Question:
Identify the product when propan-2-ol heated with Cu at \(573\) K.
Identify the product when propan-2-ol heated with Cu at \(573\) K.
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Hot copper dehydrogenates alcohols. A secondary alcohol loses hydrogen to form a ketone.
Updated On: Nov 12, 2026
- propanol
- Propene
- Propanone
- propanoic acid
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
When alcohol vapour is passed over copper at 573 K, dehydrogenation takes place. Primary alcohols give aldehydes and secondary alcohols give ketones.
Step 2: Reaction
Propan-2-ol is a secondary alcohol: \(\text{CH}_{3}\text{CH(OH)CH}_{3}\xrightarrow{\text{Cu, 573 K}}\text{CH}_{3}\text{COCH}_{3}+\text{H}_{2}\).
Step 3: Compare the options
Propene would need dehydration with acid at 443 K or above. Propanol and propanoic acid do not result from this reaction. The product is propanone.
Final Answer:
The product is propanone (acetone). \[ \boxed{\text{(C) }\text{Propanone}} \]
When alcohol vapour is passed over copper at 573 K, dehydrogenation takes place. Primary alcohols give aldehydes and secondary alcohols give ketones.
Step 2: Reaction
Propan-2-ol is a secondary alcohol: \(\text{CH}_{3}\text{CH(OH)CH}_{3}\xrightarrow{\text{Cu, 573 K}}\text{CH}_{3}\text{COCH}_{3}+\text{H}_{2}\).
Step 3: Compare the options
Propene would need dehydration with acid at 443 K or above. Propanol and propanoic acid do not result from this reaction. The product is propanone.
Final Answer:
The product is propanone (acetone). \[ \boxed{\text{(C) }\text{Propanone}} \]
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