The vapours of butan-2-ol are passed over heated copper at 573 K. Which of the following correctly gives the type of reaction and the main organic product?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Dehydrogenation; butanone
Heated copper at 573 K removes a molecule of hydrogen from a primary or secondary alcohol: the H of the O–H group and the H on the carbon carrying –OH. Butan-2-ol is a secondary alcohol, so a C=O group forms on C-2 and the product is a ketone:
\( \displaystyle \ce{CH3-CH(OH)-CH2-CH3} \xrightarrow[\text{573 K}]{\ce{Cu}} \ce{CH3-CO-CH2-CH3 + H2} \)
Why the other options are incorrect
Option (2) Dehydrogenation; butanal — Incorrect. An aldehyde comes from a primary alcohol (butan-1-ol would give butanal). The –OH of butan-2-ol is on C-2, so the carbonyl group forms on C-2.
Option (3) Dehydration; but-2-ene — Incorrect. Over heated copper only tertiary alcohols are dehydrated to alkenes. But-2-ene would come from butan-2-ol on heating with an acid such as concentrated H2SO4.
Option (4) Oxidation with C–C bond cleavage; propanoic acid — Incorrect. Breaking of C–C bonds needs strong oxidising agents such as KMnO4 at elevated temperatures; copper at 573 K only removes hydrogen and keeps the carbon chain.
NCERT (2026-27 edition):
7.4.4 Chemical Reactions – “When the vapours of a primary or a secondary alcohol are passed over heated copper at 573 K, dehydrogenation takes place and an aldehyde or a ketone is formed while tertiary alcohols undergo dehydration.”