Oxidation of Alcohols to Aldehydes, Ketones and Acids
Chemistry · Alcohols, Phenols And Ethers · NEET
When you oxidise an alcohol, you remove hydrogen. A primary (1°) alcohol becomes an aldehyde, and with a strong oxidiser it goes further to a carboxylic acid. A secondary (2°) alcohol becomes a ketone (it cannot go further). A tertiary (3°) alcohol has no H on the C-OH carbon, so it does not oxidise easily. Memory hook: "1° = two steps (aldehyde then acid), 2° = one step (ketone), 3° = no step."
Primary alcohols oxidise in two steps (aldehyde with a mild agent like PCC, then acid with a strong agent). Secondary alcohols give a ketone and stop. Tertiary alcohols have no H on the C-OH carbon, so they do not oxidise easily.
Your doubts, answered
Does PCC oxidise a primary alcohol to an aldehyde or to a carboxylic acid?
PCC (pyridinium chlorochromate) is a MILD oxidiser. It stops a primary alcohol at the aldehyde stage. It does not add water, so it cannot push the aldehyde on to the acid. This is a favourite NEET trick: PCC = aldehyde only. Use PCC (or CrO3 with acetic anhydride) when you want to STOP at the aldehyde.
Which oxidising agents take a primary alcohol all the way to a carboxylic acid?
Strong oxidisers do this: acidified KMnO4 (KMnO4/H+), acidified K2Cr2O7 (K2Cr2O7/H2SO4), and CrO3-H2SO4 (Jones reagent). They are strong enough to oxidise the aldehyde further to the acid. So R-CH2OH -> R-CHO -> R-COOH. Remember: strong = acid, mild = aldehyde.
What product do you get when a secondary alcohol is oxidised?
A secondary alcohol always gives a KETONE. The carbon bearing -OH has only one H, so one oxidation step forms C=O and the reaction stops there. A ketone cannot be oxidised further easily because there is no H on the carbonyl carbon. Example: propan-2-ol (CH3)2CHOH -> acetone (CH3)2C=O.
Why does a tertiary alcohol not get oxidised?
Oxidation of an alcohol needs an H atom on the carbon that carries the -OH group. In a tertiary alcohol that carbon has THREE carbon groups and NO hydrogen. So there is no H to remove and no C=O can form. Tertiary alcohols resist oxidation. Under very harsh acidic conditions they only dehydrate and then the small pieces get oxidised, which is not the exam answer.
What is the difference between oxidation and dehydrogenation of an alcohol?
Both remove hydrogen, but the pathway differs. Oxidation uses an oxidising agent (KMnO4, K2Cr2O7, PCC) in solution and can reach the acid for a 1° alcohol. Dehydrogenation passes the alcohol vapour over hot copper at 573 K and removes only H2, giving an aldehyde (from 1°) or a ketone (from 2°) but never the acid. Both give the same aldehyde/ketone, but only oxidation can go to the acid.
⚠️ The NEET trap ✗ PCC oxidises benzyl alcohol (C6H5CH2OH) to benzoic acid, just like KMnO4 and K2Cr2O7 do. ✓ PCC is a mild oxidiser and stops at the aldehyde, so it gives benzaldehyde (C6H5CHO), NOT benzoic acid. Only strong oxidisers (KMnO4/H+, K2Cr2O7/H2SO4, CrO3-H2SO4) give benzoic acid. 🧠 See PCC = STOP at aldehyde. See KMnO4 or K2Cr2O7 = GO to acid.
Real NEET questions
NEET 2019 (Odisha)
The reaction that does NOT give benzoic acid as the major product is:
A · C6H5CH2OH with K2Cr2O7
B · C6H5COCH3 with (i) NaOCl (ii) H3O+
C · C6H5CH2OH with PCC (pyridinium chlorochromate) ✓
D · C6H5CH2OH with KMnO4/H+
Solution: PCC is a mild oxidising agent, so it oxidises the primary (benzylic) alcohol only up to the aldehyde stage, giving benzaldehyde C6H5CHO and stopping there. The strong oxidisers K2Cr2O7 (A) and KMnO4/H+ (D) take benzyl alcohol all the way to benzoic acid. Acetophenone C6H5COCH3 (B) undergoes the haloform reaction with NaOCl and, after acidification, also gives benzoic acid. So the exception is (C).
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 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: Only strong oxidisers take a 1° alcohol all the way to the acid. CrO3-H2SO4 (Jones), acidified K2Cr2O7, and KMnO4 (alkaline then acid work-up) are all strong systems that give carboxylic acids. Cu at 573 K only dehydrogenates to an aldehyde/ketone, and CrO3 with acetic anhydride is a MILD reagent that stops at the aldehyde. So only (A), (B) and (C).
NEET 2024
Identify the correct reagents to convert an alkene to an aldehyde via the anti-Markovnikov primary alcohol (e.g. C6H5CH2CH=CH2 -> C6H5CH2CH2CHO):
A · (i) BH3, (ii) H2O2/OH-, (iii) PCC ✓
B · (i) BH3, (ii) H2O2/OH-, (iii) alk. KMnO4, (iv) H3O+
C · (i) H2O/H+, (ii) PCC
D · H2O/H+
Solution: Hydroboration-oxidation (BH3 then H2O2/OH-) adds water across the double bond in an anti-Markovnikov way, giving the primary alcohol R-CH2CH2OH. To STOP at the aldehyde you must use the mild oxidiser PCC. Option (B) uses KMnO4, a strong oxidiser, which would over-oxidise to the carboxylic acid, so it is wrong. Hence (A).
Solved Alcohols, Phenols And Ethers NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Not easily. A ketone has no hydrogen on the carbonyl carbon, so mild and normal oxidisers leave it alone. Only very strong oxidisers under harsh conditions can break the C-C bonds to give a mixture of smaller carboxylic acids, which is not asked in NEET single-step questions.
What is the correct order of oxidation for a primary alcohol?
Primary alcohol -> aldehyde -> carboxylic acid. R-CH2OH first loses hydrogen to give R-CHO (aldehyde), and a strong oxidiser then adds oxygen to give R-COOH (acid). A mild oxidiser like PCC stops at R-CHO.
Which alcohol gives a ketone on oxidation?
A secondary (2°) alcohol gives a ketone. The C-OH carbon has one H, so one oxidation step forms C=O and the reaction stops. For example, propan-2-ol gives acetone.
Why is this topic important for NEET?
NEET almost every year asks a reagent-matching question: which reagent stops at the aldehyde and which goes to the acid. Knowing PCC/CrO3-acetic anhydride = mild (aldehyde) and KMnO4/K2Cr2O7/CrO3-H2SO4 = strong (acid) lets you solve these in seconds and avoid the classic trap.
Is copper at 573 K the same as oxidation?
No. Copper at 573 K is dehydrogenation. It only removes H2 and gives an aldehyde (from 1°) or a ketone (from 2°). It can never give a carboxylic acid. True oxidation with a strong agent can reach the acid for a primary alcohol.
