I was trying to determine the reaction mechanism behind the production of diethyl ether. Looking at the ethyl alcohol + sulphuric acid method, I have come to the following conclusions for the reaction steps:

The ethanol nucleophilic oxygen gets protonated by the addition of sulphuric acid. This is more favourable because of the high density of electrons surrounding the oxygen. The bond broken between the carbon and oxygen is also needed to form the bond with hydrogen.

Water is then released and the ethyl alcohol conjugate base is formed (H3C-CH2+). The remaining ethyl alcohol molecules can either react with the HSO4- or the ethyl conjugate... Since HSO4- is more stable than the ethyl conjugate, ethyl alcohol reacts with the ethyl conjugate. Carbon is more electronegative than hydrogen and is an electrophile. It is attracted to the oxygen from the ethyl alcohol and replaces the H, thus forming diethylether, H+ and H2O where the sulphuric acid essentially acts as a catalyst!

Am I correct in presenting the mechanism this way? If the sulphuric acid still remains in the solution, why does it not protonate the diethyl ether? Is it because the stability of the diethyl ether conjugate base is unstable?



2 Answers 2


Your mechanism would create a primary alkyl cation, namely the ethyl cation ($\ce{CH3CH2+}$). Primary carbocations are extremely unstable and only generated if there are stabilising effects present such as in the benzylic or allylic case — and these are then often not considered primary cations but rather allylic or benzylic ones.

Rather, the protonation of ethanol ($\ce{CH3-CH2-OH}$) creates an oxonium ion as you suggested in the first step:

$$\ce{CH3-CH2-OH + H2SO4 -> CH3-CH2-OH2+ + HSO4-}$$

The $\ce{CH2}$ carbon is then attacked from behind in a nucleophilic $\ce{S_{N}2}$ reaction liberating water in the same step. It is important to note that the attack of another ethanol molecule and the liberation of water happen synchronously; there is no first or second.

$$\ce{CH3-CH2-OH2+ + CH3-CH2-OH -> CH3-CH2-OH+-CH2-CH3 + H2O}$$

Finally, the oxygen which is now part of a protonated ether loses the proton to give the final product diethyl ether.

In the case of a nucleophilic substitution on a substrate like tert-butanol, or even better, tert-butyl bromide, one would consider the leaving group to be liberated first and a carbocation to be created; however, tertiary carbocations are orders of magnitude more stable than primary ones.

  • $\begingroup$ in case of sn1 reaction will unsymmetric ether be formed? $\endgroup$
    – Pole_Star
    Aug 25, 2018 at 14:47
  • $\begingroup$ @starunique2016 In the case of $\mathrm{S_N1}$ reactions you are well in the area where sterics play an enormous role and you may not get any ether formation at all. But if you do and you have two different alcohols then you will likely get a distribution symmetrical and unsymmetrical ethers. If you have a case like tert-butanol and isopropanol, the unsymmetrical ethers will probably be the main product by a significant margin. $\endgroup$
    – Jan
    Oct 16, 2018 at 13:58

You got it all wrong. First of all the reaction written by you concisely is

$$\ce{diethyl~ ether + H2SO4 -> CH3-CH2-OH + CH3-CH2-HSO4}$$

Your first step is correct that oxygen gets protonated due to acid. Then you said that water is then released and the ethyl alcohol conjugate base is formed $\ce{H3C-CH2+}$. This is the point where you got it wrong. The reaction goes under SN2 mechanism( I am not going into the details of it) and there is no formation of intermediates as you gave it as $\ce{H3C-CH2+}$ So no carbocation inter mediates are formed only transition stage is formed. Your next question was for what if $\ce{H2SO4}$ is in excess. Here it goes like this

  1. Ethanol formed is again protonated by $\ce{H2SO4}$ by the same reason you stated.

  2. Hence now there are $\ce{CH3CH2OH2+}$( where +ve charge is on oxygen) and $\ce{HSO4-}$ anions.

  3. These $\ce{HSO4-}$ again attack on $\ce{CH3CH2OH2+}$ by SN2 mechanism.
  4. Now you will get salt of the 3rd step and water as by product.
  • 2
    $\begingroup$ I don't see how the HSO4- would attack the protonated ethyl alcohol. If that were true, you wouldn't have diethyl ether. $\endgroup$
    – jakerz
    Aug 22, 2015 at 3:03
  • $\begingroup$ According to wiki, the unprotonated ethyl attacks the protonated ethyl. From the looks of it, even though the protonated ethyl alcohols are positive, the oxygen still forces the C to be the electrophile. Thus the SN2 attack is of unprotonated ethyls on the carbon back with an H2O and H+ leaving group. $\endgroup$
    – jakerz
    Aug 22, 2015 at 3:16
  • $\begingroup$ I am not getting what is your question. Please send me that link and I think that I will help you out $\endgroup$ Aug 22, 2015 at 9:39
  • 5
    $\begingroup$ @NukulParmar, your "You got it all wrong" comment isn't exactly constructive to the original poster of the question. The reason people post their questions on stack exchange is because they want answers/clarification, not so someone can tell them how wrong they are... this entire community works because people feel able to do this. If you take issue with the original question, you're free to down vote it, just as people are free to down vote on your 'answer'. $\endgroup$
    – NotEvans.
    Dec 22, 2015 at 12:33

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