CHEM331L
Organic Chemistry Laboratory
Revision 1.2
Visible Spectrum of Lycopene
In this laboratory exercise we will examine the UV-VIS spectrum of the Lycopene we isolated
from Tomato Paste in a previous laboratory. We will then isomerize the Lycopene molecule and
re-examine its spectrum.
As noted previously, Lycopene is the pigment present in the Tomato that is responsible for the
Tomato’s characteristic red color.
Lycopene Crystals
http://en.wikipedia.org/wiki/File:Lycopene_powder.jpg
The color of ogranic pigments such as Lycopene (red), -Carotene (orange) and the xanthophylls
(yellow) is due to their ability to absorb electromagnetic radiation in the visible region of the
spectrum.
P age |2
Color
Red
Orange
Yellow
Green
Blue
Indigo
Violet
Approximate Wavelengths [nm]
750 – 620
620 – 590
590 – 570
570 – 495
495 – 450
450 – 420
450 – 380
In many cases, the color a pigment appears is the complement of the color its molecules absorb.
So a molecule, such as Lycopene, that absorbs light in the Blue-Green region (max = 470 nm) of
the visible spectrum will appear as Orange-Red. Complementary colors are provided by the
standard Color Wheel:
http://www.tigercolor.com/color-lab/color-theory/color-theory-intro.htm
Colors on opposite sides of the Color Wheel are complementary.
The absorbance of electromagnetic radiation by organic molecules in the Ultraviolet-Visible
region of the electromagnetic spectrum is typically due to electronic transitions within the
molecule; the absorbed photon provides sufficient energy to cause valence electrons to transition
from an energetically lower quantum state to one that is higher.
P age |3
The wavelength () of the absorbed photon is related to its speed (c) via:
c = 
(Eq. 1)
where  is its frequency. In a vacuum the speed of a photon is c = 2.99792 x 108 m/sec. Its
energy (E) is then related to its frequency via:
E = h
(Eq. 2)
where h represents Planck’s constant; h = 6.62608 x 10-34 Jsec. The energy of the photon will
then equal the energy of the molecular transition.
To understand why organic compounds absorb radiation in the UV-Visible region of the
spectrum, consider the chemical bonding in a molecule of Ethene. The two Carbon atoms are
bound together via a sigma () and a pi () bond.
In the Valence Bond Theory of chemical bonding, the loosely held  electrons are the result of
an overlap of p-orbitals on the adjacent Carbon atoms.
P age |4
However, this picture is a bit simplistic. We have to admit that orbitals are merely
representations of Wavefunctions, mathematical waves, which, like other wave pulses, can have
a positive or negative Amplitude.
If two wave pulses of positive amplitude interact, they will constructively interfere:
However, if the pulses are of opposite amplitude, the interference will be destructive:
Likewise, interaction of Atomic Orbitals, such as the 2p orbitals of our Carbon atoms, can be
between orbitals of similar amplitude, which will lead to constructive interference, or between
orbitals of opposite amplitude, giving rise to destructive interference.
P age |5
Constructive interference produces an energetically favorable  bonding molecular orbital.
Destructive interference produces an unfavorable * antibonding orbital.
We see both 2p valence electrons will wind-up in the  bonding orbital, giving rise to the  bond
between the Carbon atoms.
When we include the more tightly held  electrons in this picture, we obtain the following
energy diagram; the highest occupied molecular orbital (HOMO) and lowest unoccupied
molecular orbital (LUMO) are indicated.
P age |6
For Ethene, a photon of 163 nm will have sufficient energy to cause an electron in the HOMO to
be promoted into the LUMO. This is referred to as a  -> * transition.
In a molecule such as 1,3-Butadiene, four valence level 2p orbitals are involved in the  bonding
scheme; 2  bonding orbitals and 2 * anti-bonding orbitals will be formed. Because of the
conjugation in the bonding, the HOMO and LUMO in Butadiene will be more closely spaced
than in Ethene. This means a longer wavelength photon is required for the transition;  = 217
nm.
In general, higher levels of conjugation will cause the HOMO and LUMO to become more
closely spaced, resulting in absorbed photons of longer wavelengths. This is why highly
conjugated molecules like Lycopene have absorbances in the longer wavelength Visible region
of the spectrum.
In addition to  -> * transitions, non-bonding electrons like those found on the Oxygen in
Formaldehyde:
P age |7
can absorb radiation and be promoted into the * anti-bonding orbital. This n -> * transition is
typically much weaker than a  -> * transition.
Measuring the wavelength of absorbed photons is accomplished in a spectrometer. The
spectrometer will also measure the Intensity of the absorbance. This intensity is then related to
the number, or concentration, of the absorbing species.
A UV-Vis Spectrometer consists of a Light Source (Tungsten Filament or Deuterium Lamp),
dispersive optics (Prism or Diffraction Grating) to separate the wavelengths of the light, a sample
compartment and a detector.
Light from the source passes through an Entrance Slit and is focused on the dispersive element,
such as a diffraction grating. This separates out the various wavelengths comprising the Light
into a rainbow of colors. The desired wavelength is selected by rotating the dispersive element
such that the desired color passes through an Exit Slit, through the Sample and then onto a
detector.
The Transmittance of the light is then defined as:
T = P / Po
(Eq. 3)
This is simply a measure of how many photons pass through the sample without being absorbed.
This is then related to the Absorbance by:
A = - log T
(Eq. 4)
The magnitude of the Absorbance will depend directly on the Concentration (c) of the absorbing
species, the Path Length (b) of the light through the cell and the ability of the species to absorb
P age |8
radiation at the given wavelength, the extinction coefficient (). These quantities are related to
the absorbance via the Beer-Lambert Law:
A = bc
(Eq. 5)
Note the Absorbance is directly proportional to each factor.
A major complication occurs if an absorbing or emitting species is in a condensed phase; a liquid
or a liquid solution. If the absorbing molecule/atom is in a solution, it is surrounded by
constantly jostling solvent molecules. Thus, each molecule/atom finds itself in a slightly
different environment than its brothers. This causes the energy gap between the quantum states
responsible for the absorbance of photons to be slightly different for each molecule/atom. This
means we will have a series of very, very closely spaced absorbance lines. Practically, this
means the Absorbance Spectrum will be a broad band, rather than a sharp line. This is as
diagramed below:
In this case, we usually identify the absorbance band by the wavelength of maximal absorbance,
max.
We will start by observing the UV-VIS spectrum of the Lycopene we isolated during last week’s
lab. Once the spectrum of Lycopene has been obtained, partial isomerization can be effected by
adding a little Iodine to the Lycopene solution. This will cause a trans-to-cis conversion about
the C13-C14 bond.
P age |9
The spectrum of the resulting mixture can then be taken. Because of the bent geometry of the cis
compound, a peak at 360 nm will appear where none appears in the spectrum of the all trans
isomer. Additionally, a hypochromatic effect (decrease in intensity) is observed upon
isomerization. This is because the isomerization is reversible and an equilibrium trans-cis mix
eventually forms. Finally, a slight hypsochromatic shift (blue shift to shorter wavelengths) is
also observed. You should compare your spectra to those reported in the literature; Barrie Tan
and David N. Soderstom, Journal of Chemical Education 66 (1989) 258.
P a g e | 10
Pre-Lab Questions
1.
What concentration, in grams per milliliter, of a substance with a MW = 200 g/mol should
be prepared in order to give an Absorbance of 0.8 if the substance has  = 16,000 M-1cm-1
and the cell to be used has a path length of 1 cm?
2.
Calculate E for the HOMO-LUMO transition for both Ethene and 1,3-Butadiene.
3.
The UV spectrum of 3,6,6-Trimethylcyclohex-2-en-1-one is shown below. The
concentration is 1.486 x 10-5 g/mL. Draw the structure of this compound. Determine max
and  at max.
Introduction to Organic Chemistry, 2nd Ed.
Andrew Streitwieser Jr. and Clayton H. Heathcock
P a g e | 11
Procedure
Dissolve your sample of Lycopene obtained from the Tomato Paste in sufficient 10% AcetoneHexanes to fill a 1 cm Quartz spectrophotometer cell. (Quartz is used because, unlike Glass, it
does not absorb in the UV region of the spectrum.) Fill the cell and obtain the spectrum. Your
laboratory instructor will assist you in obtaining the spectrum. Scan the spectrum from 250 nm
to 600 nm.
We will measure Absorbance spectrun using a double beam Shimadzu UV-2550 scanning UV-VIS
spectrometer. This spectrometer uses a double beam arrangement to correct for stray radiation problems
associated with the sample cells, etc.
This instrument’s spectral range is from 190 – 900nm; operating with a resolution of 0.1nm. It
uses 50W Halogen (VIS) and Deuterium (UV) lamps as lights sources. Wavelength dispersal is
via a double-blazed holographic grating monochrometer. Radiation is detected using an R-928
Photomultiplier. The instrument itself is interfaced to a computer data acquisition system and is
under control of the acquisition software.
After completing the spectrum, transfer your Lycopene solution to a small beaker. Mix in a drop
of 0.025% Iodine to effect the isomerization. Allow the solution to stand in strong sunlight for
about 15 minutes. Record another spectrum over the same wavelength region as before.
Estimating the percentage trans-Lycopene in your sample can be accomplished via the following
procedure:
P a g e | 12



Draw a straight, horizontal line tangent to the bottom of the valley between the last
(highest wavelength) two peaks on the spectrum.
Measure the height of both peaks from that line, and call the heights a and b.
Divide the smaller height (b) by the larger (a), subtract 0.40, multiply by 100% a, and
divide the product by 0.40, as shown in the following equation.
Operational Organic Chemistry by John W. Lehman
P a g e | 13
Post Lab Questions
1.
What is the concentration of Lycopene in your spectrometer cell?  = 1.86 x 104 M-1cm-1 at
471 nm for Lycopene.
2.
Draw the structure of 13-cis-Lycopene.
3.
If Lycopene is treated with large amounts of Iodine, the color of its solution fades and may
dissappear. Explain this, and give an equation for a possible reaction.
4.
Based on the highly predictable nature of the max, Woodward and Fiesser developed
“rules” for predicting said max for dienes. According to these rules, the max can be
calculated according to:
Homoannular
(cisoid)
Heteroannular
(transoid)
Parent
253 nm
214 nm (217 if acyclic)
Increments for:
Double Bond
Extending Conjugation
30
30
Alkyl Substituent or
Ring Residue
5
5
Exocyclic Double Bond 5
5
Use these rules to predict the max for each of the following two hydrocarbonds.
Experimentally this is found to be 235 nm and 275 nm respectively.
#1
#2