Q.Give simple chemical tests to distinguish between the following pairs of compounds.
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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 key is that aldehydes are easily oxidised to carboxylic acids (they reduce Tollen’s reagent and Fehling’s solution), while ketones are not. Also, methyl ketones give the iodoform test, and phenols give a colour with neutral FeCl₃.
Reasoning steps:
- Propanal vs Propanone: Propanal (aldehyde) reduces Tollen’s reagent (silver mirror) and Fehling’s solution (red precipitate); propanone does not.
- Acetophenone vs Benzophenone: Acetophenone (methyl ketone) gives a yellow precipitate of iodoform with I₂/NaOH; benzophenone does not.
- Phenol vs Benzoic acid: Phenol gives a violet colour with neutral FeCl₃; benzoic acid does not.
- Benzoic acid vs Ethyl benzoate: Benzoic acid reacts with NaHCO₃ to evolve CO₂ gas; ethyl benzoate does not.
- Pentan-2-one vs Pentan-3-one: Pentan-2-one (methyl ketone) gives iodoform test positive; pentan-3-one gives negative.
- Benzaldehyde vs Acetophenone: Benzaldehyde reduces Tollen’s reagent (silver mirror); acetophenone does not. …
The key idea is to use specific chemical tests that exploit differences in functional groups, oxidation behaviour, or structural features. For each pair, a single reagent gives a distinct observable change (colour, precipitate, gas) for one compound but not the other.
(i) Propanal and Propanone
Concept: Both are carbonyl compounds, but propanal is an aldehyde (easily oxidised) while propanone is a ketone (resistant to mild oxidation). The Tollens' test (ammoniacal silver nitrate) is perfect here — aldehydes reduce Ag⁺ to a silver mirror, ketones do not.
Step-by-step:
- Prepare Tollens' reagent: add dilute NaOH to AgNO₃ solution, then add just enough NH₄OH to dissolve the brown precipitate.
- Add a small sample of each compound to separate test tubes containing the reagent.
- Warm gently in a water bath (do not boil).
- Observation: Propanal gives a bright silver mirror on the tube wall. Propanone gives no change. …
Here is the clear solution method for distinguishing the given pairs of compounds using simple chemical tests.
Method Name: Functional Group & Oxidation State Differentiation via Specific Reagent Tests
Core Concept: Different functional groups (aldehydes, ketones, carboxylic acids, phenols, esters) react differently to specific oxidizing agents or nucleophiles. The key is to identify a reagent that gives a positive test (color change, precipitate, gas evolution) with one compound but no reaction with the other.
(i) Propanal and Propanone
Method: Tollens' Test (Silver Mirror Test)
- Reagent: Ammoniacal silver nitrate ([Ag(NH3)2]+OH−).
- Steps:
- Take a small amount of each compound in separate clean test tubes.
- Add Tollens' reagent to each.
- Warm gently in a water bath (do not boil).
- Observation:
- Propanal: A silver mirror forms on the inner wall of the test tube (aldehyde is oxidized to carboxylate; Ag+ is reduced to Ag).
- Propanone: No reaction (ketones are not easily oxidized by Tollens' reagent).
(ii) Acetophenone and Benzophenone
Method: Iodoform Test (Methyl Ketone Test)
- Reagent: Iodine (I2) in the presence of sodium hydroxide (NaOH).
- Steps:
- Dissolve a small amount of each compound in a little ethanol or water.
- Add I2 solution and then NaOH solution dropwise until the brown color of iodine just disappears.
- Warm gently.
- Observation:
- Acetophenone: A yellow precipitate of iodoform (CHI3) forms (it has a CH3CO− group).
- Benzophenone: No yellow precipitate (it lacks the CH3CO− group).
(iii) Phenol and Benzoic acid
Method: Neutral FeCl3 Test (Ferric Chloride Test)
- Reagent: Neutral ferric chloride solution (FeCl3).
- Steps:
- Dissolve a small amount of each compound in water (or a little alcohol if insoluble).
- Add a few drops of neutral FeCl3 solution.
- Observation:
- Phenol: A violet/blue color appears (due to formation of a colored iron-phenolate complex).
- Benzoic acid: A buff-colored precipitate (or no color change) forms (carboxylic acids give a different, less intense color or precipitate).
(iv) Benzoic acid and Ethyl benzoate
Method: Sodium Bicarbonate Test (NaHCO3 Test)
- Reagent: Aqueous sodium bicarbonate (NaHCO3).
- Steps:
- Take a small amount of each compound in separate test tubes.
- Add a few mL of NaHCO3 solution.
- Observe for effervescence.
- Observation:
- Benzoic acid: Brisk effervescence of CO2 gas (acid reacts with bicarbonate).
- Ethyl benzoate: No effervescence (ester is not acidic enough to react with NaHCO3).
--- …
Here are the common mistakes students make when tackling these distinction tests in oxidation reactions, along with how to avoid each.
General Mistake: Confusing "Chemical Test" with "Physical Property"
- The Mistake: Trying to distinguish compounds by smell, colour, or solubility (e.g., "Propanal smells different").
- Why it fails: Exam questions demand a reagent and a visible observation (precipitate, colour change, gas evolution). Physical properties are not considered "chemical tests."
- How to Avoid: Always ask yourself: "What reagent do I add, and what do I see?" If you can't name a reagent, your answer is incomplete.
(i) Propanal vs. Propanone
Common Mistake: Using Tollen’s reagent or Fehling’s solution but forgetting that both are aldehydes/ketones.
- The Error: Students write "Propanal gives silver mirror, Propanone does not" — which is correct, but they often mislabel which is which.
- How to Avoid: Remember: Only aldehydes reduce Tollen’s reagent. Propanal (CH3CH2CHO) is an aldehyde; Propanone (CH3COCH3) is a ketone.
- Correct Test: Add Tollen’s reagent (ammoniacal silver nitrate) and warm.
- Propanal: Silver mirror forms.
- Propanone: No reaction.
(ii) Acetophenone vs. Benzophenone
Common Mistake: Assuming both are ketones, so no test works.
- The Error: Students forget that methyl ketones (like Acetophenone, C6H5COCH3) give the iodoform test, while Benzophenone (C6H5COC6H5) does not.
- How to Avoid: Check the structure: Does the carbonyl have a CH3 group attached? If yes → iodoform test positive.
- Correct Test: Add iodine (I2) and NaOH (or NaOI) and warm.
- Acetophenone: Yellow precipitate of CHI3 (iodoform).
- Benzophenone: No yellow precipitate.
(iii) Phenol vs. Benzoic acid
Common Mistake: Using NaHCO3 test but forgetting that phenol is a weaker acid.
- The Error: Students think both react with NaHCO3 because both are acidic. Phenol does not react with NaHCO3 (it needs NaOH).
- How to Avoid: Remember the acid strength order: Benzoic acid (pKa≈4.2) > Carbonic acid (pKa≈6.4) > Phenol (pKa≈10). Only benzoic acid displaces CO2 from NaHCO3.
- Correct Test: Add aqueous NaHCO3.
- Benzoic acid: Effervescence of CO2 gas.
- Phenol: No effervescence.
(iv) Benzoic acid vs. Ethyl benzoate
Common Mistake: Using NaHCO3 test but forgetting that esters are neutral.
- The Error: Students think Ethyl benzoate (C6H5COOC2H5) might react because it has a carbonyl group. It does not — esters do not release H+ ions.
- How to Avoid: Only free carboxylic acids (−COOH) react with NaHCO3 to give CO2. Esters are not acidic.
- Correct Test: Add aqueous NaHCO3.
- Benzoic acid: Effervescence of CO2.
- Ethyl benzoate: No reaction.
(v) Pentan-2-one vs. Pentan-3-one
Common Mistake: Using Tollen’s or Fehling’s test — both are ketones.
- The Error: Students forget that both are ketones, so neither reduces Tollen’s reagent. They need a test that distinguishes a methyl ketone from a non-methyl ketone.
- How to Avoid: Check the position of the carbonyl. Pentan-2-one (CH3COCH2CH2CH3) has a CH3CO− group → iodoform positive. Pentan-3-one (CH3CH2COCH2CH3) does not.
- Correct Test: Iodoform test (iodine + NaOH).
- Pentan-2-one: Yellow precipitate of CHI3.
- Pentan-3-one: No precipitate.
(vi) Benzaldehyde vs. Acetophenone
Common Mistake: Using Tollen’s test but forgetting that Benzaldehyde is an aldehyde and Acetophenone is a methyl ketone.
- The Error: Students often use only one test and miss the fact that both can give positive results with different reagents. …
- 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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