Mass Spectrometer and Infra-red
Spectroscopy
Aims:
To review year 12 analytical
techniques!
What do we use the mass spectrometer for and
how does it work?
What is it used for?
• To identify unknown compounds
• To determine the abundance of each isotope
in an element
• To gain information about the structure and
properties of molecules
How it works
The sample is first
Vapourised.
Then it is Ionised by
having high energy
electrons fired at it.
M(g) + e- → M+(g) + 2e-
VIA
These ions are then
Accelerated by a
strong electric field
How it works
A good vacuum is
essential in the whole
of the apparatus
How it works
The accelerated ions are Deflected by the magnetic field, bigger ions
are deflected less.
The ion beams
can be
focused on
the Detector
We can use the mass spectrometer to work
out the Ar of a sample containing different
isotopes.
Antimony has two isotopes 121-Sb and 123Sb. In a bullet at a crime scene there was a
sample of antimony containing 57.3% 121-Sb
and 42.7% 123-Sb.
Calculate the relative atomic mass of the
sample of antimony from the bullet.
Identifying organic compounds
• Organic molecules do not hold charge well.
• The ions break up or fragment.
• They break into smaller ions and neutral fragments.
• They do this in regular patterns.
i.e. the same molecule always fragments in the same way.
• This allows us to analyse them.
Fragmentation
• M(g) + e- → M+(g) + 2e• M+ is called the molecular ion, M+. Since M+ has lost an
electron, it is less strongly held together than the
original molecule. So, some of the molecular ions fall
apart (fragment) in the ion source to give fragment ions
(F+) and neutral fragments (N).
• M+(g) → F+(g) + N*(g)
• Analysis of the charged particles in the mass
spectrometer gives the MASS SPECTRUM of the
molecule.
Mass Spectrum of ethanol
Molecular ion:
tells you the
molecular
mass of the
whole
compound.
Mass Spectrum of ethanol: Identify the
fragments responsible for the following
peaks:
46
45
31
29
15
Possible fragments
+
+
45
31
M+
+
46
++
29
+
+
15
Mass Spectrum of an alcohol.
Identify the alcohol and draw structures for the
fragment ions represented by the peaks at: 60,
59, 43, 31 and 29.
Possible fragments
+
+
59
31
M+
+
60
+
+
43
+
+
29
IR spectroscopy
What do you remember?
How IR spectroscopy works…
•Shine a range of IR frequencies, one at a time
through a sample of organic compound and at
some of the frequencies the energy will be
absorbed.
•A detector on the other side of the sample
would show that some frequencies are
absorbed whilst others are not.
• If a particular frequency is being absorbed as
it passes through the compound being
investigated, it must mean that its energy is
being transferred to the compound.
• Energies in infra-red radiation correspond to
the energies involved in bond vibrations.
Bend and Stretch
• In covalent bonds, atoms aren't joined by rigid
links;
• the two atoms are held together because both
nuclei are attracted to the same pair of
electrons.
• The two nuclei can vibrate backwards and
forwards - towards and away from each other
- around an average position.
All molecules absorb infrared radiation;
absorbing energy makes the bonds vibrate.
STRETCHING
BENDING
Every bond vibrates at its own frequency, dependant upon bond
strength, bond length and the mass of atoms involved in bond.
Fingerprint regions
• You must be able to identify the following
peaks.
– C-H in alkanes/alkenes/aldehydes
– O-H in alcohols/carboxylic acids
– N-H in amines
– C=O in aldehydes and ketones
– C-X in halogenoalkanes
• You will have data books in exam, if required.
http://www.rsc.org/learnchemistry/resource/res00000059/in
fra-red-spectrometer
C-O in alcohols, esters & carboxylic acids
1000-1300
C=O in aldehydes and ketones
1640-1750
C-H in alkanes/alkenes/aldehydes
2850-3100
O-H in carboxylic acids
2500-3300 (very broad)
N-H in amines/amides
3200-3500
O-H in alcohols/carboxylic acids
3200-3550 (broad)
C-O in alcohols, esters & carboxylic acids
1000-1300
C=O in aldehydes, ketones & carboxylic acid
1640-1750
C-H in alkanes/alkenes/aldehydes
2850-3100
O-H in carboxylic acids
2500-3300 (very broad)
N-H in amines/amides
3200-3500
O-H in alcohols/carboxylic acids
3200-3550 (broad)
C-O in alcohols, esters & carboxylic acids
1000-1300
C=O in aldehydes and ketones
1640-1750
C-H in alkanes/alkenes/aldehydes
2850-3100
O-H in carboxylic acids
2500-3300 (very broad)
N-H in amines/amides
3200-3500
O-H in alcohols/carboxylic acids
3200-3550 (broad)
C-O in alcohols, esters & carboxylic acids
1000-1300
C=O in aldehydes, ketones & carboxylic acids
1640-1750
C-H in alkanes/alkenes/aldehydes
2850-3100
O-H in carboxylic acids
2500-3300 (very broad)
N-H in amines/amides
3200-3500
O-H in alcohols/carboxylic acids
3200-3550 (broad)
Lesson 2
Nuclear Magnetic Resonance
Used to study molecular structures in
detail.
NMR relies on the fact that protons and neutrons
have a property called spin and can be in one of
two opposite directions. In the nucleus opposite
spins pair up and cancel out.
Some nuclei have odd numbers of nucleons. This
produces a small net nuclear spin which generates
a small magnetic field.
Nuclei with an overall spin can be
thought of as tiny magnets. They line up
in a strong magnetic field and are either
with the field or opposed to it.
Nuclei that oppose the magnetic
field have a higher energy than
those that align. This creates an
energy gap.
Resonance
A nucleus in the low spin state can be ‘excited’ to the upper spin
state by providing energy that matches the energy gap.
This is done by applying low-energy radiofrequency waves.
Resonance
A nucleus in the low spin state can be ‘excited’ to the upper spin
state by providing energy that matches the energy gap.
This is done by applying low-energy radiofrequency waves.
The excited nucleus will drop back down to its low-energy state
emitting the same amount of energy: this is called relaxation.
The cycle of resonance continues as long as the frequency of
radiation supplied matches exactly the energy gap.
The magnetic field felt by a nucleus depends on 2
factors
1) The applied strong magnetic field
2) Weaker magnetic fields from electrons
surrounding the nucleus and in surrounding atoms.
These electrons have a shielding effect.
Different environments result in different resonance
frequencies and so different chemical shifts– this
can give us clues about the structure of a molecule.
Chemical shift
Chemical shift, δ = place in NMR spectrum at which a
nucleus absorbs energy.
Measured in parts per million and is measured relative to
a reference signal from a standard compound –
tetramethylsilane or TMS.
TMS has 12 equivalent protons and gives a sharp signal
that is easy to identify. The chemical shift of TMS is
defined as δ=0
Solvents
The sample is dissolved in a deuterated solvent. This
is because hydrogen atoms cause a signal which
would interfere with the spectrum.
To avoid this Deuterium is used as it has an even
number of nucleons and so produces no signal.
CDCl3 is often used as a solvent and evaporated off
afterwards.
C-13 NMR Spectra
Carbon-13 has an odd number of nucleons and
so has a residual magnetic spin.
This property allows the identification of
carbon atoms in an organic molecule.
How it works
• Carbon-13 makes up 1.1% of all naturally
occurring carbon atoms.
• Carbon-13 has 13 nucleons, an odd number so
it has a residual magnetic spin.
• i.e. it shows up on NMR.
• Carbon -13 atoms in different environments
have different chemical shifts.
Carbon-13 is sensitive to nuclear
shielding and gives a large range of
chemical shifts that show up as
separate peaks on the spectrum.
Carbon-13 chemical shifts: see page 86
Interpreting the spectrum
The number of peaks tells us the number of
carbon environments
The chemical shifts tell us the types of carbon
environment
The size of the peak is not useful in this case (in
contrast to proton NMR – next lesson)
Propan-1-ol
3 peaks: 3 C atoms in 3 different environments.
Propan-1-ol
Propan-2-ol: why are there only 2
peaks?
Propan-2-ol
2-bromopropane
Your tasks
Read pages 86-87
Answer questions 1 and 2 on page 87
Read through worked examples 1 & 2 on page
88-89
Answer Q1 on page 89
Prep: read pages 90-91 and 98-99.
Make notes on both pages and
answer the questions. Due next
lesson.
Lesson 3
Prep: Hand in your prep from last
lesson.
What do you remember about
NMR?
Proton NMR
Why does Hydrogen give an NMR signal?
A 1H nucleus is a single proton and so has a residual
spin.
The spectrum can be interpreted in a
similar way as carbon-13 spectra
- The number of peaks = number of proton
environments
- The chemical shift tells us about the type of
proton environment
In addition:
The relative peak area tells us the
proportions of protons in each
environment (sometimes added onto
the spectrum as an integration trace, see
page 91)
Spin-spin coupling gives information
about adjacent protons
Predict and number the environments
H
H
O
O
H
H
H
O
H
3-hydroxypropanoic acid
H
H
C
C
H
H
H
C
C
H
H
(2E)-but-2-ene
H
Predict the number of proton environments in
each of these examples
H
H
H
H
H
H
H
H
H
H
H
H
H
O
H
H
H
H
H
H
OH
H
H
H
H
H
H
H
H
H
CH3
Predict Each Proton’s Chemical Shift
1
e.g.
Proton
H
H2
O
Relative Area
Functional
Group
δ / ppm
H
H
3
O
2
1
1
O-H
1-12
2
2
O-CH
3-4.5
3
2
O=CCH
2-3
4
1
COOH
11-12
H3
O
H 4
Did we get it right?
Spin-spin coupling
On proton NMR you may see little clusters
rather than a single peak.
This arises due to the interaction with the spin
states of protons on adjacent C atoms.
These ‘clusters’ can help you identify the
number of protons in the environment next to
it.
Possible arrangements of
neighbouring protons
With 1 H there are
2 possible fields, 1
up or 1 down.
With 2 H there are 3
possible fields, both
up, 1 up, 1 down or
both down.
• There is always one more field than the
number of adjacent protons: these are seen
on the spectrum as sub-peaks
• This tells us about the number of protons on
the ADJACENT carbon.
The N+1 rule
Look at this CH3 group.
It has a CH2 group next
to it. The CH2 group
contains 2 protons.
2+1 = 3
Therefore we expect to
see 3 miniature peaks,
or a triplet.
We can use the N+1 rule to identify how many
protons are in the environment next to it.
Look at this CH group. It has
Example.
no protons next to it.
3
0+1 = 1, therefore we expect
Look
at this
to see
oneCH
peak,
or a singlet.
2 group,
it has a CH3 group next
to it, with 3 protons.
3+1 = 4,
Therefore we expect to
see 4 mini peaks next
to it, or a quartet.
The N+1 Rule
This CH3 group has a group next
to it with one proton.
1 + 1 = 2, therefore we
see a doublet.
This aldehyde proton has a CH3
group next to it.
3 + 1 = 4,
Therefore we see a quartet group.
Note: Spin-spin coupling only happens between non-equivalent
protons.
So the three protons in CH3 have the same chemical shift and are
equivalent (you only see one peak that represents all three). They
do not couple with each other.
The N+1 Rule
N
N+1
Multiplicity
0
0+1 = 1
Singlet
1
1+1 = 2
Doublet
2
2+1 = 3
Triplet
3
3+1 = 4
Quartet
The relative intensities of the lines can be determined by
Pascal’s triangle.
i.e. a doublet has a relative intensity of 1:1
a triplet has an intensity of 1:2:1
a quartet has an intensity of 1:3:3:1
The gap between the peaks also stays the same and is
determined by the adjacent proton environment, this is
called the coupling constant.
OH and NH protons
These peaks can be difficult to identify
They appear over a wide range of chemical
shift values, they can be broad and there is
usually no splitting pattern.
Splitting from OH or NH
OH and NH peaks show up as a singlet
which may be broad.
This is due to the sample Hydrogen
bonding to small traces of water that are
difficult to avoid.
Splitting from OH or NH
NMR peaks from OH or NH are not split.
Protons on adjacent carbons are also not
split by these groups either
So basically ignore OH and NH when it
comes to splitting
Use of D2O
Heavy water can be used in NMR because Deuterium does
not generate an NMR signal (why not?)
D2O is used as follows:
-A proton NMR spectrum is run
- D2O is added and the mixture shaken
- A second proton NMR spectrum is run. Any peak due to OH
or NH disappears
-The deuterium in D2O exchanges with the OH or NH proton
-These groups can be easily identified by comparing the two
spectra
On Your Whiteboard
Predict the chemical shift for each proton in
Proton
Relative Area
Functional
Group
δ / ppm
H
6
R-CH
1-2
2
4
R-CH
1-2
H
H
H
H
H
H
1
H
H
H
Did you get it right?
Chromatography
In pairs:
Use this diagram to write a
summary of what
chromatography is and how
it works
Key words: mobile phase,
stationary phase, separation,
affinity.
Chromatography
•Chromatography is a technique in which substances in a
mixture are separated between a mobile phase and a
stationary phase.
•Substances are separated as
they have different affinities for
the mobile and stationary phases
•Some substances stay dissolved
in the solvent and move further.
•Others are more attracted to the
stationary phase and so they are
slowed down.
Analysis
Two types of chromatography
Thin-layer chromatography:
Mobile phase is a liquid
Stationary phase is a solid
Gas chromatography:
Mobile phase is a gas
Stationary phase is a liquid or a solid on a
support.
Separation
Solid stationary phase separates by
adsorption: as the mobile phase
passes over the stationary phase
some molecules bind to the surface
of the stationary phase. The
stronger this interaction the more
the molecules are slowed down.
Liquid stationary phase separates by
solubility. The more soluble a
substance is in the liquid stationary
phase the more it will be slowed
down.
Thin layer chromatography
•Thin layer chromatography can be used to
monitor the progress of a reaction or to
check the purity of compounds.
• Stationary phase = glass, plastic, or
aluminum foil. Coated with a thin layer of
silica gel, aluminium oxide, or cellulose
• The mobile phase is a liquid solvent that
moves vertically up the plate.
Quantitative Chromatography - calculating Rf value
Rf = distance moved by sample
distance moved by solvent
Solvent Front
Worked examples:
Rf = 5.5 = 0.92
6.0
Rf = 3.0 = 0.50
6.0
Sample start point
Solvent start point
Analysis
Rf values can be compared with those of pure
substances to help identify unknown compounds.
Analysis
Limitations of TLC
- Rf values can be similar for similar
compounds
- Unknown compounds may have no Rf
value for comparison
- It can be difficult to find a solvent that
works for all components of a mixture.
Gas Chromatography
http://my.rsc.org/video/55
Gas Chromatography
• Mobile phase is an unreactive
gas (helium, nitrogen)
• Stationary phase is a solid or
liquid layer on the inside surface
of a column
•Sample is injected into machine
•Sample is vaporised and then
carried through the column
• Computer records exact
quantities of each part of the
mixture as it moves through the
column.
Organic compounds have
known retention times, they
can be identified by looking
up the values of the
unknowns
Interpreting Gas Chromatographs
•The position of the peak (retention time) identifies the compound: compared to
known retention times.
•The area of the peak is used to calculate the quantity of material in the sample
•eg. There is a lot more linoleic acid than arachidic acid. There is very little
linolenic acid in this sample.
Analysis
Limitations of GC
Similar compounds have similar retention times so it is
difficult to distinguish between them
Unknown compounds have no reference retention
times
Not all substances will necessarily be separated and
detected
Gas Chromatography-mass spectrometry
By combining GC with Mass spectrometry we can obtain far more
information than with either technique alone.
GC can separate substances but cannot identify them conclusively
MS can identify compounds but cannot separate a complex
mixture.
Gas Chromatography-mass spectrometry
This combined technique can be used in:
-Forensics: to identify substances found at the scene of a crime
- Environmental analysis: e.g. To monitor the quality of drinking
water
- Airport security: detecting explosives
- Space probes: analysing material from other planets/moons
Thin Layer Chromatography
Thin layer chromatography
•Make a flow chart of the steps in TLC using page
78
Analysis