Chemical Shifts

The chemical shift is a measure of the degree of shielding or deshielding of a nucleus
with respect to a reference nucleus.
The chemical shift is defined as
 which has units ppm.

The applied magnetic field will induce currents in the electron density surrounding the
nucleus. These current in turn produce magnetic fields opposite in direction to the
applied field, thus reducing the all-over magnetic field experienced by the nucleus. This
property is known as shielding, or more specifically diamagnetic shielding. The degree
to which a nucleus is shielded depends strongly on the structure of the electron density
and thus chemical structure.
Diamagnetic Shielding By the Electrons Surrounding A Nucleus

As a first approximation the degree of shielding is directly proportional to the electron
density around the nucleus. That is why 19F nuclei have a much larger chemical shift
range (ca. 250 ppm) than do 1H (14 ppm).

Substituents near the nucleus of interest can remove or add electron density, as a
consequence of their inductive effect, this would thus affect the degree of shielding
accordingly. In other words an electron withdrawing substituent would remove electrons
and thus decrease the degree of shielding. The converse is true for electron donating
groups.

The degree of hybridization has a profound chemical shielding, the degree of s-character
is inversely related to the degree of shielding. An sp2 hybridized carbon has 33% s
character while that of an sp3 hybridized carbon is 25%, thus the sp2 cabon is more
electronegative and hence deshields a bond hydrogen with respect to a sp2 carbon. In
other words the chemical shift of a sp2 bond H is larger than that of a sp3 bound H. The
sp bound H is deshielded with respect to that of an sp2 bound H due some other factor.
i.e. 1 ppm
=CH2 4.5 to 7 ppm

The sp3 hybridized H chemical shift depend in the degree of substitution on the carbon.
The higher the degree of substitution the higher the degree of shielding.

Sometimes certain bonding structures allow electrons to circulate in a manner otherwise
not possible. For instance in highly conjugated systems, the electrons can move freely
about between nuclei. For instance in aromatic systems the electron actually circulate in
the  systems giving rise to a ring current, which has profound effect on the chemical
shifts. (7 ppm for 1H, and 120 ppm for 13C)
Deshielding Caused by the Ring Current in an Aromatic

Notice than the degree of shielding will be different depending on the position of the
nucleus with respect to the ring current. A proton on the ring will be deshielded, whilst
one placed nearer the center would be greatly shielded. This phenomenon is known as
diamagnetic anisotropy. It serves to explain the substituent effect of many -bonded
system of neighboring nuclei.

The diamagnetic currents in a carbonyl group, of an aldehyde, lead to the proton being
strongly deshielded. However a nucleus along an axis perpendicular to the C=O bond
axis bisecting the bond, would experience shielding. For similar reasons the protons in
an alkyne system are more shielded than their alkene counterparts.
Diamagnetic Shielding Anisotropy in Alkynes
Diamagnetic Shielding Anisotropy in Carbonyl Groups

The diamagnetic shielding effects, inductive effects, etc. of neighboring substituents are
remarkably consistent between analogues systems. These effects can be thought of as
being transferable and cumulative. Consequently one can compile tables of substituent
effects that allow one to predict the chemical shift of a nucleus given the neighboring
substituents.
Substituent effect on the chemical shift of methylene hydrogens

In summary the electronic environment around the nucleus determines its chemical shift
and hence its frequency.

The NMR spectrum is resolved into separate lines corresponding to the different
electronic environments for the nuclei in a given chemical species. The frequency of
these lines is determined by its chemical shift.
Indirect Spin-Spin Coupling

In solution state direct spin-spin interactions, like bar magnets, are averaged out due to
molecular motion. However they can interact with each indirectly through the electron
density. This is known as indirect spin-spin coupling or also a scalar coupling.

The magnetic moment of the nucleus polarizes the spin of its surrounding electrons, these in
turn, through various electron-electron interactions, transfer this spin polarization to electron
surrounding another nucleus, polarizing the spin of that nucleus. These effects are very small
and are independent of the applied field, since the applied field only serves to provide a
direction of polarization.

Spin polarization can be propagated through the electronic system through more than on path
way. Consequently the coupling constant is often discussed in terms of contributions from
the different pathways. The most common path is through the sigma bonded network. This
would be through the s and spn orbitals connecting the two nuclei in question, and is referred
to as the sigma contribution.

As a rule the spin polarization propagated through the sigma bonded network changes sign
for every bond it passes through and is attenuated by approximately an order of magnitude
for every bond.
For example:
In alkanes
JHH = 210 Hz, 2JH,H = -10 to –20, and 3JHH = 2-14 Hz
JCH = 120 Hz, 2JCH = -4.5, and 3JCH= 3 Hz.
1
1
Significant other contributions make the two-bond coupling smaller than expected to be.

The alternating sign is a consequence of the Pauli-exclusion principle. The sign of the
polarization within orbital on the same atom are preserved as a consequence of the Hunds
rule.
Propagation of Spin Polarization Via the Bonded Network

Sometimes spin polarization will take an alternative route through the delocalized  system,
when the opportunity arises, thus two contributions to coupling would have to be considered.
Propagation of Spin Polarization Via the Bonded Network

The  contribution depends heavily on the geometry of the system, in order for it to be
significant the orbital must overlap in a favorable way.

One bond-coupling has a positive sign. It ranges between 100 to 300 Hz between C and H
depending on the degree of hybridization. For an sp3 Carbon the coupling is 115-125 Hz, an
sp2 carbon it ranged between 150 to 170 Hz, and for sp carbon it occurs around 240 to 270
Hz.

The two-bond coupling, often referred to as the geminal coupling, are most often negative.
Between two hydrogen they occur between –9 to –15 Hz for unstrained sp3 hybridized
systems. The magnitude of the geminal coupling depends strongly on the angle of the two CH bond involved, as the angle decreases the coupling constant increases. This trend is the
result of increasing in overlap between the sp3 bonding orbitals involved in the coupling
interaction..

Consider the three-bond vicinal coupling constant. It has both significant  and 
contributions, the former is mostly independent of the dihedral angle between the C-H bonds,
the later is dependent on the overlap between the Pz orbitals on both carbons, and hence it is
very dependent on the dihedral angle between the C-H bonds.

As a consequence it is expected to attain a maximum at 0 and 180o and a minimum at 90o.
Such a relationship is found and is known as the Karplus relationship.
 and  Coupling Mechanism for the Vicinal 3JHH

As a consequence it is expected to attain a maximum at 0 and 180o and a minimum at
90o. Such a relationship is found and is known as the Karplus relationship.

The Karplus relationship has the following form:
3
J H, H ( )  A cos 2   B cos   C
where A, B and C are parameters which are empirically determined. These have been
optimized for various systems, including the rings, peptides, alkanes, amides, carboxylic
acids, etc. Additional terms can be incorporated to empirically account for substituent
effects.
The Karplus Curve

In general the curve starts at around 10 –12 Hz at  = 0o, goes to 0-2 Hz at  = 90o and finally
back to 12-14 Hz  = 180o. The rotationally averaged couplings tend to fall between 6- 8 Hz.
Question
a)
What are the 3JHH in the three rotamers of 1,1-dichlo-2,2-dibromoethane.
b)
What are all the 3JHH’s in the three rotamers of 1-chloro-2-bromoethane
c)
What are the 3JHH’s in the chair and boat conformation of 1-chorocyclohexane, between
the proton’s on carbon 1 and 2.

The geometrical dependence of coupling constants can be attributed largely to changes in
the  contribution due to variation in  orbital overlap with bond angles, these include
valence and dihedral angles.

In general long-range couplings follow the zigzag rule, where in an all trans conformation
a coupling will exist, whilst if there is any intervening cis conformation the coupling will
disappear. This hold in most situations where the coupling extends to 4 bonds or beyond.
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
Splitting Patterns

Indirect coupling between spins gives rise to splittings in the line of each nucleus
involved. The nature of the splitting patterns are dependent on then the number of spins
involved, their symmetry, and the coupling constants.

Two spins that differ in frequency are coupled indirectly at a value of J. Thus the nucleus
experiences two slightly different filed depending on J and the orientation of the other
spin with respect to the applied field.
The Total Magnetic Field Experience by Nucleus a and b When I) Parallel II) Antiparallel
I
Electron density
BJb
Bo
Transfer
a
of Spin
b
BJa
Bo
polarization
J
II
BJb
Bo
Transfer
a
of Spin
b
BJa
Bo
polarization
J

For every coupling interaction with another nucleus the nucleus experiences two fields
(that is for spin-1/2 system). Thus for N interactions there are 2N fields and thus 2N lines.
General Tree Diagram
N=0
N=2
N=3
N=4
Chemical shift and Magnetic Equivalence
 When two nuclei have an identical chemical shifts they are said to be chemical shift
equivalent.

When two nuclei have the same chemical shift and are coupled equally to all their mutual
interacting partners, they are said to be magnetically equivalent.

A spin system is named according to the number of spins in each group that are magnetically
equivalent.
For example:
Ethanol contains 3 groups of nuclei two of which have more than one spin that are chemical
shift equivalent, the CH3 has three equivalent protons and the CH2 has two. Thus this system
is called an A3B2C system.

Chemical and magnetic equivalency is often the result of symmetry in the spin system and/or
rapid motion.
For example:
In alkanes chemical and magnetic equivalence exists between proton on the same carbon due
to rapid interconversion between the three rotamers of each C-C bond. The coupling
constants for protons between two different carbons are thus averaged.
In 1,2,3-tribromo-benzene the two meta proton are magnetically equivalent since they are
coupled identically to the para proton as a result of the C2V symmetry in the molecule. It is
referred to as a A2B spin system.
On the other hand in 1,2-dibromo-benzene, protons 3 and 6 are chemical equivalent but no
magnetically equivalent since 3 has a different coupling to 4 as 6 has to 4, a similar situation
exists for proton 5. Thus this system is referred to as an AA’BB’ where the prime indicates
magnetic non equivalence between the A and A’, and B and B’.

Often these coupling interactions involve several nuclei of the same type where the
interaction are the same. In this case the coupling patterns follow the N + 1 rule, where N is
the number of neighboring equivalent nuclei involved. The intensity of the corresponding
line follows the binomial distribution (i.e. Pascal’s Triangle)
Tree Diagram for equivalent Nuclei
1
1
2
1
N=1
1
3
1
3
N=2
1
4
N=3
6
4
1
N=4
Predict the spectrum of ethyl chloride.
i)
ii)
iii)
How many lines are there for the methyl group?
How many lines are there for the methylene group?
Which is more shielded the CH2 or the CH3?
What coupling patterns would you expect for the CH group of an isopropyl group, when the
methyl groups are the only nuclei coupled to it?
Strong Coupling

When the frequencies difference between the interacting nuclei becomes of the same order as
the coupling interaction, distortions appear in the spectrum. These manifest themselves as
shifts in the frequency and changes in the relative intensity of the spectral lines. This can
happen to such an extent the above splitting rules no longer apply. These are known as
second order effects.
1
The Effects of Strong Coupling on an AB system.
J/
50
2
1
0.5
0.25
0.05
Frequency

The rule in naming spins systems have to be amended to accommodate strong coupling. The
basic idea is that, the closer the letters are in the alphabet, the closer they are in frequency.
For instance two strongly coupled spins-1/2 spins are referred to as an AB system however
weakly coupled they would be referred to as AX.
Name the following spin systems: 1) 13CH3 2) CHClCH2 3) CF2CH2 4) o-dichlorobenzene 5)
trans-1,2-dichlorocyclopropane 6) 1-chloro-2-bromobenzene 7) 2-chloro-ethanol . 8) 1-13Cbenzene
Explain why the two methylene protons of 1,2-dichloroethanol are non-equivalent?
Alkanes
General formula:
CnH2n+2
Chemical shifts:
Indirect Coupling Constants:
-CH3
-CH2-CRH-
3
Example
JH,H = 7 Hz for –CH-CH-
0.7 to 1.3 ppm
1.2 to 1.4 ppm
1.4 to 1.7 ppm
1
H Spectrum of n-Octane
Alkenes
General formula:
RR’C=CR’’R’’’
Chemical Shifts:
Indirect Coupling Constants
C=C-CH3
C=C-CH2C=C-CRH=CHR-
1.6 ppm
2.05 ppm
2.27 ppm
4.5 to 6.5 ppm
H-C=C-H
Example:
1
3
JH,H (trans)
JH,H (cis)
2
H-C-H
JH,H (gem.)
4
-CH=C-C-H JH,H
3
H Spectrum of 2-methyl-pentene
11 – 18 Hz
6- 15 Hz
0 – 1 Hz
0 – 3 Hz
Aromatics 
General formula:
C6H6-nXn
Chemical Shifts
Indirect Coupling Constants
-CHPhCH-
6.5 to 8.0 ppm
2.3 to 2.7 ppm
3
Example:
1
Jom = 3Jmp = 7 to10 Hz
4
Jop =4Jmm = 1 to 3 Hz
5
Jip= 0 to 1 Hz (across the ring)
6
J = 0 to –1 Hz (benzyl proton to para-proton)
H Spectrum of chloro-p-xylene.
Alkynes
General Formula:
R-C≡C-R’
Chemical shifts
Indirect Coupling Constants
HC≡
H-C-C≡
1.7 to 2.7 ppm
1.6 to 2.6 ppm
H- C≡C-C-H
Example:
1
H Spectrum of 1-pentyne
4
JHH = 2-3 Hz
Alkylhalides
General Formula:
RR’R’’CX
Chemical Shifts
Indirect Coupling Constants
-CHX- for X=
-CH-F
-CH-CF
F
Cl
Br
I
2
JHF = -50 Hz
JHF = 20 Hz
3
4.2 to 4.8 ppm
3.1 to 4.1 ppm
2.7 to 4.1 ppm
2.0 to 4.0 ppm
Example: 1H spectrum of 1-chlorobutane
Alcohols
General Formula:
RR’R’’COH
Chemical Shifts
Indirect Coupling Constants
C-OH
CH-OH
CH-OH
0.5 to 5.0 ppm
3.2 to 3.8 ppm
Example: 1H Spectrum of 2-methyl-1-propanol
3
JHH = 5 Hz (in aprotic solvent)
Ethers
General Formula:
R-O-R’
Chemical shifts:
Indirect Coupling Constants:
R-O-CH-
-CH-O-CH 4JHH = O Hz or 2.5 to 4 Hz when
constrained to a fixed geometry.
3.2 to 3.8 ppm
Example: 1H Spectrum of Butyl-methyl-ether
Amines
NRR’R’’
Chemical Shifts
Indirect Coupling Constants
R-N-H
-CH-NPh-N-H
0.5 to 4.0 ppm
2.2 to 2.9 ppm
3.0 to 5.0 ppm
-N-H
-N-CH
H-C-N-H
Example:
1
H Spectrum of Propylamine
1
JNH = 50 Hz
JNH < 1 Hz
3
JHH often lost due to chemical
Exchange.
2
Nitriles
General Formula: RR’C-CN
Chemical Shifts
-CH-CN
2.1 to 3.0 ppm
Example: 1H Spectrum of Valeronile
Aldehydes
General Formula:
R-CH=O
Chemical Shifts:
Indirect Coupling Constants
R-CHO
R-CH-CHO
9.0 to 10.0 ppm
2.1 to 2.4 ppm
-CH-CHO
Example:
1
3
JHH = 1 to 3 Hz
H Spectrum of 2-methylbutyralaldehyde
Ketones
General Formula:
R-C(=O)-R’
Chemical Shifts:
R-CH-C(=O)-R’
Example:
1
2.1 to 2.4 ppm
H Spectrum of 3-Methyl-2-pentanone
Esters
General Formula:
R-C(=O)-O-R’
Chemical Shifts
R-CH-C(=O)-O-R’
R-C(=O)-O-CH-R’
Example:
1
2.1 to 2.5 ppm
3.5 to 4.8 ppm
H Spectrum of Isobutyl acetate
Carboxylic acids
General Formula:
R-C(=O)O-H
Chemical Shifts:
R-COOH
CH-COOH
11.0 to 12.0 ppm
2.1 to 2.5 ppm
Example:
1
H Spectrum of ethylmalonic acid
Amides
General formula:
R-C(=O)-NR’R’’
Chemical Shifts
R(CO)-N-H
-CH-CONHR(CO)-N-CH
Indirect Coupling Constants
5.0 – 9.0 ppm
2.1 to 2.5 ppm
2.2 to 2.9 ppm
1
-N-H
JNH = 50 Hz often not observed*
2
-N-CHJNH = 0*
3
-NH-CHHHH= 0*
*signal is lost due to rapid quadrupolar relaxation
of N or chemical exchange of N-H.
Example:
1
H Spectrum of butyramide