Author Topic: (iso)propyliodide synth  (Read 4888 times)

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Bandil

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(iso)propyliodide synth
« on: May 05, 2002, 10:38:00 PM »
Swim was looking at the

https://www.thevespiary.org/rhodium/Rhodium/chemistry/methyliodide.html

document and found it interesting!

Can the same proc be used on propanole and isopropanole, or is the some inhibition of the reaction because of the bigger alkyl chain?

if it can be done, what reactiontimes should be used? Swim supposes that the amounts of chemicals should just be scaled up with regards to moleweight?

Thanks!

Regards
Peter

Rhodium

  • Guest
I would say that the only thing you need to ...
« Reply #1 on: May 06, 2002, 12:14:00 AM »
I would say that the only thing you need to adjust for is the molar weight of the reactants, and don't forget to look up boiling points op propyl iodide and isopropyl iodide.

Here are two other methods for iodinating alcohols, should work just as well with propanol/IPA.

- - - -

Preparation of Ethyl Iodide

Method 1

A mixture of 500 g (3.94 mol) of iodine, 800 mL of 84% (w/v) ethanol, and 60 g (2.224 mol) of aluminum foil in small pieces is warmed gently. Once started, the reaction increases progressively in vigor, but can be controlled by cooling; it subsides in about 10 min, and is complete in an hour. The product is distilled until deep red fumes appear. A cold mixture of 700 mL of 84% ethanol and 400 mL of 85% sulfuric acid is added to the cooled residue. After 15 min, distillation is begun, and continued until no more oily drops form in the water in the receiver. The yield of crude ethyl iodide is 260 mL (504 g; approx. 80% yield).

Method 2 [7]

1 mole of alcohol and 2 moles of KI (or equimolar amount of NaI) are added to 2.96 moles of 95% phosphoric acid. Reflux the mixture 3-4 hrs. A two-phase mixture will result. Add water and ether to the reaction mixture. Seperate the ether layer, decolorise it with Na-thiosulfate, wash it with saturated NaCl solution and dry it with Na2SO4. Evaporate the ether and distill the remaining alkyl iodide to get ~90% yield.


Bandil

  • Guest
That looks like a very nice reaction.
« Reply #2 on: May 07, 2002, 03:13:00 PM »
That looks like a very nice reaction. Do you have a reference on the reaction? Is the Al foil to be used "as is", or should it be amalgated first?

Thanks!

Regards
Peter

Rhodium

  • Guest
Al for alkyl iodides
« Reply #3 on: May 07, 2002, 04:20:00 PM »
No amalgam. The references are found towards the end of this document:

https://www.thevespiary.org/rhodium/Rhodium/chemistry/nitroalkane.html



UTFSE and you will find people trying out the above method in practice.

halfapint

  • Guest
Stronger Iodine Solution
« Reply #4 on: May 08, 2002, 03:09:00 AM »
The salient point about making isopropyl iodide, is that the farmer's "Stronger Iodine Solution", a standard article of commerce, is not composed of the same ingredients as the familiar 2% drugstore tincture, with its water, ethanol, iodine and sodium iodide.

Instead, the 7% iodine consists of 7% iodine in dry isopropanol. It's a ready-made reagent for this, just add aluminum glitter and heat. (If you do use aluminum flake, bee prepared with a cooling bath to moderate its initial intensity.) I haven't run this yet, but plan to do so with the following end in mind: to react i-propyl iodide with fused sodium acetate to produce the ester i-propyl acetate water free, less drying and purification labor to make use of such ester in anhydrous reactions.

A further incentive, is that I dislike the loss and mess of pulling elemental iodine out of the isopropanol, so I'm looking for a clean way to get a handle on that iodine in a less lossy way, eventually in its salt. I expect there may bee some elimination to propylene, but this shouldn't prove a major factor.

a half a pints a half a pound a half a world a half a round
Sidearm n. Flask neck tube.

Osmium

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Isolation of I2 dissolved in an alcohol: Add ...
« Reply #5 on: May 08, 2002, 11:01:00 AM »
Isolation of I2 dissolved in an alcohol:

Add slight excess of NaOH solution:

I2 + 2 NaOH -----> NaIO + NaI

Evaporate away the alcohol. Add acid to the solid inorganic residue:

NaIO + NaI + 2HCl -----> 2 NaCl + I2 + H2O

Has been posted before by me, but of course nobody cares.

I'm not fat just horizontally disproportionate.

catseye1

  • Guest
Actually I do
« Reply #6 on: May 08, 2002, 07:29:00 PM »
Osmium. Haven't before, but that's cuz I hadn't seen this specific thread topic before. Thanks.
Concise, information laden thread. Thanks everyone, This one goes in my hive thread bookmarks.