Detonationology
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Free Radical Halogenation
Lately I've been reading some literature regarding the free radical chlorination of chloroform to produce carbon tetrachloride. Magpie preformed the
experiment using a 400w Hg vapor lamp. Is the bond energy of Cl2 used to calculate the energy required to make chlorine radicals, and at what
bandwidth? Is there any equations or sources that can help me better comprehend this subject? I realize that in this situation, overkill is always
better. Is a grow lamp the best source of cheap, efficient UV? Or would something on the smaller side work, such as a nail polish curing unit,
containing 4x9w UV bulbs? I understand that the amount of time and electricity necessary to carry out the chlorination to completion is dependent on
the quantity of reactants, but I would just like to have a more lucid understanding before purchasing anything. Here is a pic of one of the nail
polish curing devices on eBay for $15.99.
[Edited on 11-5-2015 by Detonationology]
“There are no differences but differences of degree between different degrees of difference and no difference.” ― William James
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AJKOER
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Per this source http://scorecard.goodguide.com/chemical-profiles/html/chloro... , to quote:
"Chloroform can be further chlorinated to carbon tetrachloride by elemental chlorine upon irradiation of the vapor"
where, I suspect, irradiation includes not just light but the use of microwave pulse radiation, although the yield may be low.
The same source also states "Chloroform can be manufactured, or can occur naturally when a chlorine molecule (Cl2) reacts with an organic substance
such as humic or fulvic acids."
A more inventive route, in my opinion, however, is by first noting that Cl2 is partially soluble in cold CHCl3 (see, for example, https://books.google.com/books?id=McQGCAAAQBAJ&pg=PA101&... ). Next, this interesting article, "Free radical formation induced by ultrasound
and its biological implications", by Peter Riesz and Takashi Kondo, link: http://www.sciencedirect.com/science/article/pii/08915849929... . To quote part of the abstract:
"The chemical effects of ultrasound in aqueous solutions are due to acoustic cavitation, which refers to the formation, growth, and collapse of small
gas bubbles in liquids. The very high temperatures (several thousand K) and pressures (several hundred atmospheres) of collapsing gas bubbles lead to
the thermal dissociation of water vapor into .OH radicals and .H atoms. Their formation has been confirmed by electron spin resonance (ESR) and spin
trapping. The sonochemistry of aqueous solutions of gases and of volatile and nonvolatile solutes is reviewed. The similarities and differences
between sonochemistry and radiation chemistry of aqueous solutions are explained. Some unusual characteristics of aqueous sonochemistry can be
understood by considering the properties of supercritical water. "
where one would replace water with CHCl3, in my proposed novel possible embodiment of CCl4 creation (a possible issue is the breakup of the CHCl3 as
well, so there could be many products formed).
Note, I have generally notice more references to ultra sound techniques in patent processes as a seemingly viable method for effecting reactions.
Also, currently relatively inexpensive ultrasound devices are available.
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[Edit] My suspicion that CHCl3 is broken down under " ultrasonic radiation ", per the source listed below, is likely given the following path on the
fate of CCl4:
CCl4 ---) .CCl3 + .Cl
.CCl3 ---) .CCl2 + .Cl
From which, for CHCl3, I would similarly expect under ultrasonic radiation:
CHCl3 ---) .CCl3 + .H
.CCl3 + Cl2 ---) CCl4 + .Cl
CHCl3 + .Cl---) .CCl3 + HCl
among other possible reactions based on the products above.
Source: "Physicochemical Treatment of Hazardous Wastes" by Walter Z. Tang, page 447, link: https://books.google.com/books?id=0pWGmybyyH4C&pg=PA447&...
For those trying to apply sonolysis to CHCl3/Cl2, please not that the chlorine must be dry (water vapor can form hydroxyl radicals that will destroy
CCl4) and pure (free of oxygen).
[Edited on 5-11-2015 by AJKOER]
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Detonationology
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Is the same amount of energy required to make chlorine free radicals with UV the same as supersonic? Assume 1 mol of Cl2 -> 2 .Cl. How do you
calculate the amount of energy required? I found this website and it has helped me better understand why energy is needed to preform the reaction, but I couldn't find much info about the
calculation.
[Edited on 11-5-2015 by Detonationology]
“There are no differences but differences of degree between different degrees of difference and no difference.” ― William James
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AJKOER
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Quote: Originally posted by Detonationology  | Is the same amount of energy required to make chlorine free radicals with UV the same as supersonic? Assume 1 mol of Cl2 -> 2 .Cl. How do you
calculate the amount of energy required? I found this website and it has helped me better understand why energy is needed to preform the reaction, but I couldn't find much info about the
calculation.
[Edited on 11-5-2015 by Detonationology] |
As I noted above for CHCl3, I would similarly expect under ultrasonic radiation (in other words, employing ultrasound devices), the following:
CHCl3 ---) .CCl3 + .H
.CCl3 + Cl2 ---) CCl4 + .Cl
CHCl3 + .Cl---) .CCl3 + HCl
This reaction chain is interesting, and one reason I posted it, as the energy required to break apart CHCl3 in the 1st initiation step (which
propagates the chain) may be less than required to break the Cl-Cl bond to form the .Cl radical directly via:
Cl2 ---) Cl. + Cl.
One can test the ability of ones ultrasonic (or photolysis) device to perform by noting its ability to degrade a like organic compound in the presence
of water (as a consequence of hydroxyl radical formation).
If it works in that scenario, I would expect some positive results also for a CHCl3/Cl2 mix (as we are just modifying the reactant mix to produce
hopefully some CCl4).
[Edited on 5-11-2015 by AJKOER]
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aga
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I really love to see a comment from AJKOER.
Always so much to see, learn and links to follow, and learn more.
Just thought a comment might be warranted, as i've read a lot of your posts and not commented on them due to lack of knowledge.
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