Ethanol in the presence of concentrated sulfuric acid (conc. H2SO4) at 443 K produces
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → CH2=CH2
Concentrated sulphuric acid dehydrates ethanol. The acid protonates the –OH group, water leaves, and the product depends on the temperature:
At 443 K, a proton is lost from the neighbouring carbon and ethene is formed (intramolecular dehydration).
At 413 K, with excess ethanol, a second ethanol molecule attacks and ethoxyethane (diethyl ether) is formed.
\( \displaystyle \ce{C2H5OH} \xrightarrow{\text{conc. }\ce{H2SO4}\text{, 443 K}} \ce{CH2=CH2 + H2O} \)
\( \displaystyle \ce{2C2H5OH} \xrightarrow{\text{conc. }\ce{H2SO4}\text{, 413 K}} \ce{C2H5OC2H5 + H2O} \)
So at 443 K ethanol gives ethene, CH2=CH2.
Why the other options are incorrect
Option (2) CH3COOH — Incorrect. Ethanoic acid needs a strong oxidising agent such as acidified \( \ce{K2Cr2O7} \) or \( \ce{KMnO4} \).
Option (3) CH3CHO — Incorrect. Ethanal is obtained by mild oxidation (PCC) or by passing ethanol vapour over hot copper at 573 K.
Option (4) CH3CH2OCH2CH3 — Incorrect. Diethyl ether forms at the lower temperature, 413 K.
NCERT (2026-27 edition):
Preparation of Ethers – …
