Sciencemadness Discussion Board

Iodine + liquid butane??

m1tanker78 - 17-3-2013 at 09:13

I performed a small experiment to see what would happen if iodine is stored under pressure from a secondary substance with a low BP and/or high vapor pressure. I placed a single iodine crystal in a vial and introduced some liquid butane then capped the vial tightly.

Immediately, the butane started to turn purple and it darkened by the minute. I let it sit overnight and it seems to have reached a maximum level of saturation(?) but the crystal is still there.

I admit that I assumed (didn't scour the literature much) that the butane should be mostly inert and that any insignificant reactions would yield colorless products. Is this simply a dispersion of iodine in the butane or what's going on?



Tank

AndersHoveland - 17-3-2013 at 09:42

The halogens (Cl2,Br2) will fairly rapidly react with alkanes (including butane) in the presence of light. The reaction proceeds through a radical cascade mechanism. First the halogen molecule is split by the light, then the halogen radical attacks the alkane.

You should have done the reaction in a darkened room. If I remember correctly, chlorine will not react in red frequency light.

Edit: I have been informed that iodine is not reactive with alkanes, even in the presence of light/UV. While surely I2 is more easily broken into radicals than the other halogens, presumably the I• radical is not a strong enough oxidizer to attak alkanes.

The halogens also react fairly rapidly with acetone, through a different mechanism, enolization. The dark color of dissolved iodine takes several minutes to dissappear in acetone at neutral pH, but at lower pH it dissappears much more rapidly. The reaction forms iodoacetone, a potent lachrymatory agent.

[Edited on 18-3-2013 by AndersHoveland]

kristofvagyok - 17-3-2013 at 09:45

Quote: Originally posted by m1tanker78  
I admit that I assumed (didn't scour the literature much) that the butane should be mostly inert and that any insignificant reactions would yield colorless products. Is this simply a dispersion of iodine in the butane or what's going on?

The butane is perfectly inert for iodine, it just dissolves it, nothing more. No reaction, no nothing, even at elevated pressure and temperatures.

What do you want to get from that? Unless you put something reactive in that flask nothing will happen, it's only an iodine solution in an alkane.

[Edited on 17-3-2013 by kristofvagyok]

AndersHoveland - 17-3-2013 at 10:03

Alkanes do tend to be inert, but not to radical attack.

m1tanker78 - 17-3-2013 at 10:03

Thanks, I did read that butane would be inert to iodine. I didn't know that alkanes would dissolve iodine, though.

My purpose was to store nice, big iodine crystals under pressure so they don't sublime. Pipe dream?

Tank

AndersHoveland - 17-3-2013 at 10:05

Bromine dissolves in carbon tetrachloride, no need to worry about photosensitive reactions in that case.

An easy way to make CCl4 is to just prepare chloroform first (from the haloform reaction), and then saturate the chloroform with plenty of chlorine and leave it in sunlight for a few days in a closed glass jar (without any water), periodically adding more chlorine in to replace the chlorine lost. (the jar will be mostly filled with Cl2 gas, with a small layer of chloroform at the bottom).

[Edited on 18-3-2013 by AndersHoveland]

simba - 17-3-2013 at 10:18

You can not judge the amount that has dissolved only by the color of the solution. Probably only a small portion has dissolved, but that was probably enough to make a huge change in color.

AndersHoveland - 17-3-2013 at 12:08

Off topic, but chlorine gas spontaneously reacts with iodine, either solid crystals or dissolved in CCl4. Iodine trichloride actually exists as a dimer, I2Cl6, which is a dull orange solid melting at 63 °C.

Another interesting fact, vicinal perchlorate esters can be formed by first by reacting iodine trichloride with acetic anhydride, with the subsequent addition of aqueous perchloric acid and an alkene. (in other words you could convert ethylene into ethylene glycol di-perchlorate, though I doubt it would be stable enough to be isolated from solution, better to use a longer chain alkene)

WARNING: do not attempt such a reaction before reading the "Ethyl Perchlorate" thread, this compound is insanely dangerous, exploding without provocation


[Edited on 17-3-2013 by AndersHoveland]

Endimion17 - 17-3-2013 at 12:53

I remember dissolving iodine in mixture of propane and butane. It did dissolve rather well as expected, however the colour was brownish, obviously because the liquefied gas contained sulphur compounds for early leak detection. It was fun, nonetheless. :)

If you use very pure liquid or liquefied alkane, no matter which one you use, you get a purple solution because iodine molecules aren't being stretched around by anything in such nonpolar solvents. Methane or hexane, it doesn't matter.

Iodine doesn't readily attack alkanes even in brightly lit conditions. It would probably take strong, direct light and some time for some reaction to occur. It's simply too sluggish, unlike chlorine.

DraconicAcid - 17-3-2013 at 12:58

Quote: Originally posted by AndersHoveland  
The halogens (Cl2,Br2,I2) will fairly rapidly react with alkanes (including butane) in the presence of light. The reaction proceeds through a radical cascade mechanism. First the halogen molecule is split by the light, then the halogen radical attacks the alkane.

You should have done the reaction in a darkened room. If I remember correctly, chlorine will not react in red frequency light, iodine is more sensitive.


Are you sure? I didn't think the free radical reaction was very fast under ordinary illumination conditions.

AndersHoveland - 17-3-2013 at 13:21

Quote: Originally posted by DraconicAcid  
Are you sure? I didn't think the free radical reaction was very fast under ordinary illumination conditions.

Try using a spiral CFL bulb. With all the UV radiation those wretched things emit, I would be surprised if you do not notice any reaction.

I think I recall reading that typical reaction times are about 2 hours.


In case you were wondering, blacklights also give off UV, but it is mostly a different type of UV, less energetic and generally harmless. Special effect blacklights use a thin coating of Wood's Glass applied to the inside of the tube which only lets lower frequencies of UV through, and also a special phosphor to convert the "hard" 184-253nm UV into safer lower frequency 350-353nm UV that can pass through the filter. In fact, spiral CFLs likely leak out more of the hard 253nm than normal blacklights because of the typical cracks in the phosphor coating formed during manufacture shaping the glass tube into a spiral. This "hard" higher frequency UV is not a good thing in terms of health, and health experts suggest minimizing your exposure to CFLs: http://www.climatechange.gov.au/what-you-need-to-know/lighti...

[Edited on 17-3-2013 by AndersHoveland]

Endimion17 - 17-3-2013 at 13:45

Quote: Originally posted by AndersHoveland  
Quote: Originally posted by DraconicAcid  
Are you sure? I didn't think the free radical reaction was very fast under ordinary illumination conditions.

Try using a spiral CFL bulb. With all the UV radiation those wretched things emit, I would be surprised if you do not notice any reaction.

I think I recall reading that typical reaction times are about 2 hours.


(Blacklights also give off UV, but it is mostly a different type of UV, less energetic and generally harmless)

[Edited on 17-3-2013 by AndersHoveland]


CFL tubes to not emit more energetic light than blacklight fluorescent bulbs. It's impossible as it would be a violation of energy conservation.
Mercury vapor inside produces mostly radiation in the UV-C spectrum, whether it's a germicidal, tanning, blacklight or regular light fixture.
Germicidal tubes are made of quartz and they are rather transparent to UV-C, so they let out pretty much everything emited by the vapor.
Tanning tubes emit mostly UV-B because UV-C hitting its special phosphors induces emission in the UV-B spectrum and some visible part, to. Their glass envelope is made transparent to UV-B, but opaque to UV-C.
Blacklights' phosphor coating absorbs UV-C and releases mostly UV-A, with some visible violet. It's opaque to UV-C, but transparent to UV-A (Wood's glass).
Light fixtures' phosphors absorb UV-C and emit mostly in the visible spectrum. Their tubes stop UV-C and strongly attenuate UV-B (possibly even stopping it totally) and UV-A.

CFL light fixtures most certainly do not emit neither higher frequencies than blacklights, and the frequencies they share are strongly attenuated by the glass tube.

They do emit more total UV than incadescent lightbulbs, but certainly less than any bulb designed to release any UV light.

Even with the cracks in the phosphor layer, the sodium glass stops or strongly attenuates what is supposed to stay inside.

[Edited on 17-3-2013 by Endimion17]

AndersHoveland - 17-3-2013 at 14:13

EDIT: sorry this post sort of turned into a long rant, I will just quickly summarize it for you so you do not have to read it all.

1. Despite what you may be inclined to think, CFLs leak out more energetic short wave UV radiation than common blacklight tubes. Blacklight tubes do emit more UV light overall, but it is the less energetic longer wave type.

2. CFL lighting emits UV that can interfere with some chemical reactions, or alternatively a CFL bulb could potentially be utilized as an improvised source of UV for other reactions.

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Quote: Originally posted by Endimion17  

Germicidal tubes are made of quartz and they are rather transparent to UV-C, so they let out pretty much everything emited by the vapor. Tanning tubes emit mostly UV-B because UV-C hitting its special phosphors induces emission in the UV-B spectrum

I was not referring to quartz tubes. Yes, obviously quartz germicidal tubes emit the most and most energetic form of UV radiation. These things can really be damaging to you eyes if you get too much exposure.
Quote: Originally posted by Endimion17  

CFL light fixtures most certainly do not emit neither higher frequencies than blacklights, and the frequencies they share are strongly attenuated by the glass tube.

They are not specifically designed to, but they do.

Quote: Originally posted by Endimion17  
Even with the cracks in the phosphor layer, the sodium glass stops or strongly attenuates what is supposed to stay inside.

The glass tube absorbs most of the UV generated by the mercury vapor, but is too thin to absorb all of it. Much of it still leaks out. The phosphor coating on the inside of the glass actually plays a bigger part in filtering out the UV.
Quote: Originally posted by Endimion17  
They do emit more total UV than incadescent lightbulbs,

Incandescent bulbs barely release any UV. What tiny ammount of UV they do release is all soft near vissible UV, mostly harmless. Halogen bulbs use a quartz capsule, which is transparent to UV, and without an outer glass cover they can indeed cause significant eye irritation. But again, the UV is still only being produced by an incandescent filament, and is not the same type of hard much more damaging UV produced by mercury discharge.
Quote: Originally posted by Endimion17  
They do emit UV, but certainly less than any bulb designed to release any UV light.

In terms of potential health hazards, how "much" UV is rather irrelative. The question is how much of the hard UV they emit, and I suggested that CFLs likely leak out more of the hard UV than special effect blacklights. Both because of the cracks in the phosphor with CFLs, and because the Wood's Glass coating inside a blacklight tube helps filter out the hard UV.

I just wanted to point out the paradox, because otherwise many people would just jump to the seemingly obvious assumption that blacklights give off "more UV". And then the next reasoning in that line of thinking would erroneously be "since black lights are mostly harmless, CFLs must be too". I wouldn't want anyone going out and wasting their money on an ordinary blacklight as a source of hard UV, when they likely already have a better more powerful source of hard UV lurking in one of their lamp sockets. Sure, quartz germocidal tubes would be the best, but in many cases one does not need that level of intensity. Amateur science is all about improvisation.

I think there should be a special thread about "CFLs and the Amateur Experimenter", because these spiral bulbs can both interfere (in several different ways) with amateur chemistry experiments, and potentially be utilized for certain reactions that require UV.

Actually, the effect of all this energetic short wave UV can be quite pronounced. It causes cross-linking in polymers. An observable manifestation of this is that over time a CFL bulb can cause the plastic lampshade to become embrittled and crack/disintegrate. It can also cause fabric colors to fade over time, and even cause weakening of the fibers.
http://www.pro-tecttint.com/preventing-furniture-fading-from...
http://www.cci-icc.gc.ca/caringfor-prendresoindes/articles/1...

Because of the incomplete spectrum they give off, CFL lighting can also make certain qualitative color change observations difficult:
Quote: Originally posted by strontiumred  
I'm with you 100% on the "Compact Fluorescent Lamp" issue Anders. It's absolutely outrageous that we are forced to use them.

You cannot do chemistry under them either. Salts all look either washed out or totally the wrong colour (Holmium Oxide looks yellow under all other lights, halogen, sunlight, incandescent Etc, but bright PINK under CFL).

They are the work of the Devil and no mistake.


[Edited on 17-3-2013 by AndersHoveland]

Heuteufel - 17-3-2013 at 18:01

kristofvagyok is completely right, when he says: "The butane is perfectly inert for iodine, it just dissolves it, nothing more. No reaction, no nothing, even at elevated pressure and temperatures. "
@AndersHoveland: You are as wrong as one can be - I remember (from my first year at university :P), that iodine does not react at all with hydrocarbons - no matter, what you do. Here, what Wiki says:
Quote:

The reactivity of the different halogens varies considerably. The relative rates are: fluorine (108) > chlorine (1) > bromine (7×10−11) > iodine (2×10−22). Hence the reaction of alkanes with fluorine is difficult to control, that with chlorine is moderate to fast, that with bromine is slow and requires high levels of UV irradiation while the reaction with iodine is practically non-existent and thermodynamically unfavorable.

And yes, if you search hard, you will find some papers, where they report a free-radical iodination of an hydrocarbon under special conditions, but generally: NO, it doesn´t work!

Here something, that might also interest you: http://forum.lambdasyn.org/index.php/topic,1451.0.html

AndersHoveland - 17-3-2013 at 19:02

Quote:

The relative rates are: fluorine (108) > chlorine (1) > bromine (7×10−11) > iodine (2×10−22).

Generally true, but it is more complex than that. Br2 is broken into radicals more easily than Cl2. I think one of the members of this forum mentioned in another thread that green frequency light was enough.

I read a patent somewhere else discussing oxidation of chlorate to FClO3 using fluorine. Apparently acetone was one of the solvents they investigated, and the reaction still worked. Yet we all know acetone will even react with iodine. Obviously the fluorine does attack the acetone (and any alkane I am sure) but the rate of reaction must be slower than the oxidation of the chlorate.
US 3,556,726 (1971)

Oxidatizing strength and "reactivity" does not alway correlate to reaction rate.
Quote:

Hence the reaction of alkanes with fluorine is difficult to control, that with chlorine is moderate to fast, that with bromine is slow and requires high levels of UV irradiation while the reaction with iodine is practically non-existent and thermodynamically unfavorable.


I question whether the radical reaction with bromine is any slower than with chlorine, and further do not think UV is really necessary; I think regular light should work just fine.

As for radical iodination, it is possible:
Quote:

A cheap and efficient iodination of hydrocarbons can be achieved by generating tert-butyl hypoiodite from iodine and sodium tert-butoxide. The alkane is reactant and solvent, and this metal-free process provides a clean solution for their direct iodination.

"Direct iodination of alkanes", Raúl Montoro, Thomas Wirth, Cardiff University, UK, Organic Letters 2003; 5(24):4729-31.


Quote: Originally posted by Heuteufel  
kristofvagyok is completely right, when he says: "The butane is perfectly inert for iodine, it just dissolves it, nothing more. No reaction, no nothing, even at elevated pressure and temperatures. "
@AndersHoveland: You are as wrong as one can be -

Perhaps I erroneously jumped to conclusions of iodine being reactive towards alkanes in light. (I suppose iodine radicals would not be a very powerful oxidizer) But I do know for certain iodine reacts with acetone, though that proceeds through a different mechanism.

[Edited on 18-3-2013 by AndersHoveland]

madscientist - 28-3-2013 at 06:11

Quote:
@AndersHoveland: You are as wrong as one can be - I remember (from my first year at university :P), that iodine does not react at all with hydrocarbons - no matter, what you do. Here, what Wiki says:


Even with alkenes the addition of I<sub>2</sub> is reversible.
Quote:
I question whether the radical reaction with bromine is any slower than with chlorine, and further do not think UV is really necessary; I think regular light should work just fine.


You're wrong, sorry.