Electra
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Photochemical weakening/breaking of Carbon-Chlorine bonds.... mechanism of action?
I haven't had any formal education in Photochemistry so the field is entirely foreign to me.
I have read much that Carbon-Chlorine bonds are broken by UV light, and that Carbon-Chlorine molecules absorb UV light..... which can lead to
breakage, but how does this play in with the wattage of the light?
If heat is typically used to accelerate/increase the disassociation of C-Cl bonds for a reaction to take place, how does a UV lamp substitute this?
With reactions, we can say, "run at 100 degrees for 3 hours to complete the reaction". How would this sort of energy input be calculated for
photochemical input of a specific wattage? The UV light catalyst has the advantage that the solution does not need to be heated, which would be ideal
in certain situations. Is there some sort of wattage-molar calculation as there is for electrochemical reactions? Or does the light just infinitely
bombard it, with the wattage determining the energy the bond is receiving?
I want to adequately determine how powerful of a light I would need for experimental photochemical reactions without going overkill on some 500w
light, and more importantly, gain an understanding of how the light power relates to the moles being displaced.
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macckone
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UV light provides the energy to break the bond. Because the energy is
better tuned to the bond it breaks it more readily (quantum mechanics).
Of course a UV light needs to be the right frequency for this to happen.
That frequency will vary for the chemical.
The unit of measure you are looking for is Lumens rather than watts.
The wattage of a UV bulb will somewhat correspond with Lumens but
not reliably enough to make an advisement on a specific wavelength,
wattage or brand.
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Electra
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I was looking into mercury vapor to be specific. I read they were suited correctly for this purpose.
When you say the bond breaks more rapidly, do you mean it is an all or nothing process? I fear the bond "breaking" when I aim for it to undergo a
simple substitution reaction could be detrimental to the reaction, though I am entirely unsure, since a substitution is more of a bond swapping.
Premature chlorine and Carbon- radicals could create unwanted reactions. Do different degrees/strengths of luminosity not weaken/activate bonds,
instead of only breaking them?
[Edited on 25-2-2014 by Electra]
[Edited on 25-2-2014 by Electra]
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DraconicAcid
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For a single bond (such as a carbon-chlorine bond), the absorption of a photon of the right energy (wavelength) will promote an electron from a
bonding orbital to an antibonding orbital, reducing the bond order to zero. This breaks the bond. The luminosity of your lamp determines how many
photons will be given off per second, and thus (if you've got the right wavelength) the number of molecules that will have their bonds broken per
second. Not every photon will be absorbed, of course, but it will be proportional. If your light is dim, you will have a slow photolysis, but it
will not weaken bonds instead of breaking them- it will just break fewer of them per second. If your wavelength is too long, it won't do anything.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Dr.Bob
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This is not trivial chemistry, these reactions tend to be substrate dependent. And UV chemistry requires quartz or other UV transparent materials
(normal glass in not OK).
If simple chemistry, better to do outside in open top dish on a sunny day than in Pyrex with a light bulb. That requires a low-volatile solvent.
UV is most often used to add chlorine to molecules in my experience, but not a trivial system, mostly used for industrial type work.
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macckone
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As a side note although glass is not totally opaque to UV but it will
severely diminish the amount reaching the reaction mixture.
In direct sunlight you may get enough UV to initiate a chlorination reaction
for example but not enough to do many other reactions.
The moral is use quartz glass or sapphire window for the UV window.
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Electra
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The reaction I was looking into doesn't requiring heating or anything fancy. I was planning to have the light/lamp shining down directly onto the
stirring mixture.
Is bond breaking what I am looking for if I wish the chlorine atoms to undergo a substitution reaction, with say, NaOH? In a typical reaction heat
would allow the chlorine to effectively leave. Would the right light have the same effect by forcing the bonds to break while the solution is
stirring, thus facilitating the reaction? If this is so, for the fastest reaction would I just aim for a very powerful light?
Or would the UV light instead of facilitating the substitution reaction, create radicals that would react to form unwanted products? If I understand
correctly, the forced breaking by UV light would not have the same effect as heating the reaction mixture up to make chlorine a better leaving group?
Or is the effect the same, simply giving more energy to chlorine so that it can leave more easily?
Quote: Originally posted by DraconicAcid  | | The luminosity of your lamp determines how many photons will be given off per second, and thus (if you've got the right wavelength) the number of
molecules that will have their bonds broken per second. |
Is there a way to calculate Luminosity -> Bonds broken per second, if I know the luminosity of the lamp?
| Quote: Originally posted by macckone | As a side note although glass is not totally opaque to UV but it will
severely diminish the amount reaching the reaction mixture.
In direct sunlight you may get enough UV to initiate a chlorination reaction
for example but not enough to do many other reactions.
The moral is use quartz glass or sapphire window for the UV window.
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I planned to shine light down clearly into the stirring mixture in a beaker. You say the UV light is used to initiate a chlorination. I am attempting
to do a Williamiamson Ether synthesis. The Chlorine technically would be chlorinating the Sodium Hydroxide, but usually this reaction would work by
abstracting the hydrogen from the hydroxyl group, accepting chlorine, so that the radical can react with the oxygen, forming the ether. Would the UV
light interfere with this process, or facilitate it?
[Edited on 25-2-2014 by Electra]
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Galinstan
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the williamson ether synthesis works using an SN2 mechanism not a radical one so attempting to make it go down a radical pathway will undoubtedly
leave you with a mess of product and almost certainly not your ether. as homolytically breaking the C-Cl bond will create a carbon and chlorine
radical which will go on to react forming any number of diffrent "products"
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Electra
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| Quote: Originally posted by Galinstan | | the williamson ether synthesis works using an SN2 mechanism not a radical one so attempting to make it go down a radical pathway will undoubtedly
leave you with a mess of product and almost certainly not your ether. as homolytically breaking the C-Cl bond will create a carbon and chlorine
radical which will go on to react forming any number of diffrent "products" |
Okay so UV light cannot be used to weaken C-Cl bonds but only to break them? So it's basically out of the question for facilitating an SN2 reaction? I
will be experimenting with Chloromethane CH3Cl, if it makes any difference.
Since a carbon-chlorine bond is electrophilic and NaOH is nucleophilic towards the chlorine, and the NaOH is nucleophilic towards a C-OH bond, then
wouldn't the chlorine instantly bond with the sodium hydroxide, allowing a hydrogen from the CH3 to take the place of the hydrogen from the OH bond,
since NaOH is mediating this process?
A phase transfer catalyst would be used to optimize yields of course, but should this work as NaOH is already trying to mediate the substitution, just
the Chlorine leaving barrier needs to be overcome? Sorry for being stubborn... I'm just trying to get a clear idea of this reaction in my head and why
it will or will not work.
[Edited on 26-2-2014 by Electra]
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AvBaeyer
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As others have already pointed out, photochemical reactions are very complex processes whose outcome depends on a variety of variables. It seems to me
that you are trying to apply "wet" chemistry ideas to reactions taking place in the excited state of a molecule. As someone who spent several years
doing organic photochemistry, I suggest that you get a better understanding of the basics, particularly with regard to types of excited states and how
chemical properties of molecules in excited states differ from those in the ground state. For example, the book "Molecular Reactions and
Photochemistry" by DePuy and Chapman (available on ABEbooks) is an excellent entry level starting point. I have also attached a paper which you might
find useful from a descriptive standpoint with respect to some kinds of chemical transformations of excited states.
Attachment: AROMATIC PHOTOSUBSTITUTION REACTIONS Pure Appl Chem 47_1_1976.pdf (290kB)
This file has been downloaded 420 times
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Electra
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Thanks AvBaeyer,
I'll check it out.
With your experience in the field, would you say the situations I have described above with UV light irradiation are more likely to generate waste
products than to enhance my reaction?
As far as I know this typical reaction can be run at room temperature if a hydrocarbon with a either bromine or ideally iodine were used. The reaction
speed is directly tied to how easily the halogen can leave the molecule so that it can subject to the Sn2 reaction
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AvBaeyer
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Electra:
I just noticed (sorry) that you propose doing photochemistry on chloromethane. Frankly, there is no way you will be able to use this compound as a
photochemical reactant without the use of very special conditions and equipment. If you wanted to run a photochemical reaction on chloromethane you
would need to excite it at or near its uv absorption maximum. For this compound, the uv absorption max is near 200 nM. You would need special
photochemical equipment and some very expensive uv source to approach this wavelength. Thus, photolysis of chloromethane and sodium hydroxide (in what
proposed solvent?) with any easily available light source you could lay your hands will lead to no reaction at best. Keep in mind that to get a
compound to react photochemically you need to excite it as close as possible to its aborption maximum. Many organic compounds (particulalry aromatic
compounds) have more than one absorption maximum and each has the possibility of leading to different photochemical reactivity and reaction products.
Photochemistry is an extremely interesting and complex field and with the understanding of some fundamental principles you could have "hours of fun."
As Dr. Bob mentioned above some very interesting photochemistry can be done in Pyrex glassware or just straight sun exposure using good old sunlight.
Another, more advanced text you might find useful is "Molecular Photochemistry" by N. Turro. It is considered a classic in the field and readily
available on ABEbooks.
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Electra
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I was never intending an easily available light source. I can afford to buy any needed light source for experimental purposes. Not exactly on a
budget... I never intended to try to pull of such a reaction with a regular light sources :p.
Anyways, after reassessing not only the above reaction and a couple of others I wanted to try, I decided the photochemical approach probably is not
ideal for a number of reasons. C-Cl bond is not the only one I would need to weaken. I will still probably look into this field for knowledge sake, as
I am always interested in alternative ways of catalyzing reactions.
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