Iodine monochloride

[pHarmacist]

I recived following letter from Anatole Zhoborov, a man located in Leningrad. Irrelevant details about him beeing drugged and forced in to what he calls "The Big House" to give information to KGB and/or him getting away to Moscow for a week in feer for his life for what he calls "The Narco Mafia" are not presented.

Iodine monochloride

In a fume cupboard, pass dry chlorine gas into 127 g (0.5 mol) of iodine in a distilling flask until the weight has increased by 34.5 g (0.49 mol). The chlorine should be fed in via a glass frit located at or below the surface of the iodine while the flask is gently shaken. Distil the iodine in an ordinary distillation apparatus; protect the reciver flask from the atmospheric moisture by a calcium chloride guard-tube. Collect the fraction, b.p. 97-105 C; the yield is 140 g (88%). Preserve the iodine monochloride in a dry, glass-stoppered bottle.

DANGER!; The product is a corrosive liquid that gives off a harmful vapour. Use dilute HCl (aq) (1:1) as an effective antidote if this nasty stuff ever comes in contact with the skin.

Why did Zhoborov send me this? Does it has anything to do with the well-known 2C-I/DOI synthesis presented in PiHKAL by Shulgin? Has Anatole Zhoborov been secreatly in contact with Shulgin all along? I might never know! Oh, just let it bee...

Refs:

TiHKAL and my imagination
Vogel's, 5:th ed, p. 440

For the details regarding the generation of Cl2 itself from KMnO4 (s) and HCl (aq), turn your attention to rhodium.ws and the document written by Eleusis.

[pHarmacist]

But first some erroneous stoichiometry:

Eleusis starts with: "Molar ratios; 37g KMnO4 to 7.1mL 31.45% HCl to .41g Cl" <- Wrong!

He goes on: "In the addition funnel you will eventually run a volume of "Muriatic" Acid (31.45% HCl) 19 times the weight of the KMnO4 (304mL for this example)." <- This also... overkill!

Corrections:

[0.367 g of KMnO4 gives 0.412 g of Cl2] Meaning; [1.000 g KMnO4 requires 6.2 mL of conc. HCl] (Vogel's)

So, if you're going to follow Eleusis document and start with 16.00 g KMnO4, it'll bee enough to use 99.20 mL of HCl (aq) and NOT 304 mL as he suggests. You should get 17.97 g of chlorine gas. Now, HCl (aq) is a cheap chemical and this dosen't matter much, but still, why waste in the long run, it's a bad practice.

Convertion factor KMnO4 <=> Cl2 is 1.123; m(KMnO4)x1.123 = m(Cl2) generated or m(Cl2)/1.123 = m(KMnO4) required; Example: 16.00 x 1.123 = 17.97 g Cl2 generated or 17.97 / 1.123 = 16.00 g KMnO4 required...
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Eleusis is a skilful ASCii-artist and his document contains cool illustrations regarding apparatus assembly, so check it out!

[Osmium]

> start with 16.00 g KMnO4, it'll bee enough to use 99.20 mL of HCl (aq)

100ml of HCl contain about 35g chloride, why isn't it all converted into Cl2?
(Should probably check the stoichometry of this reaction, th rest will probably end up as KCl and MnCl2 etc. WTF nevermind, I think I answered my own question)

Is it possible to dissolve that much KMnO4 in such a little bit of solvent? I don't think so, and even if that isn't necessary you might still end up with quite some syrup, or crystal mush in your flask.

[pHarmacist]

well... calculations were based on the experiment found in Vogel's so it should run smoothly with that KMnO4/HCl ratio...