4. 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
5. 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
did nmr
C–H functional
going to look at uv so that
leaves ir and...
groups
6. determining structure
O CH3
H3C N
N
H
O N N
CH3 Mass:
caffeine
194.08
C8H10N4O2
we’ll also look at mass
spectrometry (not a form
of spectroscopy, but still
useful!)
8. 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
9. 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
17. ...this means we can use
Hooke’s law on extension
and elasticity...
Hooke’s
law
18. ...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
19. 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
20. 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
21. strong (short) bond
requires...
...so it is easy to see
multiple bonds in IR...
more energy
22. light (hydrogen) atom
vibrates faster
...and bonds to
hydrogen (C–H, O–H
etc)
more energy
24. ...they will only
absorb energy of the
same frequency or
wavelength...
λ
bond will only absorb
energy of same frequency
25. energy / light
long wavelength short wavelength
low frequency high frequency
low energy high energy
26. 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
28. 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
29. infrared spectroscopy
high low
energy energy
υ1 υ1
υ1 υ2 υ3
O
υ2
EtO CH3
υ3 υ3
high low
the real version
wave wave
number number
30. infrared spectroscopy
high low
energy energy
υ1 wavenumber is the
υ 1
inverse (1/λ) of the wavelength
in cm and measures υ1 υ2 υ3
O energy...
υ2
EtO CH3
υ3 υ3
high low
the real version
wave wave
number number
31. more on the theory of
IR can be found at:
www.massey.ac.nz/~gjrowlan
‘
in ‘Introduction to organic
and bioorganic molecules and
reactions’
32. interpreting IR spectra
energy to cause vibration
4000 3000 2000 1500 1000 cm-1
O H C C C C C O
change in scale
N H C N C O C F
C H C O C Cl
bonds to triple double single
hydrogen bonds bonds bonds regions of
four
the spectrum are
important...
33. interpreting IR spectra
energy to cause vibration
4000 3000 2000 1500 1000 cm-1
O H C C C C C O
change in scale
N H C N C O C F
C H C O C Cl
bonds to triple double single
hydrogen bonds bonds bonds
light
atoms
(H)
34. interpreting IR spectra
energy to cause vibration
4000 3000 2000 1500 1000 cm-1
O H C C C C C O
change in scale
N H C N C O C F
C H C O C Cl
bonds to triple double single
hydrogen bonds bonds bonds
strong strong
bonds bonds
35. interpreting IR spectra
energy to cause vibration
4000 3000 2000 1500 1000 cm-1
O H C C C C C O
change in scale
N H C N C O C F
C H C O C Cl
bonds to triple double single
hydrogen bonds region isn’t that
actually fingerprint bonds bonds
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
36. 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
37. 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
40. functional group absorptions
can predict roughly where
most functional groups will come...you’ll
be given this in an exam if
you need it...
41. 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!)
42. examples of IR spectra
H3C H
N
N–H
3443
N-methylbenzenamine
N-methylaniline
43. examples of IR spectra
Ph
O–H
3224 O O
H H
H H
O O
Ph Ph
phenol - H-bonding
44. examples of IR spectra
Ph
O–H
3224 O O
hydrogen bonding causes H
H
the peak to be very broad as the H H
strength of H-bonds varies O O
depending on factors like
distance... Ph Ph
phenol - H-bonding
45. examples of IR spectra
O–H
3627 H
O O
H
2,6-di-tert-butyl-4-methylphenol
46. examples of IR spectra
O–H
3627 H
no H-bonding so O–H has
specific strength bond and sharp
O O
peak (there is no H-bonding as
the large tert-butyl groups
H
prevent the two molecules
getting close to each other...)
2,6-di-tert-butyl-4-methylphenol
47. examples of IR spectra
O
C=C
1642
C=O
hex-5-en-2-one 1718
48. examples of IR spectra
C=C
C=O
1634
O 1674
note how putting the two
groups in conjugation makes the
bonds weaker (and hence have a
lower wavenumber)
(E)-hex-4-en-3-one
49. examples of IR spectra
C=C
C=O
1634
O 1674
why does conjugation make the
bonds weaker? Think about the
resonance forms...
(E)-hex-4-en-3-one
50. examples of IR spectra
O–H
3010
C=O
1712 O
H
O
butanoic acid
51. examples of IR spectra
N–H C=O
3356 1662 O
3184 1634
H
N
H
butanamide
52. 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
54. ‘a mass spectroscopist is
someone who figures out what
something is by smashing it with a
hammer & looking at the pieces
JEOL (manufacturer) website
56. 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
57. molecular mass
molecular ion
NH3 + – +NH + –
e 3 2e
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)
58. molecular mass
molecular ion
NH3 + – +NH + –
e the radical cation 3 2e
(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)
59. 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)
NH3 + – +NH + –
e 3 2e
17.031 5.5 x 10–4 17.030
mass-to-charge ratio (m/z)
60. cyclohexane
+
C6H12 + e– [C6H12 ] + 2e–
m/z 84
C = 6x12
H = 12x1
a simple example m/z 85
showing the mass due to 13C
of cyclohexane... isotope
61. cyclohexane
+
C6H12 + e– [C6H12 ] + 2e–
m/z 84
C = 6x12
H = 12x1
...all these other
peaks are
useful...but lets
ignore
m/z 85
due to 13C
isotope
62. 31
isotopes
35Cl 37Cl
chlorine exists as two
isotopes (same element
different mass due to
number of neutrons)
63. 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)
64. isotopes
C6H5Cl + e– [C6H5Cl]+ + 2e–
the mass spectrum has two
peaks...one for each isotope and
the relative size of these peaks will m/z 112
be 3 : 1... due to
35Cl
Cl m/z 114
due to
37Cl
C6H535Cl Mr = M+ = 112
C6H537Cl Mr = M+2 = 114
65. 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...
66. isotopes
C6H5Br + e– [C6H5Br]+ + 2e–
...will have two peaks of equal
intensity with a mass 2 apart m/z 156
due to
79Br
m/z 158
due to
Br 81Br
C6H579Br Mr = M+ = 156
C6H581Br Mr = M+2 = 158
67. what have
....we learnt?
• to determine molecular
structure
• thebasics of infrared
spectroscopy
• the basics of mass
spectrometry
