Proton NMR.pdf

7/29/2019 Proton NMR.pdf

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NMR SPECTROSCOPY

Handout 2

Felipe Garcia

[email protected]


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Proton NMR

The scale of the chemical shift1H NMR is smaller than that in

13C NMR. Proton NMR spectra get

complicated really quickly due to the coupling. However,1H NMR is incredibly useful since a lot of

information can be derived from the spectra.

Most organic compounds resonate between 0 and about 14 ppm. As we saw for13

C NMR, the

chemical shift depends on the local magnetic field a nucleus experiences. All the explanations and

principles applied to 13NMR are also applicable to 1H NMR:

For example if we have a look at the proton NMR of ethylbenzene

As we can see there are signals at very distinctive areas. The chemical shifts tent to be difficult to

predict and interpret. However, we can make certain generalizations about where certain groups and

resonances will appear in the spectrum.

The magnetic field experienced by a proton is influenced by various structural factors. Since electrons

shield the nuclei from the external magnetic field, the less electron density around an atom, the less

shielded it is, and hence it resonates a higher frequencies with a larger chemical shift.


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However some shifts seem counterintuitive, such as the chemical shifts of protons on a benzene rings

are found at higher chemical shifts than protons on a double bond, but the the ones attached to a triple

bond are found at much lower chemical shift. This is the way the electron move in a system.

Electronegative or electropositive substituents affect the chemical shifts, the presence ofelectropositive atoms (donating) or electron withdrawing atoms affect the local electrondensity of the nucleus of study

X C H X C HLi -14 -1.94 NH2 26.9 2.47SiMe3 0 0 OH 50.2 3.39H -2.3 0.23 F 75.2 4.27Me 8.4 0.86 SMe 19.3 2.09Et 15.4 0.91 Cl 24.9 3.06

Hydrogen is more electropositive than carbon , with the result of every replacement of hydrogen by an

alkyl group causes a downfield shift in the resonance of that carbon and any remaining hydrogens on

it.

CH4 MeCH3 Me2CH2 Me3CH Me4C

H 0.23ppm 0.86 1.33 1.68C -2.3ppm 8.4 15.9 25.0 27.7

For example, in the case of H3COOCH2CH3

CH3 protons (A) not deshielded. But next to a CH2 group..slightlyShielded..upfield

CH3 protons (B) 3 bonds away from Oxygens..less deshielded than C..downfieldfrom A

CH2 protons (C) next to a electronegative O, just two bonds awayhighly deshielded -downfield

Aromaticity and (magnetic anisotropy): There are some types of nuclei whose chemicalshifts are not easily explained by simple considerations of the electronegativity of the groups

attached. The anomalous shift is due to the presence of an unsaturated system (one with -

electrons) in the vicinity of the nuclei in question. All groups in a molecule that have -

electrons generate an anisotropic field. In electromagnetic terminology an isotropic field is


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one of either uniform density or spherically symmetric distribution; and anisotropic field is

not isotropic (not uniform). All the groups in a molecule that have -electrons generate

secondary anisotropic fields.

For example, in the previously mentioned

ethyl benzene the aromatic protons are

found between 6.0-9.5

For example the protons on a benzene ring typically have larger magnetic fields than protons

on a C=C double bond, but protons on a triple bond have much smaller shifts. This is due to

the way the electrons in the system of the aromatic ring move in a magnetic field.

When the benzene ring is placed in a magnetic field the electrons in the -system of the

benzene ring move in such a way as to generate a magnetic field that opposes the field applied

at the centre of the ring. This field is said to be due to what is called ring current. Thus a

proton attached to a benzene ring in influenced by three magnetic fields:

i) the strong field applied by the electromagnets of the NMR spectrometer

ii) a weak magnetic field by the shielding by the valence electrons around the proton

iii) a weak magnetic field due to the anisotropy generated by the ring-system -electrons

This field is opposed to the applied field in the centre of the ring, but is aligned with the

applied field for protons on the outside of the ring. Is this anisotropy effect ring current -

that gives the benzene protons a chemical shift higher than expected.


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CH2H2C

H2C CH2

H2C

ringcurrent6electrons

= 1.0ppm

= 2.0ppm

H

H

H

H

H

H

H

HH

H

H

H

H

H

H

H

= 1.8ppm

= 8.9ppmringcurrent18electrons

bonds (magnetic anisotropy).Similar ring currents are set up by the -electronsof an alkyne; but is this case the proton

attached to an alkyne is in the region where the magnetic field is decreased slightly.

The same effect discussed for benzene - aryl protons are found between 6.0-9.5- also happens

for vinylic protons-deshielded to a lesser extend than aryl protons.

Vinylic, R2C=CH2 4.6 - 5.0ppm

Vinylic, R2C=CRH 5.2 - 5.7ppm

In general the anisotropy could be represented as follows :

(-) (-)

(+)

(+)

(-) (-)

(+)

(+)

C C C C(-) (-)

(+)

(+)

C O HH

(-)

(-)

(+) (+)

The -bonded electrons of a -bonds create a magnetic field around them that causes a

downfield shift of the signals from trigonal (sp2) and diagonal (sp) carbons and of the signals

from the protons attached to them. All the protons falling into the conical areas are shielded

(lowed chemical shifht), the ones falling outside the conical areas are deshielded (higherchemical shift).


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A few examples of chemical shifts for carbon and proton are given in the table below,

C H C HCH3H -.23 0.23 CH3CHO 31.2 2.20CH3CH=CH2 22.4 1.71 CH3COCH3 28.1 2.09CH3CCH 5.9 1.80 CH3CN 1.30 1.98CH2=CH2 123.3 5.25 CH3CHO 199.7 9.80CH3CCH 66.9 1.80 CH3COCH3 206.0CH3CCCH3 79.2 CH3CN 117.1

Conjugation. A alkyl group is donating and has not effect on the framework. Whereas amethoxy group is donating and inductively withdrawing. The effect on the chemical shift is

similar, with the MeO group more powerful than the Me group

OMe OMe

inductiveeffect-withdrawingconjugativeeffect

-donating

H Me OMe

H Me OMe

H

H

H

H

H

H

H

H

H

H3.85

HH5.73 H6.38

H4.88

C123 C133.9 C152

C115.4 C84.4

Both groups induce similar effects but the influence of the -overlap is greater that the inductive

effect in the -framework.

C C

H

H

H

X

X Electron nature C C H HH Reference compound 123.3 123.3 5.28 5.28

Me Weak and donor 115.4 133.9 4.88 5.73

OMe donor and -aceptor 84.4 152.7 3.85 6.38

Cl -aceptor, weak donor 117.2 125.9 5.02 5.94

CH=CH2 Single conjugation 130.3 136.9 5.06 6.27

SiMe3 aceptor, donor 129.6 138.7 5.87 6.12

COMe aceptor, aceptor 129.1 138.3 6.40 5.85

Exchange and hydrogen bonding.The shifts for protons connected to oxygen or nitrogen are very variable and can come out almost

anywhere. Some appropriate shifts are shown below,


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Despite of being so variable the NH or OH signals are very easy to spot. They are much broader

than the resonances for protons attached to carbons. The more hydrogen bonding there is the more

the proton is deshielded. Another reason why these signals are easy to identify is that on shaking

with a little deuterium oxide, D2O, the signals from O-H and N-H disappear.

Why would a signal disappear?

Consider the alcohol case for example: R-OH + D2O R-OD + HOD

During the hydrogen bonding, the alcohol and heavy water can "exchange" -H and -D amongst

each other, so the alcohol becomes R-OD. Although D is NMR active, it's signals are of different

energy and are not seen in the H-NMR, hence the signal due to the -OH disappears. (Note that

the HOD will appear...).

Spin-spin coupling:

In the case of protons NMR the coupling follows the same rules as seen before, but in this case there

is coupling with other proton. We have already seen 13C-1H coupling what happens when two nuclei

with I=1/2 are coupled. Much the same is true for proton-proton coupling, except that we are now

observing longer range coupling principally two bond (2J) and three bond coupling (

3J)

HA

C

HB

A

C C

B

Consider the 1H-NMR of ethanol; CH3CH2OH,


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The multiplicity of a multiplet (whether it is a triplet or doublet and so on..) is given by the number of

equivalent protons in neighboring atoms plus one, i.e. the n + 1 rule (where n is the number of

neighboring atoms)

Hb

Hb

Cl

Ha

Cl

Cl

Hb

Hb

Cl

Ha

Cl

Cl

two neighbours

give a triplet

(n+1 = 3)(Area= 1)

one neighbours

give a doublet

(n+1 = 2)

Hb

Hb

Cl

Ha

Ha

Ha

Hb

Hb

Cl

Ha

Ha

Ha

two neighbours

give a triplet

(n+1 = 3)(Area=3)

Three neighbourgs

give a quartet

(n+1 = 3)(Area=2)

Ha

Hb

CN

Hc

The relative intensities of the lines in a coupling pattern is given by a binomial expansion or more

conveniently by Pascal's triangle as already discussed for13

C NMR

As seen before eequivalent nuclei do not interact with each other. The three methyl protons in

ethanol cause splitting of the neighboring methylene protons but they do not cause splitting among

themselves. The relative rations follow what we saw previously using Pascals triangle.


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So for H-NMR a proton with zero neighbours, n = 0, appears as a single line, a proton with one

neighbour, n =1 as two lines of equal intensity, a proton with two neighbours, n = 2, as three lines of

intensities 1 : 2 : 1, etc.

For example, (CH3)2CHCl

There are two different environments: environment a (composed by one Hproton)environment b (composed by six Hprotons)

(CH3)2CHCl: the environment composed by H couples to six protons H in environment bgiving rise to a septet.

(CH3)2CHCl: the environment composed by H couples to one proton H in environment agiving rise to a doublet

(CH3CH2)2CO


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There are two different environments: environment a (composed by three Hprotons)environment b (composed by two Hprotons)

(CH3CH2)2CO: the environment composed by Hprotons couples to two protons H inenvironment b giving rise to a triplet.

(CH3CH2)2CO: the environment composed by H protons couples to one three H inenvironment a giving rise to a quartet.

More complicated coupling patters

More complex splitting patterns can be seen when a nuclei couples with more than when Hb isadjacent to nonequivalent Ha on one side and Hc on the other, the resulting coupling gives rise to a

doublet of doublets.

In the proton nmr of acrylonitrile, all the hydrogens are different - Ha is on the same side as (orcis to)

rhe cyanide group, whereas Hb is opposite (or trans). The orange integrals indicate that there are three

different environments (Ha, Hb and Hc)

Ha

Hb

CN

Hc


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But unlike in previous examples, Hc couples to Ha and Hb (which are different, one cis and one trans

to Hc). Therefore two different coupling constants will be observed Ha-Hc and Hb-Hc. Thefore rather

then a triplet, the signal will split into a doubletdue to Ha-Hc and consequently into anotherdoublet

Hb-Hc; this leads into the formation of a doublets of doublets.

If we construct a tree diagram for the doublet for doublets the resulting signal patter will be

independent of the order in which we consider the coupling constants

Another example will be ethyl propenoate, an unsymmetrical terminal alkene, the three vinylic

hydrogens are nonequivalent due to the restricted rotation about the double bond.

a tree diagram for the complex coupling of the three vinylic hydrogens in ethyl propenoate


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Identifying multiplets - coupling constants

Multiplets can be easily distinguished from closely spaced chemical shift signals. They will have the

same separation between the signals - same coupling constant

To calculate J we can use the simple formula

Operating Frequency (MHz) x Splitting (ppm) = Coupling Constant (Hz)

For a quartet like the one on the left you there are 3

splitting values (x-y), (y-z) and (z-g) and final value is

taken from the average , final J value expressed usually

expressed with just one decimal place

Splitting is a scalar quantity so there is no sign

Remember - Coupled protons have same J values

The coupling between two atoms

The coupling between two nuclei appears on the spectrum as the separation between two lines and is

measured in Hz. The coupling constant between two atoms coupled to each other is the same.

Whether is the same spectrum or a different one. Remember the example for fluoroacetic acid

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Another example where we can observe the splitting between adjacent proton environments is

CH3CH2Br

We can observe how under every peak there is a number, this number indicates what is called the

integral of the peak

Equivalent hydrogens:

have the same chemical environment a molecule with 1 set of equivalent hydrogens gives 1 NMR

signal

a molecule with 2 or more sets of equivalent hydrogens gives a different NMR signal for each set


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Integration of signals

The NMR spectrum not only distinguished how many different types of protons in a molecule but also

reveals how many of each type are composing each environment. In NMR the area under the peak is

proportional to the number of hydrogens generating that peak.

For example, in the case of 2-butanone

The ration showed are 1.99 : 3.04 : 2.96 that approximates 2:3:3.

If we increase the number of carbon atoms and we move from bromo-ethane to bromo propane, the

spectra will look like the following:

-----------------------------------

Further reading:


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Spectroscopic methods in organic chemistry, 5th edition, Dudley H. Williams and Ian Fleling,

McGraw-Hill (1995)

Structure determination of organic compounds, 3rd

edition, E. Pretsch, P. buhlmann, C. Affolter and

M. Badertscher, Springer-Verlag (2001).

Structural Metods in Inorganic chemistry, 2nd

edition, E. A. V. Ebdsworth, David W. H. Rankin and

Stephen Cradocl, CRC Press (1991).

Spectrometric Identification of Organic Compounds, 5th edition, R. M. Silverstein, G. C. Bassler and

T. C. Morrill, Wiley (1991).

Spectrometric identification of organic compounds, 7th

edition, Robert M. Silverstein, Francis X.

Webster, David J. Kiemle, Chapter 3

Essential NMR for scientists and engineers Blumich, Bernhard.

Modern NMR spectroscopy. Oxford U.P

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