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Author: Subject: Ethyl Iodide
paccman278
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[*] posted on 30-11-2004 at 01:53
ethyl iodide


hope this helps. helped me once, form rhodiums site....



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).
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[*] posted on 30-11-2004 at 07:17


Maybe it would have helped, had it not already been posted a few posts upthread. Redundant information is not particularily helpful -- read the thread first.
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[*] posted on 6-12-2004 at 21:57


About the monoethylamine from ethyl bromide and excess NH3, could anyone post reaction details, neither vogel's practical o-chem nor mann's practical o-chem had any details on this reaction. They merely confirmed that an excess of amine was necessary, but how is the reaction carried out, is it a 5 mole ammonia excess or a 50 mole ammonia excess? Is it carried out in an aqueous solution or in alcoholic solution? Does it need to be dry?



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[*] posted on 6-12-2004 at 22:57


In the reference given in Thorpe, 90% EtOH was saturated with NH3. Water bad. Higher concentration of alkyl halide would give higher percentage of sec. and tert. amines, and quat. ammoniums. There was 16X excess ammonia and a 34% yield. Then you have to separate the amines.

So there is a reason why this isn't in many books.
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[*] posted on 7-12-2004 at 02:09


Quote:
Originally posted by Mendeleev
About the monoethylamine from ethyl bromide and excess NH3, could anyone post reaction details, neither vogel's practical o-chem nor mann's practical o-chem had any details on this reaction. They merely confirmed that an excess of amine was necessary, but how is the reaction carried out, is it a 5 mole ammonia excess or a 50 mole ammonia excess? Is it carried out in an aqueous solution or in alcoholic solution? Does it need to be dry?


Not a good route. For amines from alkylhalides use azide as an nucleophile rather than amonia. You can obtain azide salt from an air-bag at the car-junk. Run a two phase reaction (alkylhalide in DCM, azide in water). In case tetraalkyl ammonium halids are unavailable as a PTC catalyst you can use Softlan (TM) wash & clean ditergent available in supermarket, otherwise they can be prepared from ammonia and alkylhalide. The azide can be reduced to amine with sodium dithionite also available OTC. If there is interest in this route I can dig up the references, I don't have the experimantal details right now.


[Edited on 7-12-2004 by Nosferatu]




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[*] posted on 28-12-2004 at 00:40


One could easily make ethyl chloride from ethanol + HCl + ZnCl2

One could also make ethyl bromide from ethanol + NaBr + H2SO4

Once either of these compounds is formed, ruflux a saturated solution of the ethyl halide with an equimolar amount of NaI in acetone for 30 minutes or so.




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[*] posted on 29-12-2004 at 19:16


Quote:

One could easily make ethyl chloride from ethanol + HCl + ZnCl2

Is that an easier route than the common one?
I've heard that its actually a pain in the arse, becasue you need to reflux, put it at about 170 degrees, and ethyl chloride being a gas and all...




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[*] posted on 18-12-2005 at 09:14


You could start with the ethylbromide then turn it into ethyliodide:

Begin by first making your hydrogen bromide from sodium bromide;
NaBr + HCl + H2O2 -> Br2 + NaCl + H2O
3Br2 + S + H2O -> 6HBr + H2SO4

then proceed to the ethylbromide:

CH3CH2OH + HBr -> CH3CH2Br + H2O
small amounts of ether, ethene, and CH3CH2SO3

then in acetone do a displacement rxn:
CH3CH2Br + NaI -> CH3CH2I + NaBr




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[*] posted on 18-12-2005 at 09:50


Quote:
3Br2 + S + H2O -> 6HBr + H2SO4
How do you expect this to work, I'll admit that I could be entirely wrong but it seems you're relying on some sulfur halogen compound as an intermediate and its subsequent hydrolysis to yield your desired sulfuric acid and hydrogen bromide (not to mention your reaction is unbalanced with the exception of the bromine and sulfur). If that is the case 1) Sulfur only reacts appreciably with chlorine at its melting point as a few members here can attest. 2) These types of compounds react to form HBr, S, thiosulfates, and sulfites, this can be found in an MSDS (although it took me some searching to find it), no H<sub>2</sub>SO<sub>4</sub>. The reaction as you wrote it just doesn't seem to work, however the reaction between elemental bromine, sulfur dioxide, and water can indeed give hydrogen bromide and sulfuric acid.



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[*] posted on 18-12-2005 at 10:52


Chochu3 the last reaction you mentioned won't work eighter.

Bromide ions are stronger nucleophils than iodide ions, so the iodide ion can't displace the bromide ion.
The reaction work only with stronger nucleophils like Cl- , F- and OH- (for example CH3CH2Br+NaCl or NaF or NaOH ==>CH3CH2X X is Cl,F,OH + NaBr)
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[*] posted on 18-12-2005 at 11:06


That's what I figured, however in acetone maybe the decreased solubility of sodium bromide with respect to sodium iodide may help force the reaction. However if iodine is present in overwhelming excess, it will also force the reaction. Still, I don't see an advantage over this method then just the regular reaction to produce ethyl iodide.



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[*] posted on 18-12-2005 at 20:29


It is unbalanced:
3Br2 + S + 4H2O -> 6HBr + H2SO4

Rxn b/w Br2 and S forms S2Br2 then it hydrolyzed by water.

This is how it works:
S2Br2 + 2Br2 + 4H2O -> 6HBr + H2SO4 + S(precipitated)

Reference for Sulfur + bromine to form hydrogen bromide
A course in inorganic preperations; by William Edwards Henderson; 1935; pg. 80
Modern Inorganic Chemistry: An intermidate text; by C. Chambers and A. K. Holliday; 1975; pg. 333
Inorganic Laboratory preparations; by shlessinger; 1962; pg. 144


Displacement of bromide with iodide proceeds by sodium bromide not being soluble in acetone which sodium iodide is.

Reference for NaI in acetone for halogen exchange
Vogel practical organic chemistry 5th ed.; by Vogel and others; pg. 572 (expirement 5.62)
displacement of bromide in acetone

[Edited on 19-12-2005 by chochu3]




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[*] posted on 18-12-2005 at 22:17


Quote:
Originally posted by hinz
Chochu3 the last reaction you mentioned won't work eighter.

Bromide ions are stronger nucleophils than iodide ions, so the iodide ion can't displace the bromide ion.
The reaction work only with stronger nucleophils like Cl- , F- and OH- (for example CH3CH2Br+NaCl or NaF or NaOH ==>CH3CH2X X is Cl,F,OH + NaBr)


Actually, iodide ions are rather strong nucleophiles, at least in protic solvents they are. They're stronger than all the other halides and OH-. This may not be true in acetone as the solvent but the displacement of Br with I occurrs anyway in acetone because NaBr has very poor solubility in acetone. It's actually a standard reagent test for alkyl bromides and alkyl chlorides to react them with NaI in acetone and see if a precipitate forms.
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[*] posted on 19-12-2005 at 14:24


chochu3, I couldn't find hide nor hair of that reaction in your first reference but the third reference was a winner. The stated reaction is:

S<sub>2</sub>Br<sub>2</sub> + 5Br<sub>2</sub> ---> 12HBr + 2H<sub>2</sub>SO<sub>4</sub>

150 g Br<sub>2</sub> with 10 g of sulfur dissolved in it, slowly hydrolyzed in a 500 ml flask with 200 grams of ice over the course of an hour and a half. The HBr being distilled off.

Sorry for maintaining my off topicness.

[Edited on 12/19/2005 by BromicAcid]




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[*] posted on 19-12-2005 at 16:19


that reference just justfies it but does not actually show no more than that.



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[*] posted on 12-2-2006 at 08:25


Quote:
Originally posted by paccman278
hope this helps. helped me once, form rhodiums site....



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).


Could this reaction work with methanol to get the methyl iodide instead of ethyl iodide??
I'm pretty sure it could, after all it's just one methyl less.
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[*] posted on 13-2-2006 at 10:38


Quote:
Could this reaction work with methanol to get the methyl iodide instead of ethyl iodide??
I'm pretty sure it could, after all it's just one methyl less.


It works but yields are lower. I'd need to check but not less than 50%.
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[*] posted on 13-2-2006 at 16:10


I remember it being more around 77% with sulfuric acid, and could be higher with phosphoric acid (no oxidation of the HI). I saw this somewhere on rhodium some time ago, I remember the yield and general procedure since it was straightforward and easy.

Sulfuric acid, ammonium iodide and methanol was >85% yield IIRC.
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[*] posted on 16-12-2012 at 17:05


I have been looking for the best procedure for making ethyl iodide now for a few days. This is what I have come up with so far:

1. The classic method. This uses red P, which I don't have.
2. phosphoric acid method. garage chemist has reported that this gives poor yields.
3. HI method. This is described on p. 102 in "Experimental Organic Chemistry," by Norris (forum library). Brauer tells how to make HI, but this requires H2S. In the Erowid archive procedure for HI a hard to remove solid phosphate is deposited on the bottom of the flask.
4. Diethyl Sulfate method. I would prefer to avoid this compound if possible for safety (risk to health) reasons.
5. aluminum foil method. This is described upthread.

I have been leaning toward the use of method 3. Please indicate your favorite method and why you chose it.

[Edited on 17-12-2012 by Magpie]

[Edited on 17-12-2012 by Magpie]

[Edited on 17-12-2012 by Magpie]




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[*] posted on 19-12-2012 at 11:47
alkyl iodide alkyl halides


Without any modification I copy here some other options from The Hive.
(chemistrydiscourse\000051160.html)

Posted by obituary
MeI methods Bookmark

different ways to make MeI for whatever you want it for:

1. MeSO4 + KI -- MeI
2. Slowly distill MeOH w/large excess of HI
3. electrolysis of an aq. solution of K acetate w/ I2 or KI
4. KI + Me-p-toluenesulfonate
5. methanol + PI5 (in MeI {of all things})
6. MeOH + PI3 or MeOH + P(yellow or red) + I2

sources:
#1. weinland and schmid, ber. 38, 2327 (1905); ger pat. 175,209 [Frd 1. 8, 17 (1905-07)].
#2. Norris, Am. Chem. J. 38, 639 (1907).
#3. Kaufler and Herzog, ber. 42, 3860 (1909).
#4. Peacock and Menon, Quart. J. Indian Chem. Soc. 2, 240 (1925); Rodionow, Bull. Soc. Chim., 39, 305 (1926).
#5. Walker and Johnson, J. Chem. Soc. 87, 1595 (1905).
#6. Dumas and Peligot, Ann. 15, 20 (1835); Ipatiew, S. Russ. Phys-Chem. Soc. 27, I, 364 (1895) [Ber. 29(R)90(1896)]

By foxy2

MeI and alkyl iodide synthesis
(Rated as: excellent) Bookmark

Okay we have Rhodiums Synths
https://www.rhodium.ws/chemistry/nitroalkane.html
https://www.rhodium.ws/chemistry/methyliodide.html

Here is a possible variation on the RedP procedure above.
Improved preparation of methyl iodide

Kizlink, Juraj; Rattay, Vladimir.
Chem. Listy (1980), 74(1), 91-2.

Abstract
MeI was prepd. in .apprx.90% yield by the red P-catalyzed reaction of MeOH with I2. The reaction rate was controlled by the rate of addn. of MeOH to a mixt. of I2 and P.


I'll add a few more.

Alkyl iodides.
Patent US3053910
Abstract
Alkyl iodides were prepd. by treating a dialkyl sulfate at pH 1-6.5 with an iodide prepd. by the action of a reducing agent, e.g., Al, Fe, Sn, SO4, oxalic acid, or N2H4, on an aq. slurry of elemental iodine. In an example, 241 g. Fe powder was added to 820 g. H2O and 1050 g. com. iodine during 2-3 hrs. at 20-60°, at pH 5.3. Et2SO4 (1540 g.) was added dropwise at 70-80°, while distg. the EtI as formed. The reaction mixt. was heated to 95° to complete the EtI distn., and the product was washed with cold H2O, 5% N2CO3, and dried (CaCl2) to give 99% EtI. Similarly prepd. were 99.1% MeI, 97.2% PrI, and 99.4% AmI.


Preparation of alkyl iodides or aryl iodides.
Patent JP62246527
Abstract
Alkyl iodides or aryl iodides, useful as methylating agents and cocatalysts for carbonylating agents, were prepd. by a reaction of group IA, IIA, IIIA, IB, IIB, or IVB metals and iodine or iodides of metals and alcs., carboxylic acid esters, dialkyl ethers, and/or diaryl ethers at 15-150° and 1-50 atm without formation of H2O and corrosive materials. Thus, a soln. of iodine in AcOMe was added dropwise to a mixt. of Al and AcOH over 60 min and then heated at 65° for 150 min to give 65.4% (based on iodine) MeI.


Methyl iodide.
Zadorozhnaya, I. E.
Khim. Reaktivov Prep. (1967), No. 15 96-7.
Abstract
MeI was prepd. in 91.5% yielded by adding 1.55 kg. PhSO3Me dropwise to a soln. of 1.35 kg. NaI in 1.35 1. H2O, first at room temp. until one third of the ester had been added, and then at 60-70° in a system which provided for distg. the product at this temp. into a receiver contg. ice and H2O.


Antoncho might find this one funny

Preparation of alkyl halides and sulfur from hydrogen sulfide, alcohols, and halides
Patent US3649197
Abstract
Alkyl halides were prepd., by reaction of H2S, Br or I, and EtOH or MeOH in the presence of an alkali metal halide and, optionally, a hydrohalic acid, in a 2-stage reaction. Thus, a mixt. of EtOH, Br, LiBr, and 12M HCl was treated with 15 psi H2S, S filtered, and the filtrate heated 2 hr at 130° to give 65% EtBr. Similarly prepd. were 89.0% EtI and 43.0% MeI.


Alkyl iodides
Patent JP50121205
Abstract
Alkyl iodides were prepd. by treating fatty acid alkyl esters with stoichiometric amts. of metal iodides at 100-250°. Thus, a mixt. of 2.02 g Me oleate (I) and 10 g CaI2 was stirred 2 hr at 200° under N to give 6 g MeI (purity 99%). PrI was similarly prepd.; MgI2, MnI2, LiI, AlI3, and SrI2 were also used in place of CaI2. AcOMe was also used in place of I.


Methyl iodide.
Patent CS164707
Abstract
Hot aq. KI was continuously treated with a 3:2 mixt. of H2SO4 and MeOH, and MeI was gradually distd. in 90% yield.


Hydrogen iodide, lithium iodide and methyl iodide.
Patent US4302432
Abstract
HI is prepd. by reaction under anhyd. conditions of H2 and I2 in a non-alc. solvent using a homogeneous Rh catalyst [e.g., Rh2(CO)4Cl2]. LiI and/or MeI are obtained by including LiOAc and/or MeOAc in the reaction medium.


Anhydrous alkyl iodide.
Patent EP46870
Abstract
Anhyd. RI were prepd. by treating R1CO2R (R = C1-4 alkyl; R1 = H, C1-8 alkyl or aryl) with iodine and H (and optionally CO) in the presence of a Pt metal compd. as catalyst and a quaternary heterocyclic N compd. or quaternary P compd. as promoter. Thus, 250 g AcOMe, 50 g AcOH, 60 g N,N-dimethylimidazolium iodide, 1.2 g [Rh(CO)2Cl]2 and 50 g iodine were pressurized with 2 bar CO and 4 bar H, then stirred 58 min at 373 K to give 99.6% MeI.



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[*] posted on 7-2-2013 at 15:22


I recently prepared some ethyl iodide (EtI) using the method found in Norris in the forum library. Since this is an alternative to the classic method using red phosphorus and iodine I thought my results would be of interest.

The Norris procedure specifies 60g of 57%HI. Since I only had 50g I scaled back accordingly. Not having a 200mL distilling flask I used a 250mL RBF. I used a small pressure-equalizing funnel with a 6mm glass tube attached as a dropping funnel. Some time ago when cleaning this funnel I accidently broke off the small inner glass tube at the tip of the funnel. This was serendipitous for this use as I could now clearly see the ethanol drops as they were added at 1drop/s as required by the procedure. The boiling HI did tend to percolate up the tube but this gave no apparent problem. As Norris indicated the EtI accumulated in the receiver drop by drop as the ethanol was added drop by drop. A photo of the reaction apparatus is shown below:

dripping ethanol into boiling HI.JPG - 92kB
dripping ethanol into boiling 57%HI

For the workup, per Norris, some water (~7mL) was used to wash the product. The washed product had a deep brown color due to dissolved iodine. At this point I switched over to the workup in Brewster. I washed with 5mL of 3% NaOH which discharged all the iodine color. I then washed the product with 12mL of water. This was removed by separatory funnel as shown in the photo below. There was a very clean separation as the density of the EtI is 1.93.

water wash of EtI.JPG - 86kB
water wash of EtI

The product was then placed in a small Erlenmeyer flask and about 15 BB's of CaCl2 were added for drying. As you can see in the photo below they floated. This was left overnight, and yes, the rubber stopper did swell. It likely would be difficult to remove if left like that for much longer.

EtI drying.JPG - 86kB
EtI drying

Today the product was placed in a 25mL RBF and distilled. The boiling point was 72C (literature 72.3C). The yield was 11.7g; the % yield on ethanol was 41.7%.

Questions, comments, and suggestions are welcomed, as usual.





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[*] posted on 7-2-2013 at 19:23


Excellent report Magpie. How did you come by the HI?
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[*] posted on 7-2-2013 at 20:38


Quote: Originally posted by Paddywhacker  
Excellent report Magpie. How did you come by the HI?


Thank you. With some reluctance I first tried the method in Brauer, ie, the use of H2S and I2. But this did not work well as my scale was so small. Also, I don't like making elemental iodine, then working with it - so messy and volatile, and stains everything. But even worse is working with H2S due to its stench and toxicity.

Also with some reluctance due to the phosphate polymer formation I then went to the method of Argox. This worked exactly as advertised. My %yield was low at 44%, but this was expected as I did not recover the HI dissolved in the gas trap, nor did I push the phosphate polymer production to the endpoint.

Production of HI (aq).JPG - 103kB
Production of 57%HI

Attachment: phpgva8rh (116kB)
This file has been downloaded 362 times
Purification of 57%HI




[Edited on 8-2-2013 by Magpie]




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[*] posted on 10-2-2013 at 23:29


Magpie, that seems to be a good yield for your method. I remember making my sample via the classic RP and iodine in situ. I had a small yield because I don't have a small distillation setup so lots of it just went to vapors. :(

Be sure to put a piece of clean copper or silver wire in your sample for any stray iodine, and store it in a dark bottle. Wrap it in aluminium foil to be sure.




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[*] posted on 24-11-2017 at 13:52
Ethyl Iodide


Hi guys, after some lurking I've decided to finally post something ;)

Having a bit of an issue understanding some of the concepts of forming Ethyl Iodide via reaction of elemental Iodine, Aluminium and Ethanol.

For a start, there is not much literature on this reaction apart from chemplayer and an erowid text I believe. Chemplayer mentions in the video that the reaction initially starts as thus:
Al + 3I --> AlI3
Am I correct in believing that since Iodine is diatomic, that it should be:
2Al + 3I2 --> 2AlI3?

Secondly, based off of this (if correct) the molar calculations are incorrect and have been calculated via simply the atomic mass of the element Iodine, 126.9 , when diatomic it should be 253.8g/mol, since a molecule, no? This is not in any way, shape or form a way of trying to slate Chemplayer at all, I just want to make sure my knowledge on chemistry is right.

Thirdly, from the reaction of Iodine and Aluminium to form Aluminium iodide, what products and bi products are formed from the reaction of Aluminium Iodide and Ethanol?
AlI3 + 3 EtOH --> 3EtI + Al(OH)3
Is this also correct as cannot find any information on this particular reaction?

Performing this reaction led to a highly undesireable yield of ethyl iodide and I'm hoping that I'm right so that I can change the moles of reactants to provide a better yield. Following on with this particular part in the reaction, would a lower percentage alcohol by volume be an advantage as the production of HI from water and Aluminium Iodide would lead to more potential for ethyl iodide to be formed?

Last, erowid mentions to add an acidic alcoholic solution (Sulphuric acid and Ethanol) to the distillate (namely EtI) to be further distilled, excuse my ignorance, but what is the purpose of this? Could someone please explain?

Hopefully not too any dumb questions or too much of a mouthful for a first post, eh? ;)


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