Unit One Part 10:
infrared spectroscopy and
mass spectrometry
my last
lecture...yippee

Unit One
Part10 spectroscopy?
What is infrared
Functional groups & IR
Mass spectrometry

determining structure
UV
uv-vis nmr ir
conjugation C–H functional
groups

determining structure
UV
three common forms of
spectroscopy, all used to
uv-vis nmr ir
look at molecules (and
determine their
structure)...
conjugation C–H functional
groups

determining structure
UV
three common forms of
spectroscopy, all used to
uv-vis nmr ir
look at molecules (and
determine their
structure)...
conjugationlast lecture...not
C–H functional
did nmr
groups
going to look at uv so that
leaves ir and...

determining structure
O CH3
H3C N
N
H
N
O N
Mass:
CH3
194.08
caffeine
C8H10N4O2
we’ll also look at mass
spectrometry (not a form
of spectroscopy, but still
useful!)

look at infrared
spectroscopy first
what is infrared
spectroscopy?

infrared spectroscopy
molecule in excited
energy state E2
energy
energy
basically we shine
energy
infrared radiation on a
molecule and some of it
is...
energy
molecule in
energy state E1

infrared spectroscopy
molecule in excited
energy state E2
energy
energy
is absorbed, exciting the
molecule...we measure what comes energy
through and note what has been
absorbed (unlike nmr when we
measured what was emitted)
energy
molecule in
energy state E1

what energy is absorbed
by the molecule?

atom 1 atom 2
bond
two atoms connected by
a bond can be thought to
be the same as...

spring
...two masses / blobs
joined by a spring...
mass 1 mass 2

bonds
vibrate
bonds vibrate
(stretching and
contracting) in the
same way...

like
springs
...a spring would...

...this means we can use
Hooke’s law on extension
and elasticity...
Hooke’s
law

...of course, I hate maths
so I’m not actually going
to show you Hooke’s law
but just tell you what it
reveals...
Hooke’s
law

energy to vibrate bond
depends on...
spring
the force needed to
vibrate a bond / spring depends
on the strength of the bond /
spring and...
mass 1 mass 2
strength

energy to vibrate bond
depends on...
spring ...the difference in
mass of the two ends
(or atoms)...this
means...
mass 1 mass 2
difference in mass

strong (short) bond
requires...
...so it is easy to see
multiple bonds in IR...
more energy

light (hydrogen) atom
vibrates faster
...and bonds to
hydrogen (C–H, O–H
etc)
more energy

bonds only vibrate
with fixed frequencies
and...
bonds vibrate with
certain frequencies

...they will only
absorb energy of the
same frequency or
wavelength...
λ
bond will only absorb
energy of same frequency

energy / light
long wavelength short wavelength
low frequency high frequency
low energy high energy

energy / light
...so, how does all
this mumbo-jumbo (I
mean physics) effect IR
spectroscopy?
long wavelength short wavelength
low frequency high frequency
low energy high energy

infrared spectroscopy
υ1 υ1
υ1 υ2 υ3
O
υ2
EtO CH3
υ3 υ3
the cartoon version

infrared spectroscopy
υ1 υ1
υ1 υ2 υ3
O
υ2
EtO CH3
υ3 υ3 ...shine IR on molecule...
certain bonds will absorb certain
wavelengths of energy and thus by
observing which wavelengths are
missing we have a clue as to what
bonds are in the molecule...
the cartoon version

infrared spectroscopy
low
high
energy
energy
υ1 υ1
υ1 υ2 υ3
O
υ2
EtO CH3
υ3 υ3
high low
the real version
wave wave
number number

infrared spectroscopy
low
high
energy
energy
wavenumber is the
υ1 υ 1
inverse (1/λ) of the wavelength
υ1 υ2 υ3
in cm and measures
O energy...
υ2
EtO CH3
υ3 υ3
high low
the real version
wave wave
number number

more on the theory of
IR can be found at:
www.massey.ac.nz/~gjrowlan
‘
in ‘Introduction to organic
and bioorganic molecules and
reactions’

interpreting IR spectra
energy to cause vibration
1000 cm-1
4000 3000 2000 1500
C C C C C O
O H
change in scale
C O
C N C F
N H
C O C Cl
C H
single
triple
bonds to double
bonds regions of
four
bonds
hydrogen bonds
the spectrum are
important...

interpreting IR spectra
energy to cause vibration
1000 cm-1
4000 3000 2000 1500
C C C C C O
O H
change in scale
C O
C N C F
N H
C O C Cl
C H
single
triple
bonds to double
bonds bonds
hydrogen bonds
light
atoms
(H)

interpreting IR spectra
energy to cause vibration
1000 cm-1
4000 3000 2000 1500
C C C C C O
O H
change in scale
C O
C N C F
N H
C O C Cl
C H
single
triple
bonds to double
bonds bonds
hydrogen bonds
strong
strong
bonds
bonds

interpreting IR spectra
energy to cause vibration
1000 cm-1
4000 3000 2000 1500
C C C C C O
O H
change in scale
C O
C N C F
N H
C O C Cl
C H
single
triple
bonds to double
hydrogen bonds region isn’t that bonds
bonds
actually fingerprint
useful...it is unique to each molecule but
almost impossible to interpret...good if
you have a computer database I guess...
1500–400cm–1
fingerprint
region

interpreting IR spectra
these two molecules have same
functional group (ketone) and
are almost identical in three
regions...
O O
1500–400cm–1
fingerprint
region

interpreting IR spectra
only really differ in fingerprint...
but I couldn’t tell you what bond
stretching caused this peak!
O O
1500–400cm–1
fingerprint
region

bad slide alert!

functional group absorptions

functional group absorptions
can predict roughly where
most functional groups will come...you’ll
be given this in an exam if
you need it...

examples of IR spectra
H H
N
NH2 benzeneamine
3480 aniline
3395 two N–H stretches visible...but
not for the reason you think
(I’m not going to tell you why
in case it confuses you!)

examples of IR spectra
H3C H
N
N–H
3443
N-methylbenzenamine
N-methylaniline

examples of IR spectra
Ph
O–H
3224 O
O H
H H
H
O
O
Ph
Ph
phenol - H-bonding

examples of IR spectra
Ph
O–H
3224 O
O
hydrogen bonding causes H
H H
H
the peak to be very broad as the
O
O
strength of H-bonds varies
depending on factors like
Ph
Ph
distance...
phenol - H-bonding

examples of IR spectra
O–H
3627 H
O O
H
2,6-di-tert-butyl-4-methylphenol

examples of IR spectra
O–H
3627 H
no H-bonding so O–H has
O O
specific strength bond and sharp
peak (there is no H-bonding as
H
the large tert-butyl groups
prevent the two molecules
getting close to each other...)
2,6-di-tert-butyl-4-methylphenol

examples of IR spectra
O
C=C
1642
C=O
hex-5-en-2-one 1718

examples of IR spectra
C=C
C=O
1634
1674
O
note how putting the two
groups in conjugation makes the
bonds weaker (and hence have a
lower wavenumber)
(E)-hex-4-en-3-one

examples of IR spectra
C=C
C=O
1634
1674
O
why does conjugation make the
bonds weaker? Think about the
resonance forms...
(E)-hex-4-en-3-one

examples of IR spectra
O–H
3010
C=O
1712 O
H
O
butanoic acid

examples of IR spectra
N–H C=O
3356 1662 O
3184 1634
H
N
H
butanamide

examples of IR spectra
weaker C=O stretch in
amide due to a resonance
form involving the nitrogen
lone pair...see IR can tell us
a lot of useful information...
N–H C=O
3356 1662 O
3184 1634
H
N
H
butanamide

mass spectrometry

‘a mass spectroscopist is
someone who figures out what
something is by smashing it with a
hammer & looking at the pieces
JEOL (manufacturer) website

a mass spectrometer
+
– –
+ +
M M
e 2e

a mass spectrometer
basically, what you need to know is that
a mass spectrometer fires electrons at your
compound knocking one electron off the compound
to form a radical cation that is then detected...
+
– –
+ +
M M
e 2e

molecular mass
molecular ion
+NH
– –
+ +
NH3 e 2e
3
because the mass
of an electron is
very very small...
17.031 5.5 x 10–4 17.030
mass-to-charge ratio (m/z)

molecular mass
molecular ion
+NH
– –
+ +
NH3 e 2e
3
the radical cation
(or molecular ion)
effectively has the same
mass as the original
compound...
17.031 5.5 x 10–4 17.030
mass-to-charge ratio (m/z)

molecular mass
molecular ion ...as a result mass
spectrometry gives us the
molecular mass...it also gives
a lot more info but thats for
another day (or course)
+NH
– –
+ +
NH3 e 2e
3
17.031 5.5 x 10–4 17.030
mass-to-charge ratio (m/z)

cyclohexane
+
e– 2e–
C6H12 + +
[C6H12 ]
m/z 84
C = 6x12
H = 12x1
a simple example m/z 85
showing the mass due to 13C
of cyclohexane... isotope

cyclohexane
+
e– 2e–
C6H12 + +
[C6H12 ]
m/z 84
C = 6x12
H = 12x1
...all these other
peaks are
useful...but lets
ignore
m/z 85
due to 13C
isotope

31
isotopes
35Cl 37Cl
chlorine exists as two
isotopes (same element
different mass due to
number of neutrons)

31
isotopes
35Cl 37Cl
there is 3 times as much Cl
mass 35 than Cl mass 37 (hence
average is 35.5 as shown on
most periodic tables)

isotopes
[C6H5Cl]+
e– 2e–
C6H5Cl + +
the mass spectrum has two
peaks...one for each isotope and
the relative size of these peaks will m/z 112
due to
be 3 : 1...
35Cl
Cl m/z 114
due to
37Cl
C6H535Cl Mr = M+ = 112
C6H537Cl Mr = M+2 = 114

11
isotopes
79Br 81Br
chlorine not the only element with
isotopes...bromine exists as two
isotopes in equal proportion so the
spectrum of a bromide...

isotopes
[C6H5Br]+
e– 2e–
C6H5Br + +
...will have two peaks of equal
m/z 156
intensity with a mass 2 apart
due to
79Br
m/z 158
due to
Br 81Br
C6H579Br Mr = M+ = 156
C6H581Br Mr = M+2 = 158

what have
....we learnt?
determine molecular
•to
structure
basics of infrared
• the
spectroscopy
• the basics of mass
spectrometry

good luck!