JEE (Advanced) · Chemistry, Class 12
Aldehydes, Ketones and Carboxylic Acids: JEE Advanced PYQ, 2007–2026
Aldehydes, Ketones and Carboxylic Acids (Chemistry, Class 12) has been asked 56 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.4 questions a paper, 7% of all Chemistry questions. It is the 2nd most-asked of 31 Chemistry chapters. The two 2026 papers had 3 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 2in 2 papers |
| 2008 | 4in 2 papers |
| 2009 | 3in 2 papers |
| 2010 | 5in 2 papers |
| 2011 | 2in 2 papers |
| 2012 | 4in 2 papers |
| 2013 | 3in 2 papers |
| 2014 | 3in 2 papers |
| 2015 | 2in 2 papers |
| 2016 | 4in 2 papers |
| 2017 | 3in 2 papers |
| 2018 | 3in 2 papers |
| 2019 | 1in 2 papers |
| 2020 | 0in 2 papers |
| 2021 | 2in 2 papers |
| 2022 | 1in 2 papers |
| 2023 | 4in 2 papers |
| 2024 | 4in 2 papers |
| 2025 | 3in 2 papers |
| 2026 | 3in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Aldol and Cannizzaro19
- Nucleophilic addition16
- Acidity of carboxylic acids12
- Tests to distinguish aldehydes and ketones9
Try it
10 Aldehydes, Ketones and Carboxylic Acids questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
Consider (I) benzoic acid, (II) methanoic acid, (III) 4-nitrobenzoic acid and (IV) ethanoic acid. What is the order of decreasing acid strength in water?
Answer: (C)
The pKa values are: 4-nitrobenzoic acid 3.41, methanoic acid 3.75, benzoic acid 4.19, ethanoic acid 4.76. The nitro group withdraws electrons and steadies the anion most; methanoic acid has no group that releases electrons; the phenyl group withdraws weakly; the methyl group releases electrons and weakens the acid. So III > II > I > IV.
Q2 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
In the Cannizzaro reaction of benzaldehyde with concentrated NaOH, which species transfers a hydride ion (H⁻) to the carbonyl carbon of a second benzaldehyde molecule?
Answer: (D)
The first step is nucleophilic addition of OH⁻ to the C=O of one benzaldehyde molecule, giving the alkoxide C6H5–CH(OH)–O⁻. In this adduct the negative oxygen pushes electron density toward carbon, and the C–H hydrogen is transferred with its bonding pair as hydride to the carbonyl carbon of a second benzaldehyde. The donor becomes benzoic acid, which is deprotonated to benzoate, and the acceptor becomes the benzyl alkoxide, which takes a proton to give benzyl alcohol. Hydroxide and enolate cannot donate hydride, and benzoate is the oxidised product.
Q3 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
Oxalic acid, HOOC–COOH, has pKa₁ = 1.25 and pKa₂ = 4.27 at 298 K, while ethanoic acid has pKa = 4.76. Which statement best accounts for pKa₁ being much lower than that of ethanoic acid and for pKa₂ being higher than pKa₁?
Answer: (A)
The acidity of a carboxylic acid depends on how stable its conjugate base is. In oxalic acid the second –COOH group is directly attached to the first and strongly withdraws electrons (–I effect). This stabilises the monoanion, so pKa₁ (1.25) is far below that of ethanoic acid (4.76). In the second step, H⁺ is removed from HOOC–COO⁻, which is already negatively charged. The Coulombic attraction and the added charge repulsion in the dianion make this less favourable, so pKa₂ (4.27) is higher than pKa₁. Only the first option gives both effects correctly.
Q4 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
Methanal, ethanal, propanone and 2,2,4,4-tetramethylpentan-3-one (di-tert-butyl ketone, (CH3)3C–CO–C(CH3)3) are each treated with HCN under identical conditions. Which sequence correctly gives their order of decreasing reactivity towards this nucleophilic addition?
Answer: (A)
Nucleophilic addition of HCN to C=O converts a planar sp2 carbon into a crowded sp3 carbon. The rate therefore depends on (i) how electrophilic the carbonyl carbon is, since alkyl groups reduce it by +I, and (ii) steric hindrance to the incoming CN⁻. Methanal has two H atoms, so it is least hindered and most electrophilic. Ethanal has one alkyl group, acetone has two, and the di-tert-butyl ketone has two very bulky groups. Both effects point the same way, so the order is HCHO > CH3CHO > CH3COCH3 > (CH3)3C–CO–C(CH3)3.
Q5 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
Benzaldehyde in which the aldehydic hydrogen is replaced by deuterium (C6H5–CDO) is heated with concentrated NaOH in ordinary water (H2O) until the Cannizzaro reaction is complete. Which pair correctly describes the products after acidification and ignoring exchange of O–H hydrogens?
Answer: (A)
Principle: the Cannizzaro reaction of an aldehyde with no α-hydrogen proceeds by hydroxide adding to C=O, then a hydride shift from this adduct to the C=O of a second aldehyde molecule. Step 1: OH– adds to C6H5–CDO to give C6H5–CD(O–)(OH). Step 2: the D leaves as D– and attacks the carbonyl carbon of another C6H5–CDO, giving C6H5–CD2–O–. The donor becomes the carboxylate C6H5COO– with no C-bound D. Step 3: protonation by water gives C6H5CD2OH, and acidification gives C6H5COOH. The alcohol carbon holds two D atoms and the water supplies no C–H.
Q6 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
Consider the acids (I) CH₃COOH, (II) ClCH₂COOH, (III) Cl₂CHCOOH and (IV) FCH₂COOH. What is the order of decreasing acid strength in water?
Answer: (C)
The electron-withdrawing inductive effect of a halogen stabilises the carboxylate ion. Two chlorine atoms do more than one halogen atom of either kind, and one fluorine does more than one chlorine. The pKa values are about 1.3 (Cl₂CHCOOH), 2.6 (FCH₂COOH), 2.9 (ClCH₂COOH) and 4.8 (CH₃COOH): III > IV > II > I.
Q7 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
5.7 g of hexane-2,5-dione (CH₃COCH₂CH₂COCH₃, molar mass 114 g mol⁻¹) is warmed with excess iodine and aqueous NaOH. Assume each CH₃CO– group undergoes complete haloform reaction and no other reaction occurs. What mass of iodoform is obtained? (H = 1, C = 12, I = 127)
Answer: (A)
The iodoform test is given by compounds containing CH₃CO–. Each CH₃CO– group is cleaved to give one CHI₃ and a carboxylate. Hexane-2,5-dione has two such groups and therefore gives two CHI₃ per molecule, along with disodium butanedioate. Moles of diketone = 5.7 ÷ 114 = 0.05 mol. Moles of CHI₃ = 2 × 0.05 = 0.10 mol. Molar mass of CHI₃ = 12 + 1 + 3(127) = 394 g mol⁻¹. Mass = 0.10 × 394 = 39.4 g.
Q8 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
Butanoic acid and its three monochloro derivatives are listed below. Which one is the strongest acid in water?
Answer: (A)
A substituent that withdraws electrons by the inductive effect stabilises the carboxylate anion and so increases acid strength. The effect decreases rapidly with distance along the carbon chain. Cl on C-2 is closest to the COOH group, then C-3, then C-4. The order of acidity is 2-chloro > 3-chloro > 4-chloro > butanoic acid, so 2-chlorobutanoic acid is the strongest.
Q9 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
The formation of the semicarbazone of cyclohexanone from semicarbazide (H₂N–NH–CO–NH₂, conjugate acid pKₐ ≈ 3.7) is carried out in buffers of different pH. The rate is very low at pH 1, highest at about pH 4–5, and falls again in strongly basic solution (pH 12). Which explanation correctly accounts for BOTH the low rate at pH 1 and the low rate at pH 12?
Answer: (A)
Addition of ammonia derivatives (Z–NH₂) to C=O needs two things. First, the nitrogen must have a free lone pair so that it can attack the carbonyl carbon. Second, some acid must be present to protonate the carbonyl oxygen, which makes carbon more electrophilic. Acid is also needed to protonate the –OH of the carbinolamine so that water is eliminated to give C=N. At pH 1, which is well below the pKₐ of 3.7, nearly all the semicarbazide is present as the protonated, non-nucleophilic ion, so the rate is low. At pH 12, hardly any acid is available for carbonyl activation and dehydration, so the rate is low again. The maximum therefore lies at weakly acidic pH (about 4–5), which is why these reactions are done in a weakly acidic medium.
Q10 · Chemistry · Aldehydes, Ketones and Carboxylic Acids
Propanone is treated with dilute Ba(OH)2, and the aldol addition product formed is then heated. What is the final organic product?
Answer: (B)
Two molecules of propanone undergo aldol addition in dilute base to give (CH3)2C(OH)CH2COCH3, 4-hydroxy-4-methylpentan-2-one. This is a β-hydroxy ketone with acidic α-hydrogens. On heating it loses water, with the α-H and the β-OH leaving, to give the conjugated enone (CH3)2C=CHCOCH3, 4-methylpent-3-en-2-one.
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All 56 real JEE Advanced questions from Aldehydes, Ketones and Carboxylic Acids.
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About Aldehydes, Ketones and Carboxylic Acids in JEE Advanced.
How many questions come from Aldehydes, Ketones and Carboxylic Acids in JEE Advanced?
Aldehydes, Ketones and Carboxylic Acids (Chemistry, Class 12) has been asked 56 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.4 questions a paper, 7% of all Chemistry questions. It is the 2nd most-asked of 31 Chemistry chapters. The two 2026 papers had 3 from it.
Which topics of Aldehydes, Ketones and Carboxylic Acids are asked most in JEE Advanced?
Aldol and Cannizzaro (19), Nucleophilic addition (16), Acidity of carboxylic acids (12), Tests to distinguish aldehydes and ketones (9), counted over the papers from 2007–2026.
Is Aldehydes, Ketones and Carboxylic Acids important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 2 of 31 in Chemistry; use that to decide how much time it gets, not whether to study it.
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On Nikaalle, free: sign up and practise every past JEE Advanced question from Aldehydes, Ketones and Carboxylic Acids with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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