Q.Give names of the reagents to bring about 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.
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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 — controlled oxidation of alcohols, alkenes, and nitriles to aldehydes or ketones using specific reagents.
Reasoning steps:
- Primary alcohol → aldehyde: Use a mild oxidising agent that stops at the aldehyde stage, avoiding over-oxidation to carboxylic acid.
- Secondary alcohol → ketone: Use a standard oxidising agent; ketones resist further oxidation.
- Methylarene → benzaldehyde: Use chromyl chloride (Etard reaction) to oxidise the methyl group directly to aldehyde.
- Nitrile → aldehyde: Use DIBAL-H (partial reduction) followed by hydrolysis.
- Allyl alcohol → α,β-unsaturated aldehyde: Use a mild oxidant like PCC that does not attack the double bond. …
The key idea is to use mild oxidizing agents that stop at the aldehyde stage without over-oxidizing to the carboxylic acid. The final reagents are: (i) PCC,
(ii) anhydrous CrO₃ (PCC or Na₂Cr₂O₇/H⁺ also work),
(iii) CrO₂Cl₂ (Etard reagent), then H₃O⁺,
(iv) DIBAL-H then H₃O⁺,
(v) PCC,
(vi) O₃ then Zn/H₂O.
Oxidation reactions in organic chemistry are all about controlling the degree of oxidation. For alcohols, a primary alcohol can be oxidized to an aldehyde or further to a carboxylic acid. The trick is to choose a reagent that stops at the aldehyde stage — typically a mild oxidant like PCC (pyridinium chlorochromate) or a controlled reduction of a nitrile. For alkenes, ozonolysis cleaves the double bond to give carbonyl compounds. Let's go through each transformation.
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Hexan-1-ol to hexanal
Hexan-1-ol is a primary alcohol. To get the aldehyde without over-oxidation, use PCC in dichloromethane. PCC is a mild, anhydrous reagent that stops at the aldehyde. Strong oxidants like K₂Cr₂O₇/H₂SO₄ would give hexanoic acid.
Reagent: PCC (pyridinium chlorochromate) in CH₂Cl₂.
-
Cyclohexanol to cyclohexanone
Cyclohexanol is a secondary alcohol. Oxidation of a secondary alcohol gives a ketone, and there's no risk of further oxidation (ketones are resistant to mild oxidants). The textbook's reagent is anhydrous CrO₃ — without water there is no scope for over-oxidation. PCC or Na₂Cr₂O₇/H₂SO₄ also work for a secondary alcohol.
Reagent: Anhydrous CrO₃ (also acceptable: PCC or Na₂Cr₂O₇/H₂SO₄).
-
p-Fluorotoluene to p-fluorobenzaldehyde
This is a side-chain oxidation of a methyl group attached to an aromatic ring. Direct oxidation with KMnO₄ would give the carboxylic acid (p-fluorobenzoic acid). To stop at the aldehyde, use the Etard reaction: chromyl chloride (CrO₂Cl₂) reacts with the methyl group to form a chromium complex, which is then hydrolysed with H₃O⁺ to release the aldehyde.
Reagent: CrO₂Cl₂ (Etard reagent) then H₃O⁺.
-
Ethanenitrile to ethanal
A nitrile (R-C≡N) can be reduced to an aldehyde using DIBAL-H (diisobutylaluminium hydride) at low temperature, followed by hydrolysis. DIBAL-H adds hydride to the nitrile carbon, forming an imine intermediate that hydrolyzes to the aldehyde. LiAlH₄ would over-reduce to the primary amine.
Reagent: DIBAL-H (in hexane or toluene, -78°C), then H₃O⁺.
-
Allyl alcohol to propenal …
Here is the clear solution method for the given transformations, focusing on the concept of controlled oxidation.
Method: Controlled Oxidation using Specific Oxidizing Agents
The core concept is that different oxidizing agents have different strengths. To stop an oxidation at the aldehyde stage (without over-oxidizing to a carboxylic acid), we must use a mild or specific reagent.
Steps for each transformation:
(i) Hexan-1-ol to hexanal
- Reagent: PCC (Pyridinium Chlorochromate) in anhydrous conditions (e.g., dry CH2Cl2).
- Why: PCC is a mild oxidizing agent. It oxidizes primary alcohols to aldehydes without further oxidizing them to carboxylic acids. Strong agents like K2Cr2O7/H2SO4 would give hexanoic acid.
(ii) Cyclohexanol to cyclohexanone
- Reagent: Anhydrous CrO3 (the textbook's choice; PCC or Jones reagent also work).
- Why: This is the oxidation of a secondary alcohol. Secondary alcohols oxidize to ketones (which cannot be further oxidized easily without breaking carbon-carbon bonds). Both mild (PCC) and strong (Jones) reagents work here, but PCC is often preferred for sensitive molecules.
(iii) p-Fluorotoluene to p-fluorobenzaldehyde
- Reagent: Chromyl chloride (CrO2Cl2) in CS2 or CCl4, followed by hydrolysis (aqueous workup). This is the Etard reaction.
- Why: Direct oxidation of a methyl group attached to a benzene ring (side-chain oxidation) usually gives benzoic acid. The Etard reaction specifically stops at the aldehyde stage by forming a stable complex with the benzylic carbon.
(iv) Ethanenitrile to ethanal
- Reagent: Diisobutylaluminium hydride (DIBAL-H) at low temperature (−78∘C), followed by hydrolysis.
- Why: Nitriles (R−C≡N) are reduced by DIBAL-H to an imine intermediate. Controlled hydrolysis of this imine yields the aldehyde. This is a reduction, not an oxidation, but it is the standard method to get an aldehyde from a nitrile.
(v) Allyl alcohol to propenal
- Reagent: PCC (Pyridinium Chlorochromate) in CH2Cl2.
- Why: Allyl alcohol is a primary allylic alcohol. PCC selectively oxidizes the −OH group to an aldehyde without attacking the carbon-carbon double bond. Stronger oxidants would break the double bond.
(vi) But-2-ene to ethanal …
Here are the common mistakes students make when tackling these specific oxidation transformations, along with the precise conceptual corrections needed to avoid them.
General Mistake: Confusing "Strong" vs. "Mild" Oxidising Agents
The most frequent error is using a strong oxidising agent (like KMnO4 or K2Cr2O7/H+) for every reaction. Strong agents often over-oxidise, turning an aldehyde into a carboxylic acid or breaking carbon-carbon bonds.
- The Concept: Aldehydes (RCHO) are easier to oxidise than primary alcohols (RCH2OH). To stop at the aldehyde, you must use a mild reagent that cannot further oxidise the product.
- How to Avoid: Memorise the "mild" list: PCC (Pyridinium Chlorochromate), Collins reagent, or Swern oxidation. For strong oxidation (to acid), use K2Cr2O7/H2SO4 or KMnO4.
(i) Hexan-1-ol to hexanal
Common Mistake: Using K2Cr2O7/H2SO4 (acidified dichromate).
- Why it's wrong: This is a strong oxidising agent. It will convert the primary alcohol (CH3(CH2)4CH2OH) all the way to hexanoic acid (CH3(CH2)4COOH), not stopping at the aldehyde.
- Correct Reagent: PCC (Pyridinium Chlorochromate) in anhydrous CH2Cl2.
- How to Avoid: Remember the rule: Primary alcohol PCC Aldehyde. PCC is the "stop sign" for oxidation.
(ii) Cyclohexanol to cyclohexanone
Common Mistake: Using PCC here when a cheaper, stronger reagent works perfectly.
- Why it's wrong: PCC is correct but expensive. Students often forget that secondary alcohols can be safely oxidised to ketones using strong agents because ketones are resistant to further oxidation (they have no H on the carbonyl carbon to lose).
- Correct Reagent: Anhydrous CrO3 — the textbook's answer. Acidified dichromate and PCC also oxidise a secondary alcohol cleanly.
- How to Avoid: Distinguish between primary and secondary alcohols. Secondary → Ketone is safe with strong oxidants. Primary → Aldehyde requires mild oxidants.
(iii) p-Fluorotoluene to p-fluorobenzaldehyde
Common Mistake: Trying to oxidise the methyl group (CH3) directly with KMnO4 or CrO3.
- Why it's wrong: Strong oxidants on a methyl group attached to a benzene ring will oxidise it all the way to p-fluorobenzoic acid (COOH). You cannot stop at the aldehyde stage with these reagents.
- Correct Reagent: Chromyl chloride (CrO2Cl2) in the Etard reaction.
- How to Avoid: For toluene → benzaldehyde (or substituted variants), the specific reagent is Etard's reagent (CrO2Cl2). Do not use general oxidisers.
(iv) Ethanenitrile to ethanal
Common Mistake: Trying to oxidise the nitrile (CH3CN) directly.
- Why it's wrong: Nitriles are not alcohols. You cannot "oxidise" a nitrile to an aldehyde. You must reduce it in a controlled way (DIBAL-H) and then hydrolyse the intermediate imine to the aldehyde.
- Correct Reagent: DIBAL-H (diisobutylaluminium hydride) at low temperature, then hydrolysis — the textbook's reagent. (Stephen reduction, SnCl2/HCl, is an older alternative route.)
- How to Avoid: Recognise that this is a controlled reduction of a nitrile, not an oxidation. The reaction is: RCN1. DIBAL-H2. H3O+RCHO.
(v) Allyl alcohol to propenal
Common Mistake: Using K2Cr2O7/H+ which might also attack the double bond.
- Why it's wrong: Strong oxidants like KMnO4 or hot HNO3 can cleave the carbon-carbon double bond (C=C) in allyl alcohol (CH2=CH−CH2OH), destroying the molecule. …
- 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.
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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.
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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. …
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- 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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