Question

Write a method of preparation of diethyl ether and write the mechanism of the reaction of diethyl ether with HI.

Hint:

Prepare diethyl ether by heating ethanol with conc. H₂SO₄ at 413 K (or Williamson synthesis). With HI: protonate the ether oxygen, then I^– does S_N2 on the less hindered carbon giving C₂H₅I + C₂H₅OH.

Answer 1

Preparation. The simplest laboratory/industrial route to diethyl ether is Williamson's ether synthesis, which reliably gives a pure ether. Sodium ethoxide (a strong nucleophile) attacks bromoethane by S_N2, displacing bromide:
\(C_2H_5ONa + C_2H_5Br \rightarrow C_2H_5\text{-}O\text{-}C_2H_5 + NaBr\).
On a large scale, ethanol is instead dehydrated with conc. H₂SO₄ at 413 K: \(2C_2H_5OH \rightarrow C_2H_5OC_2H_5 + H_2O\).

Reaction with HI, step by step.
Step 1: Ethers are weak bases through the oxygen lone pair. The first thing HI does is donate a proton to that oxygen, forming a dialkyloxonium ion \([C_2H_5\text{-}O(H)^+\text{-}C_2H_5]\) and freeing an iodide ion. Protonation converts –OR into the far better leaving group –OH…R.
Step 2: The freed iodide, a strong nucleophile, now looks for the easiest carbon to attack. Both carbons are primary, so a back-side S_N2 displacement occurs: I^– bonds to one ethyl carbon while that carbon's bond to oxygen breaks, expelling ethanol. Products of this stage: ethyl iodide + ethanol.
Step 3: If HI is in excess and the mixture is hot, the ethanol is itself converted to a second ethyl iodide (\(C_2H_5OH + HI \rightarrow C_2H_5I + H_2O\)), so the net result is two moles of ethyl iodide and one of water. (With a bulky/tertiary group the leaving alcohol would instead be the more substituted one, because iodide attacks the primary carbon.)