Unit 11:Data processing and analysis.
A.
B.
C.
D.
Infrared spectroscopy
Mass spectrometry
X-ray diffraction/crystallography
H NMR
A. Infrared spectroscopy
What is infrared?
An electromagnetic wave
Energy of electromagnetic radiation is
carried in discrete packets of energy
called photons or quanta.
E = hν



E = energy of a single photon of radiation
h = 6.63 x 10-34 Js (Plank’s constant)
ν = frequency of the radiation
Example: Calculate the energy of a photon of
visible light with a frequency of 3.0 x 1014 s-1.
Express in kJ mol-1.
E = hν
E = (6.63 x 10-34 Js)(3.0 x 1014 s-1)
E = 1.989 x 10-19 J
1.989  10 19 J 6.02  1023 photons 1kJ


photon
mol
1000J
 120 kJ  mol
1
Type of em radiation
Typical f (s-1) Typical  (m)
Radio waves (low energy)
3 x 106
102
Microwaves
3 x 1010
10-2
Infrared (IR)
3 x 1012
10-4
Visible (ROYGBIV)
3 x 1015
10-7
Ultraviolet (UV)
3 x 1016
10-8
X-rays
3 x 1018
10-10
Gamma rays (high energy)
> 3 x 1022
< 10-14
Note:  . ν = c = 3.0 x 108 ms-1
Thus, ν= c/ 
Wavenumber

In IR spectroscopy, the frequency of radiation
is often measured as number of waves per
centimeter (cm-1), also called wavenumber.

Example: Calculate the wavenumber in cm-1
for an IR wave with a frequency of 3 x 1013 s-1.
c=λν  1/λ = ν/c
1 n
3´1013s-1
5 -1
-1
= =
=
3´10
m
=1000
cm
l c 3´108m ×s-1
Infrared spectroscopy
A bond will absorb radiation of a frequency similar to its vibration
Light atoms vibrate at higher freq. than heavier atoms;
they absorb IR radiation of shorter wavelength ( more energy)
Multiple bonds vibrate at higher freq. than single bonds;
they absorb IR radiation of shorter wavelength ( more energy)
normal vibration
vibration having absorbed energy
By measuring the IR spectrum of a molecule we can determine what
kind of motion the molecule has and therefore what kind of bonds are
present in the molecule.
Infrared has the right energy to be absorbed by the polar
bonds of a molecule .
The IR radiations ( specific energy or frequency ) make
the bonds stretch, bend, or vibrate.
Symmetrical
stretching
Antisymmetrical
stretching
Scissoring
Rocking
Wagging
Twisting
Using IR to excite molecules

Not all vibrations absorb IR.

For absorption, there must be a change in
bond polarity (dipole moment) as the
vibration occurs.

Thus, diatomic gas molecules such as H2,
Cl2 and O2 do not absorb IR.
Vibrations of H2O, SO2 & CO2
Molecule
H 2O
Asymmetrical
stretching
Symmetrical
stretching
-
-
-
O
O
O
+
H +
H
+
+
O -
O
-
CO2
O
+
C O
IR active
IR active
+
+
S
O - - O
IR active
-
-
O
H +
H
IR active
O
IR active
-
+
S
S
-
H +
H
IR active
SO2
Symmetrical
bending
+
IR active
-
C O
IR inactive
O -
-
O
+
-
C O
IR active
Matching wavenumbers with bonds
Data Booklet -table 26
“fingerprint region”
lots of overlap, so
not very useful
very strong
broad and strong
broad and strong
Usually sharper
than OH
https://www.youtube.com/watch?v=DDTIJgIh86E
IR spectrum of ethanol, CH3CH2OH
IR spectrum of ethyl ethanoate,
CH3COOCH2CH3
C-H
C=O
“fingerprint region”
FINGERPRINT REGION
The 1400 cm-1 to 800 cm-1 range is the
“fingerprint” region : its the pattern is
characteristic of a particular compound
IR SPECTRUM OF A CARBONYL COMPOUND


carbonyl compounds show a sharp, strong absorption between 1700
and 1760 cm-1
this is due to the presence of the C=O bond
IR SPECTRUM OF AN ALCOHOL


alcohols show a broad absorption between 3200 and 3600 cm-1
this is due to the presence of the O-H bond
IR SPECTRUM OF A CARBOXYLIC ACID


carboxylic acids show a broad absorption between 3200 and
3600 cm-1
 this is due to the presence of the O-H bond
they also show a strong absorption around 1700 cm-1
 this is due to the presence of the C=O bond
Practice some more spectra

http://undergrad-ed.chemistry.ohiostate.edu/anim_spectra/index.html
IR spetra of
butanal and
butanone
B. Mass
spectrometry
B.Mass
spectrometry
The Mass Spectrometer
http://www.youtube.com/watch?v=J-wao0O0_qM&feature=related
How it works:
The sample is bombarded with a stream of high energy electrons.
The collision is so energetic that it causes the molecule to break
up into different fragments (ions).
The fragments ( + ions) of a particular mass are detected and
a signal is sent to a recorder. The strength of the signal is a
measure of the number of ions with that charge/mass ratio that
are detected.
Fragmentation Patterns :
evidence for the structure of the compound
The largest mass peak corresponds to a parent ion passing through the
instrument unscathed, but other ions produced as a result of this break up
are also detected.
For each fragmentation, one of the products keeps the + charge and will be
detected.
Generally the most stable + ion is formed
CH3-CH2-OH 
CH3-CH2+
peak 29
+
OH
no peak 17
Example: ethanol
100
relative abundance
31
45
15
29
46
0
0
30
mass/charge
60
Example: ethanol
Example: ethanol
Note: This fragmentation will
yield either CH3+ and CH2OH
or CH3 and CH2OH+, yielding
peaks at both 15 and 31
Data table 28-Mass spectral of fragment lost.
Mr
loss of…
15
CH3
17
OH
29
C2H5 or CHO
31
CH3O
45
COOH
C . X-ray diffraction / X-ray crystallography
http://www.theguardian.com/science/video/2013/oct/09/100-years-x-ray-crystallography-video-animation
http://www.rsc.org/learn-chemistry/resource/res00000020/computational-chemistry#!cmpid=CMP00001685
When X-rays shine on a crystal (orderly structure), they are
reflected and produce an ordered diffraction pattern.
The diffraction pattern produced by X-rays helps determine
the electron density of the crystal.
As the electron densities are related to the element’s electron
configuration, we can also determine the identity of the atoms.
Note: H atoms have a very low electron density ( 1e) and are not visible
on the X-ray diffraction pattern.
magnetic
resonance
D. Nuclear
Nuclear magnetic
resonance.
H NMR spectroscopy.
(NMR) spectroscopy
Nuclear magnetic resonance (NMR)
spectroscopy
When an external magnetic field is applied to a molecule the
energy of its spinning protons splits in 2 separate levels. The
spin aligned with the magnetic field will be at a lower energy
state. The spin aligned against the magnetic field will be at a
higher energy state.
We can then get a proton change its spin from with the
magnetic field to against the magnetic field by providing the
right amount of energy.
The moment the proton returns to its lower energy level and
reverses its spin, it gives out the energy that shows as a peak resonance- in the NMR spectrum.
The HNMR spectrum:
its interpretation
https://www.youtube.com/watch?v=gaGUpACXijE
Chemical shifts vs TMS
standard (tetramethylsilane)
Different H groups ( types)
Integration
Number of H in the group
= area under the peak
Multiplicity of peak
(splitting)
Number of adjacent H
See Data Table 27:
H NMR data
The TMS standard
-All 12 H’s are in identical chemical environments, so one
signal is recorded = zero point on the scale
-The H are very shielded by the electrons of C that is more
electronegative that silicon. Therefore they experience the
strength of the magnetic field the least.
-The signal doesn’t interfere with the signal given by H bonded
to Carbon. The less the H will be shielded in a group, the
greater the shift will be.
1H-NMR
spectroscopy: 1-bromopropane
CHEMICAL SHIFTS
1
2
3
3 environments = 3 signals
Triplet
Sextet
Triplet
d = 3.4
d = 1.9
d = 1.0
Signal for H’s on carbon 3
is shifted furthest
downfield from TMS due
to proximity of the
electronegative halogen
TMS
5
4
3
2
1
0

1H-NMR
spectroscopy: 1-bromopropane
INTEGRATION
1
2
3
Area ratio from relative
heights of integration lines
= 2:2:3
2
3
TMS
Carbon 1
Carbon 2
Carbon 3
3
2
2
2
5
4
3
2
1
0

1H-NMR
spectroscopy: 1-bromopropane
SPLITTING
1
1
2
3
SPLITTING PATTERN
Carbon 1
TMS
Chemically different
hydrogen atoms on
adjacent atoms = 2
2+1 =3
The signal will be a
TRIPLET
5
4
3
2
1
0

1H-NMR
spectroscopy: 1-bromopropane
SPLITTING
2
1
2
3
SPLITTING PATTERN
Carbon 2
Chemically different
hydrogen atoms on
adjacent atoms = 5
TMS
5+1 =6
The signal will be a
SEXTET
5
4
3
2
1
0

1H-NMR
1
spectroscopy: 1-bromopropane
2
3
TMS
SUMMARY
4
Peaks
Shift
Integration
Splitting
3
2
1
Three different signals as there are three chemically different protons.
Signals are shifted away from TMS signal, are nearer to the halogen
( H is deshielded more by Br ).
The integration lines show that the ratio of protons is 2:2:3
Signals include a triplet (d = 1.0) sextet (d = 1.8) triplet (d = 3.4)
The signals due to the protons attached to carbon ...
C1 triplet
C2 sextet
C3 triplet
0
(d = 1.0)
(d = 1.8)
(d = 3.4)
coupled to the two protons on carbon C2
coupled to five protons on carbons C1 and C3
coupled to the two protons on carbon C2
( 2+1 = 3 )
( 5+1 = 6 )
( 2+1 = 3 )
