[lutesium]

There are numereous ways to Prepare Diethyl Ether the most known method is H2SO4 dehydration + condensation of EtOH (Ethyl Alcohol) and after that using an Ethyl iodide + Sodium Ethoxide which is expensive hard building properties and dangerous (because of the high volatility and carcinogenic properties of Ethyl Iodide) but this reagent can be easily prepared by replacing the above Ethyl Iodide simply by the Sulphuric Acid Ester of EtOH which is easily prepared by adding Sulphuric Acid to EtOH!!!

[djay]

1)EtOH + NaOH -> EtONa + H2O
2)EtONa + X-CH2CH3 -> EtOEt + NaX
X= F, Cl, Br

IF X-CH2CH3 is expensive (which is usually really cheap)
you can make it by mixing CH2CH2 (ethene) with HX.
However, I'm not quite sure if heating is necessary

another way to make ether (probably more expensive and less effective is)

1)CH3(CO)CH2CH3 + MCPBA (or any peracid R(CO)OOH) -> CH3(CO)OCH2CH3
2)CH3(CO)OCH2CH3 + C activated or Zn(Hg) -> CH3CH2OCH2CH3

The second step needs to be heated to reduce the ester.
For the first step I think it's dangerous to heat since it's a peroxyd.

[lutesium]

Ok but MeBr is so volatile that leave its storage using it is nearly impossible!!! Leave MeCl or MeF cuz they are gases!!! And obtaining HI and Iodinating ethene is a very hard process cuz its a gas (ethene)!!! So if you have sodium or potassium in your laboratory (these are a must for many university labs) I think that the easiest way is as I've suggested!!!

[sjb]

...And obtaining HI and Iodinating ethene is a very hard process cuz its a gas (ethene)!!! So if you have sodium or potassium in your laboratory (these are a must for many university labs) I think that the easiest way is as I've suggested!!!

To be fair, if a lab has access to sodium and or potassium metal, I'd be very surprised if it did not also have easy acess to diethyl ether as well - probably cheaper too when all's said and done and labour etc is accounted for.

S

[djay]

Unless it's for an evaluation that he has to do that reaction. (ether is very cheap and very used especially in hospitals) The second method can be used if you dont want any gas.

[lutesium]

Of course you are right to say that if a lab has access to Na or K metal it has also access to Et2O but for the price issue things change from country to country!!! In the place where I got my Private Laboratory I used to buy Et2O as a reagent but it was priced as $15 for a litre ~600 (8 moles)grms but Na metal was $55 for a kg (~50moles) (surprising price) so its not much of a difference for money for a litre but for 20 lts of Et2O it makes a difference like $100 a good money that'll be transferred back to new research!!!

[xc]

Hi.

Just one thing, Lutesium. Which is the difference between using the first method you mentioned (EtOH + H₂SO₄) and the new method you propose (EtOH + H₂SO₄)?

Do you isolate this sulfuric acid esther intermediate?!

In my opinion, the first method,consisting just in adding sulfuric acid to ethanol is simple enough!

[lutesium]

No xc there's a great difference!!! In the first rxn the rxn is a condensation rxn but the rxn that I proposed is the addition of an alcoholate to an ester of sulphuric acid!!!

And I favor this method over the others because there's no need to use distillation apparatus!!! Just add the chemicals reflux them and then seperate them using a seperatory funnel!!!

No we don't isolate anything...


Lutesium...

[Butyllithium]

Hi,lutesium,

Just one doubt, how to avoid Elimination Reaction(due to sodium ethoxide as strong base)?How to purify your compounds? I think distillation will be efficient. Ah.

Butyllithium

[lutesium]

If you use an excess of ethanol while making the sulpuric acid ester I don't think an elimination reaction will occur.

No there's no need to distill as the formed salts will be left at the water phase!!! Just reflux and then seperate. So easy...


Lutesium...

[xc]

Lutesium, maybe the problem is I'm not understanding you correctly. So correct me:

- This is how I understand you propose to do it:

1^st: Mix ethanol and sulphuric acid to make the sulfuric esther. So: 2 EtOH + H₂SO₄ -> (EtO)₂SO₂ + 2 H₂O
You obtain the sulfuric esther of ethanol and water, I guess you use ethanol as the solvent, so your limiting reagent is the sulfuric acid.

2^nd: Add into this mixture two equivalents of sodium ethoxide (that you prepared previously from sodium and ethanol), so: (EtO)₂SO₂ + 2 NaOEt -> 2 Et₂O + Na₂SO₄

- This is the traditional method:

Mix ethanol with sulfuric acid and warm it up. So: 2 EtOH + H₂SO₄ -> 2 Et₂O + H₂SO₄ + H₂O
The solvent is ethanol and the sulfuric acid is catalitic.

Now, if how I described your method is correct, I explain my objections to it.

1. You need 0.5 moles of sulfuric acid per mole of ether prepared (this is crazy!!). In the other method, you just need catalytic amount of sulfuric acid.

2. You also need 1 mole of sodium ethoxide (or 1 mole of sodium) per mole of ether prepared. But well, you say you can afford this, so I have nothing to say here, it's your problem.

3. You produce 0.5 moles of sodium sulfate per mole of ether prepared. This is a lot of waste, you have to do something with it now. (Please don't throw it into the river) In the other method you don't produce any waste

4. Your method has two steps (three if you have to prepare the sodium ethoxide), what means that you have to make batches (unless you have a continuous flow system). So you have to make one reaction, the other one, and then purify. Also I don't think you use a continous flow system for the extractions, so the amount of ether you can prepare is certainly limited.
In the other method, you just have to mix ethanol and sulfuric acid, warm up to 40ºC, place a condenser and collect the pure ether as it's formed (it's evaporated at this temperature, so you isolate it at the same time the reaction is working). If you need more ether, you just have to add ethanol to the reaction mixture, because the sulfuric acid is not consumed. You can have it working the whole day, if you want. Or more.

5. The more important question I'm trying to explain: In the traditional method, when you mix ethanol and sulfuric acid, they react to form ether. Why in your method when you react ethanol and sulfuric acid you form diethyl sulfate??

I'm sure, the only problem is that I didn't understand what you were trying to explain.

[adam]

I completely agree with you xc,
I'm sorry, in my last post, in the 5th line, instead of (EtO)2SO2H, it should say (EtO)2SO2

[Borek]

I'm sorry, in my last post, in the 5th line, instead of (EtO)2SO2H, it should say (EtO)2SO2

You can edit your posts, click on Modify button right to the subject.

[xc]

Thanks Borek

[djay]

are you sure there wont be an elimination?