Section B 2 UV-visible (electronic) spectroscopy

Question 1

• Typical chromophores include :
conjugated polyunsaturated systems : -C=C-C=C- etc : carotene
polyaromatic systems : chlorophyll
systems containing C=O, C=S, C-P species etc
molecules containing transition metal ions (Fe3+, Co2+ etc) :
e.g. haemoglobin

Answer 1

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Question 2

Important things to determine :

Answer 2

1. Number and position(s) of the peaks: quoted in tables as λmax values : (but also as 1/λ in cm-1 units)
2. Relative peak intensities - this helps work out what kind of electronic transitions are involved.
3. Absolute peak intensities - you can do this by measuring the Absorbance values for several concentrations
   of a samples and using the Beer-Lambert law to determine the characteristic ε for the electronic transition
4. All this information can be used to help identify the molecule and the electronic transitions involved. It gives
   useful structural information - including the extent of polyunsaturated; aromatic systems, the presence of O,
   N, S, P etc, or if a transition metal ion is present, its coordination geometry, oxidation state etc
5. Can also be used for quantitative or quantitative analysis - important in forensics, biomedicine etc : is a
   given compound present ? How much of it is present ?

Question 3

• The energy jump is related to the frequency and wavelength of light absorbed by Planck’s
relation : ΔE = h ν = hc / λ%

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Question 4

• The UV-vis absorptions are due to electronic transitions between these orbital energy levels

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Question 5

• Transitions between σ , π, π, σ, n levels give rise
to characteristic UV-vis spectra in organic
molecules

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Question 6

ππ* transitions usually lie in the UV range (200-300
nm) for “simple” organic molecules
πσ* transitions usually lie at too high energy (ΔE = hν =
hc/λ) to be observable in laboratory UV-vis spectra

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Question 7

ethylene (C2H4)

Answer 7

λmax = 163 nm

π - π* transition

Question 8

• The π and π* orbitals interact along the =C-C=C- chain ….
%
• The π - π* transition shifts to longer wavelength as the chain length increases (red shift)
usually by ~30 nm per additional C=C bond unit (start from ethylene : λmax = 163 nm)

Answer 8

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Question 9

e.g., β-carotene : C40H56 : 11 conjugated -C-C=C- bonds : strong absorption in blue-green range (480 - 510 nm) :
gives rise to yellow-red colour component : present in carrots, butter, egg yolks, green leaves and vegetables :
precursor to vitamin A in liver (C20H30O : present in animal fats, egg yolk, fish oils) : necessary for growth and weight
gain ; resistance to infection; night vision

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Question 10

carotene -

Answer 10

absorb blue-green (400-500 nm)

Question 11

benzene

Answer 11

π - π* for benzene : λmax = 203 nm
Substituents : -CH3, -Cl, -NO2, -OH, -CO2H etc change this value between 203 - 295 nm
Polycyclic aromatic systems : anthracene, phenanthene etc : move λmax to longer
wavelength (π - π* at lower energy)

This peak is the π-π* transition for
C6H6 :

“spikes” on this peak are due to
vibrational transitions occurring along with
the electronic transition

Question 12

Chlorophyll -

Answer 12

is a polyunsaturated/aromatic ring system containing -C=C- and -C=N- bonds. Used to
capture sunlight and initiate photosynthesis process. The electronic absorptions at 400-450 (blueviolet)
nm and 600-700 nm (red) result in the green-yellow colour reflected from plant leaves

λmax = 428 nm
(blue-violet)

λmax = 660 nm
(red)

Question 13

Quantitative Spectroscopy - the Beer-Lambert Law

Answer 13

Absorbance A = (log10 Io/It) = ε . c. L

concentration c: moles per litre (moles dm-3)
ε is the molar absorptivity (extinction coefficient) : units L mol-1 cm-1

Question 14

• If ε is known for a compound, we can use the Beer-Lambert law relation to determine concentrations.
• For unknown compounds, we can determine ε by graphing measured absorbance vs known concentration.
• The value of ε determined for a compound can then be used to determine the type of transition involved -
e.g., σ-π, π-π* transition etc : this helps us work out the electronic structure and bonding

Answer 14

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Question 15

1. If there is no observable UV-vis spectrum, there are : (a) no multiple bonds ; aromatics ; O, N, S
   containing groups; transition metal ions. We are likely dealing with fully saturated hydrocarbons.

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Question 16

2. Small unsaturated hydrocarbons and aromatics (ethylene, benzene etc) have their electronic absorptions
   entirely in the UV (i.e., they are not coloured).

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Question 17

3. The π-π* transitions of aromatics and unsaturated hydrocarbons (-C=C- groups) usually give rise to the
   strongest UV-vis absorption bands (largest ε values)

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Question 18

4. Molecules containing O, N, S, Cl etc contain non-bonded electrons on the hetero atom : these give rise to n-
   π* transitions in the 200-400 nm range. Here the UV-vis absorption is usually ~10-100 times weaker than for
   π-π* transitions.

Answer 18

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Question 19

5. In a detailed study using UV-visible spectra to help determine the structure of an unknown compound, we
   can use several sets of “rules” that have been established for identifying or predicting the spectra of different
   chromophores : e.g., reduce λmax by ~30 nm for each -C=C- bond in a conjugated system, etc…. Other rules
   concern the effect of different substituents etc ……

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Question 20

In fluorescence, the excited molecule re-emits light directly as the electron goes back down to the lower level.

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Question 21

In phosphorescence, the emission can continue for a long time after the exciting light is switched off (this is usually
because the electron drops down via several intermediate levels …)

Answer 21

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