generalchem_ls_17.pdf

8/10/2019 GeneralChem_LS_17.pdf

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NMR spectroscopy: Shielding

BITSPilani, Pilani Campus

Electrons surrounding the nucleus also produce magnetic field.

Badd

= -

B

Blocal

= B + Badd

= B(1-

)

N Blocal / 2 N B / 2

Resonant frequencies being dependent on strength ofexternal magnetic field, are inconvenient to remember.

How to get rid of field-dependence?


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Chemical Shift


Resonant frequency recorded for a reference compound (denotedby

0).

Si(CH3)

4 (Tetramethyl silane) is commonly used reference in NMR

spectra of organic compounds. (Highly shielded, single intense peak,

volatile)

Shift in the resonant frequencies of sample with respect to reference

is recorded.

Chemical shift:

= {(

)/

}

106ppm


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Chemical Shift: Ethanol


TMS


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Factors affecting Chemical

Shift


Reference (TMS) protons are highly shielded. Hence, larger chemical

shift indicates greater extent of deshieldinglower shielding

constant.

Shielding constant: = local

neighbour

solvent

local

: due to the shielding from the electrons immediately

surrounding the nucleus.

neighbour

: due to the shielding from the neighbouring

environment

solvent

: the contribution from solvent molecules.


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Chemical Shift: Ethanol


TMS

Intensity of peak proportional to number of protons corresponding to

the peak: Integrate to calculate area under the peak 1:2:3


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Phenylacetone: PMR


CH2C

O

CH3

The number of signals = number of distinct sets of equivalent protons


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Benzene: PMR spectrum


Single peak: (all protons equivalent)

Chemical shift indicates high

deshielding

How will the specta of methane,

ethane, ethene, ethyne look like?

Position of the peak in the spectrum?


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Number of peaks, intensities


Number of peaks equals the number of distinct sets of

equivalent protons.

Equivalent protons: Two protons are equivalent if the

compounds resulting from replacement of either of them by

identical substituent will be chemically identicalsamemolecular formula, structural formula and geometry

Intensity of a peak directly proportional to the number of

equivalent protons responsible for the peak.


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Equivalent protons


Two protons are equivalent if the compounds resulting from

replacement of either of them by identical substituent will be

chemically identicalsame molecular formula, structural formula and

geometry.

Homotopic hydrogen atoms: If in making these replacements, we

get the same compound, the hydrogen being replaced are said to bechemically equivalent or homotopic. Homotopic atoms (or groups)

are chemical shift equivalent.

In H2C=C(CH3)2 replacement of H with Cl gives

H2C=CCH2Cl(CH3) + ClHC=C(CH3)2Six methyl hydrogens form one set; replacing any one of them with

chlorine, for example, leads to the same compound. Two vinyl

hydrogens form another set; replacing either of these leads to same

compound. So 2-metylpropene gives two 1H NMR signals


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Equivalent protons


Enantiotopic hydrogen atoms: If replacement of each of two

hydrogen atoms by the same group yields compounds that are

enantiomers, the two hydrogen atoms are said to be enantiotopic.

Enantiotopic hydrogen atoms have the same chemical shift and give

only one 1H NMR signal.

In CH3CH2Br, two hydrogen atoms of -CH2Br group are enatiotopic.

Diastereotopic hydrogen atoms: If replacement of each of two

hydrogen atoms by the same group gives compounds that are

diastereomers, the two hydrogen atoms are said to be diastereotopic.Diastereotopic hydrogen atoms do not have the same chemical shift

and gives rise to different 1H NMR signals.

The two protons of =CH2group of ClHC=CH2 are diastereotopic.


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Examples to solve:


How many peaks will be present in the PMR spectra of the

following molecules? What will be the ratio of intensities?

HCHO 1 -

CH3COCH

3 1 -

CH3CHO 2 3:1CH

3COCH

2CH

3 3 3:2:3

CH3CH=CHCH

3 2 3:1

(CH3)

2CH-CH

2CH

3 4 6:1:2:3

CH3COOCH3 2 1:1CH3CH

2CHClCH

3 4 3:2:1:3


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Factors affecting Chemical

Shift


Reference (TMS) protons are highly shielded. Hence, larger chemical

shift indicates greater extent of deshieldinglower shielding

constant.

Shielding constant: = local

neighbour

solvent

Deshielding due to electronegative elements

Anisotropic effect : Shielding/deshielding due to ring currents

Other effects: Hydrogen bonding, solvent effects, etc.


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Deshielding due to

electronegative elements


CHCl3 CH2Cl2 CH3Cl

7.27 5.30 3.05 ppm

-CH2Br -CH2CH2Br -CH2CH2CH2Br

3.30 1.69 1.25 ppm

No. of electronegative atoms Distance from the electronegative atom


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Anisotropic effect: ethene, ethyne


deshielded

shielded


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Solvent contribution (solvent)


A proton on the solute molecule is shielded because of the anisotropiceffect produced as a result of ring current of benzene as solvent


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Typical Chemical Shift ranges



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Fine structure (Spin-spin coupling)


Ethanol

Each magnetic nucleus contributes to the local magnetic field of

other nuclei.

Spin- spin coupling constant J (Hz) is independent of the applied

magnetic field.

Consider AX where A and X are spin nuclei.


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Consider the protons A and X (or in general spin nuclei)

In absence of any interaction between A and X, the total energy of thesetwo protons in a magnetic field

B

(neglecting spin-spin coupling) is

E = -N(1-A)BmA- N(1-X)BmX

Where,

Aand

Xare the shielding constants of A and X.

If the nuclei A and X are close enough in the molecule (separated by oneto three bonds, for example), their magnetic moments will perturb eachother. The spin-spin interaction energy will be

Espin-spin= hJAXmAmX

Etotal

= E + Espin-spin

= - NB{(1-A)mA+ (1-X)mX} + hJAXmAmX

Four energy levels are obtained.


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In absence of spinspin coupling

In presence of spinspin coupling


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21/23



AX2

1 : 2 : 1

The two X nuclei may

have the 22= 4 spinarrangements. Themiddle twoarrangements areresponsible for the

coincident resonance ofA.


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AX3

1 :3 : 3 : 1

There are 23= 8arrangements of thespins of the three X

nuclei, and theireffects on the Anucleus gives rise tofour groups ofresonances.


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N equivalent spin-1/2 nuclei split the resonance of a nearby spin

or group of equivalent spins into N+1 lines with an intensitydistribution. The relative intensities are given as coefficients of

the binomial expansion (1+x)n

This is given as Pascals triangle1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

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