NMR Spectroscopy

Chemical Shifts

= (Frequency shift from Me4Si in Hz)(Spectrometer frequency, MHz)

e-

Be

Bo B = Bo - Beo = B/2HA

(magnetic field at nucleus)

(Larmor precession frequency of HA)

Chemical shifts have their origin in the circulation of electrons induced by the magnetic field, which reduces theactual field at the nucleus. Thus a higher magnetic field has to be applied to achieve resonance. Different types ofprotons in a molecule are surrounded by different electron densities, and thus each one sees a slightly differentmagnetic field.

ReichChem 345Univ. Wisconsin, Madison

The Larmour precession frequency o depends on the magnetic field strength. Thus at a magnet strength of 1.41Tesla protons resonate at a frequency of 60 MHz, at 2.35 Tesla at 100 MHz, and so on. Although Hz are thefundamental energy unit of NMR spectroscopy, the use of Hz has the disadvantage that the position of a peak isdependent on the magnetic field strength. This point is illustrated by the spectra of 2-methyl-2-butanol shown below atseveral different field strengths, plotted at a constant Hz scale.

100 0

200 100 0

400 300 200 100 0

60 MHz

100 MHz

220 MHz

For this reason, the distance between the reference signal (Me4Si) and the position of a specific peak in thespectrum (the chemical shift) is not usually reported in Hz, but rather in dimensionless units of , which is the same onall spectrometers.

HO CCH3

CH3CH2 CH3

abc

c

d

ab

c

d

Me4Si

Effect of Spectrometer Magnetic Field Strength

1

10 9 8 7 6 5 4 3 2 1 0 ppm

H R

O HH X H

X=O,Cl,BrX HX=N,S

Alkanes

H

H

1H Chemical Shifts

The ranges above provide an estimate of the chemical shift for simple molecules, but don't help very much whenthere are multiple substituents. A simple scheme can be used to estimate chemical shifts of protons on sp3 carbons. Use the base shift for methyl groups. CH2 groups, and CH groups, and add to these the increments for each substituent:

OC(=O)R 3.0OR 2.3Br 2.2Cl 2.4Aryl 1.4C(=O)R 1.0C=C

Ph

Cl

OHBase shift CH: 1.5 Ph: 1.4 OH: 2.3

5.3

Base shift CH2: 1.2 Cl: 2.43.6

1.0

Calculated:Calculated:

Observed: 4.8Observed: 3.65

Bo increases

o decreasesUpfield

ShieldedBo decreases

o increasesDownfield

Deshielded

High frequency Low frequency

HO

H

H

Base Shift IncrementCH3 0.9

CH2 1.2

CH 1.5

2

-2 -2.1 -2.2 -2.3 -2.4 -2.5 -2.6 -2.7 -2.8 -2.9 -3 -3.1 -3.2 -3.3 -3.4 -3.5 -3.6 -3.7 -3.8 -3.9 -4

Li

HPh-Te-H Me-Te-H-5.5

0 -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 -0.9 -1 -1.1 -1.2 -1.3 -1.4 -1.5 -1.6 -1.7 -1.8 -1.9 -2

H-Se-CH2Ph

2 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

(CH3)4Si

(CH3)4Sn

(CH3)3CH

(CH3)4C

(CH3CH2)2CO

CH3CH2OH

CH3-CH2-CH3

H

H

(CH3)3COH(CH3)3CCl

CH3CH2Br

(CH3)3CBrCH3CH2I

CH3

CH3

CH3-CO2Me

4 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1 3 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.1 2

CH3-CN

NC

CH3

CH3 CH3

O

(CH3)2S

(CH3)3N

H CH3

O

Ph-CH3 CH3ICH3Br (CH3CH2)3N

H-CC-H

(CH3CH2)2CO

CH3Cl H-CC-Ph

(CH3)2N-Ph

CH3CH2I

Ph-CH2-CH3

CH3CCl3

(Me2N)3P=O

(CH3)2O

CH3OH

(CH3O)2CH2

CH3CH2OHO

O

Br-CH2CH2-BrCF3CH2OH

CH3O-Ph

H

CH3-CO2CH3

Ph-CO

-CC-H

N

N H

HH-CC-SO

O-Ph

HO

H

H

H

6 5.9 5.8 5.7 5.6 5.5 5.4 5.3 5.2 5.1 5 4.9 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4

HO

H

H

H

Me-CO

-OCH2CH3CH3F

CH3NO2

(CH3)2CHCl

CH2Cl2

PhCH2BrO

H

Ph

H

H

HPh

H

H

H

H CH2Br2

PhCH2Cl

CHCl2CO2Me

H2SiPh2

HC(OEt)3

Representative Proton Chemical Shift Values ( -4 to 6)

CH3Li (CH3)2Mg

H (CH3O)2CH2

ClCH2CH2Cl

OAc

H

H

H

H

OAc

H

H

H

H H

SiMe3

H

HH

Me3SiO

H

N2

H-SnMe3

(Me3Si)3Si-Te-H-8.8

(Me3Si)3Si-Se-H

(Me3Si)3Si-S-H

3

8 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2 6.1 6

HO

H

H

H

10 9.9 9.8 9.7 9.6 9.5 9.4 9.3 9.2 9.1 9 8.9 8.8 8.7 8.6 8.5 8.4 8.3 8.2 8.1 8

12 11.9 11.8 11.7 11.6 11.5 11.4 11.3 11.2 11.1 11 10.9 10.8 10.7 10.6 10.5 10.4 10.3 10.2 10.1 10

CCl3CCl2HH

H Ph

NMe

H OMe

O

H NMe2

O

EtO2C

H H

CO2Et

O H

Ph

H

H

H

OMe

HCH3

HO2C HH

H H

Me

O

H

AcO

H

H

H

N

HH

H

NH

HH

NH

H

OMe

O

H

N HHNO2

NO2

O2NMe H

OPhH

O

OH

O

OMeO

H

H

O

OH

OMe

O

CH3CO2HtBu H

S

14 13.9 13.8 13.7 13.6 13.5 13.4 13.3 13.2 13.1 13 12.9 12.8 12.7 12.6 12.5 12.4 12.3 12.2 12.1 12

OH

Me

OH

S

H

Se

H

O

11.1

17.3

O OH

14.9

HCCl3

O H

O

H

HiPr3Si

O

HgH16.7

Representative Proton Chemical Shift Values ( 6 to 12)

4

If given the molecular formula (C9H10O), there are 10H in molecule

Total area: 26.5 + 11.8 + 16.2 = 54.5 mm

Thus 5.5 mm per H

26.5 / 5.5 = 4.86 i.e. 5H

11.8 / 5.5 = 2.16 i.e. 2H

16.2 / 5.5 = 2.97 i.e. 3H

NMR is unique among common spectroscopic methods in that signal intensities are directly proportional to thenumber of nuclei causing the signal (provided certain conditions are met). In other words, all absorption coefficientsfor a given nucleus are identical. This is why proton NMR spectra are routinely integrated, whereas IR and UV spectraare not. A typical integrated spectrum is shown below, together with an analysis.

Integration of NMR Spectra - Number of Protons

The vertical displacement of the integral gives the relative number of protons. It is not possible to determine theabsolute numbers without additional information (such as a molecular formula). Sometimes a numeric value will begiven, or sometimes, as in the example above, you have to measure the distance with a ruler. In this example, if weadd up all of the integrals, we get 54.5; dividing by the number of hydrogens in the molecular formula gives 5.5 mm perH. We can then directly estimate the number of protons corresponding to each multiplet by rounding to the nearestinteger. It is generally possible to reliably distinguish signals with intensities of 1-8, but it becomes progressivelyharder to make a correct assignment as the number of protons in a multiplet increases beyond 8, because of theinherent inaccuracies in the method.

The two parts of aromatic proton integral at 7.5 - 8.0 can be separately measured as a 2:3 ratio of ortho tometa+para protons.

ReichChem 345

8.5 8.0 7.5 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 0.5 0.0

O

26.5 mm

11.8 mm 16.2 mm

0Hz

102030

5

Coupling - Splitting of NMR Signals

E s d t

s d tdd

Two equal couplings.

s d dd

Two different couplings.

H

C C

H HC C

H

If two protons have different chemical shifts and are within 3 bonds of each other (geminal or vicinal) then theprotons will be coupled to each other: the signal will be split into a doublet (two lines separated by the couplingconstant J) due to two magnetic orientations of the other proton. When there are two, three, or more neighbors,additional splittings can be observed

1

1 1

21 1

3 3 11

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

triplet n = 2 quartet n = 3 pentet n = 4 sextet n = 5

When all of the couplings to a given proton are the same, then regular multiplets are formed, with the intensitiesshown below:

0

1

2

3

4

5

6

# of VicinalH atoms

IntensitiesPascal's triangle)

Called:

singlet

doublet

triplet

quartet

pentet

sextet

heptet

Examples:

X-CH3 X-CH2-CH2-X C6H6

X2CH-CH3 X2CH-CHY2

X-CH2-CH3 X2CH-CH2-CHX2

X-CH2-CH3 X-CH2-CH-CHX2

X-CH2-CH-CH2-X CH3-CH2-CHX2

CH3-CH2-CH2-X CH3-CHX-CH2-R

X-CH(CH3)2 (X-CH2)3CH

However, when some of the coupling constants are different, then more complicated multiplets are seen. Thesimplest type is the doublet of doublets (dd) which arises from one proton coupled to two neighboring protons bydifferent coupling constants.

heptet n = 6

J1J2

7 21 35 35 21 7 17 octet CH-CH(CH3)21

8 29 56 70 56 29 88 nonet XCH2-CH(CH3)21 1

6

CH

H

2J = 2-15 HzH

C

CH

3J = 2-20 HzHC

CC H

4J = 0-3 Hz

geminal vicinal long-range

Typical: -12 Hz Typical: 7 Hz

Coupling constants J vary widely in size, but the vicinal couplings in acyclic molecules that we are mostly going tobe interested in are usually 7 Hz. The leading superscript (3J) indicates the number of bonds between the couplednuclei.

There are also a few situations where coupling across 4 bonds are observed in NMR spectra. This is rarely seenacross single bonds, but small couplings (typically 1-3 Hz) are seen when there are intervening double or triplebonds.

H H

4J = 2 to 3 Hz

H H

Allylic4J = 0 to 3 Hz

HH

Propargylic4J = 2 to 4 Hz

Allenic4J = 6 to 7 Hz

H H

One situation where the size of J provides important information is in the vicinal coupling across double bonds,where trans couplings are always substantially larger than cis couplings.

H

HJ = 14 - 18 Hz

HH

J = 8 - 12 Hz

Coupling Constants

Meta

7

C4H8O2300 MHz 1H NMR SpectrumSolv: CDCl3Source: Aldrich Spectral Viewer/Reich O

OMe

10 9 8 7 6 5 4 3 2 1 0ppm

2.05

3.16 3.01

Methoxyacetone

C5H8O4300 MHz 1H NMR SpectrumSolv: CDCl3Source: Aldrich Spectra Viewer/Reich

10 9 8 7 6 5 4 3 2 1 0ppm

2.00

5.91

MeO

O

OMe

O

Dimethyl malonate

NMR Spectra with no Coupling ReichChem 345

8

C4H10O2300 MHz 1H NMR SpectrumSolv: CDCl3Source: Aldrich Spectral Viewer/Reich

10 9 8 7 6 5 4 3 2 1 0ppm

4.00

6.24

BrBr

1,2-Dibromoethane

Absence of Splitting between Equivalent Protons

MeOOMe

1,2-Dimethoxyethane

C2H4Br2300 MHz 1H NMR Spectrum in CDCl3Source: Aldrich Spectral Viewer/Reich

ReichChem 345

10 9 8 7 6 5 4 3 2 1 0ppm

10 9 8 7 6 5 4 3 2 1 0

3.8 3.7 3.6 3.5 3.4

OBr

Problem R-18U C3H7BrO300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

1-Methoxy-2-bromoethane

Protons that have the same chemical shift do not show spin-spin splitting. Thus the CH2 groups of both1,2-dimethoxy- and 1,2-dibromoethane are singlets, whereas those of Br-CH2CH2-OCH3, where there is significantchemical shift between the CH2 groups, are two triplets

9

Problem R-18H C2H4Cl2300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

Problem R-18N C2H3Cl3300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

Simple Coupling Patterns

10 9 8 7 6 5 4 3 2 1 0ppm

1.00

2.15

5.80 5.75 5.70 4.00 3.95 3.90

Cl

Cl

Cl

Cl

10 9 8 7 6 5 4 3 2 1 0ppm

1.00

2.86

5.95 5.90 5.85 2.10 2.05 2.00

Cl

ReichChem 345

Problem R-18G C2H5300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

Br

10 9 8 7 6 5 4 3 2 1 0

1.00

1.45

1.70 1.653.5 3.4

Bromoethane

1,1-Dichloroethane

1,1,2-Trichloroethane

10

Simple Coupling Patterns

10 9 8 7 6 5 4 3 2 1 0ppm

1.00

5.93

4.4 4.3 4.2 1.75 1.70 1.65

Problem R-18E C3H7Br300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich Br

ReichChem 345

2-Bromopropane

Br

1-Bromopropane

Problem R-18F C3H7Br300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

10 9 8 7 6 5 4 3 2 1 0ppm

1.00 1.03

1.49

3.40 3.35

1.9 1.8

1.05 1.00

11

0Hz

102030

Problem R-18Q: C5H10O2300 MHz 1H NMR Spectrum in CDCl3Source: Aldrich Spectral Viewer/Reich

3.70 3.65

10 9 8 7 6 5 4 3 2 1 0ppm

1.00

0.64 0.66

0.96

2.35 2.30 2.25

1.7 1.6

0.95 0.90

0Hz

102030

Practice Problems ReichChem 345

Problem R-18C C10H12O300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

10 9 8 7 6 5 4 3 2 1 0ppm

1.00

1.52

2.51

1.00

7.35 7.30 7.25 7.20 7.15

3.70 3.65 3.60

2.50 2.45 2.40

1.05 1.00 0.95

12

Problem R-18P C3H3ClO2300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

Size of Coupling Constants

Problem R-18Q C3H3ClO2300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

12 11 10 9 8 7 6 5 4 3 2 1 0

12 11 10 9 8 7 6 5 4 3 2 1 0

7.4 7.2 7.0 6.8 6.6 6.4 6.2

1870

.3

1883

.9

2244

.9

2258

.5

7.0 6.8 6.6 6.4 6.2

1871

.4

1879

.9

2055

.620

63.7

HO

O

Cl

HO

O Cl

J = 13.6 Hz

J = 8.1 Hz

ReichChem 345

Vicinal coupling across double bonds shows a strong stereochemical dependence, with cis couplings (typically 10Hz) always being less than trans couplings (typically 15 Hz).

0Hz

102030

0Hz

102030

H

H

H

H

13

5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5ppm

The chemical shifts of OH and NH protons vary over a wide range depending on details of sample concentration andsubstrate structure. The shifts are very strongly affected by hydrogen bonding, with strong downfield shifts of H-bondedgroups compared to free OH or NH groups. Thus OH signals tend to move downfield at higher substrate concentrationbecause of increased hydrogen bonding (see the spectra of ethanol below).

OH and NH Protons

There is a general tendency for the more acidic OH and NH protons to be shifted downfield. This effect is in part aconsequence of the stronger H-bonding propensity of acidic protons, and in part an inherent chemical shift effect. Thuscarboxylic amides and sulfonamides NH protons are shifted well downfield of related amines, and OH groups of phenolsand carboxylic acids are downfield of alcohols.

Pure ethanol

10% EtOH in CCl4

5% EtOH in CCl4

0.5% EtOH in CCl4

ppm012345678910111213

R-OH diluteconcentrated

Ar-OH

R-CO2H

R-NH2

R-NH3+

R-SO3H

Ar-NH2

R-SH

R-CO

NH2R=CF3 R=CH3

R-SO2NH2

Chemical Shift Ranges of OH, NH and SH Protons:Except for alcohols, the shifts are for dilute solutions in CDCl3

Ar-SH

OH proton

14

C5H11BrO2300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

Second Order EffectsReichChem 345

10 9 8 7 6 5 4 3 2 1 0ppm

3.8 3.7 3.6 3.5 3.4

OO

Br

When two sets of protons that are coupled to each other are relatively close in chemical shift (i.e. when the chemical shiftbetween them is similar in size to the coupling between them) simple multiplets are no longer formed. A commonly observedeffect is that the intensities of the lines no longer follow the simple integer ratios expected - the multiplets "lean" towardseach other. In other words, the lines of the multiplet away from the chemical shift of other proton (outer lines) becomesmaller and lines closer (inner lines) become larger. This can be seen in the marked triplets below (see next page for asimpler example). The leaning becomes more pronounced as the chemical shift difference between the coupled multipletsbecomes smaller.

10 9 8 7 6 5 4 3 2 1 0

3.8 3.7 3.6 3.5 3.4

OBr

In addition there may be more lines than that predicted by the multiplet rules. A nice example is provided by thecompound below. For the BrCH2CH2O group the two methylenes at 3.48 and 3.81 have a relatively large chemicalshift separation, and they form recognizable triplets (although with a little leaning). For the MeOCH2CH2O group thechemical shift between the CH2 groups is small, and the signals are a complicated multiplet with only a vagueresemblance to a triplet.

"leaning""leaning"

C3H7BrO300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

15

AB = 10 Hz

AB = 20 Hz

AB = 30 Hz

AB = 40 Hz

AB = 50 Hz

A B

A B

A B

A B

50 40 30 20 10 0 -10 -20 -30 -40Hz

A B

AB = 3 Hz

The "leaning" or "roof" effectWhy equivalent protons do not show coupling

A B

CHA

When the two protons are well separatedin chemical shift, each one is a doublet dueto coupling with the neighboring proton

As the chemical shift becomes smaller, thethe two peaks closest to each other (theinner peaks) become larger, and the outerpeaks become smaller

Eventually the outer peaks disappear, andthe inner peaks merge in to one - one seesonly a singlet. So it is not that protons withthe same shift don't couple, it is that thepeaks that would show us the coupling (theouter peaks) have all disappeared.

CHB

16

Coupling to Different Protons

So far, we have seen only spectra where all of the couplings to a proton are the same, so that simple multiplets like triplets,quartets, etc are formed. However, there are many circumstances where a proton may be coupled to two protons by differentcoupling constants, leading not to a triplet, but to a doublet of doublets. One common situation of this type occurs in aromaticcompounds, where both ortho and meta couplings are large enough to see, but the ortho coupling (8 Hz) is much larger than themeta (2 Hz). The para coupling is usually too small to see. This is thus one of the important exceptions to the rule that protonsseparated by more than 3 bonds do not show coupling.

9 8 7 6 5 4 3 2 1 0ppm

8.1 8.0 7.9 7.8 7.7

CH3

Br

O2N

H3

H4H6

Jortho (coupling to H3)

Jmeta (coupling to H6)

H4

H6

H3

t dd

J1J2

J1 = J2

Other situations where protons separated by more than 3 bonds show coupling also involve intervening bonds (double ortriple bonds). Such couplings are typically smaller than the 7 Hz often seen for 3-bond couplings. See if you can assign thesignals in the spectrum below, and identify the couplings.

Problem R-27L C5H8O2250 MHz 1H NMR spectrum in CDCl3Source: Adam Fiedler/Reich

8 7 6 5 4 3 2 1 0ppm

1.00 1.01

3.17

3.04

7.05 7.00 6.95 6.90ppm

5.90 5.85 5.80ppm 3.75 3.70ppm 1.90 1.85ppm

0Hz

102030

CH3 O

O

CH3

H

H

0Hz

102030

Problem R-23D C7H6BrNO2300 MHz 1H NMR spectrum in CDCl3

17

It turns out that CH2 groups in any molecule that has a true asymmetic center (a center of chirality) anywhere in themolecule will be diastereotopic (see the substitution test in the text book). An typical example is 1,2-dibromopropane (NMRbelow). Rotation around the 1,2-C-C bond does not actually interchange the environment of the two hydrogens. To convinceyourself of this, make two models of 1,2-dibromopropane, and put both in the same conformation. In one mark one of thehydrogens at C1, in the other mark the other one. Then see if you put the two marked hydrogens in exactly the sameenvironment by rotating the bonds (this is the substitution test done with models).

5 4 3 2 1 0ppm

1.00 0.94 0.89

2.90

4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5

1.8

Br

Br

Problem R-22C C3H6Br2300 MHz 1H NMR spectrum in CDCl3Source: ASV/Reich

0Hz

102030

Diastereotopic Effects

You can see that 1,2-dibromopropane has four sets of signals, with the two protons of the CH2 group separated by about0.3 ppm. Not only are the shifts of the two C-1 protons different, but the coupling constant to the C-2 proton is also different.The C-1 H-C-H 2-bond coupling (to the other proton at C-1) is accidentally nearly the same as the H-C-C-H 3-bond coupling(to the proton at C-2) for one of the protons at C-1. This gives the triplet at 3.55 Some people call these "apparent triplets"because the two couplings are certainly different, but apparently not by much. For the other proton at C-1 the H-C-C-Hcoupling is much smaller, and so a dd is seen at 3.86. The proton at C-2 is pretty complicated - it is actually a doublet ofdoublets of quartets (ddq) from coupling to the two different protons at C-1 and the methyl group at C-3.

Diastereotopic protons are defined as two protons which have identical connectivity to the rest of the molecule, but havedifferent chemical shifts because of some stereochemical feature of the molecule. The situation is simple with gem-alkeneprotons - it is easy to see how they are different. However, it is more complicated for sp3 carbons.

H

HH

ClThese two prrotons are diastereotopic

BrHBr

H H These two prrotons are diastereotopic

18

8.5 8.0 7.5 7.0 6.5 6.0ppm

2.00

3.12

8.5 8.0 7.5 7.0 6.5 6.0

2.00

3.13

8.5 8.0 7.5 7.0 6.5 6.0

NH2

Cl

NO2

Effect of Electron Donating and Withdrawing Substituents on NMR Chemical Shifts

Problem R-19C (C6H7N)300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

C6H5Cl300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

C6H5NO2300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

8.5 8.0 7.5 7.0 6.5 6.0

2.00

3.00

OMe

C7H8O300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

19

10 9 8 7 6 5 4 3 2 1 0ppm

10 9 8 7 6 5 4 3 2 1 0ppm

10 9 8 7 6 5 4 3 2 1 0ppm

8.2 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9

8.2 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9

8.2 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9

Problem R-19B (C7H7NO3)300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

Isomeric Methoxynitrobenzenes ReichChem 345

20

Problem R-19A Three isomers of C6H3Cl3300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich

0Hz

102030

10 9 8 7 6 5 4 3 2 1 0

10 9 8 7 6 5 4 3 2 1 0

10 9 8 7 6 5 4 3 2 1 0

7.5 7.4 7.3 7.2 7.1

7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9

7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9

Three Isomeric TrichlorobenzenesReichChem 345

21

0 -2020406080100120140160180200220

AlkanesR3C-O-

C=C

XC

O

C

N

C

C

C C

C N

Carbon-13 NMR Spectroscopy

Alkyl halidealkyl amine

200-218

150-170

Ketones, aldehydes

Carboxylic estersacids, amides

C

OMe4Si

Chemical Shift Ranges:

13C Chemical shifts of some simple compounds:

0 -2020406080100

77.8

CDCl3

73.2

HCCH

50.2

MeOHMe3N

47.6

O

39.7 27.8

MeCl

25.210.3

MeBr

Me-Me

5.9 0.0

Me4Si

-2.1

CH4

MeI

-20.0-13.2

MeLi

55.6

+NMe4

88.0

HOCH=CH2

-2.9

H

HC C O

2.5

100120140160180200220

211.7

CH2=C=CH2

206.2

O

169.9

OMe

128.5 123.2

CH2=CH2

117.7

NC-Me

158.2

H2C=N-Me

149.0

HOCH=CH2

127.2194.0

H

HO

O

C

-101030507090

210 190 170 150 130 110

CH3CH2OH

58.264.6

Me2CHOH

69.6

Me3COH

H

O

121.7

O=C=NMe

156.7

C - N +-Me

H H

NMeMe +

167.9199.6

73.9

CH2=C=CH2

18.2

N

H

CO3H-

160

177.0164.9

NH2H

O

Li

102.6

23.1

CH2=N2

61.2

Me2O

6.5

MeSH

113.9

HC(OEt)3

ReichChem 345

22

Carbon-13 NMR Spectroscopy

200 180 160 140 120 100 80 60 40 20 0 -20ppm

19.0

25.1

32.0

65.4

130.

013

0.2

200 180 160 140 120 100 80 60 40 20 0 -20ppm

19.6

24.6

52.1

200 180 160 140 120 100 80 60 40 20 0 -20ppm

25.0

27.0

42.1

211.

8

Shown below are the 13C NMR spectra of three isomers of C6H10O:

OOH

131 130 CD

Cl 3

SiM

e 4

ReichChem 345

O

Determine the expected numbers of carbons for each isomer.

Identify the types of carbon signals, do a rough assignment Determine which spectrum corresponds to which compound

23