Aldehydes, ketones and carboxylic acids for NEET

Almost everything in this chapter follows from one fact: the carbonyl carbon is electron-poor and flat, so nucleophiles attack it from above or below. Aldehydes react faster than ketones because the carbon is less crowded and less shielded by alkyl groups. The hydrogens next to the carbonyl are acidic, which is what makes the aldol reaction possible. When there are no such hydrogens, concentrated base does something different: the Cannizzaro reaction.

The 2026 NMC syllabus lists the nature of the carbonyl group, nucleophilic addition (HCN, NH₃ and its derivatives, Grignard reagent), oxidation, reduction (Wolff–Kishner and Clemmensen), acidity of α-hydrogen, aldol, Cannizzaro, the haloform reaction, tests to distinguish aldehydes from ketones, and the acidic strength of carboxylic acids. NEET questions are usually "which compound gives this test", "product of this named reaction", or an order of reactivity or acidity.

The carbonyl group

The carbonyl carbon is sp², with its three σ bonds in a plane at about 120°. Oxygen is more electronegative, so the C=O bond is polar, which resonance shows as C⁺–O⁻. The carbon is the electrophile; the oxygen is a weak Lewis base.

Reactivity towards nucleophiles: HCHO > CH₃CHO > CH₃COCH₃.

• Steric reason: two alkyl groups crowd the ketone carbon more than one alkyl group and an H.
• Electronic reason: alkyl groups push electrons (+I) and reduce the positive charge on the carbon.
• Benzaldehyde is less reactive than propanal, because the ring's resonance reduces the positive charge on the carbonyl carbon.

Preparation in brief

Nucleophilic addition

The nucleophile attacks the carbonyl carbon, which changes from sp² to sp³ and becomes a tetrahedral alkoxide. The alkoxide then picks up a proton.

The reactions with ammonia derivatives (Z–NH₂) give C=N–Z and water. They are run in weakly acidic solution: enough acid to activate the carbonyl, but not so much that the nitrogen nucleophile is protonated.

Reduction and oxidation

Reduction:

• NaBH₄ or LiAlH₄, or H₂ with a catalyst: aldehyde to 1° alcohol, ketone to 2° alcohol.
• Clemmensen: Zn–Hg and concentrated HCl turn C=O into CH₂.
• Wolff–Kishner: hydrazine gives the hydrazone, which KOH in ethylene glycol on heating converts to CH₂.

Oxidation. Aldehydes are oxidised even by mild reagents to carboxylic acids; ketones resist and are oxidised only under vigorous conditions, with breaking of C–C bonds.

α-Hydrogen: aldol and Cannizzaro

A hydrogen on the carbon next to C=O is acidic. The C=O group pulls electrons (−I), and the carbanion left behind (the enolate) is stabilised by resonance with the oxygen.

Aldol condensation. Aldehydes and ketones with at least one α-hydrogen, in dilute alkali, give a β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol). On heating, these lose water to give an α,β-unsaturated compound.

• 2CH₃CHO → CH₃CH(OH)CH₂CHO (3-hydroxybutanal) → CH₃CH=CHCHO (but-2-enal).
• 2CH₃COCH₃ → 4-hydroxy-4-methylpentan-2-one → 4-methylpent-3-en-2-one.
• Cross aldol: two different aldehydes that both have α-hydrogens give a mixture of four products. If one partner lacks α-hydrogen, as benzaldehyde does, the mixture is simpler: benzaldehyde and ethanal give mainly 3-phenylprop-2-enal (cinnamaldehyde).

Cannizzaro reaction. Aldehydes with no α-hydrogen (HCHO, C₆H₅CHO, (CH₃)₃CCHO) in concentrated alkali disproportionate: one molecule is reduced to the alcohol and one is oxidised to the carboxylate salt.

• 2HCHO + conc. KOH → CH₃OH + HCOOK
• 2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa
• Crossed Cannizzaro: HCHO + C₆H₅CHO → HCOONa + C₆H₅CH₂OH. Methanal is the one oxidised.

The haloform reaction works on methyl ketones: the three α-hydrogens of the CH₃ group are replaced by halogen, and the CX₃ group is then cleaved off as haloform. CH₃COR + 3NaOI → CHI₃ + RCOONa + 2NaOH. The acid salt has one carbon fewer. Ethanol is the only primary alcohol and ethanal the only aldehyde that give it, because NaOI first oxidises CH₃CH(OH)– to CH₃CO–.

Acidity of carboxylic acids

Carboxylic acids are stronger acids than phenols. In the carboxylate ion the negative charge is shared equally by two oxygen atoms; in phenoxide, it is spread onto less electronegative ring carbons. Carboxylic acids release CO₂ from NaHCO₃; phenols do not.

Factors that change acid strength:

• Electron-withdrawing groups (–NO₂, –CN, halogens, –CF₃) stabilise the anion and increase acidity. Electron-donating alkyl groups decrease it.
• Number of halogens: CF₃COOH > CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH.
• Electronegativity of the halogen: FCH₂COOH > ClCH₂COOH > BrCH₂COOH > ICH₂COOH.
• Distance: 2-chlorobutanoic > 3-chlorobutanoic > 4-chlorobutanoic acid. The inductive effect fades with distance.
• Aromatic and simple acids (NCERT pKa): HCOOH 3.75, C₆H₅COOH 4.19, CH₃COOH 4.76. So HCOOH > C₆H₅COOH > CH₃COOH.

Worked NEET-style questions

Example 1: the aldol product of propanal

• The α-carbon of CH₃CH₂CHO is the CH₂. Its carbanion, CH₃CH⁻–CHO, attacks a second propanal molecule.
• Aldol: CH₃CH₂CH(OH)CH(CH₃)CHO, 3-hydroxy-2-methylpentanal.
• On heating: 2-methylpent-2-enal.

Example 2: which give iodoform?

Pentan-2-one, pentan-3-one, propan-2-ol, methanol, acetophenone.

• Pentan-2-one (CH₃CO–): yes. Pentan-3-one (C₂H₅COC₂H₅): no.
• Propan-2-ol (CH₃CH(OH)–): yes. Methanol: no.
• Acetophenone (C₆H₅COCH₃): yes.
• Three: pentan-2-one, propan-2-ol and acetophenone.

Example 3: distinguish propanal and propanone

• Tollens' reagent: propanal gives a silver mirror; propanone does not.
• Fehling's solution: propanal gives the red-brown precipitate; propanone does not.
• Iodoform: propanone gives it; propanal does not.

Practice MCQs

1. The most reactive towards nucleophilic addition is: (a) HCHO (b) CH₃CHO (c) CH₃COCH₃ (d) C₆H₅COCH₃
2. Which does not undergo aldol condensation? (a) CH₃CHO (b) CH₃COCH₃ (c) C₆H₅CHO (d) CH₃CH₂CHO
3. C=O is converted to CH₂ using Zn–Hg and conc. HCl in the: (a) Wolff–Kishner reduction (b) Clemmensen reduction (c) Rosenmund reduction (d) Stephen reduction
4. Which gives a positive iodoform test? (a) propanal (b) pentan-3-one (c) ethanol (d) benzaldehyde
5. Benzaldehyde does not give a positive: (a) Tollens' test (b) Fehling's test (c) 2,4-DNP test (d) reaction with NaHSO₃
6. In the crossed Cannizzaro reaction of HCHO and C₆H₅CHO, the alcohol formed is: (a) methanol (b) benzyl alcohol (c) both (d) neither
7. The strongest acid is: (a) CH₃COOH (b) ClCH₂COOH (c) Cl₂CHCOOH (d) Cl₃CCOOH
8. A reagent that distinguishes benzoic acid from phenol is: (a) NaOH (b) NaHCO₃ (c) Na (d) bromine water

Answers

1. (a) No alkyl groups: least crowded and most positive carbon.
2. (c) It has no α-hydrogen.
3. (b) Clemmensen uses zinc amalgam and acid.
4. (c) NaOI oxidises ethanol to ethanal, which has CH₃CO–.
5. (b) Aromatic aldehydes do not respond to Fehling's solution.
6. (b) Methanal is oxidised to formate; benzaldehyde is reduced.
7. (d) Three chlorines give the strongest −I effect.
8. (b) Only the carboxylic acid releases CO₂.

What to do next

• Draw the mechanism of cyanohydrin formation and of aldol condensation for ethanal.
• Make a table of every test with reagent, result and which compounds pass. Add tricky negatives: benzaldehyde with Fehling's, methanol with iodoform.
• Solve ten aldol and Cannizzaro "predict the product" questions, checking for α-hydrogen first each time.
• Revise alcohols, phenols and ethers, then move to amines and biomolecules.