Oxidation Reduction Carbonyl

8 MCQs9-step worked example
Source: NCERT BiomoleculesPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The trap that costs marks in oxidation-reduction of carbonyls is confusing mild and strong oxidising agents for primary alcohols. A question names a reagent — PCC, Jones reagent, KMnO₄ — and the correct product depends entirely on whether that reagent stops at the aldehyde or pushes through to the carboxylic acid.

The reagent-selectivity hierarchy (NCERT Class 12 Chemistry, Chapter 12, page 12):

Mild oxidising agents — PCC (pyridinium chlorochromate), PDC (pyridinium dichromate), Swern oxidation, DMP (Dess-Martin periodinane) — oxidise a primary alcohol to an aldehyde and stop. They work in anhydrous, non-aqueous conditions that prevent over-oxidation.

Strong oxidising agents — KMnO₄, K₂Cr₂O₇ (acidified), CrO₃/H₂SO₄ (Jones reagent) — oxidise a primary alcohol all the way to the carboxylic acid. These reagents operate in aqueous acidic media; the aldehyde intermediate is further oxidised before it can be isolated.

Secondary alcohols are oxidised to ketones by both mild and strong reagents. The ketone resists further oxidation because breaking the C–C bond would be required.

Tertiary alcohols are not oxidised by conventional reagents (no α-hydrogen on the carbon bearing –OH).

Reduction of carbonyls: Wolff-Kishner reduction (hydrazine/KOH, strong base, high temperature) and Clemmensen reduction (Zn-Hg/conc. HCl, strongly acidic) both convert the C=O of an aldehyde or ketone to –CH₂–. The key distinction is reaction conditions: Wolff-Kishner is basic, Clemmensen is acidic. Choose the reduction method compatible with other functional groups in the substrate — base-sensitive groups survive Clemmensen; acid-sensitive groups survive Wolff-Kishner.

Watch-out: When a stem says "oxidation of 1-butanol with PCC," the answer is butanal, not butanoic acid. Reagent identity determines the product — not the alcohol class alone.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which of the following reagents oxidises a primary alcohol to an aldehyde without further oxidation to the carboxylic acid?

MCQ 2Easy RecallPractice

Clemmensen reduction converts a carbonyl compound (aldehyde or ketone) to a hydrocarbon using:

MCQ 3Easy RecallPractice

Wolff-Kishner reduction is carried out under which conditions?

MCQ 4Direct ApplicationPractice

1-Propanol is treated with Jones reagent (CrO₃/H₂SO₄). The major product is:

MCQ 5Direct ApplicationPractice

2-Butanol is oxidised with acidified K₂Cr₂O₇. The product is:

MCQ 6Direct ApplicationPractice

A substrate contains both a ketone group and an acid-sensitive functional group (e.g., an acetal). To reduce the ketone to a methylene (–CH₂–) without cleaving the acetal, which method is appropriate?

MCQ 7Concept TrapPractice

2-Methyl-2-propanol (tert-butyl alcohol) is treated with acidified KMnO₄. What is observed?

MCQ 8CalculationPractice

An unknown primary alcohol X is treated with PCC to give compound Y. Compound Y is then treated separately with (i) Wolff-Kishner reduction and (ii) Clemmensen reduction. Which statement is correct?

Worked Example

  1. 1

    Given

    1-Pentanol is treated first with PCC in CH₂Cl₂, and the product is then treated with acidified KMnO₄.

  2. 2

    Required

    Identify the final product after both steps.

  3. 3

    Concept

    Oxidation of primary alcohols depends on reagent strength. PCC is a mild oxidant (stops at aldehyde). KMnO₄ (acidified) is a strong oxidant (oxidises aldehydes further to carboxylic acids). NCERT Class 12 Chemistry Chapter 12, page 12.

  4. 4

    Formula/Rule

    - PCC + primary alcohol → aldehyde (mild, anhydrous, stops here) - KMnO₄ (acidified) + aldehyde → carboxylic acid (strong, aqueous, continues)

  5. 5

    Substitution

    - Step A: 1-Pentanol + PCC → Pentanal - Step B: Pentanal + acidified KMnO₄ → Pentanoic acid

  6. 6

    Calculation

    No numerical calculation required. This is a reagent-identification and product-prediction problem. The reasoning is sequential: identify the intermediate (pentanal) from the mild oxidant, then apply the strong oxidant to that intermediate.

  7. 7

    Final answer

    The final product is **pentanoic acid** (CH₃CH₂CH₂CH₂COOH).

  8. 8

    Common trap

    The high-frequency error is selecting pentanal as the final product — forgetting that step B uses acidified KMnO₄ (strong oxidant), which pushes the aldehyde intermediate to the carboxylic acid. This is trap trap: oxidation reagent selectivity: reagent strength determines where oxidation stops.

  9. 9

    Similar NEET-style question

    "1-Butanol is heated with K₂Cr₂O₇/H₂SO₄. The organic product formed is: (A) Butanal (B) Butanone (C) Butanoic acid (D) 1-Butene." Answer: (C) Butanoic acid — K₂Cr₂O₇/H₂SO₄ is a strong oxidant, so the primary alcohol is fully oxidised. ---

Before solving, remember these

Key Fact

Wolff-Kishner reduction

C=O + NH₂NH₂ + base/heat → CH₂. Reduces aldehydes/ketones to alkanes via hydrazone. Useful when acid-sensitive groups present.

-- NCERT, p. 12
Key Fact

Clemmensen reduction

C=O + Zn(Hg) + concentrated HCl → CH₂. Reduces aldehydes/ketones to alkanes. Useful when base-sensitive groups present.

-- NCERT, p. 12

Formulas

pKa of carboxylic acid

Stronger acid than phenol due to more effective resonance over two equivalent oxygens. EWG substituents (Cl, NO2) increase acidity.

SymbolQuantitySI Unit
pKa-log Ka-

Valid when

  • Aqueous solution
  • α-substituent effects strongest

pKa of phenol vs aliphatic alcohol

Phenols ~10⁶× more acidic than aliphatic alcohols due to resonance stabilisation of phenoxide ion. Electron-withdrawing substituents lower pKa further.

SymbolQuantitySI Unit
pKa-log Ka-

Valid when

  • Aqueous solution
  • Substituent effects shift values

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Oxidation product depends on reagent strength

Category: Organic Reaction Conditions

1° alcohol: PCC/PDC → aldehyde (stops). KMnO4/K2Cr2O7 in acidic → carboxylic acid (continues). 2° alcohol: any oxidiser → ketone. 3° alcohol: not oxidised by ordinary reagents.

When it triggers

Question gives 1° alcohol oxidation with specified reagent.

How to avoid

PCC, PDC, Swern, DMP: mild → stop at aldehyde. KMnO4, K2Cr2O7, CrO3, jones: strong → carboxylic acid. Reagent choice matches desired product.

Past Year Questions

20 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2024Revised key

Given below are two statements: Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II : Aniline cannot be prepared through Gabriel synthesis . In the light of the above statements, choose the correct answer from the options given below:

1Both statement I and Statement II are true
2Both Statement I and Statement II are false
3Statement I is correct but Statement II is false
4Statement I is incorrect but Statement II is true
NTA Answer: Option 1(revised_final)
NEET 2024Revised key

Fehling’s solution ‘A’ is

1aqueous copper sulphate
2alkaline copper sulphate
3alkaline solution of sodium potassium tartrate (Rochelle’s salt)
4aqueous sodium citrate
NTA Answer: Option 1(revised_final)
NEET 2022

Given below are two statements: Statement I: In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with conc. HCl + ZnCl , known as Lucas Reagent. 2 Statement II: Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas Reagent. In the light of the above statements, choose the most appropriate answer from the options given below:

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 4(final)
NEET 2021

Match List-I with List-II List-I List-II CO, HCI (a) (i) Hell-Volhard-Zelinsky Anhyd. AlCl/ Cu3Cl reaction O R — C — CH + (b) (ii) Gattermann-Koch 3 NaOX reaction R — CH — OH (c) (iii) Haloform reaction 2 +R′ COOH Conc. HSO 2 4 (d) R — CH COOH (iv) Esterification 2 ( II) X /Red P 2 (ii) HO 2 Choose the correct answer from the options given below.

1(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
2(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
3(a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
4(a) - (i), (b) - (iv), (c) - (iii), (d) - (ii)
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 12 Chemistry Chapter 12, p.12

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