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Author: Subject: Isopropyl iodide from iPrOH/Red P/I2. Unexpected product
Heptylene
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[*] posted on 18-7-2018 at 09:58
Isopropyl iodide from iPrOH/Red P/I2. Unexpected product


I tried making isopropyl iodide today by the method used by Chemplayer to make methyl and ethyl iodide, but I didn't get the product I expected and can't find what it is. I couldn't find a procedure to make isopropyl iodide by the red phosphorus/iodine method so I improvised the following:

To a 100 ml round bottom flask there was added iodine (31.72 g, 0.25 mol, 1 eq.) and isopropyl alcohol (30.05 g, 0.5 mol, 2 eq.). The mixture was stirred without complete dissolution of the iodine and red phosphorus (3.23 g, 0.104 mol, 0.42 eq.) was added in one portion, which caused the temperature to rise enough for the mixture to start boiling. The mixture was further heated and stirred on an 80°C water bath for 1 hour then the volatiles were distilled. A single pale orange fraction came over between 67-71°C. The liquid was washed with dilute Na2S2O3 solution, then with water and was dried over anhydrous Na2SO4. Finally there was obtained about 20 ml of a clear liquid with a density of 1.57 g/ml and a sickly-sweet aroma reminescent of methyl iodide or DCM.

Now, isopropyl iodide boils at 88-90°C and has a density of 1.7 g/ml (Sigma Aldrich). My product boils below this temperature. I first thought it was isopropyl ether (BP = 68-69°C. The conditions seemed favorable: used excess alcohol, polar protic conditions, secondary alkyl iodide, reflux... but the density of the ether is 0.725 g/ml.

A few other possibilites are n-propyl ether (BP = 90°C, d = 0.75 g/ml), n-propyl iodide (BP = 101-102 °C, d = 1.74 g/ml) but those don't match.

I could not light the product on fire, but decomposition from the heat of a match formed some yellow liquid, similar in color to dilute iodine in ethanol.

Currently my money is on an azeotrope of isopropyl iodide and isopropyl alcohol, but I haven't found any data on that.

Any ideas?

EDIT:Of course sodium iodide in acetone does not react with the product! I don't know what I was thinking.

[Edited on 19-7-2018 by Heptylene]
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Chemi Pharma
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[*] posted on 18-7-2018 at 10:49


No idea what can it be, but I strongly suggest you follow the H3PO4 + KI process to make alkyl iodides from alkyl alcohols with high yields and low grade impurities.

The process were covered by me as an attachment at: http://www.sciencemadness.org/talk/viewthread.php?tid=2727&a...
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TheRealKP
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[*] posted on 18-7-2018 at 14:38


I tred a similar reaction and recieved a similar mystery product

H3PO4 + KI + i-PrOH —> K3PO4 + i-PrI

I isolated isopropyl iodide, but also a mystery product. I strongly suspect it is a mixture of isopropyl alcohol and isopropyl iodide, which should have a lower density than the iodide. My bet is that you have an azeotrope of iodide and alcohol, which is concentrated enough to be immiscible with water. Try washing with water 3 times, then check denisty. I did not try doing this because I had enough isopropyl iodide, and didn’t care enough.
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Heptylene
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[*] posted on 19-7-2018 at 11:51


So I followed your suggestion TheRealKP and washed the product three times with water (ca. 3x25 ml), dried with sodium sulfate and measured the density. It went up from 1.57 to 1.65 g/ml. I am now convinced that the product is indeed isopropyl iodide mixed with isopropanol.

Assuming that the density of the mixture is the sum of the densities of each constituent weighted by their respective mass fractions, this puts the composition of the product at around 94 % isopropyl iodide by mass. The final yield after all those washings and losses is about 62 % (adjusted for pure isopropyl iodide).

I'll probably try to run the reaction again using 1 eq. of alcohol instead of 2, seeing as the sepearation from excess alcohol is so difficult.

I used the product as is in a reaction with butylamine in an attempt to prepare diisopropylbutylamine.

I will probably open a new thread on that if I can get anything from the reaction.
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TheRealKP
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[*] posted on 19-7-2018 at 15:11


Quote: Originally posted by Heptylene  
I'll probably try to run the reaction again using 1 eq. of alcohol instead of 2, seeing as the sepearation from excess alcohol is so difficult..


This is a good idea for your system. You will see higher exotherm and some alcohol may evaporate, so maybe use 1.2 eq. of alcohol instead of just 1. Slight excess of alcohol is always a good idea for this protocol to maintain good reaction rates. If you choose to use 2 eq. anyway, then try adding extra red P/I2 after the exotherm dies down to further increase the amount of product and to reduce alcohol contamination.
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[*] posted on 20-7-2018 at 06:21


Try testing it with chlorine bleach. (add a drop of your product to chlorine bleach) That should at least tell you if your product is primarily an alkyl iodide. (there would be several color changes in that case)



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