I have an old copy of a book called "Organicum, practical handbook of organic chemistry" (I think this book is known to many of you)
there are some parts I don't quite understand/raise questions for me.
Therefore, I want to reach out to you guys and ask my questions in this thread
For the start there is a "Paneth's mirror method" which is mentioned in the D-1.4 section of the previously mentioned book. I don't know what
"Paneth's mirror method" is, could anyone shine light on it?
-Sarinoxsarinox - 11-9-2024 at 04:41
Hello folks,
another question:
In another part of the book "Organicum, practical handbook of organic chemistry" it is mentioned that coiled condensers cannot be used in tilted
positions and the book asks "Why?" is there anyone who knows the answer for this question?
Condensor type "e" is under question
[Edited on 11-9-2024 by sarinox]Dr.Bob - 11-9-2024 at 10:39
Condenser "E" is a Graham type condenser, and since the liquid condenses in the inner coil, the liquid has to drain from the coil via gravity. If you
fill a section of the coil with liquid, it will pool and prevent vapors from moving up the condenser in a smooth flow, this building pressure up,
which can cause the joints to leak vapors. Like a sink trap, which is meant to prevent gas flow, the coil will stop vapors from moving well.
Condenser "b" and "D" (Allihan condensers) are also poor using at an angle, as they hold liquids, and create a non-smooth flow of your product, and
also prevent vapors from seeing the cold part of the condenser. The other condensers all have the condensed liquid flowing on a smooth surface where
gravity can bring it to the outlet OK, except i, which is meant as a trap and used to collect material with itself. But they have limited capacity.
sarinox - 11-9-2024 at 11:33
Hello and thank you for your reply,
So, I cannot use d in tilted position or I can but I would lower down the yield?
You know currently I don't have equipmentss to use the condenser d in vertical position.
So, I cannot use d in tilted position or I can but I would lower down the yield?
You know currently I don't have equipmentss to use the condenser d in vertical position.
-Sarinox
I have used "d" (Allihn) condenser in tilted postion without problem. Although some "pools" of destilate will form (thus reducing the cold surface),
if you compare it to "straight" - C - Liebig condenser, for the same lenght, I think that the Allinh condenser has more cold surface, because a sphere
has more surface than the wall of cylinder (or tube) of the same height.
But maybe Im wrong. - because is the only condenser Ive got....
here is some more information:
The Liebig condenser is used only when the vapor rate is relatively low or used diagonally downward for distillation. The Allihn and Dimroth
condensers are more favored in reflux or redistillation, and the Dimroth condenser could offer better condensation efficiency.
[Edited on 11-9-2024 by RU_KLO]Rainwater - 11-9-2024 at 17:19
I believe "E" & "F" are graham style and are best used vertically unless you have a method to force the distillate through the condencer.
A pressure gradient caused by applying a vacuum and using a cappallery as a bubbler to prevent bumping is the best example I can think of.
Since learning how to use a graham to control reflux takeoff, I mostly use this in conjunction with a "G" for really hard to seperate mixtures.
Graham's are very useful when you have a compound that likes to supercool before condensing. Like butane, or when there is the need to maintain a
large thermal load in the distillate, like in a stripping column.
Sometimes, it is undesirable to use coolant far below the boiling point of a compound and a coolent will be just a few degrees below the distillate
temperature. Thermal transfer is directly proportional to the temperature difference between the two mediums, resulting in degraded performance. A
graham makes up for this with a larger surface area.
[Edited on 12-9-2024 by Rainwater]
Iodine
sarinox - 12-9-2024 at 01:01
Tanx for your replies on condensers.
Here is another question raised by reading the aforementioned book:
Why DEA has put a ban on Iodine?
I remember when I was a kid we had access to Iodine and it was cheap! but nowadays it is very expensive!
However, in a section of that book "Organicum" it is mentioned that Iodine could be precipitated using the following reaction:
KI + Peroxides + acetic anhydride
1-Is there a better way - cheaper, higher yield & high purity products- to get Iodine?
2-How to get rid of the acid and increase purity of Iodine?
1-Is there a better way - cheaper, higher yield & high purity products- to get Iodine?
I just use KI + HCl + H2O2. Muriatic acid and 3% H2O2 works fine. The iodine precipitates out and the H+ turns to water, leaving a solution of KCl. sarinox - 12-9-2024 at 09:44
@Metallophile,
Tanx for your reply, but why Muriatic acid? Also, the Iodine is retrievable using just filter? or you do a post processing of desublimation?
And do u happen to know why selling of Iodine is heavily restricted?
Edit: Also will it work for Bromine as well?
[Edited on 12-9-2024 by sarinox]Metallophile - 12-9-2024 at 13:01
I used muriatic acid because it's cheap and I can get it at the hardware store. I just filtered the iodine, but sublimation would probably improve the
purity. Yes, it does work with bromine too! In this case, I used NaBr and distilled the bromine out of the aqueous solution. I think iodine was
restricted because it's sometimes used in meth cookery. sarinox - 12-9-2024 at 14:25
Ok I got it why u used Muriatic acid. :-)
by the way I just think the restrictions should be lifted because the person who knows how to cook meth then S/He probably knows how to get Iodine! bnull - 12-9-2024 at 17:06
Dissolve copper sulfate in hydrochloric acid. If you don't have HCl, use NaCl and sulfuric acid. Dissolve potassium iodide in the least amount of
water possible and mix the two solutions. The result is a rust-brown sludge, which after some minutes separates into a rust precipitate and a reddish
liquid. On heating, iodine sublimates from the solution.
It goes like this:Copper (ii) is reduced to copper (i) and iodide is oxidized to iodine,$$2 Cu^{2+}(aq) + 4 I^−(aq) \rightarrow 2
CuI(s) + I_2(aq);$$
Hydrochloric acid reacts with copper (i) iodide to form copper (i) chloride and hydriodic acid,$$2CuI(s)+2HCl(aq) \rightarrow 2CuCl(s)+HI(aq);$$
The hydriodic acid just formed reacts with copper (ii) ions on the solution, as in the first reaction.
The process repeats itself until all copper (ii) is converted to copper (i) or all iodide is oxidized to iodine, whichever is in excess. The reddish
color comes from triiodide (I3-), which is formed when iodine dissolves in solutions containing free iodide ions; that's why I wrote
I2(aq) rather than I2(s) in the first equation. On heating, iodine vapors are released from the solution (sublimation or
evaporation, I don't know which) and can be condensed.sarinox - 13-9-2024 at 11:06
