Q.Methyl alcohol on oxidation with acidified K2Cr2O7 gives
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🔒 Start your 14-day free trial to unlock the full solution →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) | — |
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−CH2OH): Has two hydrogens on the carbon attached to –OH.
- Secondary alcohol (R2CHOH): Has one hydrogen on that carbon.
- Tertiary alcohol (R3COH): 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−CH2OH) is oxidized, the first product is an aldehyde (R−CHO).
Why? The oxidizing agent (like K2Cr2O7 / H2SO4 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−CH2OH[O]R−CHO+H2O
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., K2Cr2O7 / H2SO4, heat), the aldehyde is further oxidized to a carboxylic acid (R−COOH).
R−CHO[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 (R2CHOH) has only one hydrogen on the carbon with –OH. Oxidation removes:
- One hydrogen from –OH
- One hydrogen from the carbon
This forms a ketone (R2C=O).
R2CHOH[O]R2C=O+H2O
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 (R3COH) 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 K2Cr2O7 / H2SO4, 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∘) | 2 | Aldehyde → Carboxylic acid | Two hydrogens available; aldehyde still has one more |
A primary alcohol is oxidised in stages by a strong oxidising agent, first to an aldehyde and then, with excess oxidant, all the way to the carboxylic acid. …
A primary alcohol is oxidised to an aldehyde and then to a carboxylic acid; methanol → HCHO → HCOOH with acidified K2Cr2O7.
Methyl alcohol (methanol, CH3OH) is a primary alcohol. Oxidation with a strong oxidising agent such as acidified potassium dichromate (K2Cr2O7/H2SO4) proceeds:
CH3OH → (oxidation) HCHO (formaldehyde) → (further oxidation) HCOOH (formic acid)
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Showing the 12 most recent of 22 on this concept.
- CBSE 2026Set A1 markMCQQ.When vapours of an alcohol are passed over hot reduced copper, it gives an alkene. The alcohol is(a) Primary(b) Secondary(c) Tertiary(d) None of these
›Reveal solutionSolution
Over hot reduced copper (573 K), a primary alcohol gives an aldehyde, a secondary gives a ketone, and a tertiary gives an alkene.
When alcohol vapours are passed over hot reduced copper the behaviour depends on the class of alcohol:
- Primary alcohol -> dehydrogenation -> aldehyde
- Secondary alcohol -> dehydrogenation -> ketone …
- CBSE 2026Set ANNUAL1 markQ.Write the name of product obtained when vapour of ethyl alcohol are passed over heated Copper at 573 K.
›Reveal solutionSolution
Passing alcohol vapours over heated copper catalyses either dehydrogenation (for 1° and 2° alcohols) or dehydration (for 3° alcohols), depending on alcohol type.
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- CBSE 2026Set ANNUAL1 markMCQQ.Dehydration of tertiary alcohols with copper at 573 K gives:(a) Aldehyde(b) Ketone(c) Alkene(d) None of these
›Reveal solutionSolution
Passing alcohol vapours over heated copper at 573 K is a classification test: 1° alcohols → aldehydes, 2° alcohols → ketones, but 3° alcohols (no α-H on the carbinol carbon available for dehydrogenation) undergo dehydration to give an alkene.
When vapours of an alcohol are passed over copper catalyst at 573 K:
- Primary alcohols are dehydrogenated (lose H2) to aldehydes: RCH2OHCu,573KRCHO+H2
- Secondary alcohols are dehydrogenated to ketones: R2CHOHCu,573KR2C=O+H2 …
- CBSE 2026Set ANNUAL1 markMCQQ.When vapour's of a compound X are passed over heated copper, the major product obtained is the acetone. The compound X is:(a) n-Propyl alcohol(b) Iso-propyl alcohol(c) Acetaldehyde(d) Propane
›Reveal solutionSolution
Vapours passed over heated copper dehydrogenate 2° alcohols to ketones; since the product is acetone (a ketone), X must be a secondary alcohol — isopropyl alcohol.
Over heated copper (573 K), alcohols are catalytically dehydrogenated based on their class:
- 1° alcohol → aldehyde
- 2° alcohol → ketone
- 3° alcohol → alkene (dehydration) …
- CBSE 2026Set ANNUAL1 markMCQQ.The most suitable reagent for the conversion of RCH2OH→RCHO is(a) KMnO4(b) K2Cr2O7(c) LiAlH4(d) PCC (Pyridinium Chlorochromate)
›Reveal solutionSolution
Selective oxidation of a 1° alcohol to an aldehyde (without over-oxidation to the acid) requires an anhydrous, mild oxidant — PCC — rather than a strong aqueous oxidant.
Why KMnO4 and K2Cr2O7 (a, b) fail: these are strong oxidising agents used in aqueous, typically acidified medium. The initially formed aldehyde reacts with water to form a geminal diol (aldehyde hydrate), RCH(OH)2, which is itself readily oxidised further by these strong oxidants to the carboxylic acid, RCOOH. So the reaction cannot be stopped cleanly at the aldehyde stage.
Why LiAlH4 (c) fails: this is a powerful reducing agent (it reduces esters, acids and other carbonyls down to alcohols) — the wrong direction entirely for an oxidation.
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- CBSE 2026Set ANNUAL1 markQ.Fill in the blank: In addition of bromine in CCl4 to an alkene resulting in disappearance of reddish brown colour of bromine constitutes, an important method for the detection of ______ in a molecule.
›Reveal solutionSolution
Decolourisation of bromine in CCl4 detects unsaturation (C=C double bond).
An alkene readily adds bromine across its carbon-carbon double bond to form a colourless dibromide:
C=C + Br2 -> Br-C-C-Br
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- CBSE 2026Set ANNUAL1 markQ.Write True or False: Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water.
›Reveal solutionSolution
True - these halogen derivatives are denser than water.
The heavy halogen atoms (Br, I) and multiple chlorine atoms greatly increase the molar mass and density of the molecule. As a result, bromo, iodo and polychloro derivatives of hydrocarbons (e.g. bromoform, iodoform, chloroform, …
- CBSE 2025Set 56/5/11 markMCQQ.CH3CH2OH can be converted to CH3CHO by : (A) catalytic hydrogenation (B) treatment with LiAlH4 (C) treatment with PCC (D) treatment with KMnO4
›Reveal solutionSolution
The key idea is that converting ethanol (CH3CH2OH) to ethanal (CH3CHO) is a controlled oxidation of a primary alcohol to an aldehyde. The correct reagent is PCC (pyridinium chlorochromate), which stops at the aldehyde stage without over-oxidizing to a carboxylic acid.
This question tests your understanding of alcohol oxidation — a core reaction in organic chemistry. Ethanol is a primary alcohol. To get an aldehyde, you need to oxidize it partially. The challenge is that many strong oxidizers will push the reaction all the way to the carboxylic acid (acetic acid, CH3COOH). So the trick is choosing a reagent that is mild enough to stop at the aldehyde.
Let’s examine each option.
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Option (A): Catalytic hydrogenation
Hydrogenation (H2 with a metal catalyst like Pd, Pt, or Ni) is a reduction process. It adds hydrogen across double or triple bonds. Ethanol has no multiple bonds to reduce — it’s already saturated. This would do nothing. So this is wrong.
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Option (B): Treatment with LiAlH4
Lithium aluminium hydride is a powerful reducing agent. It reduces carbonyl compounds (aldehydes, ketones, acids, esters) to alcohols. Using it on ethanol would be pointless — ethanol is already an alcohol. It cannot oxidize anything. So this is also wrong.
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Option (C): Treatment with PCC
PCC (pyridinium chlorochromate, C5H5NH+CrO3Cl−) is a mild oxidizing agent specifically designed for the conversion of primary alcohols to aldehydes. It works in anhydrous conditions (typically in dichloromethane) and stops cleanly at the aldehyde stage.
The reaction:
CH3CH2OHPCCCH3CHO
This is the textbook method. So this is correct.
- Option (D): Treatment with KMnO4 …
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- CBSE 2025Set 56/6/11 markMCQQ.Which one of the following amines gives an alcohol on reaction with HNO2 ? (A) C6H5NH2 (aniline) (B) C2H5NH2 (C) (C2H5)2NH (D) (C2H5)3N
›Reveal solutionSolution
The key idea is that primary aliphatic amines react with nitrous acid (HNO2) to give alcohols via a diazonium intermediate that decomposes. Among the options, only C2H5NH2 (ethylamine) is a primary aliphatic amine, so it yields ethanol. The correct option is (B).
The reaction of an amine with nitrous acid (HNO2) is a classic test to distinguish between primary, secondary, and tertiary amines. Nitrous acid is unstable and is prepared in situ by reacting sodium nitrite (NaNO2) with a mineral acid like HCl or H2SO4. The outcome depends entirely on the class of the amine.
For primary aliphatic amines (like ethylamine), the reaction proceeds through an unstable alkyldiazonium salt. This salt spontaneously decomposes to give a carbocation, which then reacts with water to form an alcohol. This is the only case where an alcohol is the major product.
For primary aromatic amines (like aniline), the diazonium salt formed is stable at low temperatures (0–5°C) and does not give an alcohol with water — it gives phenol only upon heating or under specific conditions. At room temperature, aniline reacts with HNO2 to give a diazonium salt that can couple or decompose to other products, but not ethanol.
For secondary amines (like diethylamine), the reaction yields a yellow, oily N-nitrosamine — no alcohol is formed.
For tertiary amines (like triethylamine), the reaction gives a nitrosamine salt or simply dissolves, again no alcohol.
So the only amine that reliably gives an alcohol under standard conditions is a primary aliphatic amine.
Let’s check each option:
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Option (A): C6H5NH2 (aniline) — This is a primary aromatic amine. With HNO2 at 0–5°C, it forms a stable benzenediazonium salt. This salt does not decompose to give an alcohol at low temperature; it requires heating with water to yield phenol. Under the usual conditions of the reaction (room temperature or slightly above), aniline gives a diazonium salt that may undergo coupling or other reactions, but not an alcohol. So this is not the answer.
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Option (B): C2H5NH2 (ethylamine) — This is a primary aliphatic amine. The reaction with HNO2 proceeds as:
C2H5NH2+HNO2→[C2H5N2+]H2OC2H5OH+N2+H+
The intermediate ethyldiazonium ion is unstable and immediately loses N2 to form an ethyl carbocation, which then reacts with water to give ethanol. This is the classic case where an alcohol is produced. So this is the correct option.
- Option (C): (C2H5)2NH (diethylamine) — This is a secondary amine. With HNO2, it forms a yellow, oily N-nitrosamine: …
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- CBSE 2025Set D1 markMCQQ.Methyl alcohol on oxidation with acidified K2Cr2O7 gives(a) CH3COCH3(b) CH3CHO(c) HCOOH(d) CH3COOH
›Reveal solutionSolution
A primary alcohol is oxidised to an aldehyde and then to a carboxylic acid; methanol → HCHO → HCOOH with acidified K2Cr2O7.
Methyl alcohol (methanol, CH3OH) is a primary alcohol. Oxidation with a strong oxidising agent such as acidified potassium dichromate (K2Cr2O7/H2SO4) proceeds:
CH3OH → (oxidation) HCHO (formaldehyde) → (further oxidation) HCOOH (formic acid)
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- CBSE 2025Set ANNUAL1 markMCQQ.Ethyl alcohol --Cu/573K--> A, 'A' is(a) Acetaldehyde(b) Propionaldehyde(c) Acetone(d) Ethanoic acid
›Reveal solutionSolution
Copper catalyses dehydrogenation (not dehydration) of a primary alcohol at 573 K, converting ethanol to the corresponding aldehyde.
CH3CH2OHCu573 KCH3CHO+H2
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- CBSE 2025Set ANNUAL1 markMCQQ.Oxidation of propan-1-ol with alkaline KMnO4 solution gives(a) propanoic acid(b) propane(c) ethane(d) propanal
›Reveal solutionSolution
Alkaline KMnO4 is a strong oxidising agent, so it oxidises primary alcohols completely to carboxylic acids (via the aldehyde stage, which cannot be isolated).
CH3CH2CH2OHalkaline KMnO4CH3CH2COOH (propanoic acid)
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