Aldehydes, Ketones and Carboxylic Acids NEET PYQs (2005–2026) with Solutions

10 solved Aldehydes, Ketones and Carboxylic Acids PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumFriedel-Crafts acylation and haloform reaction

   For the following reaction sequence, choose the correct option. $\text{Benzene} \xrightarrow[\text{ii. NaOCl}]{\text{i. }\mathrm{CH_3COCl,\ AlCl_3}} \mathrm{P + Q}$

   1. AOption A: If P is the sodium salt of a carboxylic acid, Q is a primary alcohol
   2. BOption B: P and Q are aromatic compounds
   3. COption C: If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light
   4. DOption D: Both P and Q are carbonyl compounds

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   Correct answer: (C) If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light

   Explanation

   Step i (Friedel-Crafts acylation): benzene + $\mathrm{CH_3COCl}$/anhydrous $\mathrm{AlCl_3}$ gives acetophenone, $\mathrm{C_6H_5COCH_3}$. Step ii (haloform reaction): a methyl ketone is cleaved by sodium hypochlorite: $$\mathrm{C_6H_5COCH_3 \xrightarrow{NaOCl} C_6H_5COONa + CHCl_3}$$ So P = sodium benzoate and Q = chloroform. P on acidification gives benzoic acid, $\mathrm{C_6H_5COOH}$. Chloroform is slowly oxidised by air in the presence of light to phosgene, a poisonous gas:$$\mathrm{2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl}$$ That is why chloroform is stored in closed dark bottles. So option (C) is correct. Option (A) is wrong because Q is chloroform, not an alcohol. Option (B) is wrong because chloroform is not aromatic. Option (D) is wrong because P is a carboxylate salt and Q is chloroform; neither is an aldehyde or ketone.

2. Question 2 (NEET 2026)

   NEET 2026HardImportant Reactions & Uses

   Compound P(C₈H₈O) gives a red-orange precipitate with 2,4-DNP reagent and does not reduce Fehling's reagent. On drastic oxidation with chromic acid, P gives an aromatic acidic product Q that produces effervescence on treatment with aq. NaHCO₃. Compound P and Q respectively are:

   1. AOption A: Compound 1
   2. BOption B: Compound 2
   3. COption C: Compound 3
   4. DOption D: Compound 4

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   Correct answer: (D) Compound 4

   Explanation

   2,4-DNP positive indicates carbonyl compound. No Fehling's reduction suggests ketone. For C₈H₈O, acetophenone fits. Strong oxidation converts side chain to benzoic acid, which reacts with NaHCO₃ giving CO₂. Hence option (D) is correct.

3. Question 3 (NEET 2026)

   NEET 2026HardCross and self aldol condensation

   The compound that CANNOT be obtained from the aldol condensation reaction shown below, is (2,2-dimethylcyclopentan-1-one) $+ \mathrm{PhCHO} \xrightarrow{\mathrm{NaOH,\ \Delta}}$

   1. AOption A: Structure A
   2. BOption B: Structure B
   3. COption C: Structure C
   4. DOption D: Structure D

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   Correct answer: (A) Structure A

   Explanation

   The options show: (A) 2,2-dimethylcyclopentanone bonded at C-5 to a 5,5-dimethylcyclopent-1-enyl group; the C=C is inside the second ring, not conjugated with C=O. (B) 5-benzylidene-2,2-dimethylcyclopentanone with Ph pointing away from the C=O (E isomer). (C) 2,2-dimethylcyclopentanone joined at C-5 by an exocyclic C=C to C-1 of a 2,2-dimethylcyclopentylidene ring (an $\alpha,\beta$-unsaturated ketone). (D) 5-benzylidene-2,2-dimethylcyclopentanone with Ph on the C=O side (Z isomer). In 2,2-dimethylcyclopentanone, C-2 has no $\alpha$-hydrogen, so only C-5 can form an enolate. PhCHO has no $\alpha$-hydrogen and acts only as the carbonyl partner. Cross aldol: the C-5 enolate attacks PhCHO. Dehydration then gives 5-benzylidene-2,2-dimethylcyclopentanone, as the E and Z geometrical isomers (B and D). Self aldol: the C-5 enolate attacks the C=O of a second ketone molecule. In the aldol, the OH sits on the $\beta$-carbon (old C=O carbon of the second ring) and the $\alpha$-carbon (C-5 of the first ring) still has one H. Base-catalysed dehydration removes this $\alpha$-H, giving the conjugated $\alpha,\beta$-unsaturated ketone with the exocyclic C=C (C). Structure A would need the double bond to form between the $\beta$ and $\gamma$ carbons inside the second ring. That product is not conjugated with C=O and is not formed in aldol condensation. So A cannot be obtained: option (A).

4. Question 4 (NEET 2026)

   NEET 2026EasyEtard reaction and Rosenmund reduction

   Consider the following reaction, and choose the correct option. $\text{Toluene} \xrightarrow[\text{ii. }\mathrm{H_3O^+}]{\text{i. }\mathrm{CrO_2Cl_2,\ CS_2}} \mathrm{P}$

   1. AOption A: On treating compound P with saturated $\mathrm{NaHCO_3}$ solution, brisk effervescence is observed
   2. BOption B: Compound P can be prepared by treating benzene with anhydrous $\mathrm{AlCl_3}$ and $\mathrm{CH_3COCl}$
   3. COption C: On treatment with bromine water, compound P gives a white precipitate
   4. DOption D: Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on $\mathrm{BaSO_4}$

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   Correct answer: (D) Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on $\mathrm{BaSO_4}$

   Explanation

   Etard reaction: chromyl chloride in $\mathrm{CS_2}$ oxidises the methyl group of toluene to a chromium complex, and hydrolysis then gives benzaldehyde. So P = $\mathrm{C_6H_5CHO}$. (A) is wrong: brisk effervescence ($\mathrm{CO_2}$) with $\mathrm{NaHCO_3}$ is a test for carboxylic acids, and benzaldehyde is not an acid. (B) is wrong: benzene with $\mathrm{CH_3COCl}$/anhydrous $\mathrm{AlCl_3}$ (Friedel-Crafts acylation) gives acetophenone, not benzaldehyde. (C) is wrong: a white precipitate with bromine water (2,4,6-tribromo product) is given by phenol (or aniline), not benzaldehyde. (D) is correct: Rosenmund reduction, $$\mathrm{C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO + HCl}$$ So the answer is option (D).

5. Question 5 (NEET 2026)

   NEET 2026MediumPolarity of oxygen-containing organic compounds

   Arrange the following compounds in the increasing order of polarity A. $\mathrm{CH_3CH_2OCH_2CH_3}$ B. $\mathrm{CH_3CH_2OH}$ C. $\mathrm{CH_3COCH_3}$ D. $\mathrm{CH_3COOH}$ Choose the correct answer from the options given below.

   1. AOption A: $\mathrm{A < B < C < D}$
   2. BOption B: $\mathrm{C < A < D < B}$
   3. COption C: $\mathrm{C < A < B < D}$
   4. DOption D: $\mathrm{A < C < B < D}$

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   Correct answer: (D) $\mathrm{A < C < B < D}$

   Explanation

   A, diethyl ether: has only weakly polar C–O–C bonds and no O–H, so it cannot hydrogen bond with itself. It is the least polar. C, acetone: has a strongly polar C=O group but no O–H, so it is more polar than the ether. B, ethanol: has an O–H group and forms extensive intermolecular hydrogen bonds, so it is more polar than acetone. D, acetic acid: has both C=O and O–H in the carboxyl group and forms strongly hydrogen-bonded dimers. It is the most polar of the four. This matches the NCERT trend in boiling points: ethers < ketones < alcohols < carboxylic acids. So the order is $\mathrm{A < C < B < D}$, option (D).

6. Question 6 (NEET 2025)

   NEET 2025MediumCarboxylic Acids

   The correct order of decreasing acidity of the following aliphatic acids is :

   1. AOption A: (CH₃)₃CCOOH > (CH₃)₂CHCOOH > CH₃COOH > HCOOH
   2. BOption B: CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH > HCOOH
   3. COption C: HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH
   4. DOption D: HCOOH > (CH₃)₃CCOOH > (CH₃)₂CHCOOH > CH₃COOH

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   Correct answer: (C) HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH

   Explanation

   Electron-donating alkyl groups decrease the acidity of carboxylic acids due to the +I effect, which destabilizes the carboxylate ion. Formic acid has no alkyl group and is therefore the strongest acid among these. Increasing alkyl substitution decreases acidity. Hence: HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH.

7. Question 7 (NEET 2024)

   NEET 2024EasyImportant Reactions & Uses

   Fehling’s solution ‘A’ is:

   1. AOption A: alkaline solution of sodium potassium tartrate (Rochelle’s salt)
   2. BOption B: aqueous sodium citrate
   3. COption C: aqueous copper sulphate
   4. DOption D: alkaline copper sulphate

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   Correct answer: (C) aqueous copper sulphate

   Explanation

   Fehling’s solution consists of two separate solutions: Fehling’s solution A → aqueous copper sulphate solution. Fehling’s solution B → alkaline solution of sodium potassium tartrate. Hence, option (C) is correct.

8. Question 8 (NEET 2023)

   NEET 2023MediumImportant Reactions & Uses

   Identify product (A) in the following reaction:

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (A) Option (1)

   Explanation

   Zn-Hg with concentrated HCl carries out Clemmensen reduction, which converts carbonyl groups (>C=O) into methylene groups (−CH2−). Both ketone groups in the substrate are reduced to alkyl groups, giving the fully reduced hydrocarbon product. Hence, option (A) is correct.

9. Question 9 (NEET 2023)

   NEET 2023MediumPhysical & Chemical Properties

   Complete the following reaction: [C] is ________.Complete the following reaction: [C] is ________.

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (D) Option 4

   Explanation

   Cyclohexanone reacts with HCN to form cyanohydrin. On hydrolysis with concentrated H₂SO₄ and heat, the nitrile group converts into carboxylic acid followed by dehydration to form cyclohexene carboxylic acid. Hence option (D) is correct.

10. Question 10 (NEET 2021)

    NEET 2021MediumNamed Reactions of Carbonyl Compounds and Carboxylic Acids

    Match List-I with List-II and choose the correct answer.

    1. AOption A: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
    2. BOption B: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
    3. COption C: (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
    4. DOption D: (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)

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    Correct answer: (A) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

    Explanation

    (a) Benzene + CO/HCl in presence of AlCl3/CuCl → Gattermann-Koch reaction (b) Methyl ketone + NaOX → Haloform reaction (c) Alcohol + carboxylic acid + conc. H2SO4 → Esterification (d) Carboxylic acid + X2/Red P → Hell-Volhard-Zelinsky reaction Correct matching: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i).