Q.Write the structures of the following 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 -- but here the focus is on nomenclature to structure, not on any reaction. The key is to interpret the IUPAC name and draw the correct skeletal formula.
- alpha-Methoxypropionaldehyde Propionaldehyde is propanal (CH3CH2CHO). alpha-position is the carbon next to the aldehyde group. Replace one H on that carbon with a methoxy (-OCH3). Structure: CH3-CH(OCH3)-CHO
- 3-Hydroxybutanal Butanal with an -OH group on carbon 3. Structure: CH3-CH(OH)-CH2-CHO
- 2-Hydroxycyclopentane carbaldehyde Cyclopentane ring with an aldehyde group (-CHO) attached to carbon 1, and an -OH on carbon 2. Structure: A pentagon with OH on C2 and CHO on C1.
- 4-Oxopentanal Pentanal with a keto group (=O) on carbon 4. Structure: CH3-CO-CH2-CH2-CHO
- Di-sec. butyl ketone …
The key is to decode each systematic name into its parent chain, functional groups, and substituent positions, then draw the correct skeletal structure. The final structures are given below.
Let’s break down each name step by step. The logic is always: identify the parent chain (the longest carbon chain containing the principal functional group), number it to give the functional group the lowest locant, then place substituents accordingly.
(i) α-Methoxypropionaldehyde
Parent: Propionaldehyde = propanal (3-carbon chain with –CHO at C1).
Substituent: α-Methoxy means a methoxy group (–OCH₃) attached to the α-carbon (the carbon adjacent to the aldehyde group, i.e., C2).
Structure:
CH₃–CH(OCH₃)–CHO
In older nomenclature, “α” refers to the carbon next to the functional group. For an aldehyde, C1 is the –CHO carbon, so α = C2.
(ii) 3-Hydroxybutanal
Parent: Butanal = 4-carbon chain with –CHO at C1.
Substituent: –OH (hydroxy) at C3.
Numbering: The aldehyde carbon is always C1, so the chain is numbered from the –CHO end.
Structure:
CH₃–CH(OH)–CH₂–CHO
Do not confuse “butanal” with “butanone”. The suffix “-al” means aldehyde, so the functional group is –CHO, not a ketone.
(iii) 2-Hydroxycyclopentane carbaldehyde
Parent: Cyclopentane carbaldehyde = a cyclopentane ring with a –CHO group attached (the ring is the parent, and the aldehyde is a substituent named as “carbaldehyde”).
Substituent: –OH at C2 of the ring.
Numbering: The carbon bearing the –CHO is C1. Then number the ring to give the –OH the lowest locant (C2).
Structure:
A five-membered ring with –CHO on C1 and –OH on C2 (cis or trans not specified; any stereoisomer is acceptable unless asked).
When –CHO is directly attached to a ring, the compound is named as “cycloalkane carbaldehyde” — the ring carbon bearing –CHO is always C1.
(iv) 4-Oxopentanal
Parent: Pentanal = 5-carbon chain with –CHO at C1.
Substituent: “Oxo” means a ketone group (=O) at C4.
Numbering: The aldehyde carbon is C1, so the ketone is at C4.
Structure:
CH₃–CO–CH₂–CH₂–CHO
This is a dialdehyde? No — only one aldehyde; the other carbonyl is a ketone. The name “4-oxopentanal” tells you it’s a pentanal with an oxo group at position 4.
(v) Di-sec. butyl ketone
Parent: Ketone — the functional group is C=O.
Substituents: Two sec-butyl groups attached to the carbonyl carbon. …
Method: Systematic IUPAC Name-to-Structure Conversion
This method works by breaking down the IUPAC name into its functional groups, parent chain, substituents, and locants, then assembling the structure step-by-step.
Steps
- Identify the parent chain (alkane, alkene, cycloalkane, or aromatic ring) and the principal functional group (highest priority for naming).
- Number the parent chain to give the principal functional group the lowest possible locant.
- Add substituents (alkyl, halo, hydroxy, oxo, alkoxy, etc.) at the correct positions.
- Draw the complete structure with all bonds and atoms.
(i) α-Methoxypropionaldehyde
- Parent: Propanal (3-carbon aldehyde chain)
- Substituent: Methoxy (−OCH3) at the α-carbon (carbon adjacent to the aldehyde group)
- Structure:
CH3−CH(OCH3)−CHO
(ii) 3-Hydroxybutanal
- Parent: Butanal (4-carbon aldehyde chain)
- Substituent: Hydroxy (−OH) at carbon 3
- Structure:
CH3−CH(OH)−CH2−CHO
(iii) 2-Hydroxycyclopentane carbaldehyde
- Parent: Cyclopentane ring with a carbaldehyde (−CHO) group attached
- Substituent: Hydroxy (−OH) at carbon 2 of the ring (carbon 1 is the carbaldehyde carbon)
- Structure:
Cyclopentane ring with −OH at C2 and −CHO at C1
(iv) 4-Oxopentanal
- Parent: Pentanal (5-carbon aldehyde chain)
- Substituent: Oxo (=O) at carbon 4
- Structure:
CH3−CO−CH2−CH2−CHO
(v) Di-sec. butyl ketone …
(i) α-Methoxypropionaldehyde
✗ Common Mistakes
- Confusing “α” with “1”: Students often place the methoxy group on carbon-1 (the aldehyde carbon). But α refers to the carbon adjacent to the functional group.
- Forgetting the aldehyde: Some draw a ketone or alcohol instead of
–CHO.
✓ How to Avoid
- Step 1: Identify the parent chain: “propionaldehyde” = 3-carbon aldehyde (propanal).
- Step 2: α-carbon = carbon-2 (next to
–CHO). Attach–OCH₃there. - Structure: CH3-CH(OCH3)-CHO
(ii) 3-Hydroxybutanal
✗ Common Mistakes
- Wrong carbon numbering: Students number from the wrong end. Aldehyde carbon must be carbon-1.
- Drawing a ketone: “butanal” means aldehyde, not butanone.
✓ How to Avoid
- Rule: For aldehydes, the
–CHOcarbon is always C-1. - Parent: Butanal = 4-carbon chain with
–CHOat end. - Position:
–OHon C-3. - Structure: CH3-CH(OH)-CH2-CHO
(iii) 2-Hydroxycyclopentane carbaldehyde
✗ Common Mistakes
- Incorrect ring numbering: The aldehyde carbon is not part of the ring — it’s a substituent. Students often number the ring carbon bearing the
–CHOas C-1, but the correct IUPAC name uses “carbaldehyde” to indicate–CHOas a suffix on the ring. - Missing stereochemistry: Not required here, but students sometimes add wedges unnecessarily.
✓ How to Avoid
- Parent: Cyclopentane ring.
- Substituent:
–CHOgroup attached to the ring (named as carbaldehyde). - Position:
–OHon carbon-2 of the ring. - Structure:
A cyclopentane ring with
–OHon C-2 and–CHOon C-1.
(iv) 4-Oxopentanal
✗ Common Mistakes
- Misplacing the oxo group: “Oxo” means
=O(ketone). Students sometimes put it on C-1 (which is already aldehyde) or on C-5. - Drawing two aldehydes: “pentanal” means one aldehyde at C-1; the oxo is a separate ketone.
✓ How to Avoid
- Parent: Pentanal (5-carbon chain, aldehyde at C-1).
- Oxo position: C-4 has a
=O. - Structure: CH3-CO-CH2-CH2-CHO
(v) Di-sec. butyl ketone
✗ Common Mistakes
- Misinterpreting “sec. butyl”: Students draw a straight butyl chain instead of the branched secondary butyl group.
- Forgetting the ketone: “Ketone” means
–CO–in the middle. Some draw an aldehyde or alcohol.
✓ How to Avoid
- “Di-sec. butyl” = two identical secondary butyl groups.
- Secondary butyl =
–CH(CH₃)CH₂CH₃(the carbon attached to the ketone is secondary). - Ketone: The carbonyl carbon is between the two groups.
- Structure: CH3CH2CH(CH3)-CO-CH(CH3)CH2CH3
(vi) 4-Fluoroacetophenone
✗ Common Mistakes …
- 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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