Which of the following reagents can be used to oxidise primary alcohols to aldehydes? (Two or more than two op

NCERT Exemplar · Q58

Q. Which of the following reagents can be used to oxidise primary alcohols to aldehydes? (Two or more than two options may be correct.)

(i)

CrO_3

in anhydrous medium

(ii)

KMnO_4

in acidic medium

(iii) pyridinium chlorochromate

(iv) heat in the presence of Cu at 573 K

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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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The key is choosing reagents that stop at the aldehyde stage without over-oxidising to the carboxylic acid. The correct options are (i) CrO3 in anhydrous medium, (iii) pyridinium chlorochromate, and (iv) heat with Cu at 573 K.

Concept

A primary alcohol can be oxidised first to an aldehyde and then further to a carboxylic acid. Mild/anhydrous oxidants and dehydrogenation methods stop at the aldehyde; strong aqueous oxidants push through to the acid.

Checking each option

Option (i), CrO3 in anhydrous medium (e.g. Collins reagent, CrO3.2Py): with no water present, the aldehyde cannot hydrate to the gem-diol that would be further oxidised, so the reaction stops at the aldehyde. Correct.
Option (ii), KMnO4 in acidic medium: a strong, aqueous oxidant that carries the alcohol all the way to the carboxylic acid. Incorrect for making an aldehyde. …

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Common Mistakes in Alcohol Oxidation (Primary → Aldehyde)

Mistake 1: Assuming all strong oxidants stop at aldehyde

The error: Students think $KMnO_4$ in acidic medium (option B) will give aldehyde because it's a common oxidising agent.

Why it's wrong:

$KMnO_4$ in acidic medium is a very strong oxidant. It does not stop at the aldehyde stage — it over-oxidises the aldehyde to a carboxylic acid.

Primary alcohol $\xrightarrow[\text{H}^+]{\text{KMnO}_4}$ Carboxylic acid (not aldehyde)

How to avoid:

Remember the oxidation ladder:

Primary alcohol → Aldehyde → Carboxylic acid

Only mild/controlled oxidants stop at aldehyde. Strong oxidants like $KMnO_4$ (acidic) and $K_2Cr_2O_7$ (acidic) go all the way to acid.

Mistake 2: Confusing anhydrous vs. aqueous conditions for $CrO_3$

The error: Students reject option (A) $CrO_3$ in anhydrous medium, thinking $CrO_3$ always over-oxidises.

Why it's wrong:

$CrO_3$ in aqueous acid (Jones reagent) over-oxidises to acid. But in anhydrous medium (e.g., pyridine), it behaves as a mild oxidant and stops at aldehyde.

$CrO_3$ + pyridine → Collins reagent (selective for aldehyde)

How to avoid:

Pay attention to the medium — anhydrous conditions make $CrO_3$ mild. Aqueous acidic conditions make it strong.

Mistake 3: Forgetting PCC is the classic "aldehyde-maker"

The error: Students overlook option (C) pyridinium chlorochromate (PCC) because they don't recognise the name.

Why it's wrong to exclude it:

PCC is specifically designed to oxidise primary alcohols to aldehydes without over-oxidation. It is the most common reagent for this purpose in organic chemistry.

PCC = $C_5H_5NH^+ ClCrO_3^-$ → stops at aldehyde

How to avoid:

Memorise PCC as the go-to reagent for primary alcohol → aldehyde. Also remember Collins reagent ( $CrO_3$ ·pyridine) and Swern oxidation as alternatives.

Mistake 4: Ignoring catalytic dehydrogenation (option D)

The error: Students think "heat with Cu at 573 K" is not a valid oxidation method.

Why it's wrong:

This is catalytic dehydrogenation — a real industrial method. The alcohol loses $H_2$ (not oxygen addition), but it is still oxidation (loss of hydrogen).

$RCH_2OH \xrightarrow[\text{573 K}]{\text{Cu}} RCHO + H_2$

How to avoid: …

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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 questionQ57 — Which of the following reaction schemes will yield phenol? (Two or mor…→Next questionQ59 — Phenol can be distinguished from ethanol by the reactions with ______.…

Q.Which of the following reagents can be used to oxidise primary alcohols to aldehydes? (Two or more than two options may be correct.)

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

Concept

Checking each option

Method: Controlled Oxidation of Primary Alcohols to Aldehydes

Step-by-step reasoning

Common Mistakes in Alcohol Oxidation (Primary → Aldehyde)

Mistake 1: Assuming all strong oxidants stop at aldehyde

Mistake 2: Confusing anhydrous vs. aqueous conditions for $CrO_3$

Mistake 3: Forgetting PCC is the classic "aldehyde-maker"

Mistake 4: Ignoring catalytic dehydrogenation (option D)

Reasoning

Q.Which of the following reagents can be used to oxidise primary alcohols to aldehydes? (Two or more than two options may be correct.)

Concept

Checking each option

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