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.

Showing the 12 most recent of 14 on this concept.

GUJCET 2025Set 031 markMCQ
$Q. Identify R', R'' and R''' for the following reaction. [FIGURE: a ketone R'R''C=O reacting via$
(i) R'''MgX
(ii) H2O to give 2-methylbutane-2-ol.] (A) $R'=C_{2}H_{5}, R''=C_{2}H_{5}, R'''=CH_{3}$ (B) $R'=CH_{3}, R''=C_{2}H_{5}, R'''=CH_{3}$ (C) $R'=C_{2}H_{5}, R''=CH_{3}, R'''=C_{2}H_{5}$ (D) $R'=CH_{3}, R''=CH_{3}, R'''=CH_{3}$
›Reveal solutionSolution
Ketone $R'R''C{=}O$ + $R'''MgX \rightarrow R'R''R'''C\!-\!OH$ ; the product's three alkyls are CH $_3$ , CH $_3$ , C $_2$ H $_5$ .

Concept — Grignard synthesis of tertiary alcohols. A ketone gives a tertiary alcohol whose carbinol carbon bears the ketone's two groups plus the Grignard's group.

Steps.

2-methylbutan-2-ol: $\mathrm{CH_3\!-\!C(OH)(CH_3)\!-\!CH_2CH_3}$ . The C–OH carbon carries CH $_3$ , CH $_3$ , C $_2$ H $_5$ .

So $\{R',R'',R'''\}=\{CH_3, CH_3, C_2H_5\}$ . …
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GUJCET 2025Set 031 markMCQ
$KMnO_{4}/H_{2}SO_{4}$
›Reveal solutionSolution
[!TLDR]

Oxidising a primary alcohol to an aldehyde requires the mild, selective reagent PCC ( $C_5H_5NH^+CrO_3Cl^-$ ).

Concept

Primary alcohols are oxidised to aldehydes and can be further oxidised to carboxylic acids by strong oxidants. To stop cleanly at the aldehyde, a mild oxidant such as PCC (in anhydrous dichloromethane) is used.

Solution

The substrate (hydroxymethyl)cyclohexane has a $-CH_2OH$ group that must become $-CHO$ (cyclohexanecarbaldehyde) — a controlled oxidation to the aldehyde.

(A) $KMnO_4/H_2SO_4$ : strong oxidant, over-oxidises to the carboxylic acid.

(B) $O_3/H_2O$ –Zn: ozonolysis, cleaves C=C double bonds — not applicable to an alcohol. …
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GUJCET 2023Set 091 markMCQ
$Q. Which of the following alcohol undergo dehydration reaction with Cu (Copper) metal at 573 K temperature? (A) Secondary and Tertiary (B) Primary & Secondary (C) Primary and Tertiary (D) Only Tertiary$
›Reveal solutionSolution
With Cu at 573 K: 1° and 2° alcohols dehydrogenate (→ aldehyde/ketone), while only tertiary alcohols dehydrate (→ alkene).

Concept. Passing alcohol vapour over heated copper at 573 K:

Primary → aldehyde (dehydrogenation, loss of $H_2$ )

Secondary → ketone (dehydrogenation)

Tertiary → has no α-H on the carbinol carbon to lose as $H_2$ , so it instead loses water and dehydrates to an alkene. …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. R'-X --Na/ether--> 2,3-dimethylbutane. Identify R'.$
(a) (CH3)2CH-
(b) (C2H5)2CH-
(c) (CH3CH2)3C-
(d) (CH3)3C-
›Reveal solutionSolution

2,3-dimethylbutane is symmetric, made by Wurtz coupling of two isopropyl (2-propyl) groups.

Wurtz reaction: 2 R'-X + 2 Na --dry ether--> R'-R' + 2 NaX. It couples two alkyl groups to give a symmetrical alkane.

…

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GUJCET 2022Set 171 markMCQ
$-CH_2-CO-OCH_3$
›Reveal solutionSolution

$NaBH_4$ is a mild reducing agent — it reduces aldehydes/ketones to alcohols but does NOT reduce esters.

Concept: Sodium borohydride selectively reduces the cyclohexanone carbonyl ( $C=O$ ) to a secondary alcohol ( $CH-OH$ ), converting the ring ketone into a ring alcohol. The methyl ester group $-CH_2-CO-OCH_3$ is unreactive toward $NaBH_4$ and is retained unchanged. Among the options, only the choic …
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GUJCET 2021Set 151 markMCQ
$CH_3-CH_2-CH_2-Mg-X$
›Reveal solutionSolution

Grignard + methanal → primary alcohol $R-CH_2OH$ ; work backwards to find R.

Concept: $R-MgX + HCHO \rightarrow R-CH_2-OMgX \xrightarrow{H_2O} R-CH_2OH$ . Methanal always adds one carbon and gives a primary alcohol. …
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GUJCET 2021Set 151 markMCQ
$O_3/H_2O$
›Reveal solutionSolution

PCC cleanly oxidises 1° alcohol → aldehyde and leaves the double bond intact.

Concept: Pyridinium chlorochromate (PCC) is a mild oxidant that stops at the aldehyde stage and does not attack C=C.

$CH_2=CH-CH_2OH \xrightarrow{PCC} CH_2=CH-CHO \ (\text{propenal})$ …
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GUJCET 2020Set 071 markMCQ
$-CH_2-C(=O)-OCH_3$
›Reveal solutionSolution

NaBH₄ reduces the ketone (→ cyclohexanol) and does not touch the ester. …

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GUJCET 2020Set 071 markMCQ
$CrO_3$
›Reveal solutionSolution
[!TLDR] Secondary alcohol → ketone needs a mild oxidant; PCC cleanly gives cyclohexanone.

Concept

Secondary alcohols are oxidised to ketones. PCC (pyridinium chlorochromate) is a mild, selective, non-aqueous oxidant that converts secondary alcohols to ketones without further oxidation. Ozonolysis reagents and DIBAL-H are not alcohol-oxidation reagents.

Solution

(A) Anhydrous CrO $_3$ : a strong Cr(VI) oxidant, generally used with acid; not the selective mild reagent intended here. …
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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. Substance A, on reaction with Cu at 573 K, gives Isobutylene. Which is the structural formula of substance A in this reaction?$
(a) CH3-CH(OH)-CH2-CH3
(b) CH3-CH2-CH2-CH2-OH
(c) CH3-CH(CH3)-CH2-OH
(d) CH3-C(CH3)(CH3)-OH (i.e. (CH3)3C-OH)
›Reveal solutionSolution
Passed over hot copper at 573 K, a TERTIARY alcohol cannot dehydrogenate (it has no H on the carbinol carbon to lose alongside the O-H), so it instead undergoes dehydration to an alkene; a primary or secondary alcohol would dehydrogenate to an aldehyde or ketone instead.

Alcohols passed over copper catalyst at 573 K behave differently by class:

Primary alcohols dehydrogenate to aldehydes (R-CH2-OH -> R-CHO + H2).

Secondary alcohols dehydrogenate to ketones. …
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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Give the IUPAC name of the product obtained when Phenol is oxidized by chromic acid. (Na2Cr2O7 + Conc. H2SO4).$
(a) Cyclohexa-2,5-diene-1,4-dione
(b) Cyclohexa-1,4-dione
(c) Cyclohexanone
(d) Cyclohexa-1,4-diene-2,5-dione
›Reveal solutionSolution

Phenol + Na2Cr2O7/conc. H2SO4 -> benzoquinone = cyclohexa-2,5-diene-1,4-dione.

Phenol on oxidation with chromic acid (from Na2Cr2O7 + conc. H2SO4) is converted to para-benzoquinone. Its IUPAC name is cyclohexa-2,5-diene-1,4-dione: a six-membered ring with C= …

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Identify Pyridinium chlorochromate from the following.$
(a) pyridine ring, N+ - H . CrO3Cl^-
(b) pyridine ring, N+ - H . CrO2Cl^-
(c) pyridine ring, N+ - CrO3Cl^-
(d) pyridine ring, N+ - H2 . CrO3Cl^-
›Reveal solutionSolution
PCC = pyridinium (C5H5N-H+) chlorochromate (CrO3Cl-), i.e. option (a).

Pyridinium chlorochromate (PCC) is a mild oxidising reagent (C5H5NH+ ClCrO3-) used to oxidise primary alcohols to aldehydes (without over-oxidation to acids). It consists of:

the pyridinium cation: pyridine protonated at nitrogen, so N carries a positive charge and an N-H bond, and …
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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)

Concept

Solution

Concept

Solution

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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