Solved Examples (Dedicated Problem Set)

Dedicated Problem Set — JEE/NEET Level

This is a curated set of 32 fully worked, exam-level problems covering every theme of Aldehydes, Ketones and Carboxylic Acids — naming carbonyls and acids, predicting nucleophilic-addition products, aldol vs cross-aldol vs Cannizzaro, oxidation and reduction routes, ordering reactivity, comparing acidity of substituted acids, the distinguishing tests (Tollens/Fehling/iodoform/2,4-DNP), and named-reaction product prediction (Rosenmund, Stephen, Etard, Gattermann-Koch, HVZ, decarboxylation). Each solution shows the reasoning step by step. Work each one before reading the solution.

Naming, Structure & Reactivity

Example 1. Give the IUPAC name of CH$_3$CH=CHCHO.

Solution: A four-carbon chain with -CHO at C-1 and a double bond starting at C-2: but-2-enal.

Example 2. Arrange in decreasing reactivity toward HCN: HCHO, CH$_3$CHO, CH$_3$COCH$_3$, CH$_3$COC$_6$H$_5$.

Solution: HCHO > CH$_3$CHO > CH$_3$COCH$_3$ > CH$_3$COC$_6$H$_5$. More/larger groups (and the ring's resonance donation in the aryl ketone) reduce the carbonyl δ+ and add steric hindrance.

Example 3. Why is chloral (Cl$_3$C-CHO) extremely reactive toward nucleophilic addition (it even adds water to form a stable hydrate)?

Solution: The three electron-withdrawing chlorines pull electron density away, making the carbonyl carbon strongly δ+ (very electrophilic), so it readily adds even weak nucleophiles like water.

Example 4. Identify A: an aldehyde C$_7$H$_6$O that gives a Tollens' mirror but is negative to Fehling's and iodoform.

Solution: C$_7$H$_6$O aldehyde with no alpha-H and aromatic = benzaldehyde (C$_6$H$_5$CHO) — aromatic aldehydes give Tollens but not Fehling, and it has no CH$_3$CO group (iodoform negative).

Nucleophilic Addition & Named Products

Example 5. Give the product of propanone + HCN and state one use of the product.

Solution: The cyanohydrin (CH$_3$)$_2$C(OH)CN (2-hydroxy-2-methylpropanenitrile). On hydrolysis it gives an alpha-hydroxy acid (2-hydroxy-2-methylpropanoic acid).

Example 6. What is formed when ethanal reacts with (i) NH$_2$OH and (ii) 2,4-DNP?

Solution: (i) The oxime CH$_3$CH=NOH (acetaldoxime); (ii) the 2,4-DNP derivative (an orange hydrazone), confirming the carbonyl group.

Example 7. Predict the product of benzaldehyde + a primary amine (aniline).

Solution: A Schiff's base (imine): C$_6$H$_5$CH=N-C$_6$H$_5$ (with loss of water).

Example 8. Why is an acetal used as a "protecting group"?

Solution: The acetal (from the aldehyde + 2 alcohol/dry HCl) is stable to base and nucleophiles, so it shields the carbonyl during another reaction and is hydrolysed back with aqueous acid afterward.

Aldol, Cross-Aldol & Cannizzaro

Example 9. Write the aldol product of propanal and its dehydration product.

Solution: Aldol = 3-hydroxy-2-methylpentanal; on heating it dehydrates to 2-methylpent-2-enal.

Example 10. Predict the major product of the crossed aldol of benzaldehyde and acetaldehyde.

Solution: Benzaldehyde has no alpha-H, so it is only the electrophile; acetaldehyde supplies the enolate → after dehydration the product is cinnamaldehyde (C$_6$H$_5$CH=CHCHO).

Example 11. A compound C$_6$H$_5$CHO is warmed with conc. KOH. Name the reaction and give the products.

Solution: Cannizzaro reaction (no alpha-H): products are benzyl alcohol (C$_6$H$_5$CH$_2$OH) + potassium benzoate (C$_6$H$_5$COOK).

Example 12. Which one undergoes aldol and which undergoes Cannizzaro: 2,2-dimethylpropanal or butanal?

Solution: Butanal has alpha-H → aldol. 2,2-Dimethylpropanal ((CH$_3$)$_3$CCHO) has no alpha-H → Cannizzaro.

Oxidation, Reduction & Tests

Example 13. Distinguish between pentan-2-one and pentan-3-one.

Solution: Iodoform test — pentan-2-one (has CH$_3$CO-) gives yellow CHI$_3$; pentan-3-one does not.

Example 14. Convert acetophenone to ethylbenzene. Name the reaction.

Solution: Clemmensen reduction (Zn-Hg / conc. HCl) reduces C=O to CH$_2$ → ethylbenzene. (Wolff-Kishner with NH$_2$NH$_2$/KOH gives the same product.)

Example 15. An aldehyde gives a silver mirror with Tollens but a ketone does not. Explain.

Solution: Aldehydes are easily oxidised (to carboxylates) and so reduce Ag(I) to a silver mirror. Ketones resist oxidation (no H on the carbonyl carbon), so they do not.

Example 16. Reduce butan-2-one with NaBH$_4$. Give the product.

Solution: The C=O is reduced to CH-OH giving the secondary alcohol butan-2-ol.

Preparation & Named Reactions

Example 17. How is benzaldehyde made from benzoyl chloride without forming benzyl alcohol?

Solution: Rosenmund reduction — C$_6$H$_5$COCl + H$_2$ over Pd-BaSO$_4$ (poisoned catalyst stops at -CHO) → benzaldehyde.

Example 18. Convert toluene to benzaldehyde. Name the reaction and reagent.

Solution: Etard reaction: toluene + chromyl chloride (CrO$_2$Cl$_2$), then hydrolysis → benzaldehyde.

Example 19. Convert ethyl bromide to propanoic acid.

Solution: C$_2$H$_5$Br + KCN → C$_2$H$_5$CN → (hydrolysis) → CH$_3$CH$_2$COOH (or via C$_2$H$_5$MgBr + CO$_2$ then H$_3$O+). Either adds one carbon.

Example 20. Give the product of acetic acid + Cl$_2$ / red phosphorus.

Solution: Hell-Volhard-Zelinsky alpha-chlorination → chloroacetic acid (ClCH$_2$COOH).

Carboxylic Acid Acidity & Reactions

Example 21. Arrange in increasing acidity: acetic acid, formic acid, chloroacetic acid.

Solution: acetic acid < formic acid < chloroacetic acid. -CH$_3$ (in acetic) donates electrons (weakest); H in formic has no +I; -Cl is electron-withdrawing (strongest of the three).

Example 22. Which is more acidic: 4-nitrobenzoic acid or 4-methoxybenzoic acid? Why?

Solution: 4-Nitrobenzoic acid. The -NO$_2$ group is electron-withdrawing and stabilises the carboxylate; -OCH$_3$ is electron-donating and destabilises it.

Example 23. What forms when sodium propanoate is heated with soda lime?

Solution: Decarboxylation: CH$_3$CH$_2$COONa + NaOH/CaO → ethane (C$_2$H$_6$) + Na$_2$CO$_3$ (one carbon fewer).

Example 24. Give the products of esterification of acetic acid with ethanol.

Solution: Ethyl ethanoate (CH$_3$COOC$_2$H$_5$) + water, catalysed by conc. H$_2$SO$_4$ (reversible).

Mixed JEE/NEET Challenge

Example 25. An organic compound A (C$_3$H$_6$O) gives a positive 2,4-DNP and a positive iodoform but a negative Tollens. Identify A.

Solution: 2,4-DNP positive ⇒ carbonyl; Tollens negative ⇒ ketone; iodoform positive ⇒ CH$_3$CO-. The C$_3$H$_6$O methyl ketone is acetone (propanone).

Example 26. Compound B (C$_2$H$_4$O) gives positive Tollens, positive Fehling and positive iodoform. Identify B.

Solution: An aldehyde (Tollens/Fehling) with a CH$_3$CO arrangement (iodoform) and formula C$_2$H$_4$O is acetaldehyde (ethanal).

Example 27. Why does formaldehyde undergo Cannizzaro but acetaldehyde does not?

Solution: Formaldehyde (HCHO) has no alpha-hydrogen, so with conc. alkali it disproportionates (Cannizzaro). Acetaldehyde has alpha-H, so it prefers aldol condensation.

Example 28. Identify the product when propanal is treated with dilute NaOH and the aldol is heated.

Solution: Aldol = 3-hydroxy-2-methylpentanal; on heating → 2-methylpent-2-enal (an alpha,beta-unsaturated aldehyde).

Example 29. An acid C$_7$H$_6$O$_2$ is formed by oxidising any alkylbenzene with hot KMnO$_4$. Identify it.

Solution: The side chain is cut to a single -COOH on the ring → benzoic acid (C$_6$H$_5$COOH, C$_7$H$_6$O$_2$).

Example 30. Why does trichloroacetic acid have a much lower pKa than acetic acid?

Solution: The three electron-withdrawing -Cl atoms strongly stabilise the trichloroacetate ion (withdrawing the negative charge by induction), so the acid ionises much more readily — a much lower pKa (stronger acid).

Example 31. Give one chemical test to distinguish benzoic acid from phenol.

Solution: NaHCO$_3$ — benzoic acid gives effervescence (CO$_2$); phenol does not.

Example 32. Predict the major product of nitration of benzoic acid.

Solution: -COOH is deactivating, meta-directing, so the major product is m-nitrobenzoic acid (3-nitrobenzoic acid).

JEE Main & Advanced Level Solved Examples

These problems extend beyond the core set into mechanism, named-reaction logic, tautomerism and structure elucidation — the discriminators that decide JEE ranks. Identify whether an alpha-hydrogen is present and which carbon is electrophilic before you answer.

Example 33: Keto-enol tautomerism and enol content [JEE Advanced]

Why does pentane-2,4-dione (acetylacetone) exist with a far higher enol content (~80%) than acetone (~10$^{-4}$ %)?

Solution: Tautomerism interconverts the keto form (C=O with an alpha C-H) and the enol form (C=C-OH). In acetone the enol is strongly disfavoured. In pentane-2,4-dione the enol is stabilised by (i) conjugation of the enol C=C with the second carbonyl and (ii) a six-membered intramolecular hydrogen bond between the enolic -OH and the other C=O oxygen. Together these push the enol to about 80%. Takeaway: 1,3-dicarbonyl compounds are far more enolic than ordinary carbonyls.

Example 34: Crossed Cannizzaro reaction [JEE Advanced]

When methanal (HCHO) and benzaldehyde (C$_6$H$_5$CHO) are heated with concentrated NaOH, which aldehyde is oxidised and which is reduced? Why?

Solution: Neither has an alpha-hydrogen, so they undergo Cannizzaro (disproportionation). HCHO is oxidised to sodium formate (HCOONa) and benzaldehyde is reduced to benzyl alcohol (C$_6$H$_5$CH$_2$OH). Reason: HCHO is the better hydride donor — its carbonyl is the most electrophilic and least hindered, so hydride is transferred from the HCHO-derived intermediate to benzaldehyde. Formaldehyde effectively sacrifices itself as the reductant.

JEE/Advanced — Named Reactions: Perkin & Benzoin

Example 35: Perkin reaction [JEE]

What forms when benzaldehyde is heated with acetic anhydride and sodium ethanoate?

Solution: The Perkin reaction condenses an aromatic aldehyde with an aliphatic acid anhydride (its sodium salt as base) to give an alpha,beta-unsaturated aromatic acid. Benzaldehyde gives cinnamic acid (3-phenylprop-2-enoic acid), C$_6$H$_5$-CH=CH-COOH. The base forms the anhydride enolate, which adds to the aldehyde; dehydration and hydrolysis then give the conjugated acid.

Example 36: Benzoin condensation [JEE Advanced]

Two molecules of benzaldehyde react in aqueous-alcoholic KCN to give benzoin. What is the special role of cyanide?

Solution: In the benzoin condensation, cyanide (CN$^-$) is a uniquely suited catalyst. It adds to one benzaldehyde to give a cyanohydrin-type anion whose former carbonyl C-H is now acidic (the -CN stabilises the resulting carbanion). This carbanion acts as an acyl anion equivalent (umpolung), attacking the second benzaldehyde; loss of CN$^-$ regenerates the catalyst and gives benzoin, C$_6$H$_5$-CH(OH)-CO-C$_6$H$_5$ (an alpha-hydroxy ketone). Only aldehydes without alpha-hydrogen (typically aromatic) work well.

JEE/Advanced — Mechanism, Aldol & Structure Elucidation

Example 37: Mechanism of Fischer esterification [JEE Advanced]

In RCOOH + R'OH ⇌ RCOOR' + H$_2$O (acid-catalysed), which bond of the acid breaks, and how is it proven?

Solution: The reaction is a nucleophilic acyl substitution: protonation of the carbonyl, addition of the alcohol, then loss of water. The acyl C-OH bond of the acid breaks (acyl-oxygen fission). Proof: with the alcohol labelled as R'-$^{18}$O-H, the label appears in the ester (RCO-$^{18}$O-R'), not in the water — so the -OH lost as water comes from the acid. Being reversible, the equilibrium is driven forward by excess alcohol or by removing water.

Example 38: Why a crossed aldol of two enolisable aldehydes is impractical [JEE]

How many aldol products can form when ethanal and propanal are mixed with dilute base, and how is a clean crossed aldol achieved instead?

Solution: Both aldehydes have alpha-hydrogens, so each can be either the enolate or the electrophile, giving four aldol products (two self-aldols + two crossed) — hard to separate, so synthetically poor. A clean crossed (Claisen-Schmidt) aldol uses one partner with no alpha-hydrogen (e.g. benzaldehyde or HCHO) as the electrophile and the other as the only enolate source, giving essentially a single product.

Example 39: Identify the compound [JEE Advanced]

A compound A (C$_9$H$_{10}$O) gives an orange precipitate with 2,4-DNP, does not reduce Tollens' reagent, and gives a positive iodoform test. Identify A.

Solution: 2,4-DNP positive → a carbonyl. Tollens negative → not an aldehyde, so a ketone. Iodoform positive → it contains a CH$_3$CO- (methyl ketone) unit. A C$_9$H$_{10}$O ketone with a methyl-ketone group is phenylacetone, C$_6$H$_5$-CH$_2$-CO-CH$_3$ (1-phenylpropan-2-one). (Acetophenone is only C$_8$H$_8$O, so it does not fit the formula.)

Example 40: Intramolecular aldol condensation [JEE Advanced]

What is the major product when hexane-2,5-dione is treated with dilute base?

Solution: An intramolecular aldol condensation occurs. An enolate at one end attacks the other carbonyl; the favoured closure gives a strain-free five-membered ring, and dehydration yields 3-methylcyclopent-2-en-1-one. Five- and six-membered rings form preferentially, so 1,4- and 1,5-diketones cyclise to cyclopentenones and cyclohexenones respectively.

JEE/Advanced — Carboxylic Acids

Example 41: Using the HVZ reaction in synthesis [JEE]

How can ethanoic acid be converted into glycine (aminoethanoic acid) using the HVZ reaction?

Solution: The Hell-Volhard-Zelinsky (HVZ) reaction alpha-halogenates a carboxylic acid having an alpha C-H: CH$_3$COOH + Cl$_2$/red P → ClCH$_2$COOH. Treating the alpha-chloro acid with excess ammonia then replaces -Cl by -NH$_2$: ClCH$_2$COOH + 2 NH$_3$ → H$_2$N-CH$_2$-COOH (glycine) + NH$_4$Cl. HVZ thus installs a reactive alpha-halogen, opening routes to alpha-amino, alpha-hydroxy and alpha-cyano acids.

Example 42: The ortho effect on acidity [JEE Advanced]

Both 2-nitrobenzoic acid and 2-methylbenzoic acid are stronger acids than benzoic acid, even though -CH$_3$ is electron-donating. Explain.

Solution: This is the ortho effect: nearly any ortho substituent — electron-withdrawing or -donating — makes a benzoic acid more acidic than benzoic acid. The dominant cause is steric: the ortho group twists the -COOH out of the ring plane, reducing conjugation between the ring and the carboxyl and so stabilising the carboxylate (steric strain relief and hydrogen bonding also contribute). Because the effect is steric rather than purely electronic, even an electron-donating ortho group raises acidity — which the simple inductive/resonance picture alone cannot explain.