Preparing Aldehydes and Ketones from Alcohols: Oxidation and Cu/573K Dehydrogenation
Chemistry · Aldehydes, Ketones And Carboxylic Acid · NEET
To make an aldehyde or ketone from an alcohol, you remove two hydrogen atoms. A primary alcohol (1 degree) gives an aldehyde and a secondary alcohol (2 degree) gives a ketone. This works two ways: mild oxidation (PCC or CrO3/anhydride) that STOPS at the aldehyde, or passing the vapours over hot copper at 573 K (dehydrogenation). Memory hook: "1 makes al, 2 makes ket, 3 quits (dehydrates)."
Passing alcohol vapour over hot copper at 573 K removes H2: a 1 degree alcohol gives an aldehyde, a 2 degree gives a ketone, and a 3 degree alcohol dehydrates to an alkene. Mild oxidants (PCC) stop at the aldehyde; strong oxidants go all the way to the carboxylic acid.
Your doubts, answered
Does hot copper at 573 K give an aldehyde or a ketone?
It depends on the alcohol, not the copper. A PRIMARY alcohol (like CH3CH2OH) gives an ALDEHYDE (CH3CHO). A SECONDARY alcohol (like (CH3)2CHOH) gives a KETONE (CH3COCH3). Copper only removes two H atoms (one from the O-H and one from the carbon holding the OH); which product forms is decided by how many H atoms sit on that carbon.
What does a tertiary alcohol give with Cu at 573 K?
An ALKENE, not a carbonyl. A tertiary (3 degree) alcohol has NO hydrogen on the carbon carrying the OH, so it cannot lose H2 to form a C=O. Instead it undergoes dehydration (loses water) to give an alkene. This is a very common NEET trap: option 'aldehyde' or 'ketone' for a tertiary alcohol is always wrong.
What is the difference between oxidation and dehydrogenation of an alcohol?
Both remove two hydrogen atoms and give the same carbonyl product, so NCERT treats them as the same net change. OXIDATION uses an oxidising agent (PCC, K2Cr2O7, KMnO4) that pulls the H atoms out as water. DEHYDROGENATION passes the alcohol vapour over hot copper at 573 K, and the two H atoms leave as H2 gas. Both convert 1 degree alcohol to aldehyde and 2 degree alcohol to ketone.
Why does PCC stop at the aldehyde but KMnO4 goes all the way to the acid?
PCC (pyridinium chlorochromate) and CrO3 in acetic anhydride are MILD oxidants. They do not have water around to over-oxidise the aldehyde, so the reaction stops at the aldehyde. STRONG oxidants like acidified K2Cr2O7, KMnO4/H+, or CrO3-H2SO4 have water present, which lets the aldehyde get oxidised further to a carboxylic acid. So for a NEET question asking to make an ALDEHYDE from a 1 degree alcohol, always pick PCC (or Cu/573K), never strong oxidants.
Can Cu at 573 K convert a primary alcohol to a carboxylic acid?
No. Cu at 573 K only DEHYDROGENATES a primary alcohol to an aldehyde and stops there. It cannot go on to a carboxylic acid. This is exactly the trap in NEET 2023: 'which reagents convert alcohol to carboxylic acid?' Cu/573K is listed as a wrong option because it stops at the aldehyde stage.
⚠️ The NEET trap ✗ Cu at 573 K converts a primary alcohol all the way to a carboxylic acid, and a tertiary alcohol to a ketone. ✓ Cu/573K only removes H2. A 1 degree alcohol stops at the ALDEHYDE (not the acid), a 2 degree alcohol gives a KETONE, and a 3 degree alcohol (no H on the carbinol carbon) DEHYDRATES to an alkene. 🧠 Copper is lazy: it only pulls off H2 once. It never reaches the acid, and it gives up on tertiary alcohols by kicking out water instead.
Real NEET questions
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: Heated copper at 573 K causes catalytic DEHYDROGENATION: it removes one H from the O-H and one H from the carbinol carbon (the carbon bearing OH). In a secondary alcohol such as (CH3)2CHOH, that carbon carries two alkyl groups and one H, so loss of H2 gives a KETONE, e.g. (CH3)2CHOH gives CH3COCH3 + H2. Primary alcohols would give aldehydes; tertiary alcohols dehydrate to alkenes. Answer: (C) a ketone.
2017
For X (C2H6O): X over Cu at 573 K gives A; A with Tollens' reagent gives a silver mirror; A with OH-/heat (aldol) gives Y; A with H2N-NH-CO-NH2 gives Z. Identify A, X, Y, Z.
A · A = methoxymethane, X = ethanoic acid, Y = acetate ion, Z = hydrazine
B · A = methoxymethane, X = ethanol, Y = ethanoic acid, Z = semicarbazide
C · A = ethanal, X = ethanol, Y = but-2-enal, Z = semicarbazone ✓
D · A = ethanol, X = acetaldehyde, Y = butanone, Z = hydrazone
Solution: C2H6O passed over Cu at 573 K is dehydrogenation of a PRIMARY alcohol. So X = ethanol (CH3CH2OH) and it loses H2 to give A = ethanal (CH3CHO). The positive Tollens' test (silver mirror) confirms A is an aldehyde. Base-catalysed aldol of ethanal followed by heating gives CH3-CH=CH-CHO (but-2-enal), so Y = but-2-enal. Ethanal with semicarbazide (H2N-NH-CO-NH2) gives the semicarbazone, so Z = semicarbazone. Answer: (C).
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 correct set:
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: To reach a carboxylic acid you need a STRONG oxidant that takes a primary alcohol past the aldehyde. CrO3-H2SO4 (Jones), acidified K2Cr2O7, and KMnO4/KOH with acid work-up all do this, so (A), (B), (C) are correct. Cu at 573 K (D) only dehydrogenates to the ALDEHYDE and stops, and CrO3 with acetic anhydride (E) is a MILD reagent that also stops at the aldehyde. So (A), (B) and (C) only. Answer: (A).
Solved Aldehydes, Ketones And Carboxylic Acid NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Which alcohols give aldehydes and which give ketones?
Primary (1 degree) alcohols give aldehydes. Secondary (2 degree) alcohols give ketones. Tertiary (3 degree) alcohols give neither; they dehydrate to alkenes because their carbinol carbon has no hydrogen to lose.
What temperature is used for the copper dehydrogenation?
573 K (about 300 degrees C). The alcohol is passed as vapour over heated copper metal, which acts as the catalyst and releases the two hydrogens as H2 gas.
Which reagent should I pick to stop at the aldehyde in NEET?
Use a MILD oxidant: PCC (pyridinium chlorochromate) or CrO3 in acetic anhydride, or Cu at 573 K. These do not over-oxidise. Strong oxidants (KMnO4/H+, acidified K2Cr2O7, CrO3-H2SO4) push a primary alcohol all the way to the carboxylic acid.
Why is this concept important for NEET?
NEET repeatedly asks reaction sequences (like the 2017 X-C2H6O question) where the first step is Cu/573K, and reagent-matching questions where Cu/573K is a trap for 'gives carboxylic acid'. Knowing 1 degree gives aldehyde, 2 degree gives ketone, 3 degree gives alkene lets you eliminate wrong options fast.
