Which product is formed when ethanol is heated with concentrated H2SO4 at 443 K?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Ethene
Heating an alcohol with a protic acid such as concentrated H2SO4 causes dehydration: a molecule of water is removed. Which product forms depends on the temperature. At 443 K, one molecule of ethanol loses one molecule of water and a C=C double bond forms:
\( \displaystyle \ce{CH3CH2OH} \xrightarrow[\text{443 K}]{\text{conc. }\ce{H2SO4}} \ce{CH2=CH2 + H2O} \)
Why the other options are incorrect
Option (1) Ethane — Incorrect. Loss of water from ethanol leaves a double bond; ethane is a saturated compound and would need hydrogen to be added, which does not happen here.
Option (2) Diethyl ether — Incorrect. Ethoxyethane (diethyl ether) is the main product at the lower temperature of 413 K, where water is removed from two molecules of ethanol.
Option (4) Ethyne — Incorrect. Ethyne has a triple bond; dehydration of ethanol removes only one molecule of water and gives the double bond of ethene.
NCERT (2026-27 edition):
1. By dehydration of alcohols – “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.”