Analyzing H spectra - Homepage | DidatticaWEB

Analyzing 1H spectra •  The number of signals in the 1H-NMR spectrum of a molecule is

equal to the number of stereo-equivalent protons.

•  Protons are stereo-equivalent, if they form similar bonds, i.e., if the geometry with respect to the remainder of the molecule is identical, and if the character of the neighboring bonds is identical.

•  The identification of equivalent protons in a molecule can be achieved by two different methods. Either you can use each proton in turn as a reference point and characterize the remainder of the molecule based on it to find protons with identical environments. Or you can look for symmetry operations that relate different protons to each other.

•  In non-rigid molecules the protons of a methyl or methylene group give rise to a single, shared signal.

Chemical shifts - Number of resonance signals

•  The first piece of information required for the deduction of an NMR spectrum is the number of resonance signals that can be expected. How can it be determined?

•  In the presence of a homogenous magnetic field, every 1H nucleus shows upon irradiation with an oscillating electro-magnetic field of the proper frequency a resonance signal.

•  The resonance signal of a 1H in a chemical compound is shifted compared to the signal of an isolated 1H, since the external magnetic field will be modulated by the secondary magnetic fields induced in the electrons in the neighborhood of the proton. Therefore 1H with identical electronic environments will show chemical shifts, i.e. signals at identical resonance frequencies.


Which of the following approaches will give you the right number of 1H signals in an NMR spectrum of a compound? The number of signals in an NMR spectrum is equal to

•  the total number of hydrogen atoms in the compound, •  the number of hydrogen containing

groups in the molecule, •  the number of sterically equivalent groups of protons.

the number of sterically equivalent groups of protons Equivalent protons generate a single, common resonance signal, while non-equivalent protons have different resonance frequencies

•  If you want to identify equivalent protons in a molecule, it is important to come up with an exact idea of its spatial structure. Here is an example: All atoms of p-bromo-nitrobenzene are in one plane.


There are now two possible approaches to identify equivalent protons: a) Using a specific proton as a reference point, characterize the remainder of the molecule. In the case of H-2, for example, the other substituents are:

With H-2 as the reference point: • ortho-substituents at the benzene ring: Br- and H-3 • meta-substituents at the benzene ring: H-6 and NO2- • para-substituents at the benzene ring: H-5

Now you can compare this set of relationships with the ones for each of the of the other protons.

With H-6 as a reference point: • ortho-substituents at the benzene ring: H-5 and Br- • meta-substituents at the benzene ring: NO2- and H-2 • para-substituents at the benzene ring: H-3

H-2 and H-6 show similar relationships to the substitutents, which means that they are equivalent

b) Look for symmetry elements that define the molecule, such as mirror symmetries, rotational symmetries or combinations of the two. As a case in point, the molecule in this example has a two-fold rotational symmetry with a rotational axis C2 through the Br and N atoms:

Through a l80° rotation around this axis H-2 is placed on H-6 and vice versa, and H-3 is placed on H-5 and vice versa. That means that H-2 and H-6 as well as H-3 and H-5 are equivalent.


•  Now follow this procedure in the following example:

A) 1 signal B) 2 signals C) 4 signals

One signal is due to the protons bound to carbon 2 and 5, the second is due to the protons bound to carbons 3 and 4. It is the number of groups of equivalent protons that determines the number of signals in a 1H NMR spectrum.


How many signals do you expect in the 1H NMR spectra of these substances?

Substance Equivalent Protons

Number of signals

a) H1 b) H2, H3, H4

2

a) H2, H6 b) H3, H5

2

all of them 1


How many signals do you expect in the 1H NMR spectra of these substances?

Substance Equivalent Protons

Number of signals

a)  H1, H4 b)  H2, H3

2

all of them 1

a) H1, H2 b) H4

2

Now consider the methyl group in acetaldehyde:

1.  The methyl group in acetaldehyde can

rotate almost freely around around the C-C single bond. Therefore the methyl protons are exposed to a shielding averaged over time.

2.  The methyl protons are equivalent if averaged over the NMR time scale, and give only a single NMR signal.

Temperature dependence of the signal number within flexible systems The methyl group within acetaldehyde CH3-CHO is able to rotate around the single bond between the two carbon atoms. The result of this rotation is an averaged identical chemical shift for all methyl hydrogens. Hence we observe one signal only. The situation changes with an rotational speed slowed down due to low temperature. In this case the most favoured conformations become "freezed". The most favoured conformation in the case of acetaldehyde are I, V and IX. In all three conformations exist two different chemical environments for the hydrogen atoms.

We would observe two signals with an integral ratio of 1 : 2. In somewhat more complex systems such measurements at low temperatures enable possibilities to investigate conformations and conformational equilibrium.

Considering what you have learned so far, would you expect in a 1H-NMR-spektrum of propionaldehyde CH3-CH2-CHO at room temperature 3 signals? 4 signals? 6 signals?

3 signals


Indeed, the three protons of the methyl group and the single proton of the carbonyl group (-CH3 und -CHO) give rise to two signals. The same reasoning holds for the two protons of the methylene group (-CH2-), since they also rotate freely, and since the two protons have mirror symmetry. They also give only one shared signal.

C1 and C3 rotate for all concerns and purposes freely around the single bonds C1-C2 and C2-C3, respectively. Therefore both methylene protons experience the same average amount of shielding over time from the CH3- and the CHO- groups.


To summarize: In systems with unrestricted movement all the protons of a

•  methyl group (-CH3) •  methylene group (-CH2-) produce only a single, shared resonance signal. Complications with regard to this rule only arise

when the rotation around the single bond produces conformations that are related through a symmetry operations (e.g., when the CH2- group is next to an asymmetrical carbon atom).


Number of signals and symmetry in flexible systems

Rule: In the case of impossibility to fit different conformations by any symmetry operation,

the protons involved show different chemical shifts.

In general you will observe this behaviour if you have CH2 groups and at least one asymmetric carbon atom together in one molecule. Example: Within the compound

Number of signals and symmetry in flexible systems

with one asymmetric carbon atom (C*) it is impossible to fit the conformations:

with any symmetry operation. Each moment of the free rotation there interact different chemical environments with both methylene protons H(1) und H(2). Both methylene protons show their individual signal, as demonstrated within the experimental spectrum.


2 Signals

Diisopropylether is a highly symmetrical molecule.

The four CH3-groups as well as the two CH-groups are equivalent to each other, and have the same electronic environment, respectively. Therefore they have identical chemical shifts. Therefore only two signals appear in an NMR spectrum.


4 Signals

Each of the groups containing protons - including the two methyl groups - is exposed to a different environment.

so that each signals occurs at a different frequency. Obviously the protons of each methyl and methylene groups will only give rise to a single, shared signal. Therefore we will observe a total of 4 signals.


Now predict the number of signals from the methyl and aromatic protons in the NMR spectra of the following isomeric benzene derivates:

Dimethylbenzene

3 4 2


Now predict the number of signals from the methyl and aromatic protons in the NMR spectra of the following isomeric benzene derivates:

Trimethylbenzene

2 6 4

In a 1H-NMR spectrum you find three closely spaced signals. Which of the following isomeric molecules could give rise to this spectrum?



one signal from

one signal from

one signal from

This example should have demonstrated that 1D NMR spectroscopy can also used for the analysis of the spatial arrangement of substituents.

H1 and H3

Positions of the resonance signals in the spectra

For the prediction of the NMR spectrum of a given molecule it is necessary to know the number of resonance signals. But it is obviously also necessary to know the position of these signals in the spectrum, that is, their chemical shift. Just as in case of the number of signals, it is also possible to make predictions about their position:

Equivalent protons give rise - due to their identical electronic environment - to

identical chemical shifts. If two protons have similar electronical environments, then their chemical shifts in an NMR spectrum will be nearly identical

For example, the signals of the protons in the two methyl groups

-CH2-CH3 and >CH-CH3

will show similar chemical shifts.

The analysis of a large number of spectra showed that the signals of aliphatic methyl protons typically have a chemical shift δ = 0,6 ... 1,9 ppm.

In contrast to this leads the completely different electronic environment of a carbonyl group -CHO als to a different value for the typical chemical shift. In that case δ = 9,l ... l0 ppm for the proton of the aldehyde group.

Type of Proton Structure Chemical Shift, ppm

Cyclopropane C3H6 0.2

Primary R-CH3 0.9

Secondary R2-CH2 1.3

Tertiary R3-C-H 1.5

Vinylic C=C-H 4.6-5.9

Acetylenic triple bond,CC-H 2-3

Aromatic Ar-H 6-8.5

Benzylic Ar-C-H 2.2-3

Allylic C=C-CH3 1.7

Fluorides H-C-F 4-4.5

Chlorides H-C-Cl 3-4

Bromides H-C-Br 2.5-4

Iodides H-C-I 2-4

Alcohols H-C-OH 3.4-4

Ethers H-C-OR 3.3-4

Esters RCOO-C-H 3.7-4.1

Esters H-C-COOR 2-2.2

Acids H-C-COOH 2-2.6

Carbonyl Compounds H-C-C=O 2-2.7

Aldehydic R-(H-)C=O 9-10

Hydroxylic R-C-OH 1-5.5

Phenolic Ar-OH 4-12

Enolic C=C-OH 15-17

Carboxylic RCOOH 10.5-12

Amino RNH2

http://wwwchem.uwimona.edu.jm:1104/spectra/nmrintro.html


Between 7 - 8 delta, is the region typically where aromatic proton signals are found.

Between 10.5 - 12 delta, protons attached to acid groups may appear, ie -CO-OH.

Between 9 - 10 delta, aldehyde protons may appear ie, -CO-H.

Between 4.5 - 7 delta, protons attached to unsaturated groups may appear, eg C=C-H.


Between 1.5 - 5 delta, methylene groups CH2 may appear.

Between 0.4 - 4 delta, methyl groups in particular those attached to carbons which are next to carbonyls may appear, eg. Me-CO, etc.

Between 1.5 - 2, methyl groups attached to double or triple bonds may appear, for example MeCN, MeCOOR etc.

Between 0 - 1 delta, protons attached to saturated cyclic rings may appear

Relative to TMS 12 11 10 9 8 7 6 5 4 3 2 1 0 -1

H3C-Alkyl

H3C-Hal

H3C-C=C

H3C-CC

H3C-Aryl,-Heteroaryl

H3C-CO

H3C-S-

H3C-SO2-

H3C-N

H3C-O-Alkyl

H3C-O-Aryl,-O-CO-

C-CH2-Alkyl

Cyclopropane

C-CH2-CO

C-CH2-O-

C-CH2-S

C-CH2-NO2

C-CH2-N

C=C-CH2-C=C

N-CH2-CO-

-O-CH2-CO-

-O-CH2-O-, -Aryl

Relative to TMS 12 11 10 9 8 7 6 5 4 3 2 1 0 -1

CH-Alkyl

CH-Hal

C-CH-O-

C-CH-N

CO-CH-C=C

CH-Aryl,-NR-,-O-

-CCH Alkine

CH=C-

Ar-H

Alkyl-, Aryl-CHO

Alkyl-OH

Aryl-OH

R-COOH

Alkyl-SH

Aryl-SH

Alkyl-NH2,Alkyl2-NH

Aryl-NH2,Aryl2-NH

R-CO-NH

-CO-NH-CO

Enter now for the following compounds the number of Signals and the chemical shift ranges.


Molecule Number of Proton Expected chemical signals groups shift δ [ppm]

CH3-CO-CH3 Br2C=CH2 (CH3)2-CH-O-CH(CH3)2 CH3-CH2-CHO

1

1

2

3

CH3-CO

>C=CH2

-CH3 >CH-O-

-CH3 -CH2-CHO -CHO

1,7 ... 2,7

4,3 ... 5,3

0,6 ... 1,9 2,5 ... 5,3

0,6 ... 1,9 1,9 ... 3,2 9,1 ... 10

The NMR-spectrum of a compound with the formula C4H8O exhibits NMR signals at δ = 1.1ppm δ = 2.4ppm δ = 9.9ppm

Which of the following isomers would give rise to this spectrum? Assign the signals!


0.6 – 1.9 1.4-2.1

9.1-10.0

Signal intensities Next to the number of signals in an NMR spectrum

and their chemical shifts, additional information about the compound in question can also be derived from the intensities of the signals.

1. The intensity is measured as the integral below the resonance signal. To determine the area, the spectrum is integrated automatically, and the cumulative integral is plotted on the spectrum.

2. The intensities are proportional to the number of protons, that contribute to the signal.

Signal intensities As an example serves the spectrum of :

Above the spectrum the integral is plotted in red. The magnitude of the steps in this curve is a direct representation of the number of protons that contribute to each signal.

The molecule can be characterized as follows, based on the number of signals and

their integrals:

Signal intensities

equivalent protons number of protons a) (CH3)2-C< 6 b) -OH 1 c) -CH2- 2 d) -CO-CH3 3

I(CH3)2 : ICH3 : ICH2 : IOH = 6 : 3 : 2 : 1

Signal intensities

-OH >CH2

-CO-CH3

>C(CH3)2

Signal intensities For the following set of molecules, determine the groups of equivalent protons

and the relative intensities of their signals in an NMR spectrum!

(CH3)2CHOCH(CH3)2

Substance Equivalent protons Relative intensities

a)  CH3-CH< 3 b)  >CH- 1 c)  -CH2- 2 d)  -CH3 3

a) 2 * (CH3)2C< 6(12) b) 2 * >CH-O 1 (2) a)  3 * -CH3 3 (9) b)  3 * Phenyl-H 1 (3)

Signal intensities Two molecules of ketene CH2=C=O react spontaneously to form the dimeric

diketene C4H4O2. There are two signals in the 1H-NMR spectrum 2 of diketene. The ratio of their

relative intensities is 1:1. Considering this, which of the following four structures is the correct one for the diketene?

•  It gives rise to two NMR signals from the protons in the CH2= and >CH2 groups. •  The ratio of the intensities is 1 : 1, since both of the two groups contain two protons.

Summary of the sections "chemical shift" and "integration"

1.  If protons have similar chemical environments, then typically their NMR signals also have similar chemical shifts. It is therefore possible to assign certain characteristic chemical shift ranges to the protons of the various functional groups in organic compounds.

2.  The intensity of a signal is determined as the area under

the curve of an NMR spectrum (Done usually automatically, using a cumulative integral across the full spectral width).

3. The intensity of a signal is proportional to the number of

equivalent protons that give rise to it. Therefore the ratio between the intensities of a set of signals is roughly equal to the number of equivalent protons that contribute to each of the signals.

Through-bond spin-spin coupling

If you want to use NMR spectroscopy in the investigation of the structure of molecules, there are additional fine structure features of the spectra that can help you to determine the conformation of the molecule in question. At high enough resolution certain signals do not appear as simple lines, but as multiplets of closely spaced lines with characteristic patterns of intensities.

Here is an example: •  The molecule we started out with CH3-CH2-CO-CH2-CH3 should have only

two signals in its proton NMR spectrum:

a) 2 -CH3 δ = 0.6 ...1.9 ppm b) 2 -CH2-CO δ = 1.9 ... 3.2 ppm.

The actual spectrum does confirm these predictions:


However, at closer inspection the signal of the methyl protons turns out to be composed of three lines (a triplet) with relative intensities of approximately 1 : 2 : 1, while the signal of the methylene protons is composed of four lines (a quartet) with relative intensities of approximately 1 : 3 : 3 : 1


1.  If the magnetic fields of two protons can interact via σ- and π- electrons in bonds connecting the two protons directly, or indirectly over additional nuclei, then their energy levels in the external magnetic field of the spectrometer will show additional splittings. Because of that additional transitions with only small differences in energy become possible.

2. Since these interactions are mediated by bond electrons, this phenomenon is referred to as through-bond spin-spin coupling.

3.  In the most basic case, when only n equivalent protons X are coupling to the proton A, the following relationship exists between the number of coupling protons and the multiplicity MA of the signal δA MA = nX + 1


Multiplicity Relative intensities

2 1 : 1

3 1 : 2 : 1

4 1 : 3 : 3 : 1

5 1 : 4 : 6 : 4 : 1


The splitting of the signals is caused by the through-bond spin-spin coupling. This name already indicates that the nuclear spins of different atoms (we are only considering protons at this point) interact with each other via their magnetic fields. This coupling betwen the nuclei is mediated by the electrons of the bonds that lie between the protons and their respective magnetic fields. An example: The two protons in dichloroacetaldehyde

Cl2CH-CHO X A

exhibit through-bond spin-spin coupling: The external field B0 leads to a splitting of the energy levels of the proton A due

to the alignment of the nuclear spin parallel or antiparallel to the external field. This gives rise to the possibility of transitions between the two energy levels - the resonance signals we observe in NMR spectroscopy.


Without With

Magnetic field Coupling With X

Since the spin of proton X also can assume two different orientations with respect to the external magnetic field (mIX = + ½ and mIX = - ½) the effective magnetic field - the external field plus the field of X mediated via the bond electrons - at the proton A can assume two values. Therefore the energy levels for proton A are split up a second time, depending on whether the spin of the proton X is oriented parallel or antiparallel to the external field. Instead of a single transition, there are now two possible transitions - a doublet. One neighboring proton therefore gives rise to a doublet. (nX = 1, therefore MA = 2). As an interesting aside - the nearly equal intensities of the two lines in a doublet illustrates how minute the differences in population between the two states mIX = + ½ and mIX = - ½ are under normal experimental conditions.


What about the X part of the spectrum? The reasoning is analogous to the one for proton A, i.e. the

signal of proton X is also a doublet. In the external magnetic field B0 the energy levels of the proton X split into two, and transitions between them become possible, giving rise to resonance signals. Due to the coupling to the two possible spin alignments of the proton A there are a total of four possible energy levels, and the number of possible transitions doubles. Since the two transitions of a doublet have almost identical energy differences, they occur with the same probabilities, and the ratio of their intensities is 1 : 1.


•  Another example: In bromo-acetaldehyde BrCH2-CHO X A

two protons X interact with the proton A In an external magnetic field B0 the energy of the proton A is initially split into

two levels. The spins of the two protons HX of the neighboring CH2Br group can form the

following spin combinations in relation to the external field:

Coupling with X


•  Due to the coupling between any of the three possible spin combinations for the pair of protons X and the proton A, each of the two energy levels of A is split up into three sublevels. Instead of just one transition between the two energy levels of A, now three closely spaced transitions are possible. (Triplet: nx = 2, MA = 3). The ratio of the intensities of the individual triplet lines is 1 : 2 : 1, since out of the four possible spin combinations for the pair X two will have a total spin of 0 (↑↓ and ↓↑).

•  The signal of the proton X is a doublet, since the two equivalent protons X are coupling with a single proton A


So far you have come up with the following answers:

Fine structure of the signal

Due to coupling with

Equivalent protons

Singulet (M = 1) none

Doublet (M = 2) one

Triplet (M = 3) two

The general rule therefore is: The signal of one or several equivalent protons is a multiplet with a multiplicity of M = n + 1, if the proton(s) couples with n other equivalent protons!


Multiplicity M = n + 1

Relative intensities of the individual multiplet lines

M = 2 1 : 1

M = 3 1 : 2 : 1

M = 4 1 : 3 : 3 : 1

The general rule therefore is: The relative intensities of the individual multiplet lines follow the same patterns as the nth binominal coefficients!


Draw a schematic NMR spectrum of the molecule R2CH-CH3


•  The appearance of a quartet for the R2CH proton can be explained as follows: In the external magnetic field B0 the nuclear spin of the proton can either align itself either parallel or antiparallel to the external field. These two alignments are equivalent to two different energy levels, between which transitions are possible. And these transitions give rise to the resonance signals. The protons in the neighboring CH3 group also align themselves with the external field. However, the individual spins of the three protons add up to the following total spins:

The magnetic fields of these total spins modify - mediated by the bond electrons - the external magnetic field B at the proton of the R2CH group, so that both of the two possible energy levels of H are split up into four levels. Therefore four different, closely spaced transitions are possible. (Quartet: nCH3 = 3, therefore MCH = 4).

Prediction of 1H NMR spectra •  Predict the 1H nmr spectrum of R2CH-CH3

1.  Number and type of groups of equivalent protons : in this example: a)>CH- b)-CH3

2. relative position of the signals in the NMR spectrum

in this example: a) δ = 1.4 ... 2.1 ppm b) δ = 0.9 ... 1.9 ppm

3. relative intensities of the signals

in this example: a) 1 b) 3

4. Multiplicity of the line splitting and the intensities of the individual lines in this example: a) M = 4 (Quartet), I = 1 : 3 : 3 : 1 b) M = 2 (Doublet), I = 1 : 1

Molecule Equivalent protons

δ[ppm]

Irel M Intensities of the individual lines

Line spectrum

R2CH-CH3 R2CH

CH3

1.4 ... 2.1

0.6 ... 1.9

1

3

4

2

1/8 : 3/8 : 3/8 : 1/8

3/2 : 3/2

R2CH-CH2R' R2CH

CH2R'

1.4 ... 2.1

0.9 ... 2.2

1

2

3

2

1/4 : 1/2 : 1/4

1 : 1

R2CH-CHR2' R2CH

CHR2'

1.4 ... 2.1

1.4 ... 2.1

1

1

2

2

1/2

: 1/2

1/2

: 1/2

R2CH-CR3' R2CH 1.4 ... 2.1 1 1 1

Prediction of 1H NMR spectra Molecule Idealized spectrum RCH2-CH3 RCH2-CH2R‘ RCH2-CHR2‘ RCH2-CR3'

Determination of coupling constants

•  The coupling constant JAX is defined as follows: JAX = Separation in Hertz of two neighboring lines of the multiplet A However, the spectra are typically labelled in the δ scale. To convert from the δ scale to the ν scale, the following relation holds:

δ = ΔνPr · 106 / ν0 ppm

•  Basically, you have to convert the separation of two neighboring lines in ppm to the difference in frequency JAX in Hertz.


•  A quartet appears in the NMR spectrum of acetaldehyde CH3-CHO at δ = 9.26 ppm (Spectrometer frequency ν0 = 300 MHz ).

The coupling constant for the quartet of CHO protons equals JCH,CH3 = 2,7 Hz (2,84 Hz). Reasoning: Separation of two neighboring lines of the quartet in a 300 MHz spectrometer: ≈ 0.009 ppm, therefore JCH,CH3 = (0,009 · 10-6 · 300 · 106 Hz) = 2,7 Hz


•  As an example, the following detail of a 1H NMR spectrum shows the doublet of the methyl protons in the molecule (CH3)3CH-O-CH(CH3)2 around δ = 1.05 ppm (Transmitter frequency ν0 = 60 MHz):

What is the value for the coupling constant JCH3,CH?

JCH3,CH = 6.1 Hz


The CH-protons of the same molecule (CH3)3CH-O-CH(CH3)2 give rise to a signal at δ = 3.56 ppm.

1.  What is the multiplicity of this signal? 2.  How large is the coupling constant JCH,CH3? 3.  List the relative intensities of the individual lines and make a table

with the distance of the individual lines from the line at the center of the multiplet (3.56 ppm) on the Δν and the Δδ scale. The proton frequency ν0 of the spectrometer used is 60 MHz.

4.  Draw the expected multiplet using the x-axis below !


1. The signal at 3.56 ppm has a multiplicity M of 7 (6 equivalent neighboring protons)

2. JCH,CH3 = JCH3,CH = 6.1 Hz, since the coupling between the methyl protons and the methine proton is mediated by the same bond electrons as the reverse coupling between the methine proton and the methyl protons.

Complementary coupling constants (i.e., JAX and JXA) are always identical!

Signal Δν [Hz] Δδ [ppm] Irel 0 0 0 20 1/1‘ 6.1 0.1 15 2/2‘ 12.2 0.2 6 3/3‘ 18.3 0.3 1


•  The coupling constant J is - in contrast to the chemical shift of a resonance signal - not dependent on the external magnetic field of the spectrometer, but rather on the magnetic fields of the neighboring nuclei.

•  Example: For the methyl protons in (CH3)2CH-O-CH(CH3)2the following coupling constants have been determined at different fields: JCH3,CH (at ν0 = 250 MHz) = 6.1 Hz (at ν0 = 600 MHz) = 6.1 Hz

Determination of coupling constants Doublets of the methyl protons in (CH3)2CHOCH(CH3)2

at ν0 = 60 MHz, J = 6.1 Hz (≈ 0.1 ppm) at ν0 = 100 MHz, J = 6.1 Hz (= 0.061 ppm)

In other words: If spectra are recorded at higher fields and - respectively - frequencies while using the δ scale, then the spacing of the multiplet lines becomes tighter, while the spacing of the signals remains constant. This effect is one of the reasons why spectra are recorded at increasingly higher frequencies (Currently up to 1000 MHz) - if there is a high degree of coupling, multiplets may start to overlap if the spectra are recorded at low fields (= frequencies). At higher fields the difference between chemical shift and spin-spin coupling effects increases, and the resolution of the spectra improves.

Coupling constants are field independent

60 MHz

200 MHz

600 MHz


•  The value of the coupling constant J depends on the number of bonds between the two coupling nuclei, as well as their orientation towards each other and character of the bonds between the two coupling nuclei

•  In the case of σ bonds spin-spin couplings typically only

be observed across three bonds or fewer (e.g., H-C-C-H). If there are four bonds between the two coupling nuclei, the coupling constant typically already drops towards zero.

•  So-called long range couplings across a larger number of bonds typically only occurs if more polarizable π-bond systems are involved. (e.g., H-C=C-C-H).

12-15 2-9

0 6,5- 7,5

5,5- 7,0 • aa 5-8 • ae 2-4 • ee 2-4

0,5- 3 7-12

13-18 4-10

0,5- 2,5 0

9- 13 2- 3

1- 3 2- 4

• o 6- 9 • m 1- 3 • p 0- 1

• 1-2 1,6-2,0 • 1-3 0,6-1,0 • 1-4 1,3-1,8 • 2-3 3,2-3,8

• 1-2 2,0-2,6 • 1-3 1,5-2,2 • 1-4 1,8-2,3 • 2-3 2,8-4,0

• 1-2 4,6-5,8 • 1-3 1,0-1,8 • 1-4 2,1-3,3 • 2-3 3,0-4,2

• 1-2 4,9-5,7 • 1-3 1,6-2,6 • 1-4 0,7-1,1 • 1-5 0,2-0,5 • 2-3 7,2-8,5 • 2-4 1,4-1,9

Coupling constants

The protons of the methyl group don't couple, since they are equivalents

CH3-CO-CH2Br The protons of the methyl and the methylene groups don't show any detectable coupling, since they are separated by more than three bonds

BrCH2-C≡C-H The methylene protons couple with the ≡C-H (JH,CH2 = 2.7 Hz), although they are connected through four bonds, since due to the C≡C triple bond the coupling becomes more polarized

Coupling constants •  Determine for these two molecules which proton groups will couple with

each other, and which ones won't

Coupling constants

No coupling J coupling

within CH3, CH2

Due to equivalency

Due to large separation

Coupling constants

No coupling J coupling

within CH3, CH2

Due to equivalency

Due to large separation

CH3-CHBr-CO-CH2-CH3 CH3-CHBr-CO-CH2-CH3 CH3-CHBr-CO-CH2-CH3 CH3-CHBr-CO-CH2-CH3

CH3-CHBr-CO-CH2-CH3 CH3-CHBr-CO-CH2-CH3

Coupling constants •  Now let's look at the following two molecules:

1.  Which proton groups in each of these molecules will couple with each other? 2.  To what types of multiplets with these couplings give rise? 3.  What are the expected values for the coupling constants J?

Molecule coupling A/X multiplets A/X JAX in Hz A) C(CH3)2/CH doublet/heptet 5.5 ... 7 B) CH2/CH3 quartet/triplet 6.5 ... 7.5

A B

Coupling constants

Under what conditions would you expect to see no coupling between two given groups of protons in a molecule?

No spin-spin coupling is observed if 1.  protons are separated by four or more single bonds i.e.,

H-C-C-C-H 2. protons are equivalent, i.e.,

Coupling constants All the examples that we have covered so far dealt only with the spin-spin coupling

between one group of equivalent protons with only one other group of equivalent protons. However, what happens if a group of protons interacts with two neighboring groups of equivalent protons?

Spin-spin coupling in neighboring protons: If proton A is connected by 3 or less bonds to: 1.  a single proton or a group of n equivalent protons X, then J-coupling between A and X leads to

a splitting of the signal from proton A with a multiplicity of

MA = nX + 1

2. several sets of equivalent protons M, X... (i.e. CHn

M- CHA=CHnX) with nM, nX, etc. protons. In that case the protons of these groups

couple independently with proton A. The splitting then occurs according to

MA = (nM + 1) * (nX + 1) ...

Coupling constants Try to determine the multiplicity of the signal for proton HA in the molecule

ClCH=CH-CH2Cl M A X

A.  M=2 B.  M=3 C.  M=4 D.  M=6 If you want to solve this problem, you have to clarify two basic questions: A.  Would you expect to detect coupling between HA and HM, or between HA and

HX? Oh yes, you would! In both cases coupling should occur, since HA and HM, as well as HX are not equivalent, and they are separated by only three bonds. We would also expect JAM to be larger than JAX, since in the former case the coupling is mediated through a π bond.

B.  Are HM and the HX equivalent to each other? Of course, they aren't! They are in very different chemical environments.

Coupling constants

The signal of the proton HA in

ClCH=CH-CH2Cl M A X

is split into six individual lines, since HM and HX are not equivalent. Therefore their spin-spin couplings to HA are independent from each other. HA gives rise to a multiplet with

MA = (nM + 1) · (nX + 1)

Continuing from here, what would be the ratio of the intensities of the individual lines in the HA signal be?

Coupling constants

For the molecule ClCH=CH-CH2Cl M A X

the number as well as the relative intensities of the multiplet lines for the signal of HA can be derived based on the coupling constants, as shown in this example: The given values for HA are:

δA = 4,86 ppm; JAM = 13,1 Hz; JAX = 7,3 Hz.

The coupling between HA and HM gives rise to a doublet (nM = 1; relative intensities 1 : 1). Due to the second, independent coupling with HX, the two doublet lines are additionally split into triplets (nX = 2; relative intensities 1 : 2 : 1). In total there are therefore six lines (MA = 6).

Coupling constants

Signal of proton A Coupling with

Coupling constants Try to predict the spectrum of CH3-CH2-CHO

1.  Determine the number and relative intensities of the signals, as well as their expected spectral position.

Number Groups Relative intensities Expected chemical shift(δ) -CH3 3 0,6 ... 1,9ppm 3 -CH2- 2 1,9 ... 3,2ppm -CHO 1 9,1...10ppm

Coupling constants Try to predict the spectrum of CH3-CH2-CHO

2. Determine the multiplicity of the signals and the relative intensities of the lines in a multiplet.

Group Coupling with J [Hz] Multiplicity Intensities CH3-CH2-CHO CH3-CH2-CHO ≈ 7 3 3/4 : 6/4 : 3/4

CH3-CH2-CHO ≈ 7 CH3-CH2-CHO + 4*2=8 see below

CH3-CH2-CHO ≈ 2 CH3-CH2-CHO CH3-CH2-CHO ≈ 2 2 1/4 : 2/4 : 1/4

Coupling constants

With CH3-

With CHO

Coupling constants

Coupling constants If a group of equivalent protons A couple with two (M, X) or more other groups of protons, then •  the number of the individual lines that are resolved in the spectrum and •  the pattern of relative intensities of the individual lines

of the multiplet signal from the protons A depends strongly on the ratio between the coupling constants JAM and JAX. Therefore every case requires careful consideration. In the following examples you will encounter some of the rules that govern these patterns. In a 1H-NMR spectrum you observe a doublet (1 : 1) and a quartet (1 : 3 : 3 : 1) of lines. Which of the following two molecules could give rise to such a spectrum?

D2N-CH2-CH2-COOD A) B)

Coupling constants

In addition to the signals from its aromatic protons in its 1H-NMR spectrum, a substituted benzene derivative of the composition C9H12 gives rise to • a doublet (1 : 1) and • a heptet (1 : 6 : 15 : 20 : 15 : 6 : 1). Can you propose a structure that is in agreement with these data?

Summary 1.  Indirect spin-spin coupling occurs between two or more groups of protons, if these protons are

a.  non-equivalent and b.  in sufficient proximity

(i.e., in σ systems they should be in general separated by three or fewer σ-bonds). 2. The multipilcity of the fine structure in the NMR signal of a group of protons A due to spin-spin coupling

a single neighboring group X with nX equivalent protons can be predicted by using the following formula:

MA = nX + 1 ,

with multiple distinct neighboring groups of protons M, X, ... with respectively nM, nX, ... equivalent protons can be predicted using the following formula:

MA = (nM + 1) · (nX + 1) ...

3.  The intensities of the individual lines of a multiplet caused by interactions with a neighboring group of protons follow a pattern similar to the factors in a n-th degree binominal equation.

4.  The coupling constant JAX is a measure of the intensity of the indirect spin-spin coupling between the two

protons A and X. It is defined as:

JAX = Distance between two adjacent individual lines in the multiplet signal of the proton group A, caused by the coupling with the group X, in Hertz.

The value of the coupling constant is independent of the working frequency of the spectrometer.

Pick the molecule that gives rise to the following 1H NMR spectrum !

a) b) c)

Reasoning: This question can already be decided by simply counting the number of resonance signals. Only the second molecule would give rise to two signals in the spectrum.

Substance Equivalent protons Number of signals

a)  - C(CH3)3 b) - CH2 – c) - CHO

3

a)  - C(CH3)3 b) –CO- CH3

2

a)  - CH2 - CH3 b) - CH(CH3)2 c) - CO - CH2 – d) > CH - CO -

4

a) CH3-CO-O-CH2-C≡C-CH2Br b) CH3-O-CO-CH2-C≡C-CH2Br c) CH3-CO-O-CH2-CH=C=CHBr


a) CH3-CO-O-CH2-C≡C-CH2Br b) CH3-O-CO-CH2-C≡C-CH2Br c) CH3-CO-O-CH2-CH=C=CHBr

Reasoning: Since molecule c) can be eliminated solely based on the number of signals in the spectrum, we only have to decide between options a) and b). The long-range coupling between the CH2-groups in both molecules a) and b) gives rise in either case to two triplets. The only difference that we can predict lies in the characteristic expectation ranges for the spectral postions of the CH3- and CH2-groups, respectively:

Molecule a) Molecule b)

CH3-group CH3 - CO - CH3 - O -

CH2-groups - CH2 - O - - CH2 - CO -


Number of signals

δ [ppm] (expected)

δ [ppm] (observed)

Multiplicity (expected)

Multiplicity (observed)

CH3-CO-O-CH2- C≡C-CH2Br

CH3 - CO - - O - CH2 - - CH2Br

3 1.7 ... 2.7 3.3 ... 4.5 2.3 ... 4.7

2.12 4.3 4.7

1 3 3

1 3 3

CH3-O-CO-CH2- C≡C-CH2Br

CH3 - O - - CO - CH2 - - CH2Br

3 3.3 ... 4.1 1.9 ... 3.2 2.3 ...4.7

2.12 4.3 4.7

1 3 3

1 3 3

CH3-CO-O-CH2- CH=C=CHBr

CH3 - CO - - O - CH2 - - CH = = CHBr

4


a) b)

a) b)


Number of signals

δ [ppm] (expected)

δ [ppm] (observed)



C6H5-CH2-CH2-O- CO-CH3

-CO-CH3 C6H5-CH2- -CH2-O-

3 1.7 ... 2.7 2.4 ... 3.3 3.3 ... 4.6

2.0 2.9 4.3

1 3 3

1 3 3

C6H5-CH2-CH2-CO- O-CH3

-CH2-CO- C6H5-CH2- -O-CH3

3 1.9 ... 3.2 2.4 ... 3.3 3.3 ...4.1

4.3 2.9 2.0

3 3 1

3 3 1


A B C D

Reasoning: All four candidates would give rise to three signals in their 1H-NMR spectra. The first and fourth molecule, however, can be eliminated right away based on the expectation ranges for their chemical shifts (their spectra contain a signal with δ > 8ppm). Besides, the relative intensities of the signals in their spectra (6 : 1 : 1) diverge widely from what we see in our spectrum (3 : 3 : 2). The decision in favour of the third molecule is based on the expected values for the chemical shifts. It is important to keep in mind that for both options b) and c) the methylene groups give rise to triplets (m=3), the neighboring methyls to quartets (m=4), and the second methyl groups to singulets (m=1).

A B C D


Number of signals

δ [ppm] (expected)

δ [ppm] (observed)



CH3-CH2-CO-O-CH3

-CH2-CH3 CH3-CH2-CO- -OCH3

3 0.6 ... 1.9 1.9 ... 3.2 3.3 ... 4.1

1.2 4.1 2.0

3 4 1

3 4 1

CH3-CO-O-CH2-CH3 -CH2-CH3 -O-CH2-CH3 CH3-CO-

3 0.6 ... 1.9 3.3 ... 4.6 1.7 ...2.7

1.2 4.1 2.0

3 4 1

3 4 1


CH3-CHBr-CO-CH3 CH2Br-CH2-CO-CH3

a) b)

Reasoning: Since both molecules have the same expected number of signals at approximately the same expected positions in the spectrum, it becomes necessary to look at the fine structure of the signals - as well as their respective relative intensities - if we attempt to assign the proper molecule to the spectrum. • The two methyl groups -CH2-CH2- would give rise to two tripletts in the spectrum, while • the >CH-CH3 group would generate a doublet and a quartet .

CH3-CHBr-CO-CH3 CH2Br-CH2-CO-CH3

Molecule Equivalent protons δ [ppm] (expected)

Integral (expected)


CH3-CHBr-CO-CH3

-CHBr-CH3 -CO-CH3 -CHBr-

0.6 ... 1.9 1.7 ... 2.7 2.8 ... 5.8

3 3 1

2 1 4

CH2Br-CH2-CO-CH3 -CO-CH3 -CH2Br-CH2-CO- -CH2Br

1.7 ... 2.7 1.9 ... 3.2 2.4 ... 4.7

3 2 2

1 3 3

Match the spectra with the proper isomer of the chemical formula!

Formula: C3H5Cl3 (A)

(B)

CH3-CHCl-CHCl2

CH3-CCl2-CH2Cl

Reasoning: Starting with the composition C3H5Cl3, we can suggest the following isomers:

1.CH3-CH2-CCl3 2.CH3-CHCl-CHCl2

3.CH2Cl-CH2-CHCl2 4.CH3-CCl2-CH2Cl 5.CH2Cl-CHCl-CH2Cl

Spectrum (A): A spectrum containing two doublets and one doublet of quartets (8 individual lines) can only be generated by a molecule of the following structure: | |

CH3-CH-CH- M A X

since only in this case will the signal of HM be split due to coupling with HA (nA = 1, and therefore MM = 2) into a doublet. The same holds true for HX. Finally, under these circumstances the proton HA will couple to both HM and HX (nX = 1 and nM = 3, which leads to MA = 4 · 2 = 8), splitting the signal into a doublet of quartets. From these considerations follows that spectrum (A) has to belong to isomer 2.

Reasoning: Starting with the composition C3H5Cl3, we can suggest the following isomers:

1.CH3-CH2-CCl3 2.CH3-CHCl-CHCl2

3.CH2Cl-CH2-CHCl2 4.CH3-CCl2-CH2Cl 5.CH2Cl-CHCl-CH2Cl

Spectrum (B): The fact that we observe only two singulet signals in this spectrum indicates that there is no coupling between the two groups of equivalent protons. This limits the possible structures to the type |

HC – C - CH |

In our case that would be the isomer 4.


Formula: Note: The group C5H11 is linear!

A

B

5

2 3

6

5

2

9



1. 2.

3. 4.

5. 6.

7. 8.



A

B

5 2

6

3

9 2

5

1. 2.

3. 4.

5. 6.

7. 8.

Spectrum (A) The characteristic multiplicity of the signals in the spectral range δ = 0.5 ... 2 ppm (a triplet and a quartet) can only be caused by a structure of the type

CH3-CH2-

This motif occurs in the isomers 1, 3, 4, 6 and 8. But only for the isomers 6 and 8 would we expect three signals from the aliphatic side chains. (in the case of isomers 1, 3, 4 we would expect signals from five non-equivalent groups of protons in this spectral range) To distinguish between the two remaining options, we will have to consider the single signal at δ = 1.25 ppm. Since this signal is a singulet (Multiplicity M = 1), only isomer 6 could give rise to it. In this molecule the methyls CH3 of the group are equivalent and couple with no other group, thus being represented by only a single signal in the spectrum.

1. 2.

3. 4.

5. 6.

7. 8.

Spectrum (B) Matching this spectrum to isomer 7 is rather straight forward based on the number of aliphatic signals (2 groups of equivalent protons) and the lack of spin-spin coupling between the signals.

Based on the 1H-NMR spectrum given below, try to propose a possible structure for a molecule!

Composition: C5 H7O2N Based on its chemistry, you also know that the molecule contains a cyano group -CN and a carboxyl group -CO-O-

5

2

3

2

Reasoning:

As you should know by now, there are two approaches to solve this type of problem:

A.  Starting with the experimental spectrum, you try to derive as many

characteristics of the molecule that gave rise to it as necessary to come up with

an unambiguous structure

B.  Based on the composition of the molecule you make up a complete list of all the

possible isomers. You then develop the theoretical 1H-NMR spectra of all the

isomers and compare them to the experimental spectrum.

We will use the present problem to demonstrate the first approach.

1.  Determine the corresponding multiplets, i.e., the signals of groups of protons that couple with each other. These pairs (or triples etc.) of multiplets should exhibit the same coupling constant J in the fine structure of their peaks. In our case that would be the quartet and the triplet.

2.  Identification the structural element that gives rise to these coupled signals. In other words, determine the number of protons in the groups of equivalent protons that couple to each other. In the case of the quartet and the triplet, that would be

-CH2-CH3

Composition: C5 H7O2N

2

3

2

3. Remove this identified element from the chemical formula. Then try to identify the signals of the remaining protons in the spectrum. After removing the cyano group -CN and the carboxyl group -COO- (Their presence has been previously determined by chemical analysis as well as the identified structural element CH2-CH3 from C5H7O2N, we are left with the group -CH2- (with a chemical shift δ = 3.42 ppm).

4. Analysis of the signal integrals to determine the number of the structural elements determined. In this case it is only necessary as a second check of our results so far.


5. Combination of the identified structural elements as well as the elements that are based on the chemical composition and behavior, but which do not show signals in the 1H-NMR spectrum. This may lead to a unique structure, or - as in our case - to two or more possible structures. Now we can use chemical shifts, intensities and coupling constants to select the correct structure from the two candidates

a) CH3-CH2-CO-O-CH2-CN b) NC-CH2-CO-O-CH2-CH3


2

3

2

The crucial information in this case is the chemical shift of the -CH2- protons in the -CH2-CH3 group.

a) CH3-CH2-CO-O-CH2-CN b) NC-CH2-CO-O-CH2-CH3

a) δCH2-CO = 1.9 ...3.2 ppm b) δCH2-O- = 3.4 ... 4.5 ppm.

The experimental value is 4.25 ppm, which is in good agreement with the chemical shift expected for an esther group.


2

3

2


Composition: C10H10O Based on its chemistry, you also know that the molecule contains a phenyl group. The remaining part of the molecule is linear.

5

2

2

1

Because of the valences, the total number of rings

and double bonds in a molecule of the formula

CxHyNzOn will be:

x – 1/2 y + 1/2 z + 1


Composition: C10H10O Based on its chemistry, you also know that the molecule contains a phenyl group. The remaining part of the molecule is linear.

n.i.c.=10-5+1=6 5

2

2

1

Solution:

Reason: With the solution of this question another possibility of the way A is to be used. The compound contains a phenyl group of C6H5. Thus C4H5O must be looked up for univalent

remainder, fitting the spectrum. The small number of H-atoms suggests that multiple bonds and thus possibly "long range" couplings.

1.  Predicates from the chemical shifts of the signals:

δ in ppm Possible groupings

2,3 (triplet) -CH3 (aromatically) CH3-CO- CH3-C≡C- -CH2-C=C -CH2-CO- -OH H-C≡C

4,1(Doublet) CH3O- -CH2O- =CH-O- -OH 4,6 =CH-O- -CH2O- -OH =C=CH2

2. Analysis of intensities and couplings:

The multiplicity of the signals with δ = 2.3 ppm and δ = 4.1 ppm shows that CH2- group with a CH-group couples: HC- · · · · -CH2- . The intensities acknowledge this findings.

From the structures possible for the first signal is applicable thus only C≡C-. Altogether we maintain:

HC≡C-CH2-O-.

The singlet signal with δ = 4.6 ppm with same intensity as the doublet can come

then only from not coupling CH 2 - group. Check on the basis the sum formula:

Since the univalent remainder contains only four C-atoms, the following structure corresponding with the spectrum comes into consideration:

HC≡C-CH2-O-CH2-


Composition: C4H10O; the molecule is linear!

6

1

2

1

Because of the valences, the total number of rings

and double bonds in a molecule of the formula

CxHyNzOn will be:

x – 1/2 y + 1/2 z + 1

In our case (C4H10O):

4 –5 + 1 = 0

Isomer Signals Groups δ [ppm] Irel. Multiplicity 1. CH3-CH2-CH2-CH2-OH 5 2.

4 (CH3)2

>CH- CH2 -OH

≈ 1 ≈1.8 ≈3.6

up to 5.2

6 1 2 1

2 9 (J =) 4(2*2)

3 3.

5

4.

2

5. CH3-CH2-O-CH2-CH3 2 6. CH3-O-CH2-CH2-CH3 4 CH3

CH2 CH3-O- CH2-O-

≈1 ≈1.5 ≈3.6 ≈3.8

3 2 3 2

3 6 (J=)

1 3

7.

3

Based on the 1H-NMR spectrum given below, try to propose a possible structure for a molecule! Composition: C5H10; and the molecule is linear! Spectrum:

2 2

3

3

Answer:

Reasoning:

Isomer Number of signals Irel. Multiplicity

1.  H2C=CH-CH2-CH2-CH3 6 2. H3C-CH=CH-CH2-CH3 5 3 : 1 : 1 : 2 : 3 2 / 8 / 6 / 8 / 3 3. 5 1 : 1 : 1 : 1 : 6 4. 5 1 : 1 : 3 : 2 : 3 2 / 2 / 1 / 4 / 3 5. 4

The isomer 4 is the best fit for the spectrum. The problem was complicated by the fact that the two protons H2C= are exposed to very similar chemical environments, and therefore show very similar chemical shifts.

n.i.c.=8-4+1=5

Exch.

7.36+0.53-0.07=7.82

7.36-0.45+0.18=7.09

7.36+0.77-0.11=8.02

7.36-0.18+0.11=7.29

n.i.c.=5-6+1=0

n.i.c.=8-9/2+1/2+1=5

Exch.

Exch.

7.36+0.60-0.22=7.74

7.36+0.10-0.62=6.84

7.36+0.20-0.22=7.34

7.36+0.10-0.71=6.75

7.36+0.60-0.62=7.34

7.36+0.10-0.22=7.24

7.36+0.20-0.71=6.85

7.36+0.60-0.71=7.25

n.i.c.=13-9+1=5

Exch.

n.i.c.=13-9+1=5

Second Order Spectra

Δδ/J =40

Δδ/J =5

Δδ/J =1.7

Δδ/J =0.3

Δδ/J =0

An example for this effect is the signal of the ethylene protons in At a spectrometer frequency of ν0 = 60 MHz they generate a spectrum that

wildly deviates from what is expected in a first order spectrum (Three doublets of doublets with relative intensities of 1:1:1:1). In a 60 MHz spectrometer, the signals appear at the following positions:

νA = 344 Hz νB = 364 Hz νC = 372 Hz JAB = 11.75 Hz JBC = 0.91 Hz JAC = 17.92 Hz

Compare this to a spectrum recorded at a higher field (Same coupling constants J, but different frequencies ν):

JAB = 11.75 Hz JBC = 0.91 Hz JAC = 17.92 Hz

"First-order"-spectra can only be expected if : ΔAXν > 6 · JAX

5. Spin-spin couplings between protons and other magnetically active nuclei (e.g., 31P, 13C or 19F) can lead to additional line splittings. An example here is the coupling between 1H and 19F in CH3-CF2-CH2Cl:

From what we have coverd of 1H-NMR spectroscopy, you might have come to the impression that this method is straight-forward. But what you have encountered so far are barely the bunny sops of NMR, with unambigious spectra that are easy to analyze. The typical NMR spectrum that you might have to face in the real world is much more complicated, and therefore much less easily interpreted. Here is a short list of complications that you might run into:

1. The individual lines of signals with high multiplicities (M > 5) typically have very low intensities. That means that sometimes not all the lines are above the noise level in a recorded spectrum. One example is the multiplet of the CH-group in (CH3)2CHCH2OH where M = 21

2. Multiplets may overlap and thus give the impression of complex line splittings.

3. "Long-range"-couplings (i.e., spin-spin couplings across more than three bonds) may show stronger effects than what you have been shown in this tutorial so far (That is especially the case if π bonds are involved!). An example for this effect is the effect of the coupling between the two CH2 groups in CH3COOCH2CH=CHCH2OH.

4. Real-world spectra are not limited to first-order spectra. Whenever the diference in frequency between two coupling groups of protons ΔABν is of similar magnitude as the coupling constant JAB of these two groups, the relative intensities of the individual lines in the multiplets changes dramatically. Under these circumstances these patterns cease to follow the simple rules we have described in this tutorial.

Still, even under conditions such as these you can use what you have learned so far about 1H-NMR spectra in the structural determination of molecules:

a.  Based on the absence or presence of signals in specific regions of

the spectra you can draw conclusions with respect to the absence or presence of the respective groups in the molecule.

b.  And the analysis of limited regions of the spectrum with first-order character may deliver answers to at least some structural questions.

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