the chemical shift - uppsala universityboc2.boc.uu.se/boc14www/compendium/lec4.pdf · the chemical...

NMR Course, Lecture 4 14.29 April 14, 1999

Lecture 4

The Chemical Shift

Chapters (pages) in 'Modern NMR Techniques for Chernistry Research' by A. E. Derorne :

.Notbing unfortunately

~ Other useful books or scientific papers related to the topics in this lecture :

.'Physical Chemistry' by P. w. Atkins : Ch. 20.1

.'Proton and Carbon-13 NMR Spectroscopy. An Integrated Approach' by R. J. Abraham and P.Loftus : Ch. 2

~ .'Nuclear Magnetic Resonance Spectroscopy. A physicochemical View' by R. K. Harris :Ch. 1.8- 1.10 and Ch. 8.1 -8.16

.'N MR Spectroscopy' by H. Gtinther : Ch. 4.1

.Many books on chemistry, spectroscopy and physical chemistry in general cover the basics ofNMR including chemical shifts and their origin

11

NMR Lecture 4 14.55 April 12. 1999

With a slight oversimplification one can say that virtually everything that produces a magnetic field willinfluence the chemical shiit (the resonance frequency) of a given nucleus.The most important influence on the resonance frequency of a nucleus :

-A nucleus is shielded by the electrons surrounding it-The electrons create a small magnetic field opposed to Bo. the external field coming from the

magnet. due to the circular motion of the electrons around the nucleus induced by BO

-Since the new field is opposed to BO the nucleus will 'feel' a magnetic field slightly smallerthan BO

Bett = BO -Bshield

Bett refers to the 'felt' magnetic field. the one which determines the Larmor frequency. Vo

-The above formula can also be written as

Beft = BO -BO * 0"

Where 0" is the shielding constant From this we can see that the Bshield is dependendent onBO. the external field

-The higher the electron densityaround our nucleus. the higher is Bshield or 0" and the moreshielded our nucleus will be

-For a hydrogen atom where the distribution of electrons is spherical (in atoms. does not

happen in molecules) since there are only s-electrons the 0" is purely due to diamagnetic effects

and therefore O"=O"d

-In C W NMR the external field Bo must be increased to compensate for Bshield. Therefore.nuclei experiencing large Bshield are said to resonate upfields. In Fr NMR this corresponds

to small ~values. Normally the right side of the spectrum

-Accordingly. the left part of our spectrum is called the downfield region. BO is small in C W

NMR and the ~values are high in Fr NMR

'--J

In molecules and in atoms with p-ele(:trons (unsyrnmetric and non-spherically distributed), the

total 0" of a nucleus is also affe(:ted by circulation of ele(:trons within the entire molecule

These 'molecular' electron movements cause a reduction of 0" for our nucleus and this is called a

paramagnetic shift, O"p

Paramagnetic effects (O"p-term) causes large downfield shifts (to higher vor ppm, to the left inour spectrum)

If we knew the nature of electronic motion in a molecule (induced by BO) we could calculate allO"and subsequently all chemical shiftsThis we cannot do (except for H2 and LiR) and approximations are made to give a model whichcan be used to predict chemical shifts

0" = O"I~Cal + O"I~Cal + 0" ,

Here d docal and ~ocal are the local contributions to 0" for a particular nucleus and 0"' can be

solvent effects, anisotropic effects, ring current effe(:ts etc.

For protons the d docal and 0"' terms are most important since paramagnetic effects can not arise

from its own valency electrons (no p-orbitals in H). The neighbouring atoms may however

contribute to the ~ocal term of a proton

Page 1 / 15

NMR Lectln"e 4 14.55 April 12, 1999

The lack of p-orbitals on a hydrogen atom explains w hy virtually all protons resonate within -10ppm while other nuclei resonate within hundreds or thousands of ppm (C -250 ppm, N 900ppm, F 800 ppm, Co 18000 ppm etc )

How do all these theories affect a real spectrum?Practical considerations

",-..,

.Reducing electron density around a nucleus obviously reduces ~ocal .This is very important for

1H resonances.In 'normal' molecules the H is however not bonded directly to a highly electronegative atom (F,

Cl, O etc). Instead the proton is bonded to a carbon which in tum is bonded to the highlyelectronegative atom.The electronegativity of the substituent affects the partia! charge on the carbon which theninfluences the shielding of the proton

The O-values (in ppm) for the serles of compounds (CH3-X) below show that a more

electronegative substituent (higher electron-withdrawing power) reduces the O'~cal for the

protons of the methyl group and accordingly we see higher chemical shifts, the signal movesdownfield

Si(CH3)4 CH3CH3 CH31 CH3Cl CH30H CH3F0.00 0.88 2.16 3.05 3.38 4.26

~

-Reverse effects, with respect to Ö(lH) versus electronegativity, can be observed forCH3-CH2-X where X={F, Cl, Br, I}.

o(CH3-CH2-F) < o(CH3-CH2-Cl) < o(CH3-CH2-Br) < o(CH3-CH2-I)

Other, geometrical, factors are responsible for thisThere is also an additive effect when several subtituents are present :

-Higher number of electronegative substituents on the <x-carbon add to the total deshielding of

thelH(""', J

I CHCl3 I CH2Cl2 I CH3Cl II 7.27 I 5.30 I 3.05 I

The effect of the electronegativity of X decreases for protons further away. Protons bonded to

the J3-carbon (3 bonds from X) or to the y-carbon (4 bonds) are less affected then protons

bonded to the a-carbon (2 bonds)

"

-CH2-Br -CH2-CH2-Br -CH2-CH2-CH2-Br3.30 1.69 1.25

-Linear relationships have been found and NMR chemical shifts have been used to measureelectronegativities of a substituent X :EX = 0.684 (ÖCH2 -ÖCH3) + 1.78

-Caution must be exercised since other effects, like the 'reverse effect' for ethyl halides above,may appear, but within a small group of similar molecules the formula can be used

Page 2115

NMR Lecture 4 14.55 April 12, 1999

Proton chemical shifts for substituted CH2XY and CHXYZ groups can be predicted from thefollowing formula and table :

~H = 0.23 + 1: contributions

~Accuracy is approx. :i:O.3 ppm for CH2 groups, less accurate for the CH group

Ex. 1: ö(CH2CI2) = 0.23+2.53+2.53 = 5.29 ppm Exp. = 5.30

Ex.2: ö(CH3CH2CH2CH2Br) = 0.23+0.47+0.67 = 1.37 ppm Exp. = 1.25

-The hybridization state of the carbon atom to which a proton is bonded affects the chemicalshift

HI,..

HI

'"',""

I

H'\ /

/c=c,

H

H-v-. H-C=C-H

H H H

-Sp3 : mix ofl s and 3 p orbitals ~ little s-character of the molecular orbital

-Sp2 : mix of 1 s and 2 p orbitals ~ more s-character of the molecular orbital

-sp : mix of 1 s and 1 p orbitals ~ much s-character of the molecular orbital

-The more s-character the closer the electrons are to the C nucleus, and the further away from theH-nucleus.This results in a deshielding of the proton

-Expected chemical shifts are therefore :

o(ethyne) > o(ethene) > O(ethane)

H

r-"

THIS IS NOT THE CASE IN REAL LIFE, HOWEVER!!!

There are other effects which change this orderThe correct ranges of chemical shifts are :

ö(ethyne) ~ 2- 3

Ö(ethene) ~ 4- 7

ö(ethane) ~ O -2

Page 3 115

NMR Lecture 4 14.55 April 12. 1999

Chemical shifts of olefinic protons

The effect of substitution for olefmic protons have been investigated and a fonnula proposed :

3H = 5.25 + Zgem + Zcis + Z trans

Where Zgem. Zcis and ~s are the contributions (in ppm) for the substituent at that position.

R .HCIS

",C'

/Rtrans

/:c"

Rgem

1'"-'-

ICOOH ! 0.97 1.41 0.71 !Some strongly electronwithdrawing groups (OR, NR, F) have a strong deshielding effect if itis in the gemmal position (bonded to the same carbon as the proton) due to inductive effectsThe same group has a rather strong shielding effect if it is in cis or trans position (the 'other'carbon compared to the proton) due to electron-donating conjugative effects

Anisotropic effects

~

Proton chemical shifts are greatly affected by magnetic dipoles at neighbouring atoms or groups-The magnetic dipoles alter the local magnetic field 'felt' by the proton.

-In our molecule (diatomic, C-H) we get a magnetic moment, ~A, due to the effect the external

field BO has on electrons around atom C. We can consider ~A as a localized point dipole at thecentre of atom C. Please note that 'atom C' is not a carbon atom.

~A(Z) is big, ~A(Y) is small, and ~A(X) is intermediate in the picture on the next page.

-When our molecule tumbles in solution, the secondary field due to the dipole ~A will increase ordecrease the local, effective, field at atom H.

-We assume that our molecule tumbles freely and no position is favoured. Then we will see theavecage effect of the secondary field.

-The effects on H of the induced dipole at atom C are clearly not the same for all molecularorientations.

C is said to be magnetically anisotropic if at least one of the ~A(X), ~A(Y) or ~A(Z) is

different from the others. If ~A(X)=~A(Y)=~A(Z) then C is said to be magnetically isotropic and

C will have no effect (on average) on ~ of the H close to it. It is a through-space efTect, it isnot transmitted via bonding electrons.

Page 4 115

NMR Lectw-e 4 14.55 Apri112, 1999

z

I

y x~

B-\- i-

"'\ 1- ~

-'\\'1"- /

f

-~-/

-/

~I \

~)

I: HI..

I

/It~1

\

\t,x

I"r" \ .'

\

.\x

Iz

\\ \4, /

\

~~"- I / /L~ "-/ I "- -"- -"" :'--. "-<-

J L -"..:: y ~ "" ~A(x)'I, ~A(z)

Here we see the effect of the secondary field on the local effective magnetic field at atom H when themolecule (and the molecular coordinate system) is tumbling in solution.(left) Berr at H is increased due to the secondary field beeing parallell to Bo (H is deshielded)

(middle) The molecule's y-axis is allgned with BO and Berris increased again (H is deshielded)

(right) The molecule's x-axis is aligned with BO and Berr is decreased at atom H (H is shielded)

-Many chemical features (atoms, bonds, functional groups etc) give rise to anisotropic effects.

z

r

For instance, even a carbon-carbon single bond is anisotropic :

"'"""

+

Ha AO'=+0.14ppm ~

0.3 nm l... 0.3 nm..1 -C-C -~

~K\ .'., Hb AO' = -0.28 ppm

'~~

+Shielding cones showing how the shielding constant is changing due to anisotropic effects.

-The effect on the shielding constant is negative (leading to higher chemical shift) along the axisof the bond while the effect on the shielding constant is positive in regions perpendicular to thebond (lower chemical shift)

-The nodal planes where there is no change in O' (AO' = 0) goes along the surface of a cone with a

54.7° angle ('magic angle') to the bond (1-3cos2 ~)-The C-C bond anisotropic effect is clearly seen in cyclohexane where axial and equatorial

protons have a -0.5 ppm chemical shift difference. At low temperature the chair-chair

interconversion is slow and the Aö between Hax and Heq can be measured

Heq

4

H2

Page 5 115

NMR LectW"e 4 14.55 Apri112, 1999

On the previous page, it can be seen that it is the shielding cones of the C2-C3 and the C5-C6 bondsthat produce the anisotropic effects on the axial and equatorial protons at C 1

-Heq is in the negative shielding cone of both C-C bonds, while "ax is in the positive region.

-"ax is shielded by -0.5 ppm compared to Heq

a.-methoxygalactose and fi-methoxygalactose can be distinguished due to the differences in chemicalshifts of the methoxyprotons

~

Other bands give rise to anisatrapic effects as weIl :

+

""""+

The shielding cones for C-C single bond,C-C double bond and C-C triple bond

~+

-

+ +

The shielding cones for the carbonyl group and the nitro group

Page 6 / 15


Another very important contributor to anisotropic effects on chemical shifts :

Ring currents

In an aromatic ring system, a circular movement of the delocalized 7t-electrons will be induced

This cUn'ent will, as we saw before, create a small magnetic field opposed to the external field BO(in the center of the cicrular path)Since the aromatic ring is planar, the created field will only be created when the plane of thearomatic ring is perpendicular to Bo. As a result we have anisotropy and the chemical shiit ofnearby protons will be affected

t B

~

Page 7 115

B due to circulation of 1t-electrons

Protons located above or below the aromatic system will be in a strong shielding zone leading toupfield chemical shiftsProtons located in the plane of the ring (the benzene protons) will experience a significantdownfield shift due to the 'reinforced' magnetic field-The 6 protons of benzene resonate at 7.27 ppm-The olefmic protons ofcyclohexa-l,3-diene resonates at 5.86 ppm

-The difference of 1.4 ppm is attributed to the ring current shift


Another interesting example :

H I-I

r

r--

II H H

[16]-annulene [18]-annulene

In [16]-annulene the 'inner' protons resonate at ö = 10.3 and the 'outer' protons at ö = 5.28

In [18]-annulene, on the other hand, the 'inner' protons resonate at ö = -4.22 and the 'outer'

protons at ö = 10.75

It is the ring current effect which is responsible for these dramatic differences in ö

This shows that [18]-annulene is aromatic while [16]-annulene is not. According to the '4n + 2'rule the same result is predicted

The ring current shifts are very important in NMR of ONA and RNA as weIl as in the NMR ofproteins

-All nucleobases in RNA and ONA (A, C, G, T, U) are aromatic and give rise to ring currentshifts

-Adenine and guanine give rise to the strongest ring current shifts, cytosine of intermediatestrength and uracil of rather low strength

r'

"

Page 8 115


The confonnation of ONA is stabilired by both 'vertical interactions' (stacking) and 'horizontalinteractions' (hydrogen bonding)

HH~N 'N-H

,~..r-< NC1' ~N-<

H

o CH3

H

'c1'Distance betweentwo base planes:3.4 Ångström

o

H o H-

N H

/H

N~H

N'--~>H>-N, C I

O 1

C1

(

The protons of a nucleobase as well as Hl' are to a large extent within the shielding cone of thenext base and experience an upfield shift

Transitions of a ONA duplex ~ randoffi coil can be ffionitored by NMR. Melting curves fromNMR are as precise as the 'normal' uv melting curve-When the ordered, stacked structure becomes destacked (randoffi coil) the aromatic and

anomeric protons are no longer in the shielding cone of the neighbouring nucleobase ~downfield shift

-In non-standard structures the (temperature dependent) ring current shifts are used todetermine which nucleobases are stacked to each other

Chemical shifts of exchangeable protons etc

"'"""

,.-,

-Protons involved in hydrogen bonds (X-R Y) are strong ly affected by the electrical dipolefield created (the R-bond is an electrostatic bond) and this leads to downfield chemical shifts(deshielding) (useful for NMR of ONA and RNA)

-In general, chemical shifts of these protons are very dependent on temperature, solvent, pR etc-In samples which can form intermolecular R-bonds (R-OR, R-NH2)' the chemical shift of

the R-bonding proton is very rnuch dependent on the concentration

-EtOR in EtOR (neat, many R-bonds) : 8 = 5.3

-EtOH in CC14 (5%, almost no R-bonds) : 8 = 2.5 (~8 = -2.8)

-Phenol in CC14 (high conc., many R-bonds) : 8 = 7.5

-Phenol in CC14 (low conc., almost no R-bonds) : 8 = 4.4 (~8 = -3.1)

-In samples which can form intramolecular R-bonds, the chemical shift of the R-bondingproton is on ly little dependent on the concentration

-salicylaldehyde (neat, many R-bonds) : 8 = 11.5

-salicylaldehyde in CC4 (5%, still many R-bonds) : 8 = 11.1 (~8 = -0.4)

Page 9 / 15

,


HI


NMR of carbon-13

The basic theories for chemica! shifts, as weIl as for NMR in genera!, are true a!so for carbonatoms

One important difference is the relative sensitivity

-The nuclear spin of 13C, I, is t hut the magnetogyric ratio, y, is only 25% of y for protons

This 1eads to a 1ower sensitivity since AB = ~27t

Another important aspect concerning sensitivity is the natural abundance

-For hydrogen atoms the natural abundance is 99.98% for lH

-For carbon atoms the natural abundance is 1.108% for 13C

~ Taking 'Y and natural abundance into account the relative sensitivity of l3C is ~ compared

to lH

-The low sensitivity of l3C is an advantage when looking at lH spectra. With a higher

abundance of l3C, the lH spectra would become very complicated due to lH-l3Ccouplings.In a l3C spectrum, all those lH-l3C couplings are observed unIess special precautions aretaken

-The low sensitivity is naturallya disadvantage when observing l3C

Chemical shifts of carbon atoms

One important difference to 1 H is that carbons have p-electrons

-Carbon resonances are spread out over 200 -300 ppm

-The local paramagnetic shielding term (~ocal ) is of great importance while

ddocal (the diamagnetic term) is ofsma11 importance contrary to the lH case

" The chemical shifts of carbon atoms follow roughly the same order as for protons

ö(aldehyde) > o(alkene) > ö(alkyne) > o(alkane) > TMS

Effects of substituents on 13C chemical shifts are similar to their effects on ör H) but notidentical

-In CH3-I the 13C resonance moves upfield (to -20 ppm) contrary to the lH case. Similarly,

the 'expected' large downfield shifts (from lH analogy) are not seen for CH3-Cl and CH3-Br

The substituent effect propagates through more bonds for 13C shifts than for lH

-For lH chemical shifts, the substituent effect is virtually only observed for protons bonded to

the a.-carbon (the carbon carrying the substituent)

-For 13C chemical shifts, substituent effects are clearly observed at the y-carbon (3 bonds), insome cases even at carbons further away

-In acyclic alkanes the effect on the ö- and e-carbons are small but in cyclic compoundsthese effects may be large

Page11/15


,-.,

r

OH 48.3 10.1 -6.0 0.3 0.2ICHO 31.4 0.7 -1.9 0.8 0.5

The table shows the differences in 13C chemical shifts (L\5c) for various l-substituted pentanes. The

chemical shifts for pentane are : Ö(C1)=Ö(C5)=13.7 , ö(CV=Ö(C4)=22.6 and ö(C3)=34.5

-The substituent effects on the a.- and j3-carbons are normally deshielding

-The substituent effect on the 'Y-carbon is shielding (the 'Y-effect)

-The 'Y-effect is very clearly seen in substituted cyclohexanes

-The 'Y-effect depends on the dihedral angle between the substituent and the 'Y-carbon. Theclassical staggered rotamers gauche- (-600), trans (1800) and gauche+ (+60°) are normallyconsidered as the only three possible states

-In acyclic compounds the rotations around C-C bonds are generally so fast that we see anaverage effect only from the three staggered rotamers

-In cyclic compounds the rotation is restricted and there will only be two possible rotamers for adihedral X-C-C-C (where X is an exocyclic substituent) one gauche and trans.orten one rotamer is highly preferred over the other (depending on the nature of the endo- andexocyclic substituents)

-In cyclohexane an axial substituent on the a.-carbon will be gauche to the 'Y-carbon of the

ring and cause an upfield shift of 4-7 ppm for the 'Y-carbon

-Accordingly, an equatorial substituent on the a.-carbon will be trans to the 'Y-carbon of the

ring and cause an upfield shift of 0-3 ppm for the 'Y-carbon

33.2

r

~CH3H3C ty 13

36.2

(\

(H3C)3C

Equatorial Axial

Page 12 115

NMR Lecture 4 12:02 PM April 14, 1999

-The 31p chemical shiit is also sensitive to the R-O-P-O and O-P-a-R torsional angles in aphosphodiester-The torsional angles and the bond angle not uncorrelated phenomena :

Changing one structural feature is accompanied by a change in the other

-This interdependence is called the 'stereoelectronic 31p effect'

C

"'

f"'

,,\\\\\ c0.."' ()

it gauche- it trans

0- p -0 0- P-O

~ gauche- rtt gauche-

,..0 ,..0C \\\\\\"' C \\\\\,"'

In practice as weIl as from semiempirica! MO ca!culations we see that phosphate esters with bothP-O torsions in gauche conformation ( -60° or +60°) resonates several ppm (3-6 ppm) upfieldfrom a phosphate ester which has at least one P-O torsion in trans conformation ( -180°)

The AÖ(31p) arising from torsiona! changes are used to monitor DNA and RNA conformation :

In a stacked conformation the ~ and a. torsions (P-O torsions) take up a gauche,gaucheconformation while in a non-stacked (random coil) a!so the gauche,trans and trans,transconformers existThese conformationa! changes lead to downfield shifts of the 31 p resonances when theconformation changes toward random coil which can be induced by increasing temperature

f'

'-"

r,

Page 14/14

the chemical shift - uppsala universityboc2.boc.uu.se/boc14www/compendium/lec4.pdf · the chemical...

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