Q.Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
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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: Aldol Condensation vs. Cannizzaro Reaction
Key idea: Aldol condensation requires at least one α-hydrogen on the carbonyl compound. Cannizzaro reaction requires an aldehyde with no α-hydrogen. Neither applies to ketones without α-H or to alcohols.
Step 1 – Identify α-hydrogens
- Methanal (HCHO): no α-H → Cannizzaro
- 2-Methylpentanal: has α-H → Aldol
- Benzaldehyde: no α-H → Cannizzaro
- Benzophenone: ketone, no α-H → Neither
- Cyclohexanone: ketone with α-H → Aldol
- 1-Phenylpropanone: ketone with α-H → Aldol
- Phenylacetaldehyde: aldehyde with α-H → Aldol
- Butan-1-ol: alcohol → Neither
- 2,2-Dimethylbutanal: aldehyde, no α-H → Cannizzaro
Step 2 – Products (every aldol product below is a β-hydroxy carbonyl compound before dehydration; dehydration removes that water to give the conjugated enal/enone)
- Aldol products (self-condensation):
- 2-Methylpentanal → 2-(1-hydroxy-2-methylpentyl)-2-methylpentanal (then dehydration to the α,β-unsaturated aldehyde) …
The key is whether the aldehyde/ketone has at least one α-hydrogen (for aldol) or no α-hydrogens (for Cannizzaro). Methanal, benzaldehyde, and 2,2-dimethylbutanal undergo Cannizzaro; 2-methylpentanal, cyclohexanone, 1-phenylpropanone, and phenylacetaldehyde undergo aldol; benzophenone and butan-1-ol do neither.
Let's first understand the two reactions at a conceptual level.
Aldol condensation requires a carbonyl compound (aldehyde or ketone) that has at least one hydrogen atom on the carbon next to the carbonyl group — that's the α-carbon. In base, that α-hydrogen is acidic enough to be removed, forming an enolate ion. The enolate then attacks another carbonyl molecule, building a new C–C bond. The product is a β-hydroxy carbonyl compound, which often dehydrates to an α,β-unsaturated carbonyl.
Cannizzaro reaction is a disproportionation of an aldehyde that has no α-hydrogens. In strong base, one molecule of aldehyde is reduced to an alcohol while another is oxidized to a carboxylic acid (or its salt). Only aldehydes like HCHO or ArCHO (with no α-H) can do this — ketones never undergo Cannizzaro.
Now, examine each compound.
1. Methanal (HCHO) — No α-carbon at all. → Cannizzaro. Products: methanol + formic acid (as formate salt in base).
2. 2-Methylpentanal — CH3CH2CH2CH(CH3)CHO. The α-carbon (C2, next to −CHO) bears one H (and a methyl branch). → Aldol possible.
Self-aldol: the enolate carbon of one molecule (C2, bearing a methyl branch and losing its remaining H to form the new bond) attacks the carbonyl carbon of a second molecule. The unreacted molecule's own −CHO and its C2 methyl branch stay in place; the new C–OH-bearing carbon (from the attacked molecule) becomes a substituent hanging off that same C2. The result, before dehydration, is 2-(1-hydroxy-2-methylpentyl)-2-methylpentanal — a β-hydroxy aldehyde. Dehydration gives the corresponding α,β-unsaturated aldehyde.
3. Benzaldehyde (C6H5CHO) — The carbon next to the carbonyl is an aromatic ring carbon, with no α-H. → Cannizzaro. Products: benzyl alcohol + benzoic acid (as salt).
4. Benzophenone (C6H5−CO−C6H5) — A ketone with no α-H (both neighbouring carbons are aromatic ring carbons) and ketones never do Cannizzaro. → Neither.
5. Cyclohexanone — The ring carbons flanking the carbonyl (C2 and C6) each carry 2 H's. → Aldol possible. Self-aldol gives 2-(1-hydroxycyclohexyl)cyclohexan-1-one (one ring keeps its ketone; the other ring's former carbonyl carbon becomes a C–OH, now a substituent on the first ring's α-carbon). Dehydration gives 2-cyclohexylidenecyclohexan-1-one.
6. 1-Phenylpropanone (C6H5−CO−CH2−CH3) — The α-carbon (the CH2) has 2 H's. → Aldol possible. Self-aldol: one molecule's enolate carbon (bearing a methyl group left over from its own ethyl side-chain, plus one remaining H) attacks the carbonyl carbon of a second molecule — which keeps ITS phenyl group and gains an -OH. Both phenyl groups therefore survive into the product: 3-hydroxy-2-methyl-1,3-diphenylpentan-1-one. Dehydration gives the conjugated enone.
It is easy to drop one of the two phenyl rings when naming this product — one phenyl comes from the molecule that keeps its ketone (C1), and the SECOND phenyl comes from the molecule that is attacked (now sitting on C3, next to the new -OH). Both must appear in the name.
7. Phenylacetaldehyde (C6H5−CH2−CHO) — The α-carbon (the CH2, benzylic) has 2 H's. → Aldol possible. Because the nucleophile's own −CHO survives unchanged, the product is still an ALDEHYDE, not a ketone: one molecule's enolate carbon (bearing a phenyl substituent and one remaining H) attacks the carbonyl carbon of a second molecule (which keeps its own phenyl group and gains an -OH). The result is a 4-carbon aldehyde chain: 3-hydroxy-2,4-diphenylbutanal. Dehydration gives a substituted cinnamaldehyde-type product. …
Method: Reactivity Classification Based on α-Hydrogen Availability
This method uses the presence or absence of α-hydrogen atoms (hydrogens on the carbon adjacent to the carbonyl group) to decide between aldol condensation and Cannizzaro reaction.
Step 1: Check for α-hydrogens
- Aldol condensation requires at least one α-hydrogen on the carbonyl compound.
- Cannizzaro reaction requires no α-hydrogens and an aldehyde (not a ketone).
- Neither applies to alcohols or ketones without α-hydrogens.
Step 2: Apply to each compound
| Compound | α-Hydrogens? | Aldehyde/Ketone? | Reaction Type |
|---|---|---|---|
| (i) Methanal (HCHO) | No (H–C=O, no α-C) | Aldehyde | Cannizzaro |
| (ii) 2-Methylpentanal | Yes (α-C has H) | Aldehyde | Aldol condensation |
| (iii) Benzaldehyde (C₆H₅CHO) | No (α-C is part of ring) | Aldehyde | Cannizzaro |
| (iv) Benzophenone (C₆H₅COC₆H₅) | No | Ketone | Neither |
| (v) Cyclohexanone | Yes (α-CH₂ groups) | Ketone | Aldol condensation |
| (vi) 1-Phenylpropanone (C₆H₅COCH₂CH₃) | Yes (α-CH₂) | Ketone | Aldol condensation |
| (vii) Phenylacetaldehyde (C₆H₅CH₂CHO) | Yes (α-CH₂) | Aldehyde | Aldol condensation |
| (viii) Butan-1-ol (CH₃CH₂CH₂CH₂OH) | N/A (alcohol) | Neither | Neither |
| (ix) 2,2-Dimethylbutanal | No (α-C is quaternary) | Aldehyde | Cannizzaro |
Step 3: Write expected products
Aldol condensation products (β-hydroxy carbonyl compounds, then dehydration)
- (ii) 2-Methylpentanal Two molecules react:
2 CH3CH2CH2CH(CH3)CHOOH−CH3CH2CH2CH(CH3)CH(OH)CH(CH3)CH2CH2CH3
Then dehydration gives an α,β-unsaturated aldehyde.
- (v) Cyclohexanone
2C6H10OOH−C6H10(OH)C6H9O−H2OC6H9–C6H7O
-
(vi) 1-Phenylpropanone
Self-condensation at the α-carbon next to the carbonyl.
-
(vii) Phenylacetaldehyde
2C6H5CH2CHOOH−C6H5CH2CH(OH)CH(C6H5)CHO
Cannizzaro reaction products (alcohol + carboxylic acid salt)
- (i) Methanal …
Here is a breakdown of the common mistakes students make when classifying and solving problems on Aldol Condensation and Cannizzaro Reactions, specifically for the compounds listed.
Mistake 1: Confusing the "No α-Hydrogen" Rule for Aldol vs. Cannizzaro
The Mistake:
Students often think that any aldehyde without α-hydrogen undergoes Cannizzaro. They forget that formaldehyde (methanal) is the classic case, but other aldehydes like benzaldehyde also lack α-hydrogens. However, they fail to check if the compound is an aldehyde at all.
How to Avoid:
- Step 1: Check for the carbonyl group (C=O). Only aldehydes (and not ketones like benzophenone) can undergo Cannizzaro.
- Step 2: Check for α-hydrogen. If an aldehyde has no α-H, it undergoes Cannizzaro.
- Step 3: If an aldehyde has α-H, it undergoes Aldol condensation.
Application to the list:
- Methanal (i): Aldehyde, no α-H → Cannizzaro.
- Benzaldehyde (iii): Aldehyde, no α-H (the α-carbon is part of the aromatic ring) → Cannizzaro.
- 2,2-Dimethylbutanal (ix): Aldehyde, no α-H (the α-carbon has three methyl groups, no H) → Cannizzaro.
Mistake 2: Forgetting that Ketones Can Undergo Aldol (Crossed Aldol) but Not Cannizzaro
The Mistake:
Students often say "ketones do not undergo aldol condensation" because they think of simple ketones like acetone. They forget that ketones with α-hydrogen can undergo aldol, though the equilibrium often favors the reactant (unless conditions are forcing). More importantly, they incorrectly apply Cannizzaro to ketones.
How to Avoid:
- Rule: Cannizzaro is only for aldehydes (no α-H). Ketones never undergo Cannizzaro.
- Rule: Aldol condensation requires at least one α-H on the carbonyl compound. Ketones with α-H can undergo aldol (e.g., cyclohexanone).
Application to the list:
- Benzophenone (iv): Ketone, no α-H (both α-carbons are part of aromatic rings) → Neither.
- Cyclohexanone (v): Ketone, has α-H → Aldol condensation.
- 1-Phenylpropanone (vi): Ketone, has α-H (on the methyl group next to carbonyl) → Aldol condensation.
Mistake 3: Misidentifying α-Hydrogens in Branched or Aromatic Compounds
The Mistake:
Students look at the carbon directly attached to the carbonyl and count hydrogens. They often miss that in 2-methylpentanal, the α-carbon has a hydrogen (it is a secondary carbon), so it does have α-H. Similarly, they think phenylacetaldehyde has no α-H because the phenyl group is attached, but the α-carbon is −CHX2−.
How to Avoid:
- Draw the structure. The α-carbon is the carbon directly attached to the carbonyl carbon.
- Count the number of hydrogens on that α-carbon. If it is CHX3, CHX2, or CH (i.e., at least one H), it has α-H.
Application to the list:
- 2-Methylpentanal (ii): α-carbon is CH(CHX3)X− → has one α-H → Aldol.
- Phenylacetaldehyde (vii): α-carbon is CHX2X− (attached to phenyl) → has two α-H → Aldol.
- 2,2-Dimethylbutanal (ix): α-carbon is C(CHX3)X2X− → no α-H → Cannizzaro.
Mistake 4: Forgetting that Alcohols (like Butan-1-ol) Do Not Undergo These Reactions Directly
The Mistake:
Students see "Butan-1-ol" and think it might undergo oxidation to butanal, then aldol. But the question asks: "Which of the following compounds would undergo..." — the compound itself must have a carbonyl group.
How to Avoid:
- Aldol condensation requires a carbonyl compound (aldehyde or ketone) with α-H.
- Cannizzaro reaction requires an aldehyde without α-H.
- Alcohols do not undergo either reaction directly. They must first be oxidized to a carbonyl.
Application:
- Butan-1-ol (viii): It is an alcohol, not a carbonyl compound → Neither.
Mistake 5: Writing Incorrect Products (Especially for Crossed Aldol)
The Mistake:
When writing the product of aldol condensation, students often forget to:
- Show the aldol addition product (the β-hydroxy carbonyl) before dehydration.
- Dehydrate correctly (losing water from α and β positions).
- For unsymmetrical ketones (like 1-phenylpropanone), they forget that two different enolates can form, leading to two possible products.
How to Avoid:
- Step 1: Identify the α-carbon and the enolate formed. …
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.
-
- 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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