Q.How will you carry out the following conversions?
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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: Diazonium Salt Reactions / Sandmeyer Reaction) The key idea is that converting an aryl-diazonium salt to a carboxylic acid while KEEPING the ring's existing methyl group intact requires replacing −N2+ with −CN first (a Sandmeyer-type reaction), then hydrolysing the nitrile — not oxidising the methyl group, which would destroy it.
Reasoning steps:
- Toluene → p-toluidine: Nitrate toluene to get a mixture of o- and p-nitrotoluene (the methyl group is ortho/para-directing). Separate the p-isomer, then reduce the nitro group (−NO2) to an amino group (−NH2) using Sn/HCl or Fe/HCl, giving p-toluidine. …
The key idea is to use the directing effects of substituents on the benzene ring. For (i), we nitrate toluene (which gives mainly p-nitrotoluene due to the methyl group’s ortho/para direction), then reduce the nitro group to an amine. For (ii), we replace the diazonium group with a carboxyl group via a Sandmeyer-type reaction using cuprous cyanide, followed by hydrolysis.
Let’s break down each conversion with the reasoning behind every step.
(i) Toluene → p-Toluidine
Concept: The methyl group is an activating ortho/para director.
When you do an electrophilic substitution on toluene, the methyl group pushes electron density into the ring, making the ortho and para positions more reactive than the meta position. The para product is usually the major one because it’s less sterically hindered than the ortho product (which also forms but in smaller amount). So, to get p-toluidine, we first introduce a nitro group at the para position, then reduce it.
- Nitration of toluene Treat toluene with a mixture of concentrated nitric acid and concentrated sulfuric acid (nitrating mixture). The nitronium ion (NO2+) attacks the para position preferentially.
C6H5CH3+HNO3H2SO4p-O2N-C6H4CH3+H2O
The product is p-nitrotoluene (along with some o-nitrotoluene, which can be separated by distillation or crystallization).
- Reduction of the nitro group Reduce the nitro group to an amino group. A common laboratory method is catalytic hydrogenation (using H2 gas with a palladium or nickel catalyst) or chemical reduction using tin and hydrochloric acid (Sn/HCl).
p-O2N-C6H4CH3+3H2Pd/Cp-H2N-C6H4CH3+2H2O
The product is p-toluidine.
A common mistake is to try direct amination of toluene. That doesn’t work because ammonia is not an electrophile for aromatic substitution. You must go through the nitro intermediate.
(ii) p-Toluidine Diazonium Chloride → p-Toluic Acid
Concept: The diazonium group is a superb leaving group that can be replaced by many nucleophiles.
Here, we want to replace the -N2+ group with a carboxyl group (-COOH). The standard route is to first replace the diazonium group with a cyano group (-CN) via a Sandmeyer reaction, then hydrolyze the nitrile to a carboxylic acid.
- Formation of diazonium chloride Treat p-toluidine with sodium nitrite (NaNO2) and excess hydrochloric acid at 0–5°C. This gives the diazonium salt. p-H2N-C6H4CH3+NaNO2+2HCl0−5∘Cp-CH3-C6H4-N2+Cl−+NaCl+2H2O …
Method: Diazotisation followed by Sandmeyer-type Hydrolysis
This is a two-step sequence that uses the diazonium group as a temporary handle to replace an amino group with a carboxyl group.
(i) Toluene → p-Toluidine
Method: Nitration followed by reduction
Steps:
-
Nitration
Treat toluene with a mixture of concentrated HNO3 and concentrated H2SO4 (nitrating mixture).
- The methyl group is ortho/para-directing.
- By controlling temperature (≈ 30°C), the major product is p-nitrotoluene.
-
Reduction
Reduce p-nitrotoluene to p-toluidine using:
- Sn+conc. HCl (tin and hydrochloric acid), or
- Fe+dil. HCl, or
- Catalytic hydrogenation (H2/Pd-C).
Overall:
C6H5CH3HNO3/H2SO4p-NO2C6H4CH3Sn/HClp-NH2C6H4CH3
(ii) p-Toluidine diazonium chloride → p-Toluic acid
Method: Diazotisation followed by hydrolysis (Sandmeyer-type)
Steps:
- Diazotisation Treat p-toluidine with NaNO2 and excess dilute HCl at 0–5°C to form the diazonium salt:
p-CH3C6H4NH2NaNO2/HCl,0−5∘Cp-CH3C6H4N2+Cl−
-
Hydrolysis (replacement of –N2+ by –OH)
Warm the diazonium salt solution with dilute H2SO4 or simply heat it in aqueous acidic medium.
- The diazonium group is replaced by a hydroxyl group (–OH), giving p-cresol as an intermediate.
-
Oxidation of –CH3 to –COOH …
Common Mistakes in Oxidation Reactions (Toluene → p-Toluidine → p-Toluic Acid)
Students often struggle with these conversions because they mix up reaction sequences and reagent selectivity. Here are the most frequent errors and how to avoid them.
Mistake 1: Trying to oxidise toluene directly to p-toluidine
The error: Students think they can replace a methyl group with an amino group in one step using an oxidising agent.
Why it's wrong:
Oxidation of toluene always attacks the methyl group — it converts -CH₃ to -COOH (benzoic acid), not to -NH₂. To introduce an amino group, you need nitration followed by reduction, not oxidation.
How to avoid:
Remember:
- Oxidation changes alkyl groups to carboxylic acids.
- Amino groups come from reducing nitro groups (
-NO₂ → -NH₂).
Correct sequence for (i):
- Nitration of toluene → p-nitrotoluene (using conc.
HNO₃/ conc.H₂SO₄, at low temperature to favour para product). - Reduction of p-nitrotoluene → p-toluidine (using
Sn/HClorFe/HCl).
Mistake 2: Using the wrong oxidising agent for (ii)
The error: Students try to oxidise the diazonium salt directly with KMnO₄ or K₂Cr₂O₇.
Why it's wrong:
Diazonium salts are unstable and decompose under strong oxidising conditions. You must first replace the diazonium group with a group that can be oxidised.
How to avoid:
Use the Sandmeyer reaction or hydrolysis to convert the diazonium group into a nitrile (-CN) or methyl group, then oxidise.
Correct sequence for (ii):
- Sandmeyer reaction:
p-toluidine diazonium chloride+CuCN→ p-tolunitrile. - Hydrolysis (acidic or basic): p-tolunitrile → p-toluic acid.
Alternative (if you want to use oxidation directly):
- Replace
-N₂⁺with-CH₃(usingCH₃OH/Cu), then oxidise the methyl group to-COOH. But this is less common.
Mistake 3: Forgetting to protect the amino group during oxidation
The error: Students try to oxidise p-toluidine directly to p-toluic acid.
Why it's wrong:
The -NH₂ group is easily oxidised itself (it forms coloured impurities or gets destroyed). Strong oxidants like KMnO₄ will attack the amino group before the methyl group.
How to avoid:
Always protect the amino group by acetylation before oxidation:
- Acetylate p-toluidine → p-acetotoluidide (using
(CH₃CO)₂O/ pyridine). - Oxidise the methyl group to
-COOH(usingKMnO₄/H⁺). - Hydrolyse the amide back to
-NH₂(using dil.HClorNaOH).
Mistake 4: Ignoring para selectivity in nitration
The error: Students assume nitration of toluene gives only the para product.
Why it's wrong:
Toluene gives a mixture of ortho and para products. The ortho isomer is also formed (about 60% ortho, 40% para at room temperature).
How to avoid:
- Use low temperature (0°C) to favour para product slightly more.
- Separate the isomers by fractional distillation or crystallisation (p-nitrotoluene has a higher melting point).
- In exam problems, assume the para isomer is the desired product unless stated otherwise.
--- …
- CBSE 2024Set D1 markMCQQ.An aldehyde on oxidation gives(a) an alcohol(b) a ketone(c) an ether(d) an acid
›Reveal solutionSolution
Oxidation of an aldehyde gives a carboxylic acid.
Aldehydes carry an H on the carbonyl carbon and are easily oxidised. With oxidising agents (or even mild reagents such as Tollen's or Fehling's), an aldehyde is converted to the corresponding carboxylic acid:
R-CHO + [O] -> R-COOH
…
- CBSE 2024Set ANNUAL1 markQ.How would you obtain the following? Benzoic acid from ethyl benzene
›Reveal solutionSolution
Vigorous oxidation (hot alkaline KMnO4) of any alkylbenzene side chain, regardless of its length, converts it entirely to a single −COOH group attached directly to the ring.
Ethylbenzene, C6H5−CH2CH3, has a two-carbon side chain with benzylic hydrogens. Strong oxidising agents like hot alkaline potassium permanganate attack the side chain at the benzylic position and progressively oxidise it, cleaving off the extra carbon(s) and leaving only the ring-attached carbon as a carboxyl group — the exact chain length beyond the first carbon does not matter, the product is always benzoic acid:
…
- CBSE 2023Set 56/1/11 markMCQQ.CH3CONH2 on reaction with NaOH and Br2 in alcoholic medium gives : (A) CH3COONa (B) CH3NH2 (C) CH3CH2Br (D) CH3CH2NH2
›Reveal solutionSolution
This is the Hofmann bromamide degradation reaction. An amide (CH3CONH2) reacts with bromine and a base to give a primary amine with one fewer carbon atom. The product here is methylamine (CH3NH2), which corresponds to option (B).
The reaction you're looking at is a classic name reaction in organic chemistry — the Hofmann bromamide degradation. It's one of the most reliable ways to convert an amide into a primary amine, and it always involves a loss of one carbon from the chain. Let's understand why.
The key idea: the amide group (−CONH2) gets "chopped" by bromine in the presence of a strong base. The carbonyl carbon (the one attached to oxygen) is lost as carbon dioxide, and the nitrogen ends up attached to the alkyl group that was originally next to the carbonyl. So the product has one carbon fewer than the starting amide.
Now let's walk through the reaction step by step for your specific compound, acetamide (CH3CONH2).
-
Identify the starting material.
Acetamide has the structure CH3−CO−NH2. The alkyl group attached to the carbonyl is a methyl group (CH3−). The amide carbon is the carbonyl carbon.
-
Recall the general outcome of Hofmann degradation.
The reaction is:
R−CONH2+Br2+4NaOH→R−NH2+2NaBr+Na2CO3+2H2O
Notice that the product R−NH2 has the same R group as the starting amide, but the carbonyl carbon is gone (it becomes carbonate). So the amine has one less carbon than the amide.
-
Apply to acetamide.
Here R=CH3−. So the amine formed is CH3−NH2, which is methylamine.
-
Check the options.
- (A) CH3COONa — this is sodium acetate, not an amine.
- (B) CH3NH2 — methylamine, matches our prediction. …
-
- CBSE 2023Set ANNUAL1 markMCQQ.In Benzaldehyde + [O] --(Air)--> A, A is(a) Benzene(b) Benzoic acid(c) Benzyl alcohol(d) None of these
›Reveal solutionSolution
Aromatic aldehydes like benzaldehyde undergo slow autoxidation in air, converting -CHO to -COOH.
C6H5CHO + [O] --(air)--> C6H5COOH
On exposure to air, benzaldehyde is slowly autoxidised at the aldehydic hydrogen, converting the -CHO group into a -COOH group and giving benzoic acid. (Th …
- CBSE 2020Set 56/1/11 markMCQQ.Iodoform test is not given by (A) Ethanol (B) Ethanal (C) Pentan-2-one (D) Pentan-3-one
›Reveal solutionSolution
The iodoform test detects the presence of a methyl carbonyl group (CHX3COX−) or a methyl carbinol group (CHX3CH(OH)X−) that can be oxidised to a methyl carbonyl. Pentan-3-one lacks this structural feature, so it does not give the test. The correct option is (D).
The iodoform test is a classic qualitative test in organic chemistry. It's not just a random reaction — it's a specific probe for a very particular structural arrangement. When you see a question about which compound gives or doesn't give this test, you're really being asked: "Which of these molecules has a methyl group directly attached to a carbonyl carbon (or to a carbon that can be easily oxidised to a carbonyl)?"
The test works because the methyl group in CHX3COX− is uniquely reactive under basic, halogenating conditions. The three hydrogens on that methyl are successively replaced by iodine, forming a triiodomethyl intermediate. This intermediate is unstable and breaks apart, yielding a yellow precipitate of iodoform (CHIX3) — that's the visible "positive" result.
Now, there's a second pathway. A primary alcohol with the structure CHX3CH(OH)−R (where R can be H or any alkyl/aryl group) can be oxidised in situ by the iodine in the basic solution to give CHX3CO−R, which then undergoes the same reaction. So ethanol and any secondary alcohol with a methyl group on the alcohol carbon also give a positive test.
Let's examine each option.
-
Ethanol (CHX3CHX2OH)
This is a primary alcohol with the structure CHX3CHX2OH. Under the reaction conditions (basic IX2), it gets oxidised to ethanal (CHX3CHO), which has a methyl carbonyl group. The test is positive.
TipEthanol is the classic example of a compound that gives the iodoform test after oxidation. Many students forget this pathway and wrongly think only carbonyl compounds respond.
-
Ethanal (CHX3CHO)
This is acetaldehyde — the simplest methyl carbonyl. It has the CHX3COX− group directly. The test is strongly positive. In fact, this is the reference compound for the test.
-
Pentan-2-one (CHX3COCHX2CHX2CHX3) …
-
- CBSE 2020Set ANNUAL1 markMCQQ.Benzaldehyde + [O] --(Air)--> A. 'A' is(a) Benzene(b) Benzoic acid(c) Benzyl alcohol(d) None of these
›Reveal solutionSolution
Aldehydes are easily oxidised even by atmospheric oxygen (autoxidation); benzaldehyde left exposed to air slowly oxidises to benzoic acid.
Benzaldehyde (C6H5CHO) has a reactive aldehydic hydrogen. On standing in air, atmospheric O2 slowly oxidises it:
C6H5CHO + [O] --(air)--> C6H5COOH
…
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