Synthesis of Cyalume and Chemiluminescence Pre Lab Questions

Do Pre lab Synthesis of Cyalume and Chemiluminescence.

Experiment 10
Spring 2020
Experiment 10 —
Synthesis of Cyalume and Chemiluminescence
_____________________________________________________________________________
Pre-lab preparation: Read the Wikipedia article on “glow sticks”. If necessary, review
recrystallization, i.e., (a) if you do remember, fine; (b) if you don’t remember, then review it!
(1) Briefly state the purpose of this experiment. (2) You may remember light sticks as
one of those magical wonders of childhood. Ahhh… It’s hard to imagine a better toy for a small
child than a pressurized plastic tube filled with chemicals and broken glass. (a) Commercial
light sticks contain a solution that surrounds a fragile glass tube containing a different solution.
What are the key components of each solution? (b) What happens chemically when the glass is
broken and the two solutions mix? (c) Why does pressure increase (just a bit) inside the tube?
(3) (a) Put the following colors of light in order from lowest energy to highest energy: yellow,
UV, orange, red, blue, IR. (b) Label one end of your series “lowest E” and the other end “highest
E”. (c) Label one end of your series “longest l” and the other end “shortest l”. (4) Look at your
TLC analysis of the Wittig product (Exp 9) — what color did you record for its fluorescence?
(5) Write the mechanism of the reaction of the first equivalent of trichlorophenol with
Et3N and oxalyl chloride. (6) Draw the structures of toluene and diethyl phthalate. Write down
any physical data for these two solvents that are relevant to this experiment. (7) Why is it
important to handle the reagents with gloves and keep them in your fume hood? (8) (a)
Calculate the quantities of trichlorophenol, oxalyl chloride solution, and Et3N that you plan to
use to make the cyalume. Keep in mind that we normally measure solids by mass and liquids by
volume. (b) Calculate the theoretical yield of the cyalume product.
As always, write concise, easy-to-read procedural notes for both parts, A and B, that you
will use in lab. Reminder: These are notes on what you plan to do — These notes are not to be
copied into your notebook! In your notebook, as you work, you’ll of course write down what you
actually did, along with relevant data and observations.
There’s a lot in this write-up — it provides an overview of excited states, absorption and
emission of light, then explains how fireflies glow, in terms of reaction energetics, before
moving on to the specifics of your reaction. Allow enough time to read this carefully before lab!
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Light and color. Many molecules absorb light in the ultraviolet (UV) and visible regions
of the electromagnetic spectrum. Molecules that absorb in the visible region appear colored
under white light. When we look at a material under white light, the absorbing substance
“removes” certain wavelengths. The color we see is determined by the wavelengths that are not
absorbed. The unabsorbed light is reflected by opaque materials or passes through transparent
things like solutions or colored glass. It’s important to note that light emission is not involved
here — in fact, most molecules do not emit light. The colors we see come from the wavelengths
of white light shining on the substance that are not absorbed.
A convenient approximation assigns each of the six basic colors to a 50-nm range of
wavelengths (above). For example, purple light extends from about 400 to 450 nm, blue from
about 450 to 500 nm, etc. That’s easy to remember. (What most people perceive as yellow
actually extends over a band of about 30 nm, and the transition from orange to red is around 620
or 630 nm, not 650 — that’s fine-tuning.) What we see is the complement of the color absorbed
(or the color in the center of a broad absorption band). Complementary pairs are purple–yellow,
blue-orange, and green–red. So if a compound absorbs orange light, it will appear blue, etc.
Light absorption, MOs, and electronic states. Absorption of light by a molecule causes
an electron to “jump” from a lower energy molecular orbital (MO) to a higher energy empty MO.
For organic molecules that means the electron must go from a bonding MO or from a lone pair
orbital into an antibonding MO.
There are lots of filled MOs and lots of empty MOs, so several electronic transitions can
result from absorption of UV and visible light. The lowest energy transition — the one that
requires the longest-l light — is the one from the highest occupied to the lowest unoccupied
MO, or HOMO to LUMO (red in the diagram). This creates an excited state of the molecule that
has 1 e– in each of these MOs, and that we refer to as S1. The other electronic transitions shown
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Spring 2020
require higher-E, shorter-l light, and these create higher-energy excited
etc
states, S2, S3, S4, etc. (The lowest-E electronic state — the ground state
— is called S0.)
LUMO
In molecules, those higher-energy excited states quickly drop
down to S1 without emitting light; the “excess energy” is lost as heat.
E
In molecules that emit light, the emission almost always comes from
HOMO
that lowest excited state, S1, falling to S0. So, while a molecule might
absorb light at several different wavelengths (S0 —> S1 or —> S2 or —
> S3, etc), it will only have one emission (S1 —> S0).
etc
What can excited states do? The S1 excited state of an organic
molecule usually lives for only about 1 ns (10–9 sec). In some
molecular
orbitals (MOs)
molecules this state can fluoresce — drop down to S0 by emitting a photon of light (remember,
S1 —> S0 happens when the e– in the LUMO falls back into the HOMO). But in the grand
scheme of things, fluorescence is not very common. Most molecules find other ways to lose that
“excess” energy. In most cases when the excited state (S1) returns to the ground state (S0) it just
turns all that energy into heat. What a waste.
Some molecules can instead use the high energy of the S1 state to do chemistry that
would have been impossible in the ground state. That’s called photochemistry. Photochemical
reactions often involve homolytic bond breaking to create free radicals, as well as a huge variety
of isomerizations and rearrangements. Promotion of an e– by light can also lead to electron
transfers. After light absorption, the high-energy e– can “hop” over a short distance to another
molecule, and/or an e– from a nearby molecule can “hop” into the empty space (the “hole”) that
the excited e– left behind. Light-initiated e–-transfers are the basis for photosynthesis.
You’ve probably also heard of phosphorescence, in which light is emitted over a much
longer timescale — ms to sec or even longer (light emission in “glow-in-the-dark” stuff). Very
few organic molecules phosphoresce. In molecules that do, one e– in the S1 state flips its spin to
create a new excited state called T1. The T1 state eventually drops to S0 with emission of light.
But that’s slow because all the molecule’s random “jiggling” and “twisting” has to give the highE e– just the right extra torque to flip its spin as it falls into the low-E orbital.
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Experiment 10
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Vibrational states. In the ground electronic
state, S0, a molecule can vibrate in different ways.
S1
Vibrations are quantized — they can have only
of vibrational energy levels of the S0 electronic state,
and an analogous series of vibrational levels for S1.
When a molecule absorbs light, it can be
excited into the lowest vibrational level of S1 or any
of the higher vibrational levels. For this reason, we
sometimes see a series of regularly spaced peaks in a
E
abs
vibrational
levels
certain energies. The diagram at right shows a series
1st exc.
state
fluorescence
S0
ground
state
MOs
UV-vis absorption spectrum. But more often, other molecular motions cause those discrete
transitions to get smushed together, so we instead see an absorption “lump”.
When the absorption creates a vibrationally excited S1 state, the molecule quickly relaxes
to the lowest vibrational level (squiggly lines in the diagram). If the molecule fluoresces, it can
fall from there to any of the vibrational levels of S0, so a series of discrete peaks is sometimes
observed in the fluorescence spectrum. Or those might get mashed together into a lump.
But whether the individual vibrational transitions show up as separate peaks or contribute
to a “lump”, we can make an important generalization about absorption and emission
wavelengths. Look at the energy gap between the heavy green lines in the diagram. This gap is
the same in both directions (S0 —> S1 or S1 —> S0). Let’s call the corresponding wavelength l r
Now let’s take the vibrational levels into account. Absorption can occur at wavelength lr and at
shorter ls (higher E); emission can occur at this wavelength (l r) and at longer ls (lower E). So
fluorescence can never occur at a shorter wavelength than the corresponding absorption.
Ground state vs excited state processes. Light emission by molecules is almost always
the result of light absorption. The vast majority of chemical reactions, like the ones that you’ve
been studying in lecture, never ever produce light. This is because normal chemistry is
exclusively ground state chemistry. Reactants, products, intermediates, and even transition states
are in their lowest energy electronic states at all times. When we draw a rxn coordinate diagram,
showing energy changes during a reaction, we are drawing a ground state energy surface. Of
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Experiment 10
Spring 2020
course, all the molecules involved have electronically excited states, but the excited states are all
much higher in energy (ca. 100 kcal/mol (400 kJ/mol) higher). So in normal reactions, there’s no
way for anything to reach an excited state, and no way for light to be produced. If you were
running around in a park, perhaps up and down some small hills, you wouldn’t suddenly take a
bad step and find yourself in the top of a tree (sort of a potential energy excited state).
Chemiluminescence. But suppose the park is wooded and has some 50-foot cliffs. If
you’re frolicking merrily about (maybe pretending to be a molecule?) and not paying attention,
you might, if you were careless enough, run off a cliff and land in a tree. If you were an actual
molecule, you could then fall to the ground and emit a photon of light.
This happens on a molecular level only very rarely because the landscape of chemical
reactions has mostly rolling hills with very few energy cliffs. In these rare chemical reactions, a
single mechanistic step is downhill in energy so much that the product of that step can be formed
in an excited state. That state can then emit light. The reaction goes up in energy just a bit (over
the transition state) and then off the energy cliff — some molecules just hit the ground, and all
the energy is released as heat; other molecules land in the tree — the excited state of the product
— from which they can fall down and emit light. Nature has figured out how to use this
chemical principle to make critters like fireflies, some jellyfish, and a few other organisms
produce light. This is called “bioluminescence” — chemiluminescence in a biological system.
Firefly chemiluminescence. Fireflies use enzymes to create a compound that decomposes
to produce a fluorescent
excited state. Firefly
N
luciferin (at right) and
S
oxygen are combined with
O
O
N
O2
OH
N
O
luciferase
S
O
N
S
Firefly luciferin
S
R
the help of an enzyme called
hν
“luciferase”, to form a
S
S
excited state
compound (R), called a
This dioxetane expels CO2
N
O
*
+
strained 4-membered ring
dioxetane (di-OX-e-tayn).
N
N
N
S
S
ground state
O
P
to form an excited carbonyl compound (P*) that then emits light.
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P*
CO2
Experiment 10
Spring 2020
At first glance, that might not look like a very unusual reaction. So how is the conversion
of the dioxetane reactant (R) to the ketone product (P) and CO2 analogous to running off a cliff
and landing in a tree (P*)? The “cliff” is the huge drop-off in energy on going from R to P and
CO2. We can estimate this energy change by considering changes in bond energies and the strain
of that 4-membered dioxetane ring.
The R —> P reaction breaks a C–C bond of the dioxetane (R) (BDE about 80 kcal/mol;
335 kJ/mol) as well as a weak O–O bond (35 kcal/mol; 145 kJ/mol) and forms two strong CO pbonds (about 90 kcal/mol each; 375 kJ/mol), one in the carbonyl group of P, and the other in
CO2. If we imagine this happening in two steps, we can say that bond breaking raises the energy
by (80 + 35 =) 115 kcal/mol (480 kJ/mol) in all, and bond making then lowers it by (2 x 90 = )
180 kcal/mol (750 kJ/mol). So these bonding changes make the overall reaction downhill, i.e.
exothermic, by a whopping 65 kcal/mol, i.e. DH° ≈ –65 kcal/mol (–270 kJ/mol).
But that’s not the whole story because the reaction also breaks a 4-membered ring. This
releases about 30 kcal/mol (125 kJ/mol) of strain energy, making the reaction that much more
downhill. So in all, we can estimate that the rxn R —> P (+ CO2) should have DH° ≈ –95
kcal/mol! (–65 + –30; or –395 kJ/mol = –270 + –125) That’s hyuuuuge. Can you think of any
other reaction you’ve learned that releases anywhere near 95 kcal/mol (400 kJ/mol) of energy in
a single mechanistic step? No, you can’t.
O
O
O
R
P*
E
ΔH° ≈ –95
kcal/mol
hν
fluorescence
O
P
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O
C
O
Experiment 10
Spring 2020
Ok, so what, you ask. Why does this matter? It matters because it’s soooooo far
downhill in energy that the excited state of the ketone product can be formed. A simple ketone
has an excited state that lies about 85 kcal/mol (355 kJ/mol) above its ground state. But the
C=O of P is part of a conjugated p-system, so its excited state energy will be lower than that.
This is why the firefly emission peaks at about 560 nm (yellow-green), which corresponds to an
energy of about 50 kcal/mol (210 kJ/mol). The energy diagram above shows the wildly
exothermic R —> P + CO2 ground state rxn coordinate in green, and the pathway leading to the
excited state of the product, P* in red.
Because the ground state reaction is so exothermic and P* lies relatively low in energy,
some of the molecules can follow a pathway (red in the diagram) that leads to P* instead of
going directly to the ground state, P. The firefly system is extremely efficient, so nearly every
molecule R that dissociates ends up making P*, and nearly every P* then emits a photon of light!
Cyalume chemiluminescence. Our chemiluminescent reaction — the same one that is
used in light sticks — operates by the same basic principle as the firefly system. A strained
dioxetane ring dissociates in a step that is tremendously downhill (on the ground state surface)
— by about 95 kcal/mol (400 kJ/mol). The main difference is that in our system the excess
energy ends up making the excited state of an added fluorescent energy acceptor, called a
fluorophore. We’ll use several different fluorophores — the fluorescent compound we made via
the Wittig reaction, 9-Anth–CH=CH–Ph, some analogs with different combinations of aromatic
bits, and a few others. Although our light stick reaction is not as efficient as the firefly system, it
has been estimated that, depending on concentration and other conditions, anywhere from 0.1%
to 35% of the decomposition events can produce a photon of light.
In our reaction an
oxalyl ester will react with
Cl
Cl
Cl
Cl
Cl
–2
Cl
Cyalume-Cl6
hν
O
O
* + C + C
Fluor
molecules of CO2 (below).
O
Fluor
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O
O
H–O–O–H
O
Cl
strained dioxetanedione (dibreaks down to form two
O Cl
O
hydrogen peroxide to form a
OX-e-tayn-DI-own), which
O
O
O
O
OH
Cl
Fluor
O
O
Fluor
O
O
Experiment 10
Spring 2020
Just like in the firefly system, this decomposition is so exothermic on the ground state surface
that enough energy is available to produce excited states. But the excited state formed in this
case is not CO2*, but rather “fluorophore*”, which then emits visible light. This energy transfer
is thought to happen via a non-covalent complex between the dioxetanedione and the
fluorophore. When the strained ring comes apart, the energy goes directly into the fluorophore,
making its excited state, “fluorophore*”, rather than “CO2*”.
You should already have
plenty of trans-9-Anth–CH=CH–Ph
from the Wittig reaction. (If you didn’t
get much, don’t worry — we have
some in reserve, thanks to former
students.) We’ll also supply some
other fluorophores, shown in the box.
Cyalume preparation. You’ll
first need to synthesize the
bis(trichlorophenyl)oxalate reactant
(aka “cyalume-Cl6”, aka “The Cyalume”). This is done by a carbonyl substitution that you’ve
recently studied in lecture — the reaction of an acyl chloride with an alcohol. Triethylamine,
Et3N, is often added to these reactions to “sop up” the HCl produced. That is, it does an acidbase rxn that trades the strong and nasty HCl for a much weaker, friendlier, water-soluble
ammonium salt, Et3NH+ Cl–.
O
R
Cl
+
R’ OH
O
Et3N
R
O R’
( + Et3NH+ Cl–)
In our case, we’ll use a “double” acyl chloride, oxalyl chloride, and we’ll allow that to
react with two equivalents of trichlorophenol, to produce our cyalume. And in our reaction, the
Et3N will get things going by deprotonating the trichlorophenol (Ar–OH) to form the Ar–O–.
The negative charge makes this an even better nucleophile.
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Experiment 10
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Cl
O
Cl
O
Cl
OH
+
2
Cl
+ 2 Et3N
Cl
PhCH3
CH2Cl2
Cyalume-Cl6
(mp 190 – 192°C)
Procedure. Overview — We’re first going to make and isolate the Cyalume-Cl6, clean it
up a bit, then purify it by recrystallization. Treating a solution of this compound with H2O2 in
the presence of a fluorophore should produce light.
Caution: Oxalyl chloride is a corrosive lachrymator and must be handled in the fume
hood. It reacts vigorously with water to produce HCl and CO gases (and CO2). The solvent,
CH2Cl2, and 2,4,6-trichlorophenol are suspected carcinogens and must be handled with care.
Triethylamine smells like rotting fish and should be handled in the fume hood.
Part A. Preparation of bis(2,4,6-trichlorophenyl)oxalate, aka Cyalume-Cl6. Calculate
quantities corresponding to 2.0 mmol of oxalyl chloride, 4.0 mmol of 2,4,6-trichlorophenol, and
5 mmol of triethylamine. The Et3N is a liquid, so be sure to calculate the volume needed. The
oxalyl chloride was purchased as a 2.0-M solution in CH2Cl2. Don’t overthink this — you need
2.0 mmol of the compound, and the solution has a concentration of 2.0 M (What are the units of
molarity? Do you need the density or MW?).
Combine the trichlorophenol and 5 mL of toluene (Ph–CH3, TOL-you-een) in a clean,
dry flask, and stir the mixture until all the solid has dissolved. Then add the Et3N, and cool the
solution in an ice bath for a few minutes. Keep stirring — this will speed up the heat flow.
Next, summon your TA or instructor, who will bring a dry pipet and the oxalyl chloride
solution and will help you dispense it. Obtain the required amount and add this dropwise with
stirring, with stirring, to your ice-cold solution. A mass of off-white or tan precipitate should
form quickly, and the mixture may become rather thick. After the addition is complete, you can
remove the ice bath and let the mixture warm to room temperature. Keep stirring — always —
you’re doing a reaction — everything needs to mixed up uniformly.
Isolate the solid by suction filtration and rinse it with a few mL of cold hexane. (The
filtrate goes in the halogenated waste bottle.) Next, combine the solid with about 10 mL of
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Experiment 10
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water and stir the mixture vigorously for a few minutes. This step should dissolve most of the
water-soluble byproducts, side-products, and other “undesirables”. Filter the suspension by
suction and rinse the solid with a little water (filtrate goes into aqueous waste).
Recrystallize the crude cyalume from toluene. The Management did this with about 3
mL per gram of crude material. (What’s the bp of toluene?) If any dark, oily material remains
undissolved, decant the hot solution into a clean flask. The oily gunk should remain stuck to the
original flask. Once you have crystals, isolate them as usual, rinse with a little cold hexane
(which will evaporate much more easily than toluene). Determine the yield and mp. (The
filtrate probably contains a lot of chlorinated cyalume, so halogenated waste.)
Part B. Let there be chemiluminescence. (1) In a little screw-cap vial, combine 30 mg
of your freshly purified cyalume with about 2 or 3 mg of your fluorescent Wittig product, trans9-Anth–CH=CH–Ph, and add 5 mL of diethylphthalate (di-ETH-il-ffthfthfthffff… no… the “ph”
is silent… di-ETH-il-THAL-ate). In your fume hood, warm the mixture for a few minutes (cap
off while heating!) until all the solids have dissolved. Put a small amount of this solution in a test
tube. Cap the vial, which should still be quite warm, and take it and the test tube sample to the
dark instrument room. (The warmer the better, as long as it’s not too hot to touch.)
Add about 0.1 mL of 30% aq. H2O2, and cap and shake the vial. You can take a quick
selfie with it, but don’t get too excited; you still need to observe carefully and record your
observations in your notebook.
Shine 365-nm light on the test tube sample. Does it emit light? If so, compare the colors
of the fluorescence from this sample and the one in the vial, which should still be glowing. Are
they the same or different?
(2) Effect of temperature. Repeat the experiment above, but this time, after you dissolve
the solids, divide the warm solution in half. Keep one warm (the warmer the better) and cool the
other to room temp. Add the hydrogen peroxide to each vial, side-by-side. Is there any
difference in the light intensity?
(3) Different fluorophores. In place of the Anth–CH=CH–Ph, test two or three other
fluorophores that emit different colors of light. Be sure to keep track of which compound you
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Experiment 10
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used in each sample! For each different sample, repeat the procedure in part 1, remembering to
set aside a bit of each solution to look at under 365-nm illumination — is the color of the emitted
light the same in each case?
(4) Try to measure a light emission spectrum with the Pasco spectrometer for at least one
of your glowing samples. Print a copy for each member of your group. Your instructor or TA
will cheerfully assist with this.
Waste disposal. All the solutions should go into the halogenated organic waste. Rinse
the vials with acetone and toss them into the glass waste box. Do not throw the solutions into
the glass waste! There will be collection bottles for your unused cyalume and trans-9-Anth–
CH=CH–Ph. Please contribute. Future students will be happy to have your left-overs.
Discussion. (1) Find three of the fluorophores that you used in the set of spectra that
were handed out in lab. For each one, list the following: (a) Its color under white light that you
recorded in your notebook (also written next to the structures in the handout), (b) based on your
answer to a, what color it absorbs most strongly. (c) Locate the S0 to S1 transition in the UV-vis
absorption spectrum and report the approximate wavelength range of the light that’s absorbed
most strongly (maybe a 50- to 80-nm range), and (d) the color(s) this l range corresponds to. (e)
Are the colors you listed in b and d close to the same? If not, explain.
(2) With apologies for anticipating your experimental result… Explain why each of the
fluorophores emitted the same color light upon 365-nm excitation and via the chemical reaction.
(3) (a) For each of the three fluorophores that you picked in question 1, what color light
was emitted? (b) Look at the printed spectra and report the approximate wavelength range of
most of the emitted light (probably a 50- to 80-nm range), and (c) the color(s) that this l range
corresponds to. (d) Are the colors you listed in a and c (close to) the same? If not, explain. (e)
Is the energy of the emitted light higher of lower than the energy of the absorbed light?
(4) Fluorescence occurs on a timescale of about 1 ns. This is why any fluorescence you
saw under the 365-nm lamp disappeared as soon as you turned the lamp off (well, about 1 ns
after you turned the lamp off). Why did the chemiluminescence continue for such a long time?
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(Think about the sequence of events that precede the light emission. The energy diagram above
might provide a hint.)
(5) You saw a big difference in light intensities in the reactions initiated at different
temperatures. But the rate of the S1 to S0 transition that causes the emission is known to be
independent of temperature — it happens in about 1 ns regardless of the temperature. So why
did you see such a difference in light intensity?
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