Q.How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
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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 (side-chain oxidation, nitration, esterification)
Reasoning steps:
- Methyl benzoate – Benzene undergoes Friedel–Crafts acylation with acetyl chloride (CHX3COCl, AlClX3) to give acetophenone. Then haloform reaction (IX2/NaOH) oxidises the methyl ketone to benzoic acid. Finally, esterify with CHX3OH/HX+.
- m-Nitrobenzoic acid – Nitrate benzoic acid (already prepared) with HNOX3/HX2SOX4; the −COOH group is meta-directing, giving exclusively m-nitrobenzoic acid.
- p-Nitrobenzoic acid – First nitrate benzene to nitrobenzene (HNOX3/HX2SOX4). Then oxidise the side chain? No — nitrobenzene has no alkyl group. Instead, use a different route: convert benzene to toluene (Friedel–Crafts alkylation with CHX3Cl/AlClX3), nitrate toluene (HNOX3/HX2SOX4) to get a mixture of o- and p-nitrotoluene, separate p-nitrotoluene, then oxidise the methyl group with KMnOX4/HX+ to get p-nitrobenzoic acid.
- Phenylacetic acid – Benzene + CHX3COCl/AlClX3 → acetophenone. Reduce the carbonyl to ethylbenzene? No — use Arndt–Eistert homologation? Simpler: prepare benzyl chloride (chloromethylation: HCHO/HCl/ZnClX2), then react with KCN to give benzyl cyanide, hydrolyse (HX3OX+) to phenylacetic acid. …
The key idea is to use benzene's electrophilic substitution to introduce functional groups, then transform them via oxidation, reduction, or hydrolysis -- all using reagents with <=1 carbon. The final compounds are prepared as follows: (i) Friedel-Crafts acylation, haloform oxidation, then esterification;
(ii) direct nitration of benzoic acid (the -COOH group is meta-directing, so this gives the meta isomer directly);
(iii) alkylate benzene to toluene, nitrate toluene (the -CH3 group is ortho/para-directing), separate the para isomer, then oxidize the methyl group to -COOH;
(iv) chloromethylation to benzyl chloride, conversion to the nitrile with KCN, then hydrolysis;
(v) alkylate benzene to toluene, nitrate toluene (methyl directs para), separate the para isomer, then selectively oxidize the methyl group to -CHO (not all the way to -COOH).
Concept and Intuition
Benzene is an aromatic ring that undergoes electrophilic substitution -- not addition -- because its pi-system is too stable to break. To attach a side chain, we use reactions like Friedel-Crafts alkylation/acylation or nitration. Once a group is attached, we can oxidize or reduce it to get the desired product. The constraint 'reagents with <=1 carbon' means we cannot use Grignard reagents with longer chains or complex organometallics; we rely on simple one-carbon units like CH3Cl, HCHO, etc.
A point that decides several of these routes: which existing group is a meta-director and which is an ortho/para-director. -COOH, -CHO and -COCH3 are all electron-withdrawing and meta-directing; -CH3 (and -NH2, after protection) are electron-donating and ortho/para-directing. Whenever the target needs a nitro group PARA to a meta-director like -COOH or -CHO, nitrating that meta-director directly will not work (it gives the meta product) -- the practical route is to build the target via a -CH3 group instead (which sends the new nitro group ortho/para), separate the desired para isomer, and only then oxidize the methyl group into the -COOH or -CHO that was actually wanted.
Step-by-Step Solutions
(i) Methyl benzoate
- Introduce a carbonyl via Friedel-Crafts acylation. React benzene with acetyl chloride (CH3COCl) and anhydrous AlCl3 to form acetophenone (C6H5COCH3).
- Oxidize the methyl ketone to a carboxylic acid. Treat acetophenone with I2/NaOH (haloform-type oxidative cleavage) or alkaline KMnO4 to oxidize -COCH3 to -COOH, giving benzoic acid (C6H5COOH).
- Esterify with methanol. React benzoic acid with CH3OH in the presence of conc. H2SO4 (Fischer esterification) to form methyl benzoate (C6H5COOCH3).
Direct Friedel-Crafts alkylation with CH3Cl to get toluene, then oxidizing the methyl all the way to -COOH, also reaches benzoic acid -- but the acylation-oxidation route above is the cleaner, standard two-step answer.
(ii) m-Nitrobenzoic acid
- First make benzoic acid (as in part i, steps 1-2).
- Nitrate benzoic acid. Treat benzoic acid with a nitrating mixture (conc. HNO3 + conc. H2SO4). The -COOH group is meta-directing, so the nitro group goes predominantly to the meta position, giving m-nitrobenzoic acid directly.
C6H5COOH + HNO3 --H2SO4--> m-O2NC6H4COOH
(iii) p-Nitrobenzoic acid
Nitrating benzoic acid directly (as in part ii) gives overwhelmingly the META isomer, because -COOH is meta-directing -- there is no useful para fraction to isolate that way. To reach the PARA isomer, the nitro group must be introduced while an ortho/para-director is on the ring, and the -COOH built afterward.
- Alkylate benzene to toluene. Benzene + CH3Cl/anhydrous AlCl3 (Friedel-Crafts alkylation) gives toluene (C6H5CH3).
- Nitrate toluene and separate the para isomer. Toluene + HNO3/H2SO4 gives a mixture of o- and p-nitrotoluene (the -CH3 group is ortho/para-directing); separate p-nitrotoluene by fractional distillation/crystallization.
- Oxidize the methyl group. Treat p-nitrotoluene with hot KMnO4/H+ (or alkaline KMnO4 then acidify) to oxidize -CH3 to -COOH, giving p-nitrobenzoic acid.
Do not nitrate benzoic acid and try to separate a 'para fraction' -- the -COOH group's meta-directing effect makes the para isomer a negligible minor product, not a genuine synthetic route. The toluene route above is the actual standard answer.
(iv) Phenylacetic acid
- Introduce a -CH2Cl group via chloromethylation. React benzene with formaldehyde (HCHO) and HCl in the presence of ZnCl2 (Blanc chloromethylation) to form benzyl chloride (C6H5CH2Cl).
- Convert to benzyl cyanide. Treat benzyl chloride with KCN (or NaCN) in aqueous ethanol to get benzyl cyanide (C6H5CH2CN). …
Here is the clear, concept-first solution for each conversion from benzene.
Core Concept: Benzene is electron-rich and undergoes Electrophilic Aromatic Substitution (EAS). To introduce a carboxylic acid (−COOH) or aldehyde (−CHO) directly, we use oxidation of a pre-attached alkyl side chain. The directing effects of substituents (activating/deactivating, ortho/para vs meta) determine the sequence of steps.
(i) Methyl benzoate from Benzene
Method: Friedel-Crafts Acylation followed by Oxidation and Esterification.
Steps:
- Friedel-Crafts Acylation: Treat benzene with acetyl chloride (CH3COCl) and anhydrous AlCl3 to form acetophenone (C6H5COCH3).
- Oxidation: Oxidize the methyl ketone (−COCH3) to a carboxylic acid (−COOH) using a strong oxidizing agent like KMnO4 / H+ (alkaline KMnO4 followed by acidification). This gives benzoic acid (C6H5COOH).
- Esterification: React benzoic acid with methanol (CH3OH) in the presence of a few drops of concentrated H2SO4 (acid catalyst) to form methyl benzoate (C6H5COOCH3).
Key Equation:
C6H6CH3COCl/AlCl3C6H5COCH3KMnO4/H+C6H5COOHCH3OH/H2SO4C6H5COOCH3
(ii) m-Nitrobenzoic acid from Benzene
Method: Nitration followed by Oxidation.
Steps:
- Nitration: Treat benzene with a nitrating mixture (conc. HNO3 + conc. H2SO4) to form nitrobenzene (C6H5NO2). The −NO2 group is a meta-directing deactivator.
- Oxidation: This is tricky because −NO2 is not an alkyl group. We cannot oxidize it directly. The correct approach is to first introduce a methyl group, then nitrate, then oxidize.
- Friedel-Crafts Alkylation: React nitrobenzene with methyl chloride (CH3Cl) and anhydrous AlCl3. Because −NO2 is meta-directing, the methyl group goes to the meta position, giving m-nitrotoluene.
- Oxidation: Oxidize the methyl group (−CH3) on m-nitrotoluene to a carboxylic acid (−COOH) using KMnO4 / H+ (alkaline KMnO4 then acid). This yields m-nitrobenzoic acid.
Key Equation:
C6H6HNO3/H2SO4C6H5NO2CH3Cl/AlCl3m-CH3C6H4NO2KMnO4/H+m-HOOC-C6H4-NO2
(iii) p-Nitrobenzoic acid from Benzene
Method: Friedel-Crafts Alkylation, then Nitration, then Oxidation.
Steps:
- Friedel–Crafts methylation: Treat benzene with CH3Cl and anhydrous AlCl3 to obtain toluene (C6H5CH3).
- Side-chain chlorination: Chlorinate toluene at the side chain (Cl2, sunlight/hv) to obtain benzyl chloride (C6H5CH2Cl).
- Cyanide substitution: Treat benzyl chloride with KCN to obtain benzyl cyanide (C6H5CH2CN) — this adds the one extra carbon.
- Hydrolysis: Hydrolyse the nitrile (H3O+, heat) to obtain phenylacetic acid (C6H5CH2COOH).
Key Equation:
C6H6ClCH2CN/AlCl3C6H5CH2CNH2O/H+C6H5CH2COOH
(v) p-Nitrobenzaldehyde from Benzene
Method: Friedel-Crafts Acylation, then Nitration, then Reduction (or use a different aldehyde synthesis).
Steps:
- Friedel-Crafts Acylation: Treat benzene with acetyl chloride (CH3COCl) and anhydrous AlCl3 to form acetophenone (C6H5COCH3).
- Nitration: Treat acetophenone with a nitrating mixture (conc. HNO3 + conc. H2SO4). The −COCH3 group is a meta-directing deactivator. The major product is m-nitroacetophenone, not the para isomer.
- Correct approach: We need the −NO2 group para to the aldehyde. So, we must first introduce the aldehyde group, then nitrate.
- Aldehyde introduction: Use Gattermann-Koch formylation: Treat benzene with CO + HCl in the presence of AlCl3 and CuCl to form benzaldehyde (C6H5CHO). …
Here are the common mistakes students make when solving this specific type of benzene conversion problem, along with how to avoid each.
General Mistake: Ignoring the “One Carbon” Rule
The Mistake: Using a reagent with more than one carbon atom (e.g., using ethanol, CH3CH2OH, or acetyl chloride, CH3COCl).
Why it’s wrong: The problem explicitly restricts you to organic reagents having not more than one carbon atom. This means you can only use:
- CH3X (methyl halides)
- HCHO (formaldehyde)
- HCOOH (formic acid)
- CH3OH (methanol)
- CO2 (inorganic, but often forgotten)
- COCl2 (phosgene — rarely needed here)
How to avoid: Before writing any step, check the carbon count of every organic reagent you plan to use. If it has 2 or more carbons, stop and find a one-carbon alternative.
(i) Methyl benzoate — Mistake: Direct Esterification
The Mistake: Trying to make benzoic acid first, then reacting with methanol and H2SO4.
Why it’s wrong: This is actually correct in principle, but students often forget that benzene does not directly give benzoic acid in one step. They write:
- Benzene → Benzoic acid (wrong — no direct oxidation of benzene to benzoic acid)
How to avoid: Remember the correct sequence:
- Friedel-Crafts acylation with CO + HCl + AlCl3 (Gattermann-Koch) to get benzaldehyde.
- Oxidize benzaldehyde to benzoic acid (using KMnO4 or K2Cr2O7).
- Esterify with CH3OH / H2SO4 to get methyl benzoate.
Key point: Benzene must first be converted to a mono-carbon side-chain compound (like benzaldehyde or benzoic acid) before esterification.
(ii) m-Nitrobenzoic acid — Mistake: Wrong Order of Substitution
The Mistake: Nitrating benzoic acid directly (expecting meta product) but forgetting that benzoic acid is meta-directing — so this is actually correct. The real mistake is trying to oxidize m-nitrotoluene but using the wrong oxidising agent or conditions.
Why it’s wrong: Students often write:
- Benzene → Nitrobenzene → m-Nitrobenzoic acid (impossible — nitrobenzene cannot be directly oxidised to benzoic acid)
How to avoid: Use the correct route:
- Friedel-Crafts alkylation with CH3Cl / AlCl3 to get toluene.
- Nitrate toluene (mixed acid) to get a mixture of o- and p-nitrotoluene. Separate m-nitrotoluene? No — nitration of toluene gives mostly o- and p-, not meta. So this route fails.
Correct approach:
- Nitrate benzene to nitrobenzene.
- Reduce nitrobenzene to aniline (Sn/HCl).
- Protect amine (acetylation with CH3COCl — but that’s 2 carbons! So use formylation? No — better: use diazotisation followed by Sandmeyer to get chlorobenzene? That’s too long.)
Simplest correct method:
- Benzene → Benzoic acid (via Gattermann-Koch + oxidation)
- Nitrate benzoic acid with mixed acid → m-Nitrobenzoic acid (since -COOH is meta-directing)
Common error: Forgetting that nitration of benzoic acid gives meta product directly.
(iii) p-Nitrobenzoic acid — Mistake: Assuming Direct Nitration of Benzoic Acid
The Mistake: Trying to nitrate benzoic acid and expecting para product.
Why it’s wrong: Benzoic acid is meta-directing. Nitration gives only m-nitrobenzoic acid, not para.
How to avoid: Use a para-directing group first:
- Nitrate benzene to nitrobenzene.
- Reduce to aniline.
- Protect amine (acetylation — but again, 2-carbon problem). Instead, use diazotisation and replace with -CN (Sandmeyer) to get benzonitrile.
- Hydrolyse benzonitrile to benzoic acid.
- Nitrate? No — now the -COOH is meta-directing again.
Better route:
- Benzene → Toluene (Friedel-Crafts alkylation with CH3Cl)
- Nitrate toluene → mixture of o- and p-nitrotoluene
- Separate p-nitrotoluene (by distillation or crystallisation)
- Oxidise the methyl group to -COOH using KMnO4 → p-Nitrobenzoic acid
Key mistake: Forgetting that separation of isomers is required and that oxidation of p-nitrotoluene works.
(iv) Phenylacetic acid — Mistake: Trying to Make it from Benzene Directly
The Mistake: Writing benzene → benzyl chloride → phenylacetic acid (via NaCN + hydrolysis). This is correct but students forget the one-carbon rule for the cyanide source.
Why it’s wrong: NaCN is inorganic — fine. But the step to get benzyl chloride from benzene requires chloromethylation:
- Benzene + HCHO + HCl + ZnCl2 → Benzyl chloride
Common error: Using CH3Cl + AlCl3 to get toluene, then free radical chlorination to get benzyl chloride — this works but gives mixtures.
How to avoid: Use chloromethylation (Blanc reaction) — it’s a one-carbon addition:
- C6H6+HCHO+HClZnCl2C6H5CH2Cl
- Then C6H5CH2Cl+NaCN→C6H5CH2CN
- Hydrolysis → C6H5CH2COOH
Key point: Remember that formaldehyde (HCHO) is a one-carbon reagent.
--- …
Showing the 12 most recent of 17 on this concept.
- AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQQ.Reaction of toluene with reagent 'A' gave X which reacts with NaHCO3 and liberates CO2. In another reaction with reagent 'B' toluene gave Y which gives 2,4 – DNP test. What are A and B respectively? (A) KMnO4/OH−,H3O+; CrO3/(CH3CO)2O,H3O+ (B) Cl2/hν,H2O; CrO3/H+ (C) CrO2Cl2,H3O+; KMnO4/OH−,H3O+ (D) KMnO4/OH−,H3O+; PCC
›Reveal solutionSolution
X (reacts with NaHCO3, liberates CO2) must be a carboxylic acid; Y (positive 2,4-DNP test) must be an aldehyde. Toluene's methyl group needs full oxidation (KMnO4/OH− then H3O+) to reach benzoic acid, and a controlled, partial oxidation (an Étard-type reagent) to stop cleanly at benzaldehyde.
Concept and Intuition
A compound that fizzes CO2 with NaHCO3 is, by definition, acidic enough to be a carboxylic acid (phenols and alcohols are too weakly acidic to do this). A compound that gives a 2,4-DNP (Brady's reagent) test forms a hydrazone, which only aldehydes and ketones do. So the two reagents must take toluene's benzylic methyl group to two different oxidation states: all the way to −COOH for X, and only partway to −CHO for Y. Strong oxidants like hot alkaline KMnO4 (followed by acidification) cannot be stopped at the aldehyde stage — they always drive benzylic CH3 groups to COOH. To stop cleanly at the aldehyde, you need a milder, more controlled oxidant — the classical choice is chromyl chloride (Étard's reaction, CrO2Cl2) or the closely related CrO3 in acetic anhydride, both of which trap the benzylic carbon as a stable diester/diacetate that only releases the aldehyde on aqueous hydrolysis, never over-oxidizing to the acid.
Step-by-Step Solution
- X reacts with NaHCO3 and liberates CO2 ⟹ X is −COOH-bearing, i.e. benzoic acid.
- Toluene → benzoic acid requires full oxidation of the CH3 group: reagent A = KMnO4/OH− (forms the carboxylate) followed by H3O+ (protonates to the free acid).
- Y gives the 2,4-DNP test ⟹ Y is an aldehyde/ketone; here it must be benzaldehyde (C6H5CHO), the partial-oxidation product of toluene. …
- AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQQ.What are X and Y in the following set of reactions? Ethylbenzene $\xrightarrow{\text{(i) } Br_2/h\nu \text{(ii) } Mg/\text{dry ether} \text{(iii) } CO_2, H_3O^+} X$ Ethylbenzene $\xrightarrow{\text{(i) } KMnO_4/OH^- \text{(ii) } H_3O^+} Y(A)\mathrm{C_6H_5-CH_2-CH_2-COOH}(3−phenylpropanoicacid);\mathrm{C_6H_5-CH_2-CH_2-OH}(2−phenylethanol)(B)\mathrm{C_6H_5-CH(CH_3)-COOH}(2−phenylpropanoicacid);\mathrm{C_6H_5-COOH}(benzoicacid)(C)4−Ethylbenzoicacid,\mathrm{HOOC-C_6H_4-CH_2CH_3}(para);\mathrm{C_6H_5-CH_2-COOH}(phenylaceticacid)(D)3−Ethylbenzoicacid,\mathrm{HOOC-C_6H_4-CH_2CH_3}(meta);\mathrm{C_6H_5-COOH}$ (benzoic acid)
›Reveal solutionSolution
Benzylic radical bromination followed by Grignard–CO2 carboxylation adds −COOH at the benzylic carbon (keeping the methyl branch), giving 2-phenylpropanoic acid, while direct KMnO4 oxidation of the whole side chain always collapses it down to benzoic acid.
Concept and Intuition
Ethylbenzene is C6H5-CH2-CH3. Radical bromination with Br2/hν abstracts the most stable radical's hydrogen — here the benzylic hydrogen (stabilized by resonance with the ring) rather than the terminal CH3 hydrogen — so bromination occurs at the benzylic carbon: C6H5-CHBr-CH3. Converting this to a Grignard and then quenching with CO2 inserts a −COOH group exactly where the MgBr was, so the methyl branch survives and the acid carbon sits on what was the benzylic carbon.
In total contrast, hot alkaline KMnO4 is a powerful, non-selective oxidant for any benzylic C–H bond: it oxidizes an entire alkyl side chain down to a single −COOH group directly on the ring, regardless of the chain's original length or branching (as long as there's a benzylic hydrogen). So ethylbenzene's whole −CH2CH3 group is destroyed and replaced by −COOH, giving benzoic acid — not a two-carbon acid.
Step-by-Step Solution
- Ethylbenzene + Br2/hν → radical benzylic bromination (secondary benzylic radical is more stable than a primary one): C6H5-CHBr-CH3.
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- Mg/dry ether → Grignard reagent: C6H5-CH(MgBr)-CH3. …
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.Which of the following gives both iodoform test and Fehling's test? (A) Acetone (B) Acetaldehyde (C) Propanal (D) Benzaldehyde
›Reveal solutionSolution
This tests the structural requirements for the iodoform test (needs CH3CO− or CH3CH(OH)−) versus Fehling's test (needs an aliphatic aldehyde). Only acetaldehyde satisfies both.
Concept and Intuition
- The iodoform test is given by any compound with a methyl ketone group (CH3−CO−R) or a secondary alcohol of the type CH3−CH(OH)−R (which is first oxidised in situ to the methyl ketone by the hypoiodite reagent). The key structural requirement is a methyl group directly attached to a carbonyl (or carbinol) carbon.
- Fehling's test detects aldehydes that can be oxidised by the cupric-tartrate complex; it works for aliphatic aldehydes but not for aromatic aldehydes like benzaldehyde (which lack the required easily-oxidisable C–H reactivity pattern and resist Fehling's oxidation) and obviously not for ketones (no reactive aldehydic H).
Step-by-Step Solution
- Acetone, CH3−CO−CH3: has a methyl directly on the carbonyl → iodoform positive. It is a ketone, not an aldehyde → Fehling's negative.
- Acetaldehyde, CH3−CHO: the carbon next to the carbonyl carbon is a methyl group → iodoform positive. It is an aliphatic aldehyde → Fehling's positive. ✓ Both tests positive. …
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.Consider the following set of reactions (A = major product) (I) C3H6H2OH+AXB (II) C6H6Yanhy. AlCl3C The product 'B' from reaction (I) gives positive iodoform test whereas product 'C' from reaction (II) does not. What are X and Y respectively? (A) H2CrO4; CO, HCl (B) KMnO4/H+; CH3COCl (C) Ag, 573 K; CO, HCl (D) PCC; (CH3CO)2O
›Reveal solutionSolution
X = H2CrO4 (gives iodoform-positive acetone) and Y = CO,HCl (gives iodoform-negative benzaldehyde), so the answer is (A).
Concept and Intuition
A positive iodoform test needs a CH3CO− group or a CH3CH(OH)− group. Propene undergoes Markovnikov hydration to give the secondary alcohol propan-2-ol; oxidation to the methyl ketone acetone gives a positive iodoform test. For benzene, the second product must NOT give iodoform, so it must not be a methyl aryl ketone — benzaldehyde (from Gattermann–Koch) fits.
Step-by-Step Solution
- C3H6+H2O/H+→ propan-2-ol (A), a secondary alcohol.
- Oxidation by X = H2CrO4 gives propan-2-one, acetone (B): a methyl ketone → positive iodoform.
- Reaction (II): C6H6 with Y = CO,HCl over anhydrous AlCl3 (Gattermann–Koch) gives benzaldehyde (C).
- Benzaldehyde has no CH3CO/CH3CH(OH) group → negative iodoform, as required.
- Options B and D use CH3COCl/(CH3CO)2O, which give acetophenone (a methyl ketone, iodoform-positive) — these fail the 'C does not give iodoform' condition.
Common Mistakes …
- AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQQ.Which of the following does not form benzoic acid on oxidation with alkaline KMnO4 followed by acidification? (A) 1-phenylpropane (B) 2-phenylpropane (C) Acetophenone (D) 2-methyl-2-phenylpropane
›Reveal solutionSolution
Alkylbenzenes are oxidised to benzoic acid by hot KMnO4 only if they have a
benzylic C–H; tert-butylbenzene has none, so it is the exception. Answer: (D).
Concept and Intuition
Hot, vigorous KMnO4 oxidation of an alkylbenzene degrades the entire side chain
down to a single −COOH group directly on the ring, regardless of chain length,
as long as there is at least one hydrogen on the benzylic carbon (the carbon bonded
directly to the aromatic ring) for the oxidant to attack and initiate the degradation.
If the benzylic carbon has no hydrogen (fully substituted, e.g. attached to three
alkyl groups), the oxidant has nothing to abstract there and the side chain survives
untouched.
- 1-Phenylpropane, C6H5−CH2−CH2−CH3: benzylic carbon is CH2 (has H) → oxidised to C6H5COOH.
- 2-Phenylpropane (cumene), C6H5−CH(CH3)2: benzylic carbon is CH (has one H) → oxidised to C6H5COOH.
- Acetophenone, C6H5−CO−CH3: under vigorous oxidation the methyl-ketone side chain is oxidatively cleaved at the carbon adjacent to the carbonyl, again collapsing to C6H5COOH.
- 2-Methyl-2-phenylbenzene (tert-butylbenzene), C6H5−C(CH3)3: the ring-attached carbon bears three methyl groups and zero hydrogens — there is no benzylic C–H for KMnO4 to attack, so this compound is characteristically resistant to oxidation and does not give benzoic acid.
Step-by-Step Solution
- Identify the benzylic carbon (directly bonded to the ring) in each option. …
- AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQQ.Which one of the following compounds does not give benzoic acid when treated with alkaline KMnO4 ? (A) Acetophenone (B) n-Propyl benzene (C) Styrene (D) t-Butyl benzene
›Reveal solutionSolution
Alkaline KMnO4 oxidises a benzene side chain to −COOH only if the carbon directly attached to the ring bears at least one hydrogen; t-butylbenzene's ring-attached carbon is fully substituted (no H), so it alone fails to give benzoic acid.
Concept and Intuition
The oxidation of alkylbenzenes by strong oxidants like hot alkaline KMnO4 proceeds by repeatedly abstracting a hydrogen from the carbon attached to the aromatic ring (the benzylic-type position) and oxidising that carbon step-by-step until only the ring-COOH remains — regardless of how long or complex the side chain is beyond that first carbon. The reaction therefore requires at least one C-H bond on the ring-attached carbon to get a foothold; if that carbon has no hydrogen at all (i.e., it's fully substituted/quaternary), the chain is inert to this oxidation.
Step-by-Step Solution
- Acetophenone C6H5−CO−CH3: even though the ring-attached carbon is a carbonyl carbon (no C-H itself), vigorous oxidants like alkaline KMnO4 can oxidatively cleave the methyl-ketone side chain (analogous to haloform-type cleavage), ultimately yielding benzoic acid.
- n-Propylbenzene C6H5−CH2−CH2−CH3: the ring-attached carbon (CH2) has hydrogens, so stepwise oxidation proceeds all the way to C6H5−COOH.
- Styrene C6H5−CH=CH2: the ring-attached vinylic carbon has a hydrogen (and the double bond is itself oxidatively cleavable), so it too gives benzoic acid on oxidation. …
- AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQQ.What are X and Y respectively in the following set of reactions ? 2-Bromobutane alc. KOH C4H8 (major); C4H8 KMnO4/H+ X; C4H8 Baeyer’s reagent Y (A) CH3COOH, CH3CH(OH)CH(OH)CH3 (butane-2,3-diol) (B) CH3CH2COOH+CO2, HOCH2CH(OH)CH2CH3 (butane-1,2-diol) (C) CH3COOH, CH3CHO (D) CH3CH2COOH+CO2, CH3CHO
›Reveal solutionSolution
2-Bromobutane with alcoholic KOH eliminates (Zaitsev) to but-2-ene; hot KMnO4/H+ cleaves this to acetic acid, while Baeyer's reagent (cold, dilute, alkaline KMnO4) simply dihydroxylates it to butane-2,3-diol.
Concept and Intuition
Dehydrohalogenation of a secondary alkyl halide with alcoholic KOH follows the Zaitsev rule, favouring the more substituted (more stable) alkene. Once you have that alkene, its fate depends on the strength/conditions of the oxidant: hot, acidified KMnO4 is a vigorous oxidant that cleaves the C=C bond entirely (oxidative cleavage), while cold, dilute, alkaline KMnO4 (Baeyer's reagent) is a mild oxidant that only adds two OH groups across the double bond without breaking the C–C bond (syn dihydroxylation) — this is also the classic test for unsaturation.
Step-by-Step Solution
- Elimination: 2-Bromobutane, CH3−CHBr−CH2−CH3, with alc. KOH undergoes E2 elimination. Zaitsev's rule favours the more substituted alkene: but-2-ene, CH3−CH=CH−CH3 (major product), over but-1-ene (minor).
- Hot acidic KMnO4 (X): this vigorously cleaves the C=C bond. Since each alkene carbon of but-2-ene bears one H and one alkyl (methyl) group, oxidative cleavage converts each half of the double bond into a carboxylic acid: CH3−CH=CH−CH3→2CH3COOH. So X =CH3COOH. …
- AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQQ.What are X and Y respectively, in the following set of reactions? CH3CH3+3O2(CH3COO)2MnΔX CH3CH=CHCH3KMnO4/H+Y (A) CH3COOH,CH3CH(OH)CH(OH)CH3 (B) CH3CH2OH,CH3CHO (C) CH3CHO,CH3COOH (D) CH3COOH,CH3COOH
›Reveal solutionSolution
Both reactions ultimately give acetic acid — one via catalytic air-oxidation of ethane, the other via oxidative cleavage of 2-butene's C=C bond by hot acidic permanganate.
Concept and Intuition
Manganese(II) acetate catalyses the industrial air-oxidation of ethane directly to acetic acid (a known route to acetic acid manufacture). Separately, hot/acidic KMnO4 is a strong oxidant that cleaves alkene double bonds completely: each doubly-bonded carbon, if it bears at least one H, is oxidised all the way to a carboxylic acid (not stopping at an aldehyde/ketone, unlike cold dilute alkaline KMnO4 which only dihydroxylates).
Step-by-Step Solution
- CH3CH3+3O2(CH3COO)2MnΔX: catalytic oxidation of ethane gives acetic acid directly, so X=CH3COOH.
- CH3-CH=CH-CH3 (2-butene) is symmetric: each alkene carbon carries one CH3 group and one H. …
- AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQQ.Toluene on reaction with the reagent X gave Y, which dissolves in NaHCO3 and when reacted with Br2/Fe gave Z. What are X and Z? (A) X:(i) KMnO4/OH−, Δ(ii) H3O+; Z = benzene ring with COOH and Br para to it (B) X:(i) CrO2Cl2(ii) H3O+; Z = benzene ring with COOH and Br para to it (C) X:(i) CrO2Cl2(ii) H3O+; Z = benzene ring with CHO and Br meta to it (D) X:(i) KMnO4/OH−, Δ(ii) H3O+; Z = benzene ring with COOH and Br meta to it
›Reveal solutionSolution
Strong alkaline KMnO4 oxidises toluene's methyl group all the way to −COOH (which dissolves in NaHCO3); since −COOH is meta-directing, subsequent bromination places Br meta to it.
Concept and Intuition
Solubility in aqueous NaHCO3 (with visible effervescence of CO2) is the classic diagnostic for a carboxylic acid — it is acidic enough (pKa about 4) to react with the weak base bicarbonate, unlike phenols or aldehydes. So Y must be benzoic acid, meaning the methyl group of toluene has been fully oxidised, not just partially oxidised to the aldehyde stage. CrO2Cl2 (Etard reaction) is famous precisely because it stops at the aldehyde — so it cannot be X here. Only vigorous hot alkaline KMnO4, followed by acidification, drives the oxidation all the way to the carboxylic acid.
Once Y = benzoic acid reacts with Br2/Fe (electrophilic aromatic bromination), the ring-substituent effect of −COOH governs regiochemistry: it is electron-withdrawing (by both induction and resonance) and therefore deactivating and meta-directing.
Step-by-Step Solution
- C6H5CH3(i) KMnO4/OH−, Δ(ii) H3O+C6H5COOH (benzoic acid, Y) — dissolves in NaHCO3 as expected. …
- AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQQ.The suitable reagent to carry out the following reaction is C6H5CH(CH3)2 (isopropylbenzene) ? C6H5COOH (benzoic acid, ring with COOH substituent) (A) RCO3H (B) PCC (C) KMnO4/KOH, H3O+ (D) dil. H2SO4
›Reveal solutionSolution
Any alkylbenzene side chain (with at least one benzylic H) is oxidised all the way to –COOH by hot alkaline KMnO₄ followed by acid workup — this converts cumene directly to benzoic acid.
Concept and Intuition
A powerful oxidant like KMnO4 under alkaline, heated conditions can oxidatively cleave ANY alkyl side chain attached to a benzene ring — regardless of the chain's length or branching — all the way down to a single carboxylic acid group directly attached to the ring, PROVIDED there is at least one hydrogen on the benzylic carbon (the carbon directly attached to the ring) for the oxidant to attack. This is a hallmark reaction used to convert various alkylbenzenes uniformly into benzoic acid.
Step-by-Step Solution
- Cumene (isopropylbenzene), C6H5CH(CH3)2, has a benzylic C–H (on the isopropyl carbon attached to the ring).
- Treating with hot alkaline potassium permanganate (KMnO4/KOH) oxidises the ENTIRE side chain (both methyl groups are cleaved off as well), leaving only the ring-attached carbon as a carboxylate.
- Subsequent acidification (H3O+) converts the potassium benzoate salt formed into the free carboxylic acid, benzoic acid, C6H5COOH. …
- AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQQ.Match the following List - I | List - II A. HC≡CHHg+,H+H2O | I. H3C−COOH B. CH4O2Mo2O3,Δ | II. CH3−CO−CH3 C. (H3C)2C=C(CH3)2O3Zn,H2O | III. H3C−CHO D. CH3−CH=CH−CH3KMnO4H+ | IV. HCHO (A) A – I, B – II, C – III, D – IV (B) A – III, B – IV, C – II, D – I (C) A – I, B – IV, C – III, D – II (D) A – III, B – II, C – IV, D – I
›Reveal solutionSolution
Each reaction is a classic named transformation (Kucherov hydration, catalytic methane oxidation, ozonolysis, oxidative cleavage) whose products are acetaldehyde, formaldehyde, acetone and acetic acid respectively, giving the mapping A-III, B-IV, C-II, D-I.
Concept and Intuition
- Acetylene hydration (Hg²⁺/H⁺ catalysed, Kucherov reaction): water adds across the triple bond to give an unstable enol which tautomerises to a carbonyl compound; unsubstituted acetylene specifically gives acetaldehyde.
- Catalytic partial oxidation of methane (over a molybdenum oxide-type catalyst) is a controlled oxidation that stops at the aldehyde stage, giving formaldehyde rather than going all the way to CO2.
- Ozonolysis cleaves a C=C double bond symmetrically; each carbon of the double bond becomes a carbonyl carbon of a new (reduced, via Zn/H₂O) fragment — a fully substituted alkene carbon (bearing two methyls, as in tetramethylethylene) becomes a ketone (here, acetone) on each side.
- Hot/acidic KMnO4 oxidatively cleaves an alkene C=C bond all the way to carboxylic acids (for a CH= carbon) or ketones (for a fully substituted carbon); 2-butene has CH= on both alkene carbons, so cleavage gives two carboxylic acid fragments — here two acetic acid molecules.
Step-by-Step Solution
- A: HC≡CHHg2+,H+H2O CH₂=CHOH (unstable enol) → tautomerises to CH3CHO (acetaldehyde) = III.
- B: CH4O2Mo2O3,Δ HCHO (formaldehyde, controlled catalytic oxidation) = IV.
- C: (CH3)2C=C(CH3)2O3Zn,H2O → two equivalents of (CH3)2C=O (acetone) = II. …
- AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQQ.What are X and Y respectively in the following reaction? CH3CH2CHO+2Cu2++OH−→X+Y+3H2O (A) CH3CH2COOH, Cu(OH)2 (B) CH3CH2COO−, Cu (C) CH3CH2COOH, Cu2O (D) CH3CH2COO−, Cu2O
›Reveal solutionSolution
Tests Fehling's-solution oxidation of an aliphatic aldehyde. Answer: propanoate ion and Cu2O (option D).
Concept and Intuition
Fehling's solution contains Cu2+ complexed with tartrate in alkaline medium. Aliphatic aldehydes (but not aromatic ones, and not ketones) reduce Cu2+ to Cu+, which precipitates as brick-red Cu2O. The aldehyde itself is oxidised — in alkaline medium the carboxylic acid product exists as its carboxylate salt, not the free acid.
Step-by-Step Solution
- Propanal, CH3CH2CHO, is oxidised by Cu2+ in alkaline (OH⁻) medium.
- The aldehyde carbon is oxidised from the +1 oxidation state to the +3 state of a carboxylic acid/carboxylate.
- Since the medium is basic (OH⁻ present), the product is the carboxylate anion, CH3CH2COO−, not the free acid. …
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