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Ethyl Iodide

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tom haggen - 25-10-2004 at 11:13

I've been trying to find a synthesis for ethyl iodide today on google, and so far I haven't come up with much. So maybe I can just take a shot in the dark. I'm thinking that you could react NaI with CH3CH2OH and get ICH3CH2 maybe using vacuum distillation? So would the reaction go like this?

CH3CH2OH + NaI --> ICH3CH2 + NaOH

This seems convenient because both precursors are available over the counter.

[Edited on 25-10-2004 by tom haggen]

S.C. Wack - 25-10-2004 at 11:41

No, you will get NaI dissolved in EtOH.

Google is not what you should be reading, a number of texts have the synthesis of ethyl iodide in them. You will help yourself and everyone else by reading them, then making some and reporting back.

kyanite - 25-10-2004 at 12:04

Heads up that theres always the highly illegal route including HI acid...

HI + EtOH => EtI + H2O

Bromine will do the same, just search it up.
Like S.C. said, alot of chem textbooks have information like this. Thats where I looked...

BromicAcid - 25-10-2004 at 12:35

HI is also a powerful enough acid to reliably cleave ethers therefore, reaction of HI with Et2O will yield EtOH and EtI respectively. A mixture of phosphorus or possibly sulfur with nearly anhydrous EtOH and I2 with distillation would yield the desired product. (I've seen sulfur substituted for phosphorus in a number of reactions similar to this, but only with chlorine and occasionally bromine, iodine does not even have the power to make any compound with sulfur and therefore it may not have any effect at all.)

Maybe something involving insitu use of carbonyl iodide?

tom haggen - 25-10-2004 at 13:42

Hydriodic acid sounds way to dangerous for my liking. Also I was kind of looking for something along the lines of easily obtainable chemicals. By the way what the hell is so illegal about the HI + EtOH route? I haven't seen ICH3CH2 listed as a watched chemical?

[Edited on 25-10-2004 by tom haggen]

neutrino - 25-10-2004 at 14:02

HI is a list I meth precursor, i believe.

BromicAcid - 25-10-2004 at 14:09

Somewhat concentrated (60%) H2SO4 w/ NaI and EtOH might feasibly be heated and the EtI obtained by distillation (in the dark of course). However the catch of the matter is if your H2SO4 is too concentrated or if your temp goes too high or any of a number of other things happen, actually it probably happens all the time, it's just the rate. The I- will be oxidized by your H2SO4 and give you sulfur dioxide, water, elemental iodine and other undesirables.

kyanite - 25-10-2004 at 14:10

It's not the route thats illegal, but the hydriodic acid. It is a controlled precursor here in North America with heavy penalties. In the US, it's schedule I and thus a watched chemical by the DEA.

Looking over the Canadian Schedules, it looks like hydriodic acid isn't on any of the lists!

Can someone verify this?...

Edit: Bromic Acid, in the method you're describing, the HI is made kind of like in situ right? Question: Why do you think light will effect the reaction?

[Edited on 25-10-2004 by kyanite]

tom haggen - 25-10-2004 at 14:21

Well that cancels out the HI method. Unless I synthesize my own from H2S and Iodine which isn't going to happen cause H2S is not my buddy. So.... I could try using H2SO4 with NaI, and EtOH, and maybe distill with a vacuum. But why do I need to do this in the dark? Oh ya and hopefully that reaction would let off any H2S thats what would scare me.

[Edited on 25-10-2004 by tom haggen]

Marvin - 25-10-2004 at 14:42

One of the more recommended methods is phosphoric acid + sodium iodide to form the HI distilling over the ethyl iodide.

Another possibility is to make ethyl bromide and react with NaI to end up with ethyl iodide and sodium bromide. The problem here is that if you jump to chloride instead you end up making gasses which are harder to handle.

Alkyl iodides tend to decompose somewhat in light turning brown from free iodine.

tom haggen - 25-10-2004 at 15:49

So, inconclusion.

H3PO4 + NaI --> HI + H2 + NaPO4

HI + EtOH --> ICH3CH2 + H20

Whats a good way to remove the sodium phosphate? Distilling the HI I assume would do the trick. that seems like a pain in the ass...

[Edited on 26-10-2004 by tom haggen]

kyanite - 25-10-2004 at 16:35

Whaaaaa.....
Sorry for being hypocritical, but the first reaction maybe should be

H3PO4 + NaI -> HI + NaH2PO4 because the acid ionizes 3 times.
You might also be able to get more HI for your phosphoric by reacting 2 more moles of NaI like you can with carbonic acid. You'll probably need to heat it though.

tom haggen - 25-10-2004 at 19:00

Ok well theres already tones of information all over the place on this subject so I guess I wasn't looking hard enough. Sorry if I took up some unecessary space.

acx01b - 27-10-2004 at 23:44

H2SO4 75%-80% (drain cleaner, toilets cleaner in all drugstores all other the world just ask for the drain cleaners you will find it) and Toluene (same place) are refluxed for 30min

organic phase is separated while still hot (>60°C) to get toluenesulfonicacid monohydrate
with 99% H2So4 the yield is 99%....
with 80% i dont know

then this is mixed with any alcohol except phenol and phenol like alcohols...
to get toluenesulfonic ester

then this is reacted with NaX...
to get sodium toluenesulfonate and R-X
last step is separation and i really dont know: i remember that toluenesulfonate are used as detergents (solubility in organics and water)
by the way toluenesulphonic acid monohydrate can be purchased

[Edited on 28-10-2004 by acx01b]

tom haggen - 28-10-2004 at 10:51

What the hell are you going on about acx? What does toluene sulfonic acid monohydrate have to do with this subject? BTW, bromic acid, how possible do you think your method might be to actually working? The only thing that I see that would actually contaminate the ethyl iodide would be SO2 and H2O which seem like they could be removed from further distillation. I just realized something, the NaI that they sell at the grocery store is NaI(aqueous) and its already a very brown color. So it probably wouldn't work for this application.

[Edited on 28-10-2004 by tom haggen]

unionised - 30-10-2004 at 05:11

I presume he is going on about making the sulphonic acid to prodce a ethyl sulphonate to react with NaI to give ethyl iodide which seems to have quite a lot to do with the topic.

The brown colour of the NaI soln is probably due to free iodine and is easy enough to remove. titration with bisulphite seems like a good start.

Tacho - 30-10-2004 at 17:19

I have produced METHYL iodide a few times. The yield varies wildly with small changes in the reactant ratios and reaction conditions. The best yield I ever got was mixing 25gKI (0,15mol), 25ml 85% Phosphoric acid (0,37 mol), 25ml of methanol (0,62mol) and 0,1g of iodine crystals. This was brought to gentle reflux for 1 hour using ice-cold water in the condenser and magnetic stirring, then the reaction was left for 24 hours in a closed flask. Finally it was fractionally distilled (with magnetic stirring!) to give 6ml of MeI (13,7g, 0,096mol) a 64% yield.

You add some water to the distilate to obtain two phases, the lower is the very volatile iodide.

Would the yield be lower without the iodine crystals? I don’t know.
Does it have to be refluxed a bit and left for 24 hours? In my experience, yes. Be my guest to try something different.

For fractional distillation I used two condensers in series, one going “up” with water at 22°C (ambient), and the other going down, with ice-cold water. The receiving flask was kept in a ice-water mix.

I believe this would work for ethyl iodide using ethanol instead of methanol.

I was always VERY CAREFUL doing this reaction. Some sources say that methyl iodide is very nasty stuff (methylate your DNA) and the leftovers of the reaction are organophosphoric poisons. Ethyl iodide is probably no better…

This reaction is one of my sentimental favorites though. I figured it out by myself before I found it in the internet. Not being a chemist, I felt quite proud of it. My ignorant-but-sucessful reasoning was: If the acid(H2SO4)-alcohol(ethanol)-salt(NaBr) mix works for making ethyl bromide, some combination of them should work for methyl iodide. After many microscale experiments, there was the little brown drop in the bottom of the testube.

If you don’t like it brown, some dilute NaOH brings it to it’s natural lack of color.

Wish you luck. Be careful though…

kyanite - 31-10-2004 at 12:37

Apparently hydriodic acid is an amazing reducer, and becasue of that it can't be made from reducable acids like sulfuric and maybe even phosporic acid without the yeilds suffering from the HI oxidizing. Thats why a lot preparations use H2S.

Good ol' Rhodium:
http://www.rhodium.ws/chemistry/hydriodic.html
http://www.rhodium.ws/chemistry/hydriodic.argox.html

Maybe isolation, then use of the HI can give more accurate information on yeild.

alchemie - 12-11-2004 at 07:54

Here´s another route, using I2, Al foil, EtOH and H2SO4, also from Rhodium´s page:

http://www.rhodium.ws/chemistry/nitroalkane.html

I suppose the HI is produced in situ from hydrolysis of AlI3 formed from I2 and Al foil.

thunderfvck - 12-11-2004 at 13:48

I remember being interested in doing this. It's outlined in Vogel and he uses red phosphorous and iodine. Both can be obtained OTC, may arrouse some suspiscious but if you're only going to be using it for ethyl iodide there should be no problem.

I can write out the procedure for it later on tonight when I'm not so late for work.

trilobite - 12-11-2004 at 15:01

Maybe the latter of these two methods could be adapted for ethyl iodide as well. Why waste phosphorus for this?

https://www.the-hive.ws/forum/showflat.pl?Cat=&Number=30...

Darkfire - 16-11-2004 at 15:57

Anyone thought of h2so4 and NaI in DMSO then the ethanol or ether?

Eclectic - 17-11-2004 at 12:33

There are some preparations of alkyl bromides in a book I have, "Bromine", that use bromine dripped into a stirred mixture of the alcohol and sulfur. Maybe a similar route would work with I2?

garage chemist - 17-11-2004 at 12:58

The method from Tacho seems to be the best.
Phosphoric acid is non- oxidizing and therefore very useful for this reaction.

The standard method for ethyl iodide production is ethanol + red P + iodine, but only useful if you have enough red P.

Tom, why does it have to be ethyl iodide? Ethyl bromide works just as well for many reactions and is easily prepared from NaBr, H2SO4 and ethanol.
What are you going to use the ethyl iodide for?

tom haggen - 17-11-2004 at 17:06

I want to use it for a diethylamine synthesis. I'm haven't taken o-chem yet so I'm kind of getting ahead of myself. I'm pretty sure you can synthesize diethylamine from Ethyl Bromide but I don't know how easily NaBr is obtained. Anyway this question was more out of curiosity than anything.

neutrino - 17-11-2004 at 20:16

NaBr is readily available in pool stores at bromine tablets/powder.

tom haggen - 18-11-2004 at 00:31

Ok I'll keep that in mind. So lets see if i'm learning any thing in school. The reaction should go like this

NaBr + H2SO4 + CH3CH2OH--> CH3CH2Br + NaSO4 +H2O +H

I'm probably way off. But let me know if I'm somewheres in the ball park. By the way isn't there an explosion hazard in this type of reaction? Won't ethyl bromide react with the water produced in this reaction? This looks to be a reduction reaction but I don't know the charge of some of these poly atomic ions off the top of my head.
I assume that the ethyl molecule is reduced by the Bromine ion, and I'm pretty sure that the sodium ion is being oxidized. How am I looking?

[Edited on 18-11-2004 by tom haggen]

[Edited on 18-11-2004 by tom haggen]

[Edited on 18-11-2004 by tom haggen]

Tacho - 18-11-2004 at 02:44

Ethyl bromide is very easy to make in high yields with ethanol, NaBr and H2SO4 (maybe even sodium bissulfate). Almost nothing can go wrong. Just make sure you work with ice-cold water in the condensor and keep the receiving flask in an ice-water bath. A search in the net will give you some recipes. When I get home I'll check my notes to see if there was any catch in my experiment.

Ethylbromide is not flamable, HOWEVER, ethyl ether may be a by-product and that's a fire hazard.

Esplosivo - 18-11-2004 at 11:44

Quote:

Originally posted by tom haggen
NaBr + H2SO4 + CH3CH2OH--> CH3CH2Br + NaSO4 +H2O +H

The reaction is mainly composed of two parts, that is, the preparation of hydrogen bromide in situ as follows:
NaBr + H2SO4 --> HBr + NaHSO4

The senond part is the reaction of HBr with the alcohol:
CH3CH2OH + HBr --> CH3CH2Br + H20

Therefore the overall reaction would be:
NaBr + H2SO4 + CH3CH2OH--> CH3CH2Br + NaHSO4 + H2O

The reaction gives a good yield but if conditions are not rigorously controlled some of the HBr will be oxidised by the excess sulfuric acid giving bromine, and while refluxing at temp. >130 deg celcius some ether is formed as mentioned previously - therefore this should be done preferabbly outdoors, with the usual precausion, that is no flames, reduce sources of ignition, etc...

tom haggen - 18-11-2004 at 20:02

Before I attempt this synthesis I'm going to buy the proper glassware, and pull a vacuum to keep temperatures down.

Quote:

. Hazardous decomposition products: Toxic and corrosive gases and vapors (such as carbon monoxide) may be released in a fire involving ethyl bromide or when ethyl bromide reacts with water or steam. Ethyl bromide readily decomposes into hydrobromic acid, particularly in the presence of hot surfaces or open flames.

The part where it says that ethyl bromide reacts with water or steam is why I was concerned about an explosion, since water is one of the by products of this reaction. I got this information off of osha's website. Any thoughts on this information?

[Edited on 19-11-2004 by tom haggen]

neutrino - 18-11-2004 at 20:23

Warnings like that are usually over exaggerated. They are only meant to warn of what might happen in a fire, not necessarily what will happen in normal lab conditions. IIRC, JTBaker’s MSDS of MgSO4 warns of SO3 in a fire, an unlikely product of heating.

tom haggen - 18-11-2004 at 21:20

Better safe than sorry. But even better to know how to bend the rules.

"Post 300"
no title change though. I'm glad my 300th post was talking about safety, thats probably a good thing.

[Edited on 19-11-2004 by tom haggen]

garage chemist - 19-11-2004 at 02:39

Tom: Did you actually consider to use a VACUUM? I hope you didn't, because everyone knows what happens when vacuum is used with low- boiling substances like ethyl bromide.

tom haggen - 19-11-2004 at 08:58

Actually this may sound retarded but I didn't know there was a problem with vacuum distillation of low boiling solvents.
I guess it only makes sense that the boiling would be very violent in the reaction flask. Though why could you just vacuum distill these type of solvents at like 25C or something? What if I just vacuum distill the hydrobromic acid first and then simple distill this reaction

HBr + CH3CH2OH --> CH3CH2Br + H2O

[Edited on 19-11-2004 by tom haggen]

garage chemist - 19-11-2004 at 10:37

The yield of ethyl bromide will be zero when you use vacuum.

It's hard enough to condense it, even at atmospheric pressure.

The ethyl bromide synthesis would actually be easier if its boiling point was a bit higher.

If you attach an aspirator to a flask containing some ether (roughly the same Bp as ethyl bromide), the ether vaporizes and the flask gets VERY cold (about -20°C, maybe even lower). To condense the ether, you would need dry ice.
The same would happen if you attempted to distill ethyl bromide in vacuum.

Can you tell me more about the diethylamine synthesis involving ethyl bromide/iodide? I haven't heard of such a reaction yet

tom haggen - 19-11-2004 at 12:15

The preparation of amines from halogenoalkanes:

It only deals with amines where the functional group is not attached directly to a benzene ring. Aromatic amines such as phenylamine are usually made differently.

The halogenoalkane (Bromoethane) is heated with concentrated solution of ammonia in ethonal. The reaction is carried out in a sealed tube. You couldn't heat this mixture under reflux, because the ammonia would simply escape up the condenser as a gas.

The reaction happens in two stages, In the first stage, a salt is formed, ethylammonium bromide. This is just like ammonium bromide, except that one of the hydrogens in the ammonium ion is replaced by an ethyl group

CH3CH2Br + NH3 ------> CH3CH2NH3 + Br-

There is then the possibility of a reversible reaction between this salt and excess ammonia in the mixture.

CH3CH2NH3 + Br- + NH3 <--------> CH3CH2NH2 + NH4 + Br-

The ammonia removes a hydrogen ion from the ethylammonium ion to leave a primary amine, Ethylamine.

Then it goes into how the same reaction occurs to between bromoethane and ethylamine to make diethylamine. Diethylamine to triethylamine....and so on.
It says if to much ammonia is used in the first reaction that the reaction will never proceed past the primary amine stage.

I've also read in other literature that the fumes from this reaction are ungoddly, and it must be done in a rural area.

-Or if your a pro, you can use a fume hood

[Edited on 19-11-2004 by tom haggen]

kyanite - 19-11-2004 at 12:59

This is a very low yeilding route usually used for making those amines, about 50 % yeild I think. Also, alot of textbooks, etc do infact say that you have to use a huge excess if you want to have mostly primary amines. Not very efficient because the first amine is a neucleophilie itself

A much better method for primary amines would be from hydration of phthalimide.
http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-2-1-2.html
Also, you dont HAVE to use hydrazine like it shows there...

Edit: Whoops, sorry, I just read that you wanted secondary amines..

Maybe adding formic acid to the amine then reducing it would be better.

[Edited on 19-11-2004 by kyanite]

tom haggen - 19-11-2004 at 14:51

When you say a huge excess, are you talking about ethyl bromide or ammonia? If you mean ammonia than that totally condraticts the synthesis that I found on the net. I'm not worried so much about yields as I am availability of reagents. However, I'm always open to suggestions.

kyanite - 19-11-2004 at 18:39

Sorry, I seem to have a problem being clear :mad:

I do beleive its the amine thats in excess.
In my text book it says that "because all products can be formed, it is not considered a useful synthetic reaction. If the amine is very inexpensive, or if ammonia is used, a large excess can be used to favour monoalkylation. In this case RX is more likely to collide with those of the disired reactant, and less likely with the alkylated product."

CH3CH2CH2CH2Br + excess NH3 -- -OH -> CH3CH2CH2CH2BNH2 + Br-
n-butylamine 45 %

Because the yeild was so low, I guessed that even with large excess, a lot of the other products are still formed.

Hope that helps.

tom haggen - 19-11-2004 at 21:58

I thought that I had mentioned that ethyl, diethyl, and triethylamines were all formed. You can obtain the diethylamine by a simple distillation at about 56C . No need to get angry man.

[Edited on 20-11-2004 by tom haggen]

garage chemist - 20-11-2004 at 05:25

Isn't Triethylamine one of the worst smelling compounds? I heard it has an overpowering smell of rotten fish.

Mendeleev - 20-11-2004 at 08:49

As do just about all amines. The stench is persistent too. I kept a solution of methylamine hcl in IPA once, in a stoppered bottle (rubber stopper unfortunately), in a small shelf. After I took the bottle out, the closet still smelled for weeks. The procedure for ethyl bromide with sulfuric acid/NaBr works for all primary aliphatic alcohols, right? Concerning the synthesis of ethylamine, my logic maybe fatally flawed, but if formaldehyde + NH4Cl gives methylamine, perhaps acetaldehyde + NH4Cl gives ethylamine? Seems right, but I have a feeling the laws of chemistry are about to bite me on the ass.

[Edited on 20-11-2004 by Mendeleev]

kyanite - 20-11-2004 at 11:56

See what I mean? I feel so mis-interpreted... I was *not* mad at you man, I was pissed off at myself!
I should go and take English again.

I don't think I've smelled an amine before, but Ive also heard that they are like fish or urinals...

[Edited on 20-11-2004 by kyanite]

JohnWW - 20-11-2004 at 18:07

"Isn't Triethylamine one of the worst smelling compounds? I heard it has an overpowering smell of rotten fish."

I thought that phosphine, and derivatives like triethylphosphine, are the ones smelling like rotten fish - and highly poisonous, too.

[Edited on 21-11-2004 by JohnWW]

Esplosivo - 21-11-2004 at 01:54

Well actually amines do have a 'fishy' smell. The fact that it is described as having this smell is that amines are fomed by bacteria degrading rotting fish. A high percentage of the fish is protein, and as one well knows amino acids making up proteins have the amine group. During the breakdown of the amino acids by the bacteria amines are produced, and their smell is therefore associated with that of rotting fish.

tom haggen - 22-11-2004 at 11:56

Does ethyl bromide form explosive peroxides like ethyl ether does? I was just browsing through the ether synthesis on this site and noticed that they recommend a copper condenser to suppress the formation of peroxides. Also is a fractional column necessary for this synthesis, or do you think you could simply use a reaction flask, a condenser and a receiving flask? How useful do you think a claisen adapter would be for this synthesis?

[Edited on 22-11-2004 by tom haggen]

BromicAcid - 22-11-2004 at 14:07

No, ethyl bromide will not form peroxides, the peroxide formation in ethyl ether is dependent on the ether linkage.

Nosferatu - 25-11-2004 at 09:57

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).

Jones & Green, J. Chem. Soc. 2760 (1926)

As an alternative, you can in a RB flask dissolve Na- or KI in minimal amount of water and set this up distillation. Make a 50:50 (V/V) mixture of sulfuric acid/ethanol and add this mixture dropvise via an add. funnel to the KI/ water mix, adition should be slow. Heat the reaction mixture to 80 celsius. As EtI forms it will distill over into the ice-water in the reciever flask. Separate, wash with water, dry over MgSO4 and store in a dark bottle, yield should be around 80%.

[Edited on 25-11-2004 by Nosferatu]

Nosferatu - 29-11-2004 at 01:51

i've done this synthesis of ethyliodide (KI/H2SO4/EtOH) today. Do it on a balcony or in some other ventilated area. There is a compious amount of H2S evolved in the beginning of the reaction. Add the EtOH/H2SO4 mix _very_ slowly and stirr vigorously..

[Edited on 29-11-2004 by Nosferatu]

tom haggen - 30-11-2004 at 01:17

Won't sulfuric acid oxidize the Potassium Iodide in that reaction, making the reaction not go any further? I thought thats why you needed to use phosphoric acid to make Hydriodic acid?

ethyl iodide

paccman278 - 30-11-2004 at 01:53

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).

Oxydro - 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.

Mendeleev - 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?

S.C. Wack - 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.

Nosferatu - 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]

Protium - 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.

kyanite - 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...

chochu3 - 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

BromicAcid - 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 H2SO4. 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.

hinz - 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)

BromicAcid - 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.

chochu3 - 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]

a123x - 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.

BromicAcid - 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:

S2Br2 + 5Br2 ---> 12HBr + 2H2SO4

150 g Br2 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]

chochu3 - 19-12-2005 at 16:19

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

new-b - 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.

wa gwan - 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%.

Chris The Great - 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.

Magpie - 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]

alkyl iodide alkyl halides

Mush - 19-12-2012 at 11:47

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.

Magpie - 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

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

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.

Paddywhacker - 7-2-2013 at 19:23

Excellent report Magpie. How did you come by the HI?

Magpie - 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 57%HI

Attachment: phpgva8rh (116kB)
This file has been downloaded 1238 times
Purification of 57%HI

[Edited on 8-2-2013 by Magpie]

Endimion17 - 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.

Ethyl Iodide

SamuelLFraction - 24-11-2017 at 13:52

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?

Texium - 24-11-2017 at 14:08

Welcome to the forum! You've come to the right place to ask such questions.

Quote: Originally posted by SamuelLFraction

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.

Yes, you're right that the true balanced equation is 2Al + 3I₂ --> 2AlI₃
However, it doesn't matter which equation you use to calculate the amounts of reagents needed since both equations are still balanced in a 1:3 ratio of Al to I. 50 grams of I and 50 grams of I₂ contain the same number of I atoms, you see?

I don't have any other comments to make though as I haven't run this reaction myself.

AvBaeyer - 24-11-2017 at 14:32

Saturated primary (eg ethanol) and secondary alcohols (eg isopropanol) are unaffected by aluminum triodide. Allylic, benzylic, and tertiary alcohols are regioselectively converted to the corresponding iodide by the reagent. There is little expectation that ethyl iodide will be formed with aluminium triodide, at least in any useful amount.

Unfortunately, the chemplayer experiment is not useful. The product was not characterized (no boiling point). The identification of the product was based on "it looks like ethyl iodide." Maybe there was some iodide formed but the experiment as it stands is non-informative and unconvincing.

For some useful information please see:

P Sarmah, NC Barua, Tetrahedron 1989, 45, 3569. "Regioselective transformation of allylic, benzylic and tertiary alcohols into the corresponding iodides wrih aluminium triiodide: deoxygenation of vicinal diols"

AvB

SamuelLFraction - 24-11-2017 at 14:57

Zts - thank you for your reply, I feel a bit silly now realising this haha!
AvB - I'm glad to have an explanation for the relatively low yield I produced. It did have a boiling point of 73 degrees so at least all hope is not lost with the questionable reaction, shame such a small amount but I'm glad to learn. I really appreciate you including the material of regioselection, I now have something interesting to read and learn, thank you!

Chemi Pharma - 24-11-2017 at 23:00

Quote: Originally posted by AvBaeyer

Saturated primary (eg ethanol) and secondary alcohols (eg isopropanol) are unaffected by aluminum triodide. Allylic, benzylic, and tertiary alcohols are regioselectively converted to the corresponding iodide by the reagent. There is little expectation that ethyl iodide will be formed with aluminium triodide, at least in any useful amount.

Unfortunately, the chemplayer experiment is not useful. The product was not characterized (no boiling point). The identification of the product was based on "it looks like ethyl iodide." Maybe there was some iodide formed but the experiment as it stands is non-informative and unconvincing.
AvB

@AVBayer, you 're completely right when you say AlI3 doesn't affect primary and secondary alcohols.

Unfortunately many "ChemPlayer" videos at you tube are not trustworthy about the results and yields he claims. In my oppinion, in some experiments he clear did work up mistakes that affected the yields and others experiments clearly suffers from conceptual matters, although his experiments appears to be very instructive and interesting for the amateur chemistry.

I think he needs urgently start to use a LC-MS to make a qualitative and quantitative sample analyze, like "Nurd Rage" does.

@SamuelLFraction, if you intend to sinthesize alkyl iodides easily and with good yields, I suggest you the use of an alkaline iodide, phosphoric acid and the alcohol (or ether, or alkene), at reflux, following the method I posted in the study attached below:

Attachment: Iodides from clevage of ethers or iodination of alcohols and alkenes with KI + H3PO4.pdf (328kB)
This file has been downloaded 582 times

JJay - 25-11-2017 at 01:43

I have occasionally questioned Chem Player's results (such as when I made p-aminophenol and experienced much more rapid decomposition than was apparent in the video), but for the most part I think Chem Player demonstrates good science, and some of Chem Player's stuff is really cutting edge. There is a tendency among YouTube chemists in general to avoid extensive characterization or to show only easy characterizations that support the chemist's conclusion without ruling out impurities or other products. (For example, do you see any YouTube sulfuric acid distillation video that shows an actual titration?) But I don't think that Chem Player is worse than NurdRage or Nile Red or any other YouTube chemist in this regard, and I don't think it's caused by untrustworthiness so much as cognitive bias. There are very good reasons for peer review, after all....

That said, do any of the big box stores carry 85% phosphoric acid?

[Edited on 25-11-2017 by JJay]

SamuelLFraction - 25-11-2017 at 02:37

@Chemi Pharma, I did originally think it would be a lot easier to avoid the Aluminium route and go with phosporic acid, however, could not think of where to accumulate it.
Luckily I have had an epiphany and realised it's easily found. (Also thank you for the material!)
@JJay phosphoric acid at 85% is available as Ph adjustment for cultivation, if you wanted to know. Perhaps I better not say product name as maybe this is not allowed?

JJay - 25-11-2017 at 05:00

It is permitted, but I don't blast that sort of information on every thread either. It is up to you based on how you feel that information will be used.

Bert - 25-11-2017 at 06:16

Quote: Originally posted by SamuelLFraction

Perhaps I better not say product name as maybe this is not allowed?

Read the FAQ.

Board Topics- Mad Science FAQ

Quote:

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If an existing thread covers the topic you're about to post about, post in that thread instead of a new one. It makes it easier for members to keep up with topics of interest. You may need to search before posting if you're unsure whether or not a thread already exists. The search engine is not very powerful and may miss relevant threads, but please make at least a cursory effort.

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S.C. Wack - 25-11-2017 at 07:44

The process is not unlike P and the alcohol is not supposed to be anhydrous, so the water makes HI.

SamuelLFraction - 25-11-2017 at 08:36

@S.C Wack, this may well explain why low yield then, since anhydrous ethanol was used. I will try the reaction again, this time with more elemental iodine and dilute anhydrous ethanol until about 85%. I will report back with some figures if all goes to plan

S.C. Wack - 25-11-2017 at 09:06

The reaction is probably unlike P as now I look it up and see that calls for anhydrous alcohol. 85% you say have you read the original literature calling for 84% ethanol.

SamuelLFraction - 25-11-2017 at 09:49

@ S.C Wack, oh well, I suppose in chemistry not everything works out to how you expect. Yes I have read the literature regarding 84% alcohol. So far, the mixture is a dark grey to black with a little undissolved aluminium. There was a sudden exothermic reaction which occurred roughly 20 minutes after mixture of Iodine, Al and EtOH and began to froth after some initial heating. This died down after 10 minutes as the literature suggested and then continued to reflux for further 1.5 hours. Smells very ether-like but this of course is no certain indication of EtI. The additional water in the mixture has changed the mixture grey rather than the deep red that anhydrous ethanol made it go before. Distillate is not very miscible with water. I wonder, does the additional water create more Aluminium oxide/hydroxide? Following this reaction's results, I will attempt the Phosphorous acid KI Ethanol reaction.

[Edited on 25-11-2017 by SamuelLFraction]

S.C. Wack - 26-11-2017 at 07:22

The procedure above with the aluminum is sort of bullshit and should not be performed, because the scale was changed by the dickface who originally posted this version that has been copied many times. The reactions may be rather too exothermic to proceed as written and deep unpleasantness beyond low yield could result. Weights and volumes were originally 20x less, and a final weight was not given, but the words otherwise remain the same. You want to test this on the smaller scale.

https://www.sciencemadness.org/whisper/viewthread.php?tid=65...
Both of these are short, thus are free to save as .gif from the RSC, until they change that

SamuelLFraction - 28-11-2017 at 03:37

Finishing the procedure calling for 84% alcohol has resulted in a cloudy distillate with a boiling point of 50-60. Bare in mind this distillate was then further distilled once more and resulted the same end product. Immiscible with water, denser than water also. Aroma of sulphur/pickles, not very pleasant. The procedure was followed exactly, however, resulted in everything but ethyl iodide. The product was immediately appropriately discarded. Perhaps adding the H2SO4 and ethanol to the residue remaining from first distillation to then further distill had ruined this reaction entirely, however, i am not going to attempt this reaction again to find out. Despite the 84% alcohol, iodine and aluminium method resulting in entire disappointment, I have yet to test the suspected ethyl iodide produced (Silver Nitrate test) from anhydrous ethanol, of which is clear and denser than water. Ether aroma present with this substance.

SLF

[Edited on 28-11-2017 by SamuelLFraction]

Melgar - 28-11-2017 at 07:18

The low yield in the reaction of iodine with aluminum and ethanol is because aluminum likes to keep some of the iodine for itself, and forms combination hydroxide-iodide species, much like iron forms iron oxide-hydroxide species in the presence of water and oxygen. If you insist on doing something more fun than H3PO4 + ethanol + iodide, you can always react iodine with elemental silicon to form SiI4, which will do just about anything to swap those iodine atoms for oxygen atoms. It's 95% iodine by mass, and can be treated as though it were anhydrous HI for most practical purposes.

SamuelLFraction - 28-11-2017 at 13:12

Thank you Melgar, Silicon would most certainly be an interesting reagent to use. I'll be adding elemental silicon to my shopping list. I don't mind the idea of using H3PO4, but as you quite rightly connotated, something more fun is... well, more fun

Following on from reaction involving anhydrous ethanol with aluminium and iodine, the distillate was tested with a silver nitrate solution of ethanol, water and NaOH and produced a yellow precipitate. At least the product is actually formed in this synthesis, so not entirely a waste of my time.

JJay - 28-11-2017 at 13:56

Nicoderm's procedure using HCl looks quite easy if you have reagent grade hydrochloric acid on hand. I wonder if potassium iodide would work as well as sodium iodide.

chemplayer... - 12-12-2017 at 06:09

If someone wants to buy us a LC-MS for Christmas then hey, just tell us where to pick it up!

Seriously though, we don't try to produce scientific papers. We do it to learn and to have fun.

arkoma - 12-12-2017 at 20:15

Quote: Originally posted by chemplayer...

We do it to learn and to have fun.

AMEN!

Corrosive Joeseph - 13-12-2017 at 03:36

Quote: Originally posted by JJay

Nicoderm's procedure using HCl looks quite easy if you have reagent grade hydrochloric acid on hand. I wonder if potassium iodide would work as well as sodium iodide.

Did I miss something........?

@JJay - Can you link me please

/CJ

[EDIT] - Pardon my laziness. 7 and a half seconds of UTFSEing yielded this -

"Iodomethane Via the HCl Method" - https://www.sciencemadness.org/whisper/viewthread.php?tid=23...

Excellent.........

[Edited on 13-12-2017 by Corrosive Joeseph]

JJay - 18-7-2018 at 11:25

I didn't have any luck getting it to work when I tried it a few months ago (and this was for methyl iodide, not ethyl iodide), but I'm not exactly sure how strong my HCl was. I produced it by gassing azeotropic HCl with HCl gas until it appeared that very little was being dissolved. The amount of iodine recovered after the experiment was far less than the amount of iodide that went into the experiment, so I think that a reaction happened, but I didn't see an organic layer. I'm reluctant to believe that the methyl iodide simply escaped. My potassium iodide was granular, like table salt; it could be that I should have used more finely powdered potassium iodide, but that doesn't explain what happened to the iodine.

Surrealist - 28-7-2018 at 10:02

Dissolve sodium iodide in acetone (399g/L, or 2.66 molar at 25 C) source: Pubchem.
Then bubble ethyl chloride through this to get the Finkelstein reaction, which is just an sn2 reaction driven by the insolubility of sodium chloride in acetone. Organic chemistry portal has a good article on it: https://www.organic-chemistry.org/namedreactions/finkelstein...
As for making the ethyl chloride itself, a reactive distillation from ethanol and azeotropic hydrochloric acid should work well since ethyl chloride boils at 12.3 C source: Pubchem. Be very careful and very patient with this setup and take very good care not to let any ignition source anywhere near ethyl chloride. The ethyl chloride may not react immediately with the sodium iodide. Gentle heat to the acetone will accelerate the reaction, but can be a safety risk. Use ethyl iodide in situ if possible. But that assumes acetone is a good solvent for your reaction.
My only experience in this reaction is the production alkyl iodides in situ to catalytically increase reaction rates, but I have never attempted to purify one.
Please attempt this on a small scale and tinker before thinking of making a bunch of ethyl iodide. And read very carefully into the storage of ethyl iodide.

I hope this helps.

[Edited on 28-7-2018 by Surrealist]

[Edited on 28-7-2018 by Surrealist]

Chemi Pharma - 28-7-2018 at 13:11

@Surrealist, your idea on how to make ethyl chloride will not work the way you are thinking. Ethyl chloride needs 99,8% ethyl alcohol, gaseous and dry HCl and the most important: anhydrous zinc chloride. Without the last the acid won't react with the alcohol. The enviroment needs to be with the minimum amount of water possible and the work up is to bubble HCl into a solution of anhydrous ZnCl2 in absolute alcohol, cleaning the ethyl chloride gas first with a water wash to remove the HCl vapors and after with a H2SO4 wash bottle, to remove the umidity.

There's a synthesis of that in the book "Preparation of organic compounds, E. de. Barry Barnett, 71, 1912", published by Prepchem.com on the NET.

"To the round bottom flask fitted with properly cooled reflux condenser and gas inlet 100 g of ethyl alcohol and 50 g of anhydrous zinc chloride are placed. The top of reflux condenser is connected to two washing bottles. The first has water and the second concentrated sulfuric acid. Finally apparatus is connected to a cooled flask for condensing the reaction product. A dry stream of hydrogen chloride gas is passed through the boiling mixture. The vapors of ethyl chloride is washed with water, concentrated sulfuric and condensed in a flask which is cooled with freezing mixture of ice and salt. As ethyl chloride boils at 12° C it must be kept in sealed glass tubes. The yield is almost quantitative."

I think it's too much work to get ethyl iodide as the final target and I'd rather to use an easier method with KI + H3PO4 and ethyl alcohol I discribed as an attachment, lines above, in my last manifestation on this thread.

[Edited on 28-7-2018 by Chemi Pharma]

Surrealist - 28-7-2018 at 19:42

I should have remembered the Lucas reagent/test. Thank you for pointing out my error Pharma. Maybe a setup like this could be useful in the case of secondary alkyl iodides, but not primary since they won't react with the Lucas reagent (anhydrous zinc chloride and very concentrated hydrogen chloride), and not tertiary since those don't undergo the Finkelstein reaction. But I'm getting ahead of myself since I don't have any projects that need alkyl iodides currently.

S.C. Wack - 29-7-2018 at 06:28

Heating the Lucas converts primary alcohols so yes you can make the chloride that way, but if you're using ethanol, KI, and HCl for ethyl iodide maybe the experiment can be more direct:
https://www.sciencemadness.org/whisper/viewthread.php?tid=12...

The 1857 date BTW is a few years after the author made large amounts of nitroglycerin, leaving imperfect directions for doing so.

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