Ethanol is treated with concentrated sulphuric acid. The product formed is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(C_2H_4\)
Concentrated sulphuric acid is a dehydrating agent. When ethanol is heated with excess concentrated \( \ce{H2SO4} \) at 443 K, the alcohol is protonated, loses water to give a carbocation and then a proton, and ethene is formed:
\( \ce{CH3CH2OH} \) → \( \ce{CH2=CH2} \) + \( \ce{H2O} \) (conc. \( \ce{H2SO4} \), 443 K)
(At 413 K, with excess ethanol, ethoxyethane is the main product.)
Why the other options are incorrect
Option (1) \(C_2H_6\) — Incorrect. Ethane would need a reduction (addition of hydrogen); \( \ce{H2SO4} \) removes water, it does not reduce.
Option (3) \(C_2H_2\) — Incorrect. Ethyne would need loss of two molecules of water/hydrogen; dehydration of ethanol removes only one \( \ce{H2O} \).
Option (4) \(CH_3CH_2.OSO_3OH\) — Incorrect. The formula as written is not a real compound; ethyl hydrogen sulphate (\( \ce{CH3CH2OSO3H} \)) forms only at low temperature as an intermediate, not as the product of dehydration.
NCERT (2026-27 edition):
Preparation of Ethers – “… Alcohols undergo dehydration in the presence of protic acids (H2SO4, H3PO4). The formation of the reaction product, alkene or ether depends on the reaction conditions. For example, ethanol is dehydrated to ethene in the presence of sulphuric acid at 443 K. At 413 K, ethoxyethane is the main product.”