Ko sponholtz NMR project

Structure
Determina-on:

Nuclear
Magne-c
Resonance(NMR)

Kwonil
Kobe
Ko,
Cushing
Academy,
Senior

Project
Supervisor:
William
R.
Sponholtz,
III,
B.S.,
M.S.,
Ph.D.

1
2

1Dartmouth
College
Department
of
Chemistry.
h7ps://sites.dartmouth.edu/
mierkelab/nmr-‐facility
(accessed
December
19,
2015).

2Dominique
Marion;
An
IntroducJon
to
Biological
NMR
Spectroscopy.
Molecular
&
Cellular

Proteomics.
[Online]
2013,
pp
3009.
h7p://www.mcponline.org/content/12/11/3006.full.pdf
+html?sid=d9388787-‐b9ae-‐4dae-‐b0e8-‐770d9d293f41
(accesed
December
28,
2015).

NMR
Spectroscopy

Nuclear
Magne-c
Resonance
(NMR)
Spectroscopy
is
an
analy-cal

chemistry
technique
to
elucidate
the
structures
of
simple
and
complex

organic
compounds
employing
1-‐D
and
2-‐D
NMR.

Degrada've
vs.
non-‐Degrada've
Techniques

Large
Quan''es
vs.
A
Few
Milligrams

Many
Years
vs.
Hours

Ambiguous
vs.
Absolute
Assignment

Modern
techniques

à
NMR(1-‐D
and
2-‐D)
and
3-‐D
Shapes

à
GC-‐MS
:
Molecular
weight

à
X-‐ray
Crystallography
:
3-‐D
shapes

àIR
:
func-onal
groups
(based
on
vibra-ons)

Oldest
techniques

à
Elemental
analysis

à
Combus-on
analysis

Classical
Structure
Determina'on
vs

Modern
techniques

Top
Five
Greatest
Discoveries
in
Science?

NMR
&
X-‐ray
crystallography

à

X-‐ray
crystallography
is
a
tool
used
for
idenJfying
the
atomic
and
molecular
structure
of

a
crystal
by
causing
a
beam
of
X-‐rays
to
diffract
into
many
specific
direcJons.
Measuring
the

angles
and
intensiJes,
a
three-‐dimensional
picture
of
the
density
of
electrons
within
the

crystal
can
be
produced.
And
from
this
electron
density,
compound’s
3-‐D
structure
can
be

elucidated,
including
chemical
bonds,
chiral
centers,
gross
connecJvity,…etc.
5

Protein crystallography involves summing the scattered X-ray waves
from a macromolecular crystal
The steps in solving a protein crystal structure at high resolution are diagrammed in Figure 5-3.
First, the protein must be crystallized. This is often the rate-limiting step in straightforward
structure determinations, especially for membrane proteins. Then, the X-ray diffraction pattern
from the crystal must be recorded. When X-rays strike a macromolecular crystal, the atoms in
the molecules produce scattered X-ray waves which combine to give a complex diffraction
pattern consisting of waves of different amplitudes. What is measured experimentally are the
amplitudes and positions of the scattered X-ray waves from the crystal. The structure can be
reconstructed by summing these waves, but each one must be in the correct registration with
respect to every other wave, that is, the origin of each wave must be determined so that they
sum to give some image instead of a sea of noise. This is called the phase problem. Phase
values must be assigned to all of the recorded data; this can sometimes be done computationally,
but is usually done experimentally by labeling the protein with one or more heavy atoms whose
position in the crystal can be determined independently. The phased waves are then summed
in three dimensions to generate an image of the electron density distribution of the molecule
in the crystal. This can be done semi-automatically or by hand on a computer graphics system.
A chemical model of part of the molecule is docked into the shape of each part of the electron
part of the electron density (as shown in Figure 5-3). This fitting provides the first picture of
the structure of the protein. The overall model is improved by an iterative process called refine-
ment whereby the positions of the atoms in the model are tweaked until the calculated
diffraction pattern from the model agrees as well as can be with the experimentally measured
diffraction pattern from the actual protein. There is no practical limit to the size of the protein
or protein complex whose structure can be determined by X-ray crystallography.
Definitions
phase problem: in the measurement of data from an
X-ray crystallographic experiment only the amplitude
of the wave is determined.To compute a structure,the
phase must also be known. Since it cannot be deter-
mined directly, it must be determined indirectly or by
some other experiment.
crystals (enlarged view) diffraction patterns
phases
electron density maps atomic models
fittingX-rays
refinement
Figure 5-3 Structure determination by X-ray crystallography The first step in structure determination by X-ray crystallography is the crystallization of the
protein. The source of the X-rays is often a synchrotron and in this case the typical size for a crystal for data collection may be 0.3 ¥ 0.3 ¥ 0.1 mm. The crystals
are bombarded with X-rays which are scattered from the planes of the crystal lattice and are captured as a diffraction pattern on a detector such as film or an
electronic device. From this pattern, and with the use of reference—or phase—information from labeled atoms in the crystal, electron density maps (shown here
with the corresponding peptide superimposed) are computed for different parts of the crystal. A model of the protein is constructed from the electron density
maps and the diffraction pattern for the modeled protein is calculated and compared with the actual diffraction pattern. The model is then adjusted—or refined—
to reduce the difference between its calculated diffraction pattern and the pattern obtained from the crystal, until the correspondence between model and reality
is as good as possible. The quality of the structure determination is measured as the percentage difference between the calculated and the actual pattern.
can be studied.
Unlike X-ray diffraction, which presents a static picture (an average in time and space) of the
structure of a protein, NMR has the capability of measuring certain dynamic properties of
proteins over a wide range of time scales.
Structure Determination Chapter 5 171©2004 New Science Press Ltd
References
Drenth, J.: Principles of Protein X-Ray Crystallography
2nd ed.(Springer-Verlag,New York,1999).
Evans, J.N.S.: Biomolecular NMR Spectroscopy (Oxford
University Press,Oxford,1995).
Markley, J.L. et al.: Macromolecular structure deter-
mination by NMR spectroscopy. Methods Biochem.
Anal. 2003,44:89–113.
Rhodes,G.:Crystallography Made Crystal Clear:A Guide
to Users of Macromolecular Models 2nd ed. (Academic
Press,New York and London,1999).
Schmidt, A. and Lamzin, V.S.: Veni, vidi, vici—atomic
resolution unravelling the mysteries of protein
function.Curr.Opin.Struct.Biol. 2002,12:698–703.
Gorenstein,N.:Nuclear Magnetic Resonance (NMR)
(Biophysics Textbooks Online):
http://www.biophysics.org/btol/NMR.html
1H chemical shift (ppm)
13C = 22.9 ppm
1Hchemicalshift(ppm)
purified, labeled protein NMR spectrometer resonance assignment and
internuclear distance measurement
protein structure
data
analysis
data
collection
Figure 5-4 Structure determination by NMR For protein structure determination by NMR, a labeled protein is dissolved at very high concentration and placed
in a magnetic field, which causes the spin of the hydrogen atoms to align along the field. Radio frequency pulses are then applied to the sample, perturbing the
nuclei of the atoms which when they relax back to their original state emit radio frequency radiation whose properties are determined by the environment of the
atom in the protein. This emitted radiation is recorded in the NMR spectrometer for pulses of differing types and durations (for simplicity, only one such record
is shown here), and compared with a reference signal to give a measure known as the chemical shift. The relative positions of the atoms in the molecule are
calculated from these data to give a series of models of the protein which can account for these data. The quality of the structure determination is measured as
the difference between the different models.
NMR:

X-‐ray

Crystallo
graphy:

5
Chemwiki,
Unviersity
of
California,
Davis.
h7p://chemwiki.ucdavis.edu/AnalyJcal_Chemistry/Instrumental_Analysis/DiffracJon/X-‐
ray_Crystallography
(accessed
December
27,
2015).

3
Gregory,
A.;
Dagmar,
R.;
Protein
Structure
and
FuncJon:
Chapter
5,
Structure
determinaJon;
New
Science
Press
Ltd,
2004;
pp
168-‐173
For
X-‐ray
descripJon

3

4

4
Gregory,
A.;
Dagmar,
R.;
Protein
Structure
and
FuncJon:
Chapter
5,
Structure
determinaJon;
New
Science
Press
Ltd,
2004;
pp
168-‐173
For
X-‐ray
descripJon

NMR
vs
X-‐ray
Crystallography

disregarded otherwise. The external magnetic field B0 induces
currents in the electronic clouds in the protein; in turn, these
circulating currents generate a local induced field Bind. As a
result, the different spins sense the vector sum of the two
fields:
B
¡
loc ϭ B
¡
0 ϩ B
¡
ind
and will thus not resonate at the same frequency. Chemical
shifts are extremely sensitive to steric and electronic effects
and thus in the case of proteins, to secondary and tertiary
structure. Unlike nOe and J-coupling, chemical shift does not
depend on a single pairwise interaction between well-identi-
fied partners: its prediction or quantitative interpretation is
thus more complex. Let us consider the chemical shifts of
backbone 15
N in proteins: the standard chemical shift range
FIG. 2. Number of structures deposited to the RCSB protein data bank (http://www.rcsb.org/pdb/) over the years. The number of
structures solved by X-ray crystallography is steadily increasing whereas the NMR-based ones have hit a plateau. In the inset, the data for the
early years of crystallography are displayed with an extended vertical scale for clarity. For the former method, the crystallization step remains
a major bottleneck but once suitable diffraction data are available, the structure can be obtained rather quickly. NMR is primarily hampered
by the limitation in protein size that can be studied: despite that resonance assignment and nOe interpretation have been automated, it still
requires more human input during these processes.
FIG. 3. Without chemical shifts, NMR structural parameters
could not be measured and interpreted at atomic level resolution.
In fact, the magnetic field at the nucleus is generally different from the
applied field B0: this additional contribution (or screening) arises from
the interaction of the surrounding electrons with the applied field. The
An Introduction to Biological NMR Spectroscopy
à
NMR
spectroscopy
and
X-‐ray
crystallography
are
the
most
widely
used
modern
techniques
in
ﬁelds
of

biology
and
chemistry
for
structure
elucidaJon.
7

à
X-‐ray
is
highly
used
when
solving
a
protein
structure
because
it
is
easy
and
quick
to
solve
a
protein

structure
once
suitable
crystals
have
been
obtained.
8

à
Unlike
X-‐ray,
which
is
highly
restricted
to
protein
structure
determinaJon,
NMR
covers
a
wider
range
of

biochemistry
including
structures
that
are
not
easily
crystalized.
This
is
especially
powerful
in
drug

discovery,
because
it
provides
us
molecular
parameters
and
chemical
kineJcs
by
following
the
Jme-‐
dependence
of
the
data.
9

1

6

6
Dominique
Marion;
An
IntroducJon
to
Biological
NMR
Spectroscopy.
Molecular
&
Cellular
Proteomics.
[Online]
2013,
pp
3009.
h7p://www.mcponline.org/content/
12/11/3006.full.pdf+html?sid=d9388787-‐b9ae-‐4dae-‐b0e8-‐770d9d293f41
(accesed
December
28,
2015).

7
Dominique
Marion;
An
IntroducJon
to
Biological
NMR
Spectroscopy.
Molecular
&
Cellular
Proteomics.
[Online]
2013,
pp
3006.
h7p://www.mcponline.org/content/12/11/3006.full.pdf
+html?sid=d9388787-‐b9ae-‐4dae-‐b0e8-‐770d9d293f41
(accesed
December
28,
2015).

8
The
University
Medical
School
of
Dbrecen.
h7p://www.cryst.bbk.ac.uk/pps97/assignments/projects/ambrus/html.htm
(accessed
December
28,
2015).

9
The
University
Medical
School
of
Dbrecen.
h7p://www.cryst.bbk.ac.uk/pps97/assignments/projects/ambrus/html.htm
(accessed
December
28,
2015).

Example
of
Impact
of
2-‐D
NMR:
Strychnine

Classical
structure
determina-on
techniques
(degrada-ve)
took
approximately
50

years
to
elucidate
the
structure
of
strychnine
employing
the
efforts
of
many,
many

collabora-ng
research
groups.
However,
with
1H
and
13C
NMRs,
several
different
2-‐
D
NMRs,
and
other
spectroscopic
data,
Dr.
Sponholtz
was
able
to
solve
this
structure

in
only
ten
hours!

The
key:
2-‐D
NMR

Isolated
1818

Elucida-on
1946

X-‐Ray
confirma-on
1956

six
chiral
centers:
64
possible

six
methylene
groups;

differen-ate
pro-‐R
and
pro-‐S

N
N
O
O
Strychnine
Showing
gross

connec-vity
only

(no
stereochemistry)

Collabora-on
of
NMR
&
X-‐ray

N
N
N
N
O
N
N
N
O
N
(a)
TIC10
or
ONC201

(b)
Corrected
Structure
for
TIC10

à
The
patented
compound,
known
as
TIC10,
was

elucidated
by
The
Penn
State
group
and
owned
by
the

biotech
firm
OncoceuJcs.

à
Several
insJtuJons
have
found
TIC10
to
be
effecJve
in

brain
cancer,
prostate
cancer,
melanoma,
and
sarcomas.

à
Thus,
OncoeceuJcs
has
iniJated
Phase
I/II
clinical
trials
of

TIC10
and
was
about
to
enter
the
human
clinical
trials.

à
However,
When
Scripps’s
Kim
D.
Janda
and
coworkers

synthesized
the
iniJal
patented
structure,
they
found
it
to
be

biologically
inacJve.

àBy
using
X-‐ray
crystallography
and
NMR,
Scripps

confirmed
that
bioacJve
TIC10
has
a
different
structure(b)

than
the
patented
one.

à
The
Scripps
research
group
concluded
that
OncoceuJcs

and
several
insJtuJons
had
been
working
on
the
bioacJve

compound
but
had
patented
the
inacJve
structure.
Thus

Scripps
applied
for
a
patent
on
the
correct
structure(b)
and

licensed
it
exclusively
to
Sorrento.

10

Stu,
Borman.;
Tug-‐of-‐War
Over
Promising
Cancer
Drug
Candidate.
Drug
Discovery:
Structure
error
threatens
exisJng
patent
and
clinical
trials.
Chemical
&
Engineering
News.
[Online]
2014,
Volume
92,
Issue

21,
7.
h7p://cen.acs.org/arJcles/92/i21/TugWar-‐Over-‐Promising-‐Cancer-‐Drug.html
(accessed
December
28,
2015).

10

Necessary
Review

I.
Basic
understanding
of
electronega<vy
diﬀerence;
i.e.,
must
be
able
to
predict
electron

density
for
atoms;
e.g.,
deshielded
or
shielded.

H F
H F H F δδ
+ -
DESHIELDED SHIELDED
Δδ ATOMS
Example:

IIIII.
Basic
understanding
of
how
atoms
are
oriented
in
three-‐dimensions;
i.e.,
hybridiza<on

theory.

2py unhybridized
orbital (one e-
per orbital)
2pz unhybridized
orbital (one e-
per orbital)
C
sp Hybridized Carbon
(1800)
C
sp3 Hybridized
Carbon (109.50)
sp2 Hybridized
Carbon (1200)
C
The
coupling
of
these
two
topics
is
the
key
to

understand
and
predict
the
chemical
shics
of
NMR.
1

Obtaining
NMR

1-D Pulse NMR
Sample
Magnet
Magnetization
Perturbation
Response
Detection
Data
Fourier
Transformation
Spectrum
3-‐5
mg
of
compound

dissolved
in
a
suitable

solvent
and
transferred

to
a
NMR
tube.

NMR
tube
placed
into

the
magne-c
ﬁeld.

0123456
PPM
11
Dartmouth
College
Department
of
Chemistry.
h7ps://sites.dartmouth.edu/mierkelab/nmr-‐facility
(accessed
December
19,
2015).

11

11

11

Informa-on
Process

FID

Spectrum

The
informa-on
that
comes

out
of
the
spectrometer
is

called
a
free
induc-on
decay

(FID),
which
is
in
the
Time

Domain.
When
the
nuclei
are

pulsed,
the
spins
of
like
nuclei

group
together
and
acer
the

pulse
the
spins
move
apart
or

decay.

The
FID
is
transformed
via
a
Fourier

transforma-on
to
yield
a
spectrum,

which
is
in
the
Frequency
Domain.

You
can
determine
how
many

hydrogens
are
afached
by
comparing

areas
under
the
curve
(integra-on)
in

the
simplest
ra-o.
Moreover,
you
can

inves-gate
the
spligng
paferns
with

high
resolu-on.

FT

Chemists’
perspec-ve

Chemists
look
for
three
things:
Chemical
ShiJ,
Intensity,
and
SpliLng

I):
Chemical
ShiJ:
By
looking
at
chemical
ships
of
1H
and
13C,
chemists
can
roughly
predict

what
group
is
a7ached
to
carbon
or
hydrogen.
For
example,
if
a
sp2
carbon
is
highly

deshielded(or
downfield)
and
thus
has
a
chemical
ship
of
200-‐220
ppm,
chemists
can
imagine

the
presence
of
electron
withdrawing
group,
such
as
oxygen,
a7ached
to
that
sp2
carbon.

2):
Intensity:
Since
the
signal
intensity
is
directly
proporJonal
to
the
number
of
hydrogens

that
give
rise
to
the
signal,
chemists
can
see
how
many
chemically
equivalent

hydrogens(same
chemical
ship)
are
a7ached
by
seeing
a
raJo
of
areas
under
the

integrated
intensity
of
signals
in
1H
NMR
spectrum.

3):
SpliLng
:
Spliqng
offers
informaJon
of
how
many
neighboring
hydrogens
exist
for

a
parJcular
hydrogen
or
chemically
equivalent
hydrogens.
Looking
at
spliqng
pa7erns
and

complex
coupling
constants,
chemists
can
draw
the
tree
diagram
of
complex
NMR
and
thus

be7er
understand
spliqng
in
various
cases
because
different
couplings
are
applied

sequenJally.

1-‐D
NMR
Proton
Chemical
Shics
(rela-ve
to
TMS
in
CDCl3)

01.02.03.04.05.06.07.08.09.010.011.012.0 ppm
R2NHPhOH
ROHAmide RCONHRCO2H
O2N-CH R2N-CH
NC-CH
Esters
RCO2-CH
RCOCHBr-CH
I-CHCl-CHF-CH
R2C CR-CHAr-CHPhO-CH
RC CHHO-CHRCH CHR
Sat alkanes
R-H
Sulfides
RS-CH
Ethers
RO-CHR2C CH2AromaticsRCOH
13.014.0
(δ)

Diﬀerent
types
of
Hydrogens
in
NMR

à
Homotopic
–
hydrogens
that
would
result
in
idenJcal
molecules
if
they

were
replaced
with
another
atom
(X).
12

IdenJcal
signals
in
H
NMR

à
EnanJotopic
-‐
hydrogens
that
would
result
in
enanJomers
if
they
were

replaced
with
another
atom
(X).
12

IdenJcal
signals
in
H
NMR

à
Diastereotopic
-‐
hydrogens
that
would
result
in
diastereomers
if
they
were

replaced
with
another
atom
(X).
12

Diﬀerent
signals
in
H
NMR

C
Cl Cl
H H
C
Cl Cl
X H
,
C
Cl Cl
H X
C
F Cl
H H
C
(R)(R)
F Cl
X H
,
C
(S)(S)
F Cl
H X
C
Cl
H
H
C
(R)(R)
Cl
F
H
C
(R)(R)
Cl
H
X
C
(R)(R)
Cl
F
H
,
C
Cl
H
H
C
(S)(S)
Cl
F
X1

12
Mater
Organic
Chemistry.
Homotopic,
EnanJotopic,

Diastereotopic.
h7p://www.masterorganicchemistry.com/
2012/04/17/homotopic-‐enanJotopic-‐diastereotopic/

(accessed
January
4,
2016)

1-‐D
NMR
Carbon
Chemical
Shic
(rela-ve
to
TMS
in

CDCl3)

200 150 100 50 0
COCRC CRAmides R-CONR2
C CCSOnRCEsters R-CO2R'
Carboxylic Acids R-CO2H RC N C-OR C-Ar
Heteroaromatics C-OH C-SR
Aromatics C-NR2
R2C CH2 C-H Saturated Alkanes
Ketones, R2C=O RHC=CHR C-NO2 C-Br
C-IC-ClC-FR2C=CH2Aldehydes, RCH=O
ppm(δ)

Chemical
Shics
for
1-‐D
proton
and
carbon
NMR

Let’s
focus
on
coupling
(J)
constants
for
proton
NMR

What
are
J
values?

à J(coupling
constant)
is
a
distance(Hz)
between
split
peaks.
When
a
proton

absorbs
energy,
it
relaxes
by
giving
that
energy
back
to
surrounding
atoms
via

the
sigma
bond
framework.
Thus,
some
energy
is
passed
along
to
adjacent

protons.
The
adjacent
protons
provide
feedback
on
the
spectrum
to
the
proton

that
absorbs
the
energy.
Moreover,
J
values
are
independent
of
the
ﬁeld

strength,
Bo
.

C
H1
C
H2
C
H1
C
H2
C
H1
C
H2
H1 absorbs
energy
H1 gives
energy back
(relaxation)
via sigma
bonds
C
H1
C
H2
H2 absorbs
some of
the energy
Within the mathematics of the spectrum
software, we can see how many neighboring
protons absorb the energy, which gives rise to
coupling (J) constants.

What
inﬂuences
the
magnitude
of
J
value?

-‐Distance
to
relaxing
proton

-‐Angle
to
relaxing
proton

Type
H
H
H H
H
H
H H
H
H
J Value (Hz)
12-15
2-9
0.5-3
7-12
13-18
Type J Value (Hz)
H H
ax-ax: 6-14
ax-eq: 0-5
eq-eq: 0-5
H
H
ortho: 6-10
meta: 1-3
para: 0-1
CH
H
0.5-3
HC CH
0-3
Note:
circled
protons
are
NOT
equivalent
1

Karplus
Curve

C' C
H'
H
φ
C
H
H'
φ
0 20 40 60 80 100 120 140 160 180
4
2
6
8
10
12
JHH'(Hz)
φ
1

Spligng
Pafern

à Peak
Spliqng
occurs
due
to
coupling
of
spins
(interacJons
between
adjacent
carbons).

à Peak
Spliqng
is
not
seen
for
H
connected
to
O,
N.
(because
of
hydrogen
bonding)

à Spliqng
is
based
on
the
number
of
H’s
on
adjacent
C.

a):
If
a
proton
has
n
protons
a7ached
to
adjacent
carbons,
it
will
split
into
n+1
peaks.

b):
Only
nonequivalent
protons
couple.

c):
If
H’s
are
on
same
C
and
they
are
homotopic
or
enanJotopic,
no
spliqng
will
occur.

d):
If
H’s
are
on
diﬀerent
C’s,
but
they
are
“chemically
equivalent”,
no
spliqng
will
occur.

General
rules

Peaks
are
classiﬁed
by
how
they
are
split

13

13

13
University
of
California,
Los
Angeles
Department
of
Chemistry.
Proton
NMR
Spectroscopy-‐Split
the
signals,
not
your
brain!
h7p://www.chem.ucla.edu/harding/ec_tutorials/tutorial37.pdf
(accessed
January
9,
2016).

13

13

Examples
for
Hydrogen
chemical
shics

01234
PPM
H
H
H O H
H
H
1.11
1.11
1.11
2.0
3.57
3.57
Ethanol

Note:
Only
three
peaks
due

to
equivalent
protons
(see

3-‐D
model
if
you
are
not

convinced)

Note:

Deshielded
protons
are
downﬁeld

01234
PPM
O
H
H
H
H
H
H
3.24
3.24
3.24
3.24
3.24
3.24
Dimethyl
Ether

Note:
All
protons
are
equivalent

Note:
All
equivalent
protons
are
deshielded
(downﬁeld)

0123456
PPM
H
H
H
H
H
H
4.97
5.03
5.70
1.71
1.71
1.71
Note:
Only
three
equivalent
protons

Note:
sp2
more
electronega-ve
than
sp3

carbon;
thus,
those
afached
protons
are

more
downﬁeld
and
both
are
not

equivalent.

Propylene
(1-‐Propene)

1

0123456
PPM
H
H
H
H
H
H
H
H
5.48
5.48
1.71
1.71
1.71
1.71
1.71
1.71
Note:
Only
two
signals
due
to

equivalent
protons
(symmetry
of

molecule)

(E)-‐2-‐Butene

1

0123456
PPM
1-‐Butene

H
H
H
H H
H
HH
4.97
5.03
5.70
2.00 2.00
1.06
1.061.06
Note:
sp2
more
electronega-ve
than
sp3

carbon;
thus,
those
afached
protons
are

more
downﬁeld
and
each
is
not
equivalent.

Note:
Two
sets
of
equivalent
protons.

1

Examples
for
Carbon
chemical
shic
(x20
rules)

Ethanol

Note:
Only
three
peaks
due

to
equivalent
protons
(see

3-‐D
model
if
you
are
not

convinced)

Note:

Deshielded
protons
are
downﬁeld

H
H
H O H
H
H
16.9 55.8
0102030405060
PPM

0102030405060
PPM
H
H
H O H
H
H
16.9 55.8
Ethanol

Note:
Deshielded
carbon
is

downﬁeld

Note:
“x20
rule”
applies
very
well

Dimethyl
Ether

Note:
both
carbons
are
equivalent

Note:
both
equivalent
carbons
are
deshielded
(downﬁeld)

O
H
H
H
H
H
H
56.1
56.1
0102030405060
PPM

Note:
“x20
Rule”
works
well

Note:
sp2
more
electronega-ve
than
sp3

carbon;
thus,
those
carbons
are
more

downﬁeld
and
both
are
not
equivalent.

Propylene
(1-‐Propene)

020406080100120140
PPM
H
H
H
H
H
H
115.9
132.7
17.2

Note:
Only
two
signals
due
to

symmetry.

(E)
-‐2-‐Butene

H
H
H
H
H
H
H
H
16.7
125.3
125.3
16.7
020406080100120
PPM
Note:
sp2
more
electronega-ve
than
sp3

carbon;
thus,
those
equivalent
carbons
are

more
downﬁeld.

1-‐Butene

Note:
sp2
more
electronega-ve
than
sp3

carbon;
thus,
those
carbons
are
more

downﬁeld
and
each
is
not
equivalent.

H
H
H
H H
H
HH
115.1
137.3
26.3
13.7
020406080100120140
PPM

Predic-ng
the
spligng
pafern
of
a
proton
signal

Example:
If
a
proton
(or
group
of
equivalent
protons)
relaxes
by
giving
oﬀ

energy
to
two
equivalent,
adjacent
protons
the
signal
will
be
split
into
a

triplet
(t)

Example:
If
a
proton
(or
group
of
equivalent
protons)
relaxes
by

giving
oﬀ
energy
to
three
equivalent,
adjacent
protons
the
signal

will
be
split
into
a
quartet
(q)

01234
PPM
H
H
H O H
H
H
1.11
1.11
1.11
2.0
3.57
3.57
Will split into a tripletWill split into a quartet

0123456
PPM
H
H
H
H
H
H
H
H
5.48
5.48
1.71
1.71
1.71
1.71
1.71
1.71
Will split into a q
Will split into a d
Example
of
spligng
paferns
for
(E)-‐2-‐butene

1

0123456
PPM
Will split into a d of d of d
H
H
H
H
H
H
4.97
5.03
5.70
1.71
1.71
1.71
Will split into a d of d of q
Examples of d of d of d
Example
of
spliLng
paTerns
for
1-‐propene
(can
get
complicated
very
quickly)

Mul-plet
(m)

Ko sponholtz NMR project

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