H

H

axial

equatorialH

H

axial


4

Comments

The protons c and c’ are chemically inequivalent due to the lack of rotation around the C=C double bond.

3

The resonance structures seem to suggest that all of the methyl groups are different but remember that a structure is the average of all of its resonance structures. So the actual structure is highly symmetric so the top (a and a) and bottom (a’ and a’) pairs of methyl groups are the same. However, there is still sufficient double bond character to stop rotation, making the methyl groups a and a’ chemically inequivalent.

6Methylene groups (b and b’) adjacent to chiral centers will be diasteretopic (different). Notice that the protons of the methyl group (d) next to the chiral center do not become chemically inequivalent.

6

This question is really asking whether the NMR spectra of a chirally pure sample or a racemic sample will be the same. The answer is yes. The NMR of enantiomers are identical. Therefore, we cannot tell the difference between enantiomers by NMR. We can also not differentiate chirally pure and racemic samples by simple NMR spectra. (There are some more complex methods that can be used to differentiate enantiomers. We will talk about those later.)

ca

b

H3C

H H

H

c'

b

∂ +∂ +a a

a' a'


cd

e

ab

OH

b'

OH

racemic


5

Comments

Although this molecule does not have a chiral center, the protons (b and b’) adjacent to the central carbon become diastereotopic. The reason is that from the point of view of the individual methylene carbons (b and b’), the adjacent central carbon appears to be a chiral center (and behaves like a chiral center).

2Due to the presence of the ring, the methylene protons (a and a’) cannot interchange by rotation. Therefore, they will be diastereotopic because one is next to a Cl and the other is next to a Br.

1As drawn, the two methyl groups appear to be chemically inequivalent. However, the actual 3D structure of the molecule actually contains a plane of symmetry making the two methyl groups chemically equivalent.


OH

ab b' c

e

b b'a

O

Br

Cl

H

Ha

a'


CC

C

Cl Br

CH3H3C

CC

C

Cl Br

CH3H3C

Chemical Shift: Shielding

He-

He-

applied B0

B0

induced,

opposing B

Electron cloud partially negates (or “shields”

nucleus from) applied field.

2. Chemical Shift = local environment of proton

Proton Spectra (δ 10 to 0 ppm)

<-- (downfield) (upfield) --> Chemical shift differences are due to:

1) Electron withdrawing substituents that deshield the nuclei ~ electrons around nuclei shield the nucleus from induced field (B0)

Chemical Shift: Inductive Effects

H

e-

B0

induced,

opposing BElectronegative group

withdraws shielding

electrons, “deshields”

nucleus, shifts ! downfield.

H

Oe-

Chemical Shift

0.90

1.31

1.29

1.91

4.41

N+

O

O-

1.372.69

3.574.13

Chemical ShiftEWG effects are additive (to the first approximation)

• can estimate from tables (see appendix in book)

Chemical Shift

2. Proximity and orientation to local induced magnetic fields

~ consider the chemical shift of the series:0.86 5.25 1.80Chemical Shift: Conjugation and

Aromaticity

B0

Acetylene protons

are shielded (! =

1.5-3 ppm).

H

Chemical Shift: Conjugation and

Aromaticity

B0H

“Ring current”

reinforces applied

B0 field; deshields

aromatic protons

(! = 6.5-8.5 ppm).

H

H

H

H

Induced current is

less effective in

olefins, deshields

less (! = 4-6 ppm).Chemical Shift: Conjugation and

Aromaticity

B0H

“Ring current”

reinforces applied

B0 field; deshields

aromatic protons

(! = 6.5-8.5 ppm).

H

H

H

H

Induced current is

less effective in

olefins, deshields

less (! = 4-6 ppm).

alkene alkyne

benzene

cyclohexane

equatorial0.5 ppmmore downfield

H

H

Job/Unit: O08663 /KAP1 Date: 27-08-08 14:58:02 Pages: 8

FULL PAPER

DOI: 10.1002/ejoc.200800663

Intramolecular !-Stacking in Isostructural Conformational Probes DependsStrongly on Charge Separation, a Proton NMR Study

Pramod Prasad Poudel,[a] Jing Chen,[b] and Arthur Cammers*[a]

Keywords: !-Stacking / Solution state / Crystalline state / Conformation / NMR anisotropy

Our solution state conformational methods used previouslyto study a dicationic molecular template for intramoleculararomatic association were applied to a neutral hydrocarbonanalogue to probe the effect of charge on conformation. Con-formational analysis of the hydrocarbon revealed modest sol-vent dependence in largely unfolded molecules. Conforma-tions found in the solid state were unfolded also corroborat-ing the findings of the solution-state study. This study alsoadds solid-state evidence for three competing solution-state

Introduction

In an approach encapsulated by the early ideas of Kara-batsos,[1] and in work by Fukazawa,[2–7] and others,[8–12] acalculation-based/1H NMR-based, multi-state mathemati-cal model for the solution-phase conformation of the dicat-ions 1a–c was devised.[10–12] The steps involved in these con-formational analyses were: 1) molecular modeling and X-ray crystallography to locate candidate conformational en-ergy minima, and grouping these conformations intoclasses, 2) ab initio calculations of the effect of anisotropyon key chemical shifts in these conformations. Figure 1 dis-plays the calculated anisotropic shielding tensor of benzenewith shielding and deshielding regions at the !-face andedge respectively. 3) experimentally measuring the chemicalshift difference between key, analogous chemical shifts ofthe template molecule and an ideal reference molecule, 4)solving a system of equations designed to output the con-formational distribution given the computed chemicalshifts, the experimental chemical shifts and mass balance,5) checking the stability of the mathematical model bygrouping conformations and omitting the expression formass balance under different conditions (temperature andsolvent) to look for negative concentrations of conformersin the output of the algorithm.

[a] University of Kentucky, Department of Chemistry,Chemistry-Physics Building, 505 Rose St., Lexington, KY40506-0055, USAFax: +1-859-323-1069E-mail: [email protected]

[b] University of Kentucky, College of Pharmacy,725 Rose Street, Lexington, KY 40536, USASupporting information for this article is available on theWWW under http://www.eurjoc.org or from the author.

Eur. J. Org. Chem. 0000, 0–0 © 0000 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim 1

conformers previously predicted by calculations. In the ab-sence of charge, the molecular template does not favor intra-molecular association of aromatic substituents. These resultsagree with the chemical literature and previous reports ofneutral hydrocarbon intramolecular association in the solu-tion state.

(© Wiley-VCH Verlag GmbH & Co. KGaA, 69451 Weinheim,Germany, 2008)

Figure 1. Above: NMR-assayable !-stacking molecular template. a:R = Ph; b: R = 2,4,6-trifluorophenyl; c: R = pentafluorophenyl; d:R = methyl. X = Br or PF6. Below: The numbers are chemicalshifts due to shielding in the vicinity of the benzene ring calculatedat rb3lyp/6-311++g(2d,2p) and plotted as a contour slice throughC6H6, shown in CPK. The calculation starts at 2 Å on the x axisand 1.5 Å on the z axis from the benzene centroid.

Molecular modeling and solid-state studies elucidatedfour conformations that could compete at room tempera-ture. Two unfolded, splayed, S, conformational classes andtwo folded, C (3-ring cluster) and F (!-face-to-face), confor-mational classes had different NMR signatures due to the

Chemical Shift: Conjugation and

Aromaticity

B0H

“Ring current”

reinforces applied

B0 field; deshields

aromatic protons

(! = 6.5-8.5 ppm).

H

H

H

H

Induced current is

less effective in

olefins, deshields

less (! = 4-6 ppm).

benzene

Chemical Shift

induced magnetic field in benzene

Chemical Shift: Aromaticity

-1.0

2.0

Chemical Shift

10 9 8 7 6 5 4 3 2 1 0

aldehydearomatic heteroaromatic

alkeneCH2-O

alkyne

CHxC=O

CH3

CH2

CH

CH2Ar

Chemical Shift

Chemical Shift ~ acidic protons

~ downfield chemical shift scales with acidity (pKa)-COOH (4.8), phenol (10), alcohol (16-18), water (15.7)

~ acidic protons have a much wider range because hydrogen bonding shifts (with solvent or intramolecular) protons further downfield

3. Integration of peaks = ratio of chemically equivalent protons

1) Make sure you get proper (flat) integrals or you manually measure integrals

2) Exchangeable (acidic) protons and aldehyde protons are often too low (not accurate)

3) Integration gives RATIOs and not total number of protons.

3.953.95

0.5410.541

11

6.0

6.06.56.57.07.07.57.58.08.08.58.59.09.0ppm

3.523.52

0.3490.349

0.6430.643

6.0

6.06.56.57.07.07.57.58.08.08.58.59.09.0ppm

Course Notes NMR 1 (first NMR lecture)shimizu-uofsc.net/Ken_D._Shimizu_Group_Website/Group... · 2018-07-28 · Purity. Evidence for documenting compound purity should include one

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