When you pass alcohol vapour over hot copper at 573 K, the copper pulls out hydrogen (this is called dehydrogenation). A primary alcohol loses H and becomes an aldehyde. A secondary alcohol becomes a ketone. A tertiary alcohol has no H on its carbon, so it cannot lose H, and instead it just dehydrates to an alkene. Memory hook: "Copper counts the H on the OH carbon" - 2 spare H gives aldehyde, 1 spare H gives ketone, 0 spare H gives alkene.
Copper at 573 K removes H2: primary alcohols give aldehydes, secondary give ketones, and tertiary (no H on the OH carbon) instead lose water to give alkenes.
Your doubts, answered
What actually happens when alcohol vapour passes over copper at 573 K?
The hot copper acts as a catalyst and removes two hydrogen atoms from the alcohol. This is called catalytic dehydrogenation. One H comes from the -OH group and one H comes from the carbon that carries the -OH (called the carbinol carbon). These two H atoms leave together as H2 gas. This only works if that carbinol carbon actually has an H on it.
Why does a primary alcohol give an aldehyde but a secondary alcohol gives a ketone?
Look at the carbon holding the -OH. In a primary alcohol that carbon has two H atoms, so after losing one H it becomes -CHO, an aldehyde. In a secondary alcohol that carbon has only one H, so after losing it you get a C=O with two carbons attached, which is a ketone. The number of H on the OH carbon decides the product.
Why does a tertiary alcohol NOT give an aldehyde or ketone over copper?
In a tertiary alcohol the carbon carrying -OH has NO hydrogen on it (it is joined to three carbons). Dehydrogenation needs an H on that carbon, so it is impossible. Instead the copper causes dehydration: an H from a neighbour carbon and the -OH leave as water, giving an alkene. So tertiary alcohol at 573 K over Cu gives an alkene, not a carbonyl.
Is Cu at 573 K the same as oxidation? Does it give a carboxylic acid?
No. This is dehydrogenation, not oxidation. No oxygen is added. Copper only removes H2. So it stops at the aldehyde or ketone stage and can never make a carboxylic acid. To get a carboxylic acid you need a strong oxidiser like acidic KMnO4 or K2Cr2O7/H2SO4 or CrO3-H2SO4. This exact point was tested in NEET 2023.
Dehydrogenation vs dehydration - how do I tell them apart quickly?
Dehydrogenation removes H2 (two hydrogen atoms) and needs an H on the OH carbon, giving a carbonyl (aldehyde or ketone). Dehydration removes H2O (water) and gives an alkene. Over Cu at 573 K, primary and secondary alcohols dehydrogenate, but tertiary alcohols dehydrate because they have no H on the OH carbon.
⚠️ The NEET trap ✗ Assuming every alcohol over Cu at 573 K gives an aldehyde, or that a secondary alcohol gives an aldehyde. ✓ Primary gives an ALDEHYDE, secondary gives a KETONE, tertiary gives an ALKENE. A secondary alcohol like propan-2-ol gives acetone (a ketone), never an aldehyde. 🧠 Count spare H on the OH carbon: 2 to aldehyde, 1 to ketone, 0 to alkene.
Real NEET questions
NEET 2019 (Odisha)
When the vapours of a secondary alcohol are passed over heated copper at 573 K, the product formed is:
A · a carboxylic acid
B · an aldehyde
C · a ketone ✓
D · an alkene
Solution: Copper at 573 K causes catalytic dehydrogenation. A secondary alcohol loses one H from -OH and one H from the carbinol carbon (which carries a single H). This removal of H2 gives a C=O with two carbon groups attached, i.e. a ketone. Example: (CH3)2CHOH gives (CH3)2C=O (acetone) + H2. A primary alcohol would give an aldehyde and a tertiary alcohol would dehydrate to an alkene, so the answer is a ketone.
NEET 2023 (Phase 2)
Reagents which can be used to convert alcohols to carboxylic acids are: (A) CrO3-H2SO4 (B) K2Cr2O7 + H2SO4 (C) KMnO4 + KOH/H3O+ (D) Cu, 573 K (E) CrO3, (CH3CO)2O. Choose the most appropriate answer.
A · (A), (B) and (C) only ✓
B · (A), (B) and (E) only
C · (B), (C) and (D) only
D · (B), (D) and (E) only
Solution: A carboxylic acid needs strong oxidation. CrO3-H2SO4 (Jones), acidified K2Cr2O7, and KMnO4 (alkaline then acid work-up) are strong oxidisers that take a primary alcohol all the way to a carboxylic acid. Cu at 573 K only removes H2 (dehydrogenation) and stops at the aldehyde, so it CANNOT give a carboxylic acid. CrO3 with acetic anhydride is a mild reagent that also stops at the aldehyde. Hence only (A), (B) and (C).
Solved Alcohols, Phenols And Ethers NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the product when ethanol vapour passes over copper at 573 K?
Ethanol is a primary alcohol, so it dehydrogenates to acetaldehyde (CH3CHO) plus H2 gas.
What does propan-2-ol give over Cu at 573 K?
Propan-2-ol is a secondary alcohol, so it gives acetone (propanone), a ketone, plus H2.
What does tert-butyl alcohol give over Cu at 573 K?
Tertiary butyl alcohol has no H on the OH carbon, so it cannot dehydrogenate. It dehydrates instead and gives 2-methylpropene (an alkene) plus water.
Is passing alcohol over copper oxidation or dehydrogenation?
It is dehydrogenation, not oxidation. No oxygen is added; only H2 is removed. That is why it never reaches a carboxylic acid.
Why is 573 K and copper important for NEET?
NEET often asks the exact product for primary, secondary and tertiary alcohols under these conditions. Remembering aldehyde/ketone/alkene for 1/2/3 alcohols directly answers those one-mark questions.
