Ethanol is heated with sulphuric acid at 413 K, and ethoxyethane is the main product. How is the ether formed in this reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → By nucleophilic attack of an ethanol molecule on a protonated ethanol molecule (SN2)
In acid, ethanol is first protonated.
A second ethanol molecule, using a lone pair on its oxygen, attacks the carbon of the protonated alcohol and pushes out water in a single step. Two molecules take part in this step, so it is a nucleophilic bimolecular substitution (SN2).
Loss of a proton then gives the ether, and the acid is set free again.
Why the other options are incorrect
Option (2) By formation of an ethyl carbocation, which is then attacked by an ethanol molecule (SN1) — Incorrect. The SN1 path is followed when the alcohol is secondary or tertiary. Ethanol is a primary alcohol, and its protonated form is attacked directly by another alcohol molecule.
Option (3) By attack of an ethoxide ion on an ethyl halide (SN2) — Incorrect. This is Williamson synthesis, in which an alkyl halide reacts with sodium alkoxide. No alkyl halide or alkoxide is present when ethanol is heated with sulphuric acid.
Option (4) By loss of a proton from an ethyl carbocation (elimination) — Incorrect. Loss of a proton from the ethyl carbocation gives ethene, not an ether. This is the dehydration of ethanol at the higher temperature of 443 K.
NCERT (2026-27 edition):
7.6.1 Preparation of Ethers – “…For example, ethanol is dehydrated to ethene in the presence of sulphuric acid at 443 K. At 413 K, ethoxyethane is the main product.”
“The formation of ether is a nucleophilic bimolecular reaction (SN2) involving the attack of alcohol molecule on a protonated alcohol, as indicated below:”
\( \displaystyle \text{(i)}\ \ \ce{CH3CH2OH + H+ -> CH3CH2OH2+} \)
\( \displaystyle \text{(ii)}\ \ \ce{CH3CH2OH + CH3CH2OH2+ -> (CH3CH2)2OH+ + H2O} \)
\( \displaystyle \text{(iii)}\ \ \ce{(CH3CH2)2OH+ -> CH3CH2OCH2CH3 + H+} \)