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Author: Subject: Ethyl Iodide
neutrino
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[*] posted on 17-11-2004 at 20:16


NaBr is readily available in pool stores at bromine tablets/powder.
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tom haggen
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[*] posted on 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]




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[*] posted on 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.
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[*] posted on 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 + H<sub>2</sub>SO<sub>4</sub> --> HBr + NaHSO<sub>4</sub>

The senond part is the reaction of HBr with the alcohol:
CH<sub>3</sub>CH<sub>2</sub>OH + HBr --> CH<sub>3</sub>CH<sub>2</sub>Br + H<sub>2</sub>0

Therefore the overall reaction would be:
NaBr + H<sub>2</sub>SO4 + CH<sub>3</sub>CH<sub>2</sub>OH--> CH<sub>3</sub>CH<sub>2</sub>Br + NaHSO<sub>4</sub> + H<sub>2</sub>O

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...




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tom haggen
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[*] posted on 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]




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[*] posted on 18-11-2004 at 20:23


Warnings like that are usually over exaggerated. They are only meant to warn of what <i>might</i> happen in a fire, not necessarily what will happen in normal lab conditions. IIRC, JTBaker’s MSDS of MgSO<sub>4</sub> warns of SO<sub>3</sub> in a fire, an unlikely product of heating.
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[*] posted on 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]




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[*] posted on 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.
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[*] posted on 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]




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[*] posted on 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
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[*] posted on 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]




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[*] posted on 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 :P

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]




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[*] posted on 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.



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[*] posted on 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.




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[*] posted on 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]




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[*] posted on 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.
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[*] posted on 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]




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[*] posted on 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]




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[*] posted on 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]
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[*] posted on 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.



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[*] posted on 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]




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[*] posted on 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.



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[*] posted on 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]




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[*] posted on 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]




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[*] posted on 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?



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