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

Part A – Preparation of diethyl ether.

Step 1: Diethyl ether is prepared industrially by the acid dehydration (intermolecular dehydration) of ethanol. Ethanol is heated with excess concentrated sulphuric acid at 413 K (about 140°C).
\[ 2\,C_2H_5OH \xrightarrow[413\,K]{\text{conc. } H_2SO_4} C_2H_5\text{-}O\text{-}C_2H_5 + H_2O \]
The acid first forms ethyl hydrogen sulphate, which is then attacked by a second ethanol molecule to give the ether. (Alternatively, Williamson's synthesis: C₂H₅Br + C₂H₅ONa → C₂H₅OC₂H₅ + NaBr.)
Note: temperature control matters; at a higher temperature (443 K) ethanol instead gives ethene.

Part B – Mechanism of diethyl ether with HI.

Step 2 (protonation): The oxygen of the ether has lone pairs and is basic. It is protonated by the strong acid HI to give a positively charged oxonium (protonated ether) ion, which makes one C–O bond a good leaving group.
\[ C_2H_5\text{-}O\text{-}C_2H_5 + HI \rightarrow C_2H_5\overset{+}{\underset{H}{O}}\text{-}C_2H_5 + I^- \]

Step 3 (nucleophilic attack, S_N2): Since both alkyl groups are primary ethyl groups, the iodide ion I^– attacks the less hindered carbon from the back side, the C–O bond breaks, and ethyl iodide plus ethanol are formed.
\[ I^- + C_2H_5\overset{+}{\underset{H}{O}}\text{-}C_2H_5 \rightarrow C_2H_5I + C_2H_5OH \]

Step 4 (with excess HI): The ethanol produced reacts further with more HI to give a second molecule of ethyl iodide.
\[ C_2H_5OH + HI \rightarrow C_2H_5I + H_2O \]
Overall, with excess HI: C₂H₅OC₂H₅ + 2HI → 2C₂H₅I + H₂O.