Which of the following compounds will give butanone on oxidation with alkaline $KMnO_4$ solution? (i) Butan-1-ol (ii) Butan-2-ol (iii) Both of these (iv) None of these

Which of the following compounds will give butanone on oxidation with alkaline KMnO₄ solution? (i) Butan-1-ol

NCERT Exemplar · Q11

KMnO_4

(i) Butan-1-ol

(ii) Butan-2-ol

(iii) Both of these

(iv) None of these

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Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.

Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.

That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.

The Precise Statement

Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the -OH group, increasing the number of C-O bonds (or decreasing C-H bonds). The outcome depends on the class of the alcohol:

Alcohol Class	Structure	Product after oxidation	Reagent example
Primary (1°)	R–CH₂–OH	Aldehyde (R–CHO) then Carboxylic acid (R–COOH)	PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid)
Secondary (2°)	R–CHOH–R'	Ketone (R–CO–R')	K₂Cr₂O₇/H⁺, CrO₃, etc.
Tertiary (3°)	R₃C–OH	No reaction (under normal conditions)	—

Watch out

A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.

Why Does Tertiary Alcohol Not Oxidize?

Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.

A Real-World Analogy

Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (acidified potassium dichromate): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones. (Not to be confused with Jones reagent, which is specifically CrO₃ dissolved in dilute aqueous H₂SO₄, often used in acetone — a related but distinct oxidant with the same general 1°→acid / 2°→ketone outcome.) …

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

Alcohol oxidation is a fundamental reaction in organic chemistry, and understanding why it proceeds as it does is crucial for Indian board exams (Class 12, JEE, NEET). Let's break it down step-by-step.

1. The Core Idea: Loss of Hydrogen

Oxidation in organic chemistry means loss of hydrogen (or gain of oxygen). For alcohols, this happens at the carbon bearing the –OH group.

Primary alcohol ( $R-CH_2OH$ ): Has two hydrogens on the carbon attached to –OH.
Secondary alcohol ( $R_2CHOH$ ): Has one hydrogen on that carbon.
Tertiary alcohol ( $R_3COH$ ): Has zero hydrogens on that carbon.

Key insight: The number of hydrogens on the carbon with –OH determines if and how far oxidation can go.

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

When a primary alcohol ( $R-CH_2OH$ ) is oxidized, the first product is an aldehyde ( $R-CHO$ ).

Why? The oxidizing agent (like $K_2Cr_2O_7$ / $H_2SO_4$ or $PCC$ ) removes two hydrogens:

One from the –OH group
One from the carbon atom

The carbon–oxygen bond becomes a double bond ( $C=O$ ), forming the aldehyde.

R-CH_2OH \xrightarrow{[O]} R-CHO + H_2O

But why stop here? The aldehyde still has one hydrogen on the carbonyl carbon. If a strong oxidant is present, it can remove that hydrogen too.

Step 2: Carboxylic acid formation

With excess strong oxidant (e.g., $K_2Cr_2O_7$ / $H_2SO_4$ , heat), the aldehyde is further oxidized to a carboxylic acid ( $R-COOH$ ).

R-CHO \xrightarrow{[O]} R-COOH

Why does this happen? The aldehyde's carbonyl carbon is electrophilic (partially positive). Water (from the reaction medium) adds to it, forming a gem-diol intermediate. The oxidant then removes two more hydrogens, giving the acid.

Exam tip: To stop at the aldehyde, use a mild oxidant like PCC (pyridinium chlorochromate) in anhydrous conditions — no water means no gem-diol formation.

3. Why Secondary Alcohols Give Ketones (and Stop)

A secondary alcohol ( $R_2CHOH$ ) has only one hydrogen on the carbon with –OH. Oxidation removes:

One hydrogen from –OH
One hydrogen from the carbon

This forms a ketone ( $R_2C=O$ ).

R_2CHOH \xrightarrow{[O]} R_2C=O + H_2O

Why does it stop here? The ketone has no hydrogen on the carbonyl carbon. Without that hydrogen, further oxidation (under normal conditions) is impossible — you'd need to break a $C-C$ bond, which requires much harsher conditions.

Key result: Secondary alcohols cannot be oxidized further than ketones under standard conditions.

4. Why Tertiary Alcohols Do NOT Oxidize

A tertiary alcohol ( $R_3COH$ ) has zero hydrogens on the carbon bearing –OH.

What happens if you try? The oxidant cannot remove any hydrogen from that carbon. The only possible reaction would be breaking a $C-C$ bond, which doesn't happen under normal oxidation conditions.

Result: Tertiary alcohols are resistant to oxidation under mild to moderate conditions. They require strong heating with powerful oxidants (like $K_2Cr_2O_7$ / $H_2SO_4$ , heat) to break carbon–carbon bonds — this is destructive oxidation, not useful for synthesis.

5. The "Why" in One Table

Alcohol Type	Hydrogens on C–OH	Product	Why?
Primary ( $1^\circ$ )	2	Aldehyde → Carboxylic acid	Two hydrogens available; aldehyde still has one more

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Secondary alcohols oxidise to ketones, while primary alcohols oxidise to carboxylic acids. Butan-2-ol is a secondary alcohol, so it gives butanone; butan-1-ol is primary and gives butanoic acid. The correct option is (ii).

Alkaline KMnO4 is a strong oxidising agent -- it doesn't stop halfway unless the alcohol is secondary.

Why secondary alcohols give ketones. When a secondary alcohol is oxidised, the carbon bearing the -OH group loses a hydrogen and becomes a carbonyl carbon. For butan-2-ol, CH3-CH(OH)-CH2-CH3, the -OH is on carbon 2, bonded to a methyl and an ethyl group; oxidation gives CH3-CO-CH2-CH3, butanone.

Why primary alcohols give carboxylic acids. Butan-1-ol (CH3-CH2-CH2-CH2-OH) first becomes butanal, then is further oxidised by the strong oxidant to butanoic acid -- never butanone.

Butan-1-ol (i) -- gives butanoic acid, not butanone. Wrong. …

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Here are the common mistakes students make on this question, along with the correct reasoning to avoid them.

Mistake 1: Forgetting the type of alcohol (Primary vs. Secondary)

The Mistake: Students see "oxidation" and assume any alcohol will give a ketone. They pick (C) Both of these.
Why it's wrong: Alkaline $\text{KMnO}_4$ $KMnO_{4}$ is a strong oxidizing agent. It oxidizes:
- Primary alcohols ( $\text{R-CH}_2\text{OH}$ ) → Carboxylic acids (not ketones).
- Secondary alcohols ( $\text{R-CHOH-R'}$ ) → Ketones.
- Tertiary alcohols → No reaction (under normal conditions).
How to avoid: Before answering, identify the functional group class.
- Butan-1-ol: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$ → Primary.
- Butan-2-ol: $\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3$ → Secondary.

Mistake 2: Confusing the product of primary alcohol oxidation

The Mistake: Students think primary alcohols give aldehydes (like butanal) when oxidized with $\text{KMnO}_4$ .
Why it's wrong: Alkaline $\text{KMnO}_4$ is too strong to stop at the aldehyde stage. It over-oxidizes the aldehyde to a carboxylic acid (butanoic acid). To get an aldehyde, you need a milder agent like PCC (pyridinium chlorochromate) in anhydrous conditions.
How to avoid: Memorize the "oxidation ladder":
- Primary alcohol → Aldehyde → Carboxylic acid.
- Strong oxidants (like $\text{KMnO}_4$ , $\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+$ ) go all the way to the acid.
- Mild oxidants (PCC) stop at the aldehyde.

Mistake 3: Misidentifying the ketone product from butan-2-ol

The Mistake: Students correctly pick butan-2-ol but write the product as butanal or another compound.
Why it's wrong: Oxidation of a secondary alcohol removes two hydrogen atoms (one from the OH group and one from the carbon bearing the OH). The carbon becomes a carbonyl ( $\text{C=O}$ $C=O$ ).
- Butan-2-ol: $\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3$ → $\text{CH}_3\text{CH}_2\text{COCH}_3$ (butanone).
How to avoid: Practice drawing the structure. The carbonyl carbon is the same carbon that originally had the OH group. For butan-2-ol, that's carbon #2, so the product is a ketone (butan-2-one, commonly called butanone).

Mistake 4: Ignoring the "alkaline" condition …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04184 marksMCQ
$KMnO_4$
›Reveal solutionSolution

Stopping at the aldehyde requires an anhydrous, mild oxidant such as $CrO_3$ in a non-aqueous medium.

Strong aqueous oxidants (acidified $KMnO_4$ , acidified $K_2Cr_2O_7$ ) oxidise ethanol all the way to ethanoic acid. …
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KEAM 2026Set eng-2026-04194 marksMCQ
$_2$
›Reveal solutionSolution

PCC is the complex of chromium trioxide with pyridine and HCl, a mild oxidant for 1° alcohols → aldehydes.

Reasoning. Pyridinium chlorochromate, $C_5H_5NH^+[CrO_3Cl]^-$ , is prepared by dissolving chromium trioxide ( $CrO_3$ ) in hydrochloric acid and adding py …
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KEAM 2026Set eng-2026-04194 marksMCQ
$Q. The reagent used for the conversion of decanol into decanoic acid is (A) Tollens's reagent (B) Jones reagent (C) Grignard reagent (D) Fehling's reagent (E) DIBAL-H$
›Reveal solutionSolution

Converting a primary alcohol to a carboxylic acid needs a strong oxidant — Jones reagent (chromic acid). …

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KEAM 2026Set eng-2026-04204 marksMCQ
$Q. Match the following reactions with the corresponding reagents Reactions / Reagents$
(a) Oxidation of secondary alcohols to ketones ;
(b) Dehydration of secondary alcohols to alkenes ;
(c) Reduction of ketones to secondary alcohols ;
(d) Oxidation of phenol to benzoquinone
(i) 85% $H_3PO_4$ , 440 K ;
(ii) $Na_2Cr_2O_7/H_2SO_4$ ;
(iii) Chromic anhydride ;
(iv) $NaBH_4$ (A) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) (B) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) (C) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv) (D) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii) (E) (a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)
›Reveal solutionSolution
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).

(a) Secondary alcohol $\to$ ketone: chromic anhydride ( $CrO_3$ ) $\to$ (iii).

(b) Secondary alcohol $\to$ alkene (dehydration): 85% $H_3PO_4$ at 440 K $\to$ (i).

(c) Ketone $\to$ secondary alcohol (reduction): $NaBH_4$ $\to$ (iv). …
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KEAM 2025Set eng-2025-04274 marksMCQ
$Q. Which of the following compound contains two primary alcoholic and one secondary alcoholic groups? (A) Ethylene glycol (B) Isopropyl alcohol (C) 3° Butyl alcohol (D) Glycerol (E) 2° Butyl alcohol$
›Reveal solutionSolution
Glycerol (propane-1,2,3-triol) has terminal $–CH_2OH$ groups (two primary) and a central $–CHOH–$ (one secondary), matching the description exactly.

Reasoning

Glycerol structure:
$CH_2OH\text{--}CHOH\text{--}CH_2OH$

The two terminal carbons each bear a $–CH_2OH$ (attached to one other carbon) → primary alcohol groups (2 of them).

The central carbon bears $–OH$ and is attached to two carbons → secondary alcohol group (1 of them).

Checking others: …
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KEAM 2024Set eng-2024-06054 marksMCQ
$NaBH_{4}$
›Reveal solutionSolution

A tertiary alcohol cannot come from carbonyl reduction, so 2-methyl-2-propanol cannot be made this way.

Reducing an aldehyde gives a primary alcohol; reducing a ketone gives a secondary alcohol. A tertiary alcohol has no C–H on the carbinol carbon and thus corresponds to no aldehyde or ketone. …
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KEAM 2024Set eng-2024-06054 marksMCQ
$H_{2}$
›Reveal solutionSolution

1-Phenylethanol is a secondary alcohol, so the statement 'It is a primary alcohol' is incorrect.

Structure: $C_6H_5{-}CH(OH){-}CH_3$ . The carbinol carbon is bonded to two carbon groups (phenyl and methyl) plus one H and the OH — that is a secondary alcohol, making statement (A) false. …
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←Previous questionQ10 — Choose the most suitable reagent to convert the methyl ketone CH3-CH=C…→Next questionQ12 — In Clemmensen reduction the carbonyl compound is treated with ______. …

Q.Which of the following compounds will give butanone on oxidation with alkaline $KMnO_4$ solution?

Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

The Precise Statement

Why Does Tertiary Alcohol Not Oxidize?

A Real-World Analogy

Key Reagents to Remember (for exams)

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

1. The Core Idea: Loss of Hydrogen

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

Step 2: Carboxylic acid formation

3. Why Secondary Alcohols Give Ketones (and Stop)

4. Why Tertiary Alcohols Do NOT Oxidize

5. The "Why" in One Table

Method: Oxidation of Secondary Alcohols to Ketones

Step 1 – Identify the functional group in each option

Step 2 – Recall the oxidation rule

Step 3 – Apply to butan-2-ol

Mistake 1: Forgetting the type of alcohol (Primary vs. Secondary)

Mistake 2: Confusing the product of primary alcohol oxidation

Mistake 3: Misidentifying the ketone product from butan-2-ol

Mistake 4: Ignoring the "alkaline" condition …

Reasoning

Q.Which of the following compounds will give butanone on oxidation with alkaline KMnO4KMnO_4KMnO4​ solution?

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Q.Which of the following compounds will give butanone on oxidation with alkaline $KMnO_4$ solution?

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