I`v had to buy several gallons of Paraffin for the greenhouse heater and thought I`d bring a little into the Lab to experiment with as well.
this morning I`v done a simple substitution reaction by putting a small amount of TCCA in the bottom of a test tube, putting 10ml of paraffin onto
that and then adding a few drops of HCl every time the bubbles stop.
I`ll not go into the color changes that took place (probably a Dye)
but what I would like to know is where does the Hydrogen go?
if I`m ripping a H off a C atom and replacing it with a Cl atom, that H has to go somewhere surely?
there are Very few bubbles that make it to the surface, most seem to "Vanish" a little way up in the paraffin after the reaction interface.
[Edited on 24-11-2007 by YT2095]
substitution reaction, H replaced by Cl
woelen - 22-11-2007 at 02:05
The other H is connected to Cl.
This is the reaction which occurs: R-H + Cl2 --> R-Cl + HCl
The HCl most likely dissolves in the water of the droplets, and may also affect the dye. That is why you see so few bubbles. Here, R stands for some
saturated hydrocarbon group (e.g. for hexane, R- will be C6H13-).
This reaction can proceed further, more than 1 H-atom can be replaced.
If your petroleum also has some unsaturated hydrocarbons, then there also will be an addition reaction. Across the double C bonds, Cl atoms are added
on each side of the double bond, and the double bond becomes a single bond. Such reactions usually are very fast and energetic and certainly do not
need bright light to proceed quickly.
A funny and more instructive experiment is to do the following:
- Make 100 ml or so of chlorine gas, and assure it is slightly humid. Usually, just making it from TCCA and HCl, and leading the gas without drying
into a dry erlenmeyer works well. The erlenmeyer should be dry, you don't want droplets of water on its inside.
- Add 0.5 ml or so of petroleum ether or some other low-boiling saturated hydrocarbon to the erlenmeyer with the humid chlorine gas. Then quickly cap
the erlenmeyer with a glass plate or something like that and swirl a little and put it in the daylight (sunlight if possible). If the chlorine was a
little humid, you'll see a very nice effect. The clear glas mix in the erlenmeyer slowly becomes foggy, due to formation of HCl-fumes with the water
vapor.
- After one hour, open the erlenmeyer. You'll see white fumes of HCl coming out of the erlenmeyer, and there also will be some sweetish particular
smell of chlorinated hydrocarbons.
[Edited on 22-11-07 by woelen]YT2095 - 22-11-2007 at 02:26
that`s working in my Favor then, as the HCl given off will further react again with the TCCA making more Cl
one of the reasons I wanted to try with Paraffin rather than something like Ligroin was the risk of Fire, Paraffin is pretty stable even when warm and
doesn`t like to burn that easily.
it`s interesting that you mention other Volatile chlorides, there is a Layer creeping up the test tube as is something is Distilling, a bit like the
Alcohol line you get when you pour a brandy in a glass at room temp.
and the orange rubber stopper that`s loosely in the test tube has gone quite Black now
I`ll stick with Paraffin for now, I feel safer, and it gives me another great reason to hit the books and see if I can make an OL and then a COOH with
it YT2095 - 24-11-2007 at 03:22
I`ve gone a little further in this experimenting and have tried Bromine as well as Iodine to halogenate this Hydrocarbon (mostly Undecane).
it would seem I was right about the Dye in this yellow paraffin, any halogen will make it turn Pink/Purple to start with and then even that fades to
near clear again.
with Chlorine gas you get a Yellow liquid eventualy
with Bromine you get a Red/Orange liquid
and with I2 you get the most surprising result of all, with a KMnO4 type Purple color
mostly all the forms of Iodine I`v seen you only get the purple as the Gas form, but in Paraffin you get to keep that Purple!
I expect with Hydrolysis of these I should get something like Undecanol, although I`m uncertain if these halogens have actually Reacted with the
alkane at all or just mixed with in it?
that would go some distance to explaining the Colors.woelen - 24-11-2007 at 08:25
The chlorine probably will have reacted with it, but if it is still green/yellow, then there also still is free chlorine. Bromine does not react in
the dark, only bright daylight (preferrably sunlight) does the job. Iodine does not react at all. Iodine-solutions are purple in purely apolar
solvents (like paraffins, CS2). They are yellow (brown in higher concentration) in polar solvents, like ethanol, methanol. A bright red solution is
obtained in solvents, which are in-between (SOCl2, CH3C(O)Cl).
You can tell whether the paraffin is chlorinated to a reasonable extent, by adding a few drops to water. Chlorinated hydrocarbons have a larger
density than water, if they are not chlorinated, then they float on the water. A peculiar sweet smell also is an indication of chlorination.YT2095 - 24-11-2007 at 08:53
I`m glad you said that actually, you have confirmed the results I`ve had today
the Bromine paraffin mix was VERY red/orange in color, I`ve had this under a Halogen lamp all day and the color has faded now to just a strong orange
and from the Top downwards, there`s also a strange "Smoke" coming off it too, esp when I swirl the test tube a little?
Bromic acid maybe????
the Is in the Paraffin has done Little under the light although up the sides of the tube there is now a Brownish color (I assumed air oxidation of
some sort), but the bulk of the liquid is still deepest purple, quite a nice color actually.
I have to ask though, at What percentage of chlorination will the alkyl halide sink in water?
I can`t see Mono chlorination doing that, it would have to be quite saturated I think, Undecane would still have a good 23 Hydrogens left and all it`s
carbons.
I`de need to do the math obviously, but I think for such a long chain it would need good chlorination to sink.
would the addition of RP help the I2 reaction along any, or will the Lamp be enough?
[Edited on 24-11-2007 by YT2095]woelen - 24-11-2007 at 09:58
Addition of RP does nothing. In the presence of water, it causes formation of HI, but this does not react with the paraffin.
In order to get the material sink in water, replacement of one out of 4 hydrogen atoms should be sufficient, probably even lower chlorination will
give material sinking in water. Keep in mind, chlorine is quite heavy, compared to C and H. Suppose you have for each group -C2H4-, one hydrogen
replaced by chlorine, to get -C2H3Cl-. The unit mass then goes from 28 to 62.5. This is an enormous increase. The size of the molecules of course also
changes somewhat, but much less than the change of molecular mass. You will have much more matter in the same volume.YT2095 - 24-11-2007 at 10:32
I don`t think I Chlorinated enough to be honest, because it Did burn although quite sooty, and a little harder to light, certainly Not impossible (as
a 30 to 70% chlorination would be).
I was a little cautious as I didn`t want excess Cl2 in the Lab, I feel that it would have taken quite a bit more had I have had the balls to try it
I will re-do this again when I get a little more free time in one go.guy - 24-11-2007 at 11:38
You better shine some UV light on the chlorination reaction. That is the only way Cl2 can react with alk<b>A</b>nes.
Iodine radical is not strong enough to abstract hydrogens of alkanes so you won't have luck making iodides.
[Edited on 11/24/2007 by guy]
[Edited on 11/24/2007 by guy]Nicodem - 24-11-2007 at 12:22
I think guy meant that Cl2 is inert to alkanes which are the major if not the only constituent of paraffin oils. There is no way for
Cl2 to react with alkanes except in conditions allowing the formation of radicals (as is written in every basic org. chemistry schoolbook). YT2095,
the only reaction you have been performing was the reduction of TCCA with HCl forming cyanuric acid and Cl2. Paraffin only served to absorb the Cl2,
hence the yellow color. The same with other halogens, red/orange for solutions of Br2 in paraffin and dark purple for I2.MagicJigPipe - 24-11-2007 at 22:53
Accident
[Edited on 25-11-2007 by MagicJigPipe]YT2095 - 25-11-2007 at 01:08
that`s really rather disappointing in that case, although I don`t understand why I don`t see any brown Br2 gas in the test tube (as I normally do when
Br2 is present) and yet the liquid under the Halogen lamp is clearing quite nicely, there is also a "Smoke" like effect too, a little like when you
open a fresh bottle of conc Nitric.
I think it`s time to build a UV lightbox, I have 9x 5mm UV LEDs with 4000mcd each, if I place those Around the test tube, is that likely to be enough
photon energy there?12AX7 - 25-11-2007 at 04:57
Given substantial time, some small amount will occur naturally.
I store the hypochlorite- and chlorine-rich solution from my chlorate cell in high density polyethylene buckets, an excellent alkane if ever there was
one. After months of contact, the solution has taken on the sweet, chlorine and solvent odor corresponding to chlorinated organics.
To my knowledge, no chlorinated phase has seperated. I wouldn't expect any to, as the amount produced is very small and will dissolve in the water or
plastic, or evaporate before much is able to collect.
TimYT2095 - 25-11-2007 at 09:36
well the Bromine test tube has clear droplets all up the inside of the tube above the paraffin layer, it`s also VERY acidic.
the color in the paraffin is also fading too.
it Has had well over 24 hours of Halogen light so far also, I think perhaps I will let this run for now, take off the top layer and heat a small
amount of it, apparently Alkyl halides go dark when heated?
also a silver nitrate test may be handy also I think.woelen - 25-11-2007 at 12:42
I do not fully agree with Nicodem. You certainly will get halogenated hydrocarbons from chlorine and paraffins, even without UV-light, but the
reaction will be very slow. But if you mix them, and allow them to stand for days, then sure, some HCl is formed and some chlorination takes place.
Most likely, the mist, observed by YT2095, is due to HCl, in combination with humidity.
The high acidity in the bromine/paraffin mix also is an indication of bromination taking place.
These reactions, however, are not the best way to make halogenated alkanes. The reaction is not specific, you'll get a mix of all kinds of substituted
alkanes, and you'll need a lot of patience. For synthesis purposes it is not interesting, as an experiment to learn something from it may be nice.YT2095 - 26-11-2007 at 01:18
I`ve been reading up on the exact process in my text books, and it makes much sense to me now, the UV is actually needed to split the Cl2 bond to make
a pair of singlet Cl radicals then the rest is pretty straight forwards. it mentioned that Cl and Br are possible with UV but for Iodine all it says
is greater activation energy?
what would it mean by that? maybe Heating and UV at the same time perhaps?
I still think I`ll write down the Iodine result anyway, it`s another nice way to show the purple color without having to make a gas out of it.
as for it being interesting, it is to me, as I`m just curious what paraffin would be like turned into an Alcohol?
I can`t imagine it as a substance, what will its properties be etc...
that`s why I`m doing it, it`s more fun that Google and Wiki
edited to add: Now that I`ve rearranged the halogen lamp (it`s now 5cm away) there is a Distinct 3 layer reaction going on and lots of this "Smoke"
there is the now Very Yellow paraffin at the top for 2.5cms under that is a deep red layer about 6mm thick and below that of course is the very Orange
K2SO4 and Br2 in water, about 3cm thick.
so moving the lamp right up close has now formed 3 layers!
[Edited on 26-11-2007 by YT2095]guy - 26-11-2007 at 19:14
Iodine is not strong enough to oxidize a C-H bondwoelen - 26-11-2007 at 23:55
YT2095, could you make pictures of the three-layer system? This sounds interesting. I imagine that the border between the bottom and middle layer is
sharp/distinct, and that the border between the middle and upper layer is more fuzzy.YT2095 - 27-11-2007 at 00:07
I did actually Try to take pics but they came out terrible, and it`s been exposed all night and has changed somewhat since, the middle layer is now
thinner about 3mm.
I have to be at college in a short while 9am start, so I`ll see what I can do when I get back home (4:30).
I`ll leave it turned off for now, so not much should change in a day.
edit: here a quick idea,
[Edited on 27-11-2007 by YT2095]woelen - 27-11-2007 at 23:35
Am I seeing it correctly, that you now have an aqueous layer in the middle and an organic blob at the bottom? If this is correct, then that blob
definitely is a fairly strongly brominated amount of paraffin. With a pasteur pipette you could try to take that out and analyse it further. Of
course, it is not a pure compound, it will be an incredibly complex mix with substitution of different number of H-atoms (including 0) and
substitution at different positions.
Nevertheless, you could try to make a mix of alcohols from this by hydrolysing at high pH. Only the non-substituted molecules will not react. Keep in
mind though, that hydrolysis of this halogenated stuff is not a fast reaction.not_important - 28-11-2007 at 01:17
If there are any tertiary bromides they will pop off fairly easily to give the alcohol, letting the halide sit in contact with a solution of NaHCO3
(to mop up the HBr formed) will do.YT2095 - 28-11-2007 at 02:10
the pictures aren`t all that good I`m afraid, what you Are seeing is the Paraffin layer on the very top, then just under that is a dark red thin
layer, then under that the Aqueous layer containing mainly water and K2SO4, the solid at the very bottom is K2SO4 crystals and a little bit of
unreacted KBrO3 with dissolved Br2, both are white/clear, the pic makes it look dark, but it isn`t, it`s just a crappy photo.
I expect the K2SO4(aq) layer is quite dense and so any brominated alkane will probably still float.
----------------------------------------------------------------------------------------------------------------
the I2 in Paraffin has changed a little over the days, there is a brown layer creeping up, and when you look close it`s made of Very tiny dropletts,
maybe 1/3 mm each.
there is also a foggy ring about 3cms from the test tube opening also, the foggy ring is on the test tube glass itself and is white.
the vast majority of the liquid is still a lovely purple color though.
UV from LEDs
-jeffB - 6-12-2007 at 12:29
Quote:
Originally posted by YT2095
I think it`s time to build a UV lightbox, I have 9x 5mm UV LEDs with 4000mcd each, if I place those Around the test tube, is that likely to be enough
photon energy there?
Probably not. UV LEDs generally emit in a narrow band around 395 or 405nm. Some reactions can be driven by this light, or even less energetic light,
but most photoreactions really want the "UV-C" that you get from a mercury bulb, down around 250nm. They also want more flux than you get from a
standard LED anyhow -- 5mm LEDs of any variety are only putting out tens of milliwatts of luminous flux at best.
There are LEDs that put out a watt or more of flux, but not at much higher energies (shorter wavelengths) than blue light. There are LEDs that put
out UV-C, but only at microwatt levels. It's going to be a while before LEDs are a practical source for this kind of UV.
Also remember that glass (and air) are largely opaque to UV-C wavelengths. Photochemical apparatus typically puts the source inside the reaction
vessel, in a quartz envelope.
I actually picked up a photoreactor once at university surplus, but it came with no power supply or documentation. I found a researcher at the
university who did photochem work, and sent him a nice email asking if he had any info on how to drive the thing. He sent me back an equally nice
email saying that that reactor wasn't actually supposed to go to surplus, and asking if they could please have it back. :-) I did get my money back,
and a nice tour of the lab to boot.
