Q.Write chemical reactions to affect the following transformations:
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Oxidation Reactions
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 (Jones reagent): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones.
- KMnO₄: similar to dichromate, but stronger — can over-oxidize. …
Why this formula?
Oxidation Reactions: Why the Key Principles Hold
Oxidation reactions are fundamental to chemistry, and understanding why they work the way they do is essential for mastering Indian board exams (Class 11, 12, JEE, NEET). Let's break down the core ideas from first principles.
1. The Core Definition: What Does "Oxidation" Really Mean?
Historically, oxidation meant "adding oxygen." But that's too narrow. The modern, exam-correct definition is:
Oxidation is the loss of electrons by a species.
This is the electronic concept (given by the ionic theory). The why behind this definition comes from the behavior of atoms during chemical reactions.
Why do atoms lose electrons?
Atoms seek stability. They achieve this by having a full outer electron shell (octet, duplet, or pseudo-inert gas configuration).
- Metals (like Na, Mg, Fe) have few valence electrons (1, 2, or 3). It's energetically easier for them to lose these electrons than to gain 5, 6, or 7.
- Non-metals (like O, Cl, F) have many valence electrons (5, 6, or 7). It's energetically easier for them to gain electrons.
So, when a metal reacts with a non-metal, the metal loses electrons (gets oxidized), and the non-metal gains electrons (gets reduced).
Example:
2Na+Cl2→2NaCl
- Na loses 1 electron: Na→Na++e− (Oxidation)
- Cl gains 1 electron: Cl2+2e−→2Cl− (Reduction)
Key takeaway: Oxidation and reduction always happen together (Redox reactions). You cannot have one without the other.
2. The Key Formula(e): Oxidation Number Rules
The oxidation number (O.N.) is a bookkeeping tool. It's not a real charge (except in ionic compounds), but it helps track electron flow.
Why do we assign oxidation numbers?
Because in covalent compounds (like CH4 or H2O), electrons are shared, not transferred. We need a way to pretend they are transferred to see which atom "owns" the electrons more.
The Rules (and why they exist)
| Rule | Statement | Why this rule? |
|---|---|---|
| 1 | O.N. of an element in its free state = 0 | No electron transfer has occurred. |
| 2 | O.N. of a monatomic ion = its charge | The atom has actually lost/gained that many electrons. |
| 3 | O.N. of H = +1 (except in metal hydrides where it's -1) | H is less electronegative than O, F, Cl, but more electronegative than metals. |
| 4 | O.N. of O = -2 (except in peroxides where it's -1, superoxides -1/2, and with F where it's +2) | O is highly electronegative (3.44 on Pauling scale). It "pulls" electrons toward itself. |
| 5 | Sum of O.N. in a neutral compound = 0 | The compound has no net charge. |
| 6 | Sum of O.N. in a polyatomic ion = charge of the ion | The ion's overall charge must be accounted for. |
The Derivation of a Key Formula: Finding O.N. of an Unknown Element
Suppose you need to find the O.N. of S in H2SO4.
Step 1: Write known O.N.s:
- H: +1 (rule 3)
- O: -2 (rule 4)
- S: let it be x (unknown)
Step 2: Apply rule 5 (neutral compound sum = 0):
2(+1)+x+4(−2)=0
Step 3: Solve:
2+x−8=0
x−6=0
x=+6
Why this works: The oxidation number is a mathematical consequence of the electronegativity hierarchy. Oxygen is more electronegative than sulfur, so it "takes" the electrons. Hydrogen is less electronegative than sulfur, so it "gives" electrons to sulfur. The net result is that sulfur appears to have lost 6 electrons.
3. The Key Formula(e): Balancing Redox Equations
Two methods are exam-critical: Oxidation Number Method and Ion-Electron Method (Half-Reaction Method).
Why do we need these methods?
Because in a redox reaction, the total number of electrons lost (oxidation) must equal the total number of electrons gained (reduction). This is the Law of Conservation of Charge.
The Ion-Electron Method (for acidic medium) — Step-by-step why
Example: Balance MnO4−+Fe2+→Mn2++Fe3+ (acidic)
Step 1: Write half-reactions.
-
Oxidation: Fe2+→Fe3++e−
Why? Fe loses 1 electron (O.N. goes from +2 to +3).
-
Reduction: MnO4−→Mn2+
Why? Mn gains electrons (O.N. goes from +7 to +2).
Step 2: Balance atoms other than H and O.
- Mn is already balanced (1 on each side).
Step 3: Balance O by adding H2O.
- Left: 4 O atoms. Right: 0 O atoms.
- Add 4 H2O to the right:
MnO4−→Mn2++4H2O
Step 4: Balance H by adding H+ (because acidic medium).
- Right: 8 H atoms (from 4 H2O). Left: 0 H atoms.
- Add 8 H+ to the left:
8H++MnO4−→Mn2++4H2O
Step 5: Balance charge by adding electrons.
- Left: 8(+1)+(−1)=+7 charge.
- Right: +2 charge.
- To make left = right, add 5 electrons to the left:
8H++MnO4−+5e−→Mn2++4H2O
Step 6: Multiply half-reactions to equalize electrons.
- Oxidation: Fe2+→Fe3++e− (×5)
- Reduction: 8H++MnO4−+5e−→Mn2++4H2O (×1)
Step 7: Add them:
5Fe2++8H++MnO4−→5Fe3++Mn2++4H2O
Why this works: Every step is driven by conservation laws:
- Mass balance: Same number of each atom on both sides.
- Charge balance: Net charge on left = net charge on right.
- Electron balance: Electrons lost = electrons gained.
4. The Key Formula(e): Electrochemical Series and Cell Potential
For a galvanic cell (voltaic cell), the cell potential Ecell∘ is:
Ecell∘=Ecathode∘−Eanode∘
Why this formula? …
Concept: Oxidation Reactions
The core idea is the controlled oxidation of alcohols and aldehydes to carboxylic acids, and the oxidative cleavage of alkenes and alkyl side-chains on aromatic rings.
- Butan-1-ol → Butanoic acid
A primary alcohol is oxidised to a carboxylic acid using a strong oxidising agent like acidified potassium dichromate (KX2CrX2OX7/HX+) or potassium permanganate.
CHX3CHX2CHX2CHX2OHKX2CrX2OX7/HX+,ΔCHX3CHX2CHX2COOH
- Benzyl alcohol → Phenylethanoic acid
Benzyl alcohol is first converted to benzyl chloride (using PClX5 or SOClX2), then treated with KCN to form benzyl cyanide, followed by acidic hydrolysis.
CX6HX5CHX2OHPClX5CX6HX5CHX2ClKCNCX6HX5CHX2CNHX3OX+,ΔCX6HX5CHX2COOH
- 3-Nitrobromobenzene → 3-Nitrobenzoic acid
The textbook's route: form the Grignard reagent in dry ether, bubble in COX2 (carboxylation), then acidify.
Br−CX6HX4−NOX2Mg,dry etherBrMg−CX6HX4−NOX21⋅ COX22⋅ HX3OX+HOOC−CX6HX4−NOX2
- 4-Methylacetophenone → Benzene-1,4-dicarboxylic acid Both the methyl group and the acetyl group are oxidised to carboxylic acids using a strong oxidising agent like alkaline KMnOX4 under heat. CHX3−CX6HX4−COCHX3KMnOX4/OHX−,ΔHOOC−CX6HX4−COOH …
The key idea is to use strong oxidizing agents (like KX2CrX2OX7/HX+, KMnOX4/HX+, or KMnOX4/OHX−) to convert primary alcohols and aldehydes to carboxylic acids, and to oxidize alkyl side-chains on aromatic rings to carboxyl groups. The final products are: (i) butanoic acid,
(ii) phenylethanoic acid,
(iii) 3-nitrobenzoic acid,
(iv) benzene-1,4-dicarboxylic acid,
(v) hexane-1,6-dioic acid,
(vi) butanoic acid.
Oxidation reactions are the backbone of converting alcohols and aldehydes into carboxylic acids. The principle is simple: a primary alcohol (−CHX2OH) can be oxidized first to an aldehyde (−CHO) and then to a carboxylic acid (−COOH). An aldehyde itself is easily oxidized to the acid. For aromatic compounds, any alkyl group attached to the benzene ring can be fully oxidized to a carboxyl group (−COOH) under strong conditions, regardless of the chain length. This is because the benzylic carbon (the one directly attached to the ring) is the most reactive, and the entire side-chain is chopped down to a single carboxyl carbon.
Let’s go through each transformation step by step.
- Butan-1-ol to butanoic acid Butan-1-ol is a primary alcohol. To get the carboxylic acid, we need to push the oxidation all the way. A common reagent is acidified potassium dichromate (KX2CrX2OX7/HX2SOX4). The reaction proceeds via butanal as an intermediate, but under these conditions, the aldehyde is not isolated — it gets further oxidized to butanoic acid.
CHX3CHX2CHX2CHX2OHKX2CrX2OX7/HX2SOX4,ΔCHX3CHX2CHX2COOH
If you want to stop at the aldehyde, you would use a milder oxidant like PCC (pyridinium chlorochromate) in anhydrous conditions. But here, we want the acid, so strong conditions are correct.
- Benzyl alcohol to phenylethanoic acid Benzyl alcohol is CX6HX5CHX2OH. Oxidation gives benzaldehyde (CX6HX5CHO) first, then benzoic acid (CX6HX5COOH). But the target is phenylethanoic acid (CX6HX5CHX2COOH), which has a two-carbon side-chain. Direct oxidation of benzyl alcohol will give benzoic acid, not phenylethanoic acid. So we need a different strategy. The classic method is to first convert benzyl alcohol to benzyl chloride (using SOClX2 or PClX5), then react with KCN to form benzyl cyanide (CX6HX5CHX2CN), followed by acidic hydrolysis to get phenylethanoic acid.
CX6HX5CHX2OHSOClX2CX6HX5CHX2ClKCN,alc⋅CX6HX5CHX2CNHX3OX+,ΔCX6HX5CHX2COOH
A common mistake is to try to oxidize benzyl alcohol directly to phenylethanoic acid. That would require adding a carbon atom, which oxidation cannot do. Oxidation only removes electrons/hydrogens; it never adds carbon. So a cyanide step is necessary here.
- 3-Nitrobromobenzene to 3-nitrobenzoic acid Here we have a bromine atom on the benzene ring, and we want to replace it with a carboxyl group. The textbook converts the aryl halide to a Grignard reagent and then carboxylates it: treat 3-nitrobromobenzene with magnesium in dry ether to form the arylmagnesium bromide, pass in carbon dioxide (the Grignard adds to COX2 to give the magnesium carboxylate), and finally acidify to release the free acid.
Br−CX6HX4−NOX2 (meta)Mg,dry etherBrMg−CX6HX4−NOX2COX2OX2N−CX6HX4−COOMgBrHX3OX+HOOC−CX6HX4−NOX2
Strictly anhydrous ("dry ether") conditions are essential when forming the Grignard reagent — any moisture destroys it. The Grignard carbon then attacks the electrophilic carbon of COX2, which is what installs the new carboxyl carbon.
- 4-Methylacetophenone to benzene-1,4-dicarboxylic acid This molecule has two oxidizable groups: a methyl group at the 4-position and an acetyl group (−COCHX3) at the 1-position. Under strong oxidation (e.g., alkaline KMnOX4 or KX2CrX2OX7/HX+ with heat), both side-chains are oxidized to carboxyl groups. The methyl becomes −COOH, and the acetyl group (which is already partially oxidized) also becomes −COOH. The product is terephthalic acid (benzene-1,4-dicarboxylic acid). CHX3−CX6HX4−COCHX3 (para)KMnOX4/OHX−,ΔHOOC−CX6HX4−COOH …
Method: Controlled Oxidation Using Strong Oxidising Agents
The core concept is that primary alcohols and aldehydes can be oxidised to carboxylic acids, while alkyl side chains on aromatic rings can be fully oxidised to carboxyl groups. The most common reagent for these transformations is acidified potassium dichromate (KX2CrX2OX7/HX2SOX4) or alkaline potassium permanganate (KMnOX4). Below are the stepwise reactions.
(i) Butan-1-ol → Butanoic acid
Reagent: Acidified KX2CrX2OX7 (or KMnOX4/HX+)
Steps:
- Primary alcohol (−CHX2OH) is first oxidised to an aldehyde (butanal).
- The aldehyde is further oxidised to the carboxylic acid.
Reaction:
CHX3CHX2CHX2CHX2OH[O]CHX3CHX2CHX2COOH
Detailed:
CHX3CHX2CHX2CHX2OH+2[O]KX2CrX2OX7/HX2SOX4CHX3CHX2CHX2COOH+HX2O
(ii) Benzyl alcohol → Phenylethanoic acid
Reagent: PClX5 (or SOClX2), then KCN, then HX3OX+ — chain extension via cyanide
Steps:
- Convert benzyl alcohol (CX6HX5CHX2OH) to benzyl chloride with PClX5 or SOClX2.
- Substitute with KCN to get benzyl cyanide (CX6HX5CHX2CN) — this adds the extra carbon.
- Hydrolyse the nitrile (HX3OX+, heat) to phenylethanoic acid.
- Direct oxidation cannot work: it gives benzoic acid, one carbon short.
Route: benzyl alcohol → benzyl chloride (PClX5/SOClX2) → benzyl cyanide (KCN) → phenylethanoic acid (acidic hydrolysis). Direct oxidation cannot work — it would give benzoic acid, one carbon short.
(iii) 3-Nitrobromobenzene → 3-Nitrobenzoic acid
Reagent: Mg/dry ether, then COX2, then HX3OX+ (Grignard carboxylation — the textbook's route)
Steps:
- The bromine atom (−Br) is not oxidised; the carboxyl group is introduced through a Grignard reagent.
- Treat 3-nitrobromobenzene with magnesium in dry ether to form the arylmagnesium bromide.
- Pass in COX2 — the Grignard adds to it to give the magnesium carboxylate; acidification then releases 3-nitrobenzoic acid.
Reaction (Grignard carboxylation):
Br−CX6HX4−NOX2Mg,dry etherBrMg−CX6HX4−NOX21⋅ COX22⋅ HX3OX+HOOC−CX6HX4−NOX2
(iv) 4-Methylacetophenone → Benzene-1,4-dicarboxylic acid
Reagent: Hot alkaline KMnOX4 (strong oxidation)
Steps:
- The methyl group (−CHX3) on the benzene ring is oxidised to −COOH.
- The acetyl group (−COCHX3) is also oxidised to −COOH.
- Both side chains become carboxyl groups at positions 1 and 4.
Reaction:
CHX3−CX6HX4−COCHX3KMnOX4/OHX−,ΔHOOC−CX6HX4−COOH
(Product: terephthalic acid)
(v) Cyclohexene → Hexane-1,6-dioic acid …
Here are the common mistakes students make on these specific oxidation transformations, along with the conceptual corrections and exam-smart strategies to avoid them.
(i) Butan-1-ol → Butanoic acid
Common Mistake: Using KMnO4 or K2Cr2O7 in neutral/alkaline medium and stopping at the aldehyde stage, or using a weak oxidising agent like PCC.
Why it’s wrong:
- Butan-1-ol is a primary alcohol.
- To get the carboxylic acid, you need a strong oxidising agent in acidic medium.
- In neutral/alkaline KMnO4, the reaction often stops at the aldehyde or gives a mixture.
Correct Approach:
Use acidified K2Cr2O7 or acidified KMnO4 (warm).
CH3CH2CH2CH2OHK2Cr2O7H+CH3CH2CH2COOH
Avoid by:
- Always check the medium (acidic vs. neutral/alkaline).
- Remember: Primary alcohol → Aldehyde → Carboxylic acid requires strong oxidant + acid.
(ii) Benzyl alcohol → Phenylethanoic acid
Common Mistake: Thinking direct oxidation of benzyl alcohol gives phenylethanoic acid.
Why it’s wrong:
- Benzyl alcohol is C6H5CH2OH.
- Direct oxidation (e.g., with KMnO4) gives benzoic acid (C6H5COOH), not phenylethanoic acid (C6H5CH2COOH).
- The carbon chain length changes — you need to increase the carbon chain first.
Correct Approach:
This is a two-step process:
- Convert benzyl alcohol to benzyl chloride (using PCl5 or SOCl2).
- Do a nucleophilic substitution with KCN to get benzyl cyanide, then hydrolyse (acidic) to get phenylethanoic acid.
C6H5CH2OHPCl5C6H5CH2ClKCNC6H5CH2CNH3O+C6H5CH2COOH
Avoid by:
- Recognising that direct oxidation of a side chain always shortens it to COOH at the benzylic carbon.
- To retain or lengthen the chain, use cyanide + hydrolysis.
(iii) 3-Nitrobromobenzene → 3-Nitrobenzoic acid
Common Mistake: Trying to oxidise the bromine atom directly.
Why it’s wrong:
- Bromine is a halogen, not an alkyl group. It cannot be oxidised to COOH.
- You need to replace the bromine with a group that can be oxidised.
Correct Approach:
- Convert bromobenzene to a Grignard reagent (using Mg in dry ether).
- React with CO2 (dry ice) to form the carboxylate salt.
- Acidify to get 3-nitrobenzoic acid.
Br-C6H4-NO2Mg/etherBrMg-C6H4-NO2CO2O2N-C6H4-COOMgBrH3O+HOOC-C6H4-NO2
Avoid by:
- Remember: Halogens are not oxidisable to acids. Use Grignard or cyanide substitution.
- Check the functional group: if it’s a halogen, think nucleophilic substitution first.
(iv) 4-Methylacetophenone → Benzene-1,4-dicarboxylic acid
Common Mistake: Oxidising only the methyl group or only the acetyl group, or using a weak oxidant.
Why it’s wrong:
- You need to oxidise both the methyl (CH3) and the acetyl (COCH3) groups to COOH.
- Weak oxidants (e.g., SeO2) only affect the methyl group.
Correct Approach:
Use strong oxidising conditions — alkaline KMnO4 followed by acidification.
CH3-C6H4-COCH3KMnO4/OH−heatHOOC-C6H4-COOH
Avoid by:
- Recognise that both alkyl and acyl groups attached to benzene can be oxidised to COOH under strong conditions.
- Use alkaline KMnO4 (not acidic) to avoid side reactions.
(v) Cyclohexene → Hexane-1,6-dioic acid
Common Mistake: Using KMnO4 in cold/neutral conditions (which gives a diol) or using ozonolysis (which gives dialdehyde).
Why it’s wrong:
- Cold KMnO4 gives cis-1,2-cyclohexanediol (not ring opening).
- Ozonolysis gives hexanedial (dialdehyde), not the diacid.
Correct Approach:
Use hot, concentrated KMnO4 (or K2Cr2O7 in acid) to cleave the double bond and oxidise both ends to COOH.
C6H10KMnO4/H+heatHOOC-(CH2)4-COOH
Avoid by:
- Remember: Hot, strong oxidants cleave alkenes to carboxylic acids.
- Cold/neutral conditions give diols — not what’s asked.
(vi) Butanal → Butanoic acid …
- COMEDK 2026Set 2026-M1 markMCQQ.Etard reaction is a method of preparation of benzaldehyde by oxidation of toluene. The oxidizing agent used in this reaction is: (A) Chromic oxide (B) Chromyl chloride (C) Potassium dichromate (D) Pyridinium chlorochromate
›Reveal solutionSolution
The Etard reaction uses chromyl chloride (CrO2Cl2) to selectively oxidize the methyl group of toluene to an aldehyde, stopping at benzaldehyde without overoxidation. The correct option is (B).
The Etard reaction is a classic, elegant method for converting a methyl group attached to an aromatic ring (like toluene) directly into an aldehyde group (like benzaldehyde). The key challenge in such oxidations is stopping at the aldehyde stage — most strong oxidizers (like potassium dichromate in acid) would push all the way to benzoic acid. The genius of the Etard reaction lies in using a specific, milder oxidizing agent that forms a stable intermediate complex, preventing overoxidation.
Why chromyl chloride?
Chromyl chloride (CrO2Cl2) is a powerful but selective oxidant. It reacts with the benzylic C–H bonds of toluene to form a solid, insoluble complex (often called the Etard complex). This complex can be isolated and then hydrolyzed (with water or dilute acid) to release benzaldehyde. The chromium is reduced, but the aldehyde is protected within the complex until workup. Other chromium(VI) reagents like chromic oxide or potassium dichromate are too aggressive in acidic media — they generate the aldehyde but immediately oxidize it further. Pyridinium chlorochromate (PCC) is milder but typically used in anhydrous conditions for alcohols, not for direct methyl-to-aldehyde conversion on toluene.
Let’s walk through the reasoning step by step.
-
Identify the goal: We need an oxidant that converts the methyl group (–CH3) of toluene to a formyl group (–CHO) without further oxidation to a carboxyl group (–COOH). This requires a reagent that either (a) forms a protective intermediate or (b) is inherently mild enough to stop at the aldehyde.
-
Evaluate option (A) – Chromic oxide (CrO3): In aqueous acidic conditions, chromic oxide is a very strong oxidizer. It would oxidize toluene first to benzyl alcohol, then to benzaldehyde, and then rapidly to benzoic acid. It does not form a stable isolable intermediate with the aldehyde. So this is not suitable for stopping at benzaldehyde.
-
Evaluate option (B) – Chromyl chloride (CrO2Cl2): This is the classic Etard reagent. It reacts with toluene in carbon disulfide or carbon tetrachloride to form a brownish-red precipitate — the Etard complex. The complex is thought to be a cyclic adduct where chromium is coordinated to the benzylic carbon and oxygen. Upon hydrolysis, this complex decomposes to give benzaldehyde. The key is that the aldehyde is “masked” in the complex until workup, preventing overoxidation. This is the correct reagent. …
-
- COMEDK 2025Set 2025-A1 markMCQQ.An Alkene " X " on reaction with hot acidified KMnO4 gave a mixture of Ethanoic acid and Propanone. Identify " X ". (A) Pent-2-ene (B) 2-Methylbut-2-ene. (C) But-2-ene (D) 2,3 -Dimethylbut-2-ene
›Reveal solutionSolution
The key idea is that hot acidic KMnO₄ cleaves alkenes at the double bond, turning each doubly bonded carbon into a carbonyl group (ketone or carboxylic acid). The products given — ethanoic acid and propanone — uniquely point to the alkene 2‑methylbut‑2‑ene, option (B).
Concept & Intuition
Hot acidic potassium permanganate (KMnO₄) is a strong oxidising agent. When it reacts with an alkene, it doesn’t just stop at a diol — it cleaves the carbon‑carbon double bond completely. Each carbon that was part of the double bond becomes a separate carbonyl compound:
- If that carbon had two alkyl groups attached, it becomes a ketone.
- If it had one alkyl group and one hydrogen, it becomes a carboxylic acid (which may further oxidise to CO₂ if it’s a terminal carbon, but here we have stable products).
- If it had two hydrogens (terminal =CH₂), it becomes CO₂ and water.
So the puzzle is: Which alkene, when cut at the double bond, gives exactly one molecule of ethanoic acid (CH₃COOH) and one molecule of propanone (CH₃COCH₃)?
Step‑by‑step reasoning
- Identify the fragments from the products Ethanoic acid is CH₃–COOH. That means one half of the original double bond must have been a carbon with a methyl group and a hydrogen:
CH3–CH=(the =C becomes –COOH after oxidation)
Propanone is CH₃–CO–CH₃. That means the other half of the double bond must have been a carbon with two methyl groups attached:
–C(CH3)2(the =C becomes >C=O)
- Reconstruct the alkene Join the two fragments at the double bond:
CH3–CH=C(CH3)2
This is 2‑methylbut‑2‑ene.
(Check the name: a 4‑carbon chain with a double bond between C2 and C3, and a methyl substituent on C2.)
- Verify with the options
- (A) Pent‑2‑ene: CH₃–CH=CH–CH₂–CH₃ → would give propanoic acid + ethanoic acid (not propanone). …
- COMEDK 2025Set 2025-E1 markMCQQ.An organic compound [X] (Molecular formula C6H12O2 ) reacts with dil. H2SO4 to form an alcohol [B] and a carboxylic acid [C]. Reaction of compound [B] with Jones reagent yielded compound [C]. When compound [C] was heated with P2O5 an Anhydride was formed. Compound [X] is ------------- -- (A) C2H5−COO−(CH2)2−CH3 (B) CH3−COO−(CH2)3−CH3 (C) CH3−COO−CH2−CH−(CH3)2 (D) CH3−COO−C−(CH3)3
›Reveal solutionSolution
[X] hydrolyses to an alcohol [B] and acid [C]; since [B] is oxidised (Jones reagent) to the same acid [C], and [C] forms an anhydride with P2O5, [X] is propyl propanoate, option (A).
Work through the clues.
- [X] (C6H12O2) dil. H2SO4 alcohol [B] + carboxylic acid [C] — so [X] is an ester.
- [B]Jones reagent[C]: the alcohol is oxidised to the acid. A primary alcohol R–CH2OH oxidises to R–COOH. For this product to be [C] itself, the acid and alcohol must share the same carbon skeleton — i.e. the ester is R–COO–CH2–R with both halves derived from the same acid.
- [C]P2O5, Δ anhydride, confirming [C] is a carboxylic acid.
Test option (A), C2H5–COO–(CH2)2–CH3 (propyl propanoate):
- Hydrolysis → propan-1-ol [B] + propanoic acid [C]. …
- KCET 2022Set B-31 markMCQQ.The test to differentiate between pentan-2-one and pentan-3-one is (A) Fehling’s test (B) Iodoform test (C) Baeyer’s test (D) Benedict’s test
›Reveal solutionSolution
Only a methyl ketone gives iodoform; pentan-2-one is one and pentan-3-one is not, so I2/NaOH separates the pair.
1. The two compounds
Pentan-2-one: CH3−C∣∣O−CH2−CH2−CH3
Pentan-3-one: CH3−CH2−C∣∣O−CH2−CH3
They are positional isomers (C5H10O) — same functional group, different position. So any test that merely detects a ketone will respond identically to both. We need a test sensitive to the position of the carbonyl.
2. Why the iodoform test discriminates
The iodoform reaction requires a CH3 group directly attached to the carbonyl carbon (a methyl ketone), or a CH3CH(OH)− group that can be oxidised to one. Mechanism: the three α-H's of that methyl group are successively replaced by iodine under basic conditions; the resulting −CI3 is an excellent leaving group and is expelled by hydroxide as CHI3 — a yellow crystalline precipitate with a characteristic antiseptic smell.
CH3COR+3I2+4NaOH⟶CHI3↓+RCOONa+3NaI+3H2O
- Pentan-2-one: the carbonyl carries a CH3 group ⇒ positive (yellow CHI3; the other product is sodium butanoate). …
- KCET 2019Set A-11 markMCQQ.Which of the following can be used to test the acidic nature of ethanol ? (A) Blue litmus solution (B) NaHCO3 (C) Na2CO3 (D) Na metal
›Reveal solutionSolution
Ethanol is too weak an acid for litmus or carbonate tests; only sodium metal is basic/reactive enough to pull off its −OH proton, releasing H2.
Step 1 — How weak is ethanol as an acid?
Ethanol ionises as C2H5OH⇌C2H5O−+H+, with pKa≈16 — it is even weaker than water (pKa≈14), and far weaker than carbonic acid (pKa≈6.4) or a carboxylic acid (pKa≈5). The alkyl group is electron-donating (+I effect), which destabilises the alkoxide ion C2H5O− and suppresses ionisation.
Step 2 — Why (A) fails.
Blue litmus turns red only for an acid strong enough to give an appreciable [H+] in water. Ethanol is neutral to litmus — a classic exam point.
Step 3 — Why (B) and (C) fail.
NaHCO3 and Na2CO3 liberate CO2 only with acids stronger than carbonic acid — that is the standard test that distinguishes carboxylic acids (which do effervesce) from alcohols and phenols (which do not). Ethanol, at pKa≈16, is nowhere near strong enough, so no brisk effervescence occurs. (Even phenol, pKa≈10, fails this test.) …
- KCET 2018Set A-11 markMCQQ.In the following reaction CHX3CrOX2ClX2HX3OX+ CSX2X⟶Z the compound Z is (A) Benzoic acid (B) Benzaldehyde (C) Acetophenone (D) Benzene
›Reveal solutionSolution
CrOX2ClX2 in CSX2 (Etard reaction) oxidises toluene's methyl group only to the aldehyde stage — the intermediate chromium complex X hydrolyses to benzaldehyde, Z.
Step 1 — Recognise the reagent.
CrOX2ClX2 is chromyl chloride; used in an inert solvent such as CSX2 or CClX4 on a methyl arene, this is the classic Etard reaction.
Step 2 — What X is.
Chromyl chloride attacks the benzylic −CHX3 group and forms a brown chromium complex (an addition complex at the benzylic carbon):
CX6HX5−CHX3+CrOX2ClX2CSX2CX6HX5−CH(OCrOHClX2)X2(= X, the Etard complex)
The key point is that the oxidation is arrested at this complex — the carbon is not oxidised further while it is tied up in the complex.
Step 3 — What Z is.
Hydrolysis of the Etard complex with HX3OX+ liberates the carbonyl compound at the aldehyde oxidation level:
CX6HX5−CH(OCrOHClX2)X2HX3OX+CX6HX5−CHO …
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