NEET Aldehydes, Ketones and Carboxylic Acids PYQs & Practice Questions

10 free NEET Chemistry questions on Aldehydes, Ketones and Carboxylic Acids, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Grignard Reaction with Nitriles and Ketones · Partial Reduction of Esters to Aldehydes · Order of Increasing Acidic Strength · Alcohol That Gives Butanone on Oxidation · Reagent Used in Clemmensen Reduction · Aldol Condensation · Cyanohydrin Formation and Dehydration · Clemmensen Reduction · Decarboxylation (Soda Lime Reaction) · Tollens' Reagent - Selective Oxidation of Aldehydes · Aldol Condensation (Claisen-Schmidt Reaction) · Friedel-Crafts Acylation Followed by Haloform Reaction

1. Q1NEET 2026Aldol Condensation (Claisen-Schmidt Reaction)

   2,2-dimethylcyclopentan-1-one is reacted with PhCHO (benzaldehyde) in the presence of NaOH and heat (a Claisen-Schmidt / crossed aldol condensation). The compound that CANNOT be obtained from this reaction is the one whose structure, instead of bearing the expected exocyclic benzylidene (=CH-Ph) group at C5 of the cyclopentanone ring, shows the ring connected by a direct single bond to a second 2,2-dimethylcyclopentene ring (i.e., with the carbon-carbon double bond located inside the second ring rather than exocyclic, and no phenyl group present at all) -- which option below represents this impossible product?

   • A. (E)-2,2-dimethyl-5-benzylidenecyclopentan-1-one (straightforward exocyclic C=CH-Ph product)
   • B. 2,2-dimethyl-5-(2,2-dimethylcyclopentylidene)cyclopentan-1-one (exocyclic alkylidene linked to a second dimethylcyclopentane ring instead of phenyl)
   • C. (Z)-2,2-dimethyl-5-benzylidenecyclopentan-1-one (same connectivity as option A, opposite alkene geometry)
   • D. 2,2-dimethyl-5-(4,4-dimethylcyclopent-1-en-1-yl)cyclopentan-1-one (single C-C bond to a ring bearing an internal double bond, no phenyl group)
   Show answer

   Answer: C — (Z)-2,2-dimethyl-5-benzylidenecyclopentan-1-one (same connectivity as option A, opposite alkene geometry)

   C2 of 2,2-dimethylcyclopentanone bears two methyl groups and has no alpha-hydrogens, so enolization and aldol addition can only occur at C5. Reaction with PhCHO (which has no alpha-hydrogens of its own and so cannot self-condense) proceeds as a crossed Claisen-Schmidt condensation: the C5 enolate attacks the carbonyl carbon of PhCHO, and base-catalyzed dehydration gives an exocyclic C5=CH-Ph double bond conjugated with the ring carbonyl. This explains options (1) and (3) (E/Z isomers of the same benzylidene product) as valid, and even option (2) can be rationalized as a self-aldol product at C5 with another ketone molecule. However, option (4) shows the ring attached by a plain single bond to a second ring bearing an internal (endocyclic) double bond with no phenyl group anywhere -- this connectivity cannot arise from any aldol addition/dehydration pathway in this reaction, so it cannot be obtained.

2. Q2NEET 2026Friedel-Crafts Acylation Followed by Haloform Reaction

   For the following reaction sequence, choose the correct option: Benzene --[i. CH3COCl, AlCl3; ii. NaOCl]--> P + Q

   • A. P and Q are aromatic compounds.
   • B. If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light.
   • C. Both P and Q are carbonyl compounds.
   • D. If P is the sodium salt of a carboxylic acid, Q is a primary alcohol.
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   Answer: A — P and Q are aromatic compounds.

   Friedel-Crafts acylation of benzene with CH3COCl/AlCl3 gives acetophenone (PhCOCH3). Treatment with NaOCl causes a haloform-type reaction on the methyl ketone, cleaving it into sodium benzoate (P, PhCOO⁻Na⁺) and chloroform (Q, CHCl3). P on acidification gives benzoic acid (a carboxylic acid). Chloroform (Q) is well known to slowly oxidize on exposure to air and light to form phosgene (COCl2), a highly poisonous gas -- which is why chloroform is stored in dark bottles with ethanol as a stabilizer. This matches option (2).

3. Q3NEET 2026Etard Reaction and Rosenmund Reduction

   Toluene is treated with (i) CrO2Cl2, CS2 (Etard reaction) and then (ii) H3O+, to give compound P. Choose the correct option:

   • A. Compound P can be prepared by treating benzene with anhydrous AlCl3 and CH3COCl
   • B. On treatment with bromine water, compound P gives a white precipitate
   • C. Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on BaSO4.
   • D. On treating compound P with saturated NaHCO3 solution, brisk effervescence is observed.
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   Answer: B — On treatment with bromine water, compound P gives a white precipitate

   The Etard reaction (CrO2Cl2/CS2, then aqueous hydrolysis) converts toluene's methyl group selectively to an aldehyde, giving P = benzaldehyde (PhCHO), stopping at the aldehyde stage without over-oxidation. Benzaldehyde is also obtained by the Rosenmund reduction: controlled hydrogenation of benzoyl chloride (PhCOCl) using Pd catalyst poisoned with BaSO4, which selectively reduces the acid chloride to the aldehyde without further reduction to the alcohol. Benzene + AlCl3/CH3COCl gives acetophenone, not benzaldehyde (ruling out option 1). Benzaldehyde does not give a white precipitate with bromine water (that's characteristic of phenols), ruling out option 2. Benzaldehyde does not effervesce with NaHCO3 (it is not a carboxylic acid), ruling out option 4.

4. Q4NEET 2025Grignard Reaction with Nitriles and Ketones

   The major product of the following reaction is: Ph-CO-CH2-CH2-CN treated with (i) CH3MgBr (excess) (ii) H3O+

   • A. Ph-C(CH3)(OH)-CH2-CN (Grignard adds only to nitrile, ketone untouched)
   • B. Ph-C(CH3)(OH)-CH2-CH2-CO-CH3 (Grignard adds to both ketone and nitrile, nitrile converted to methyl ketone after hydrolysis)
   • C. Ph-CH(CH3)-OH...C(CH3)2-OH (double addition at both carbons, no selectivity)
   • D. Ph-CO-CH2-CH2-CO-CH3 (nitrile converted to ketone, original ketone untouched)
   Show answer

   Answer: A — Ph-C(CH3)(OH)-CH2-CN (Grignard adds only to nitrile, ketone untouched)

   Excess CH3MgBr first adds to the more electrophilic ketone carbonyl, forming a tertiary alkoxide with a pendant nitrile; the nitrile then also reacts with the Grignard reagent to form a magnesium ketimine salt. Aqueous acidic workup (H3O+) hydrolyzes the ketimine to a methyl ketone and protonates the alkoxide to a tertiary alcohol, giving Ph-C(CH3)(OH)-CH2-CH2-CO-CH3.

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5. Q5NEET 2025Partial Reduction of Esters to Aldehydes

   Identify the suitable reagent for the following conversion: Methyl benzoate (PhCOOCH3) to benzaldehyde (PhCHO).

   • A. (i) LiAlH4, (ii) H+/H2O
   • B. (i) AlH(iBu)2 [DIBAL-H], (ii) H2O
   • C. (i) NaBH4, (ii) H+/H2O
   • D. H2/Pd-BaSO4
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   Answer: A — (i) LiAlH4, (ii) H+/H2O

   DIBAL-H (diisobutylaluminium hydride) is a mild reducing agent that partially reduces esters to aldehydes at low temperature, stopping at the aldehyde stage before over-reduction to alcohol. LiAlH4 would over-reduce the ester all the way to the alcohol.

6. Q6NEET 2023Cyanohydrin Formation and Dehydration

   Complete the following reaction: Cyclohexanone [A] reacts with HCN to give 1-hydroxycyclohexane-1-carbonitrile [B], which on treatment with conc. H2SO4 and heat (Δ) gives [C]. [C] is ________.

   • A. Cyclohexanecarboxylic acid (saturated ring, -COOH)
   • B. Cyclohex-1-ene-1-carbaldehyde (-CHO on cyclohexene)
   • C. Cyclohex-1-ene-1-carboxylic acid (-COOH on cyclohexene)
   • D. Cyclohexanol (-OH)
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   Answer: B — Cyclohex-1-ene-1-carbaldehyde (-CHO on cyclohexene)

   Cyclohexanone reacts with HCN to form the cyanohydrin [B] (1-hydroxycyclohexane-1-carbonitrile). On treatment with concentrated H2SO4 and heat, the nitrile is hydrolyzed to a carboxylic acid and the molecule simultaneously undergoes dehydration (loss of the OH as water), forming the α,β-unsaturated acid cyclohex-1-ene-1-carboxylic acid [C].

7. Q7NEET 2023Clemmensen Reduction

   Identify product (A) in the following reaction: A cyclohexane ring bearing an acetyl group (COCH3) and a meta-substituted phenyl ring which itself bears another acetyl group (COCH3), treated with Zn-Hg / conc. HCl, gives (A) + 2H2O.

   • A. The diol formed by reducing both C=O to CH(OH) groups
   • B. The diol formed by reducing both C=O to CH2OH groups
   • C. The compound with both carbonyls reduced to CH3 groups directly on the rings
   • D. The compound with both carbonyls fully reduced to CH2 (ethyl-linked), i.e. an ethyl group on the cyclohexane ring and an ethyl group on the benzene ring
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   Answer: C — The compound with both carbonyls reduced to CH3 groups directly on the rings

   Zn-Hg/conc. HCl is the Clemmensen reduction, which reduces a carbonyl group (C=O) of a ketone completely to a methylene group (CH2), not just to an alcohol. Applied to both acetyl (COCH3) groups in this diketone, each is reduced to an ethyl group (CH2CH3), giving the product with an ethyl substituent on the cyclohexane ring and an ethyl substituent on the benzene ring, releasing 2H2O. Note: the solution's worked diagram shows this diethyl product even though the printed answer-key digit reads '2' — the diagram (matching option with ethyl groups) is taken as authoritative here since Clemmensen reduction converts C=O fully to CH2, not to CH-OH.

8. Q8NEET 2023Decarboxylation (Soda Lime Reaction)

   Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide, is:

   • A. 32
   • B. 30
   • C. 18
   • D. 16
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   Answer: —

   Heating sodium ethanoate (CH3COONa) with soda lime (NaOH + CaO) causes decarboxylation to give methane (CH4), which has a molecular weight of 16 g/mol. The weight of two moles of CH4 is 2 x 16 = 32 g.

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9. Q9NEET 2023Tollens' Reagent - Selective Oxidation of Aldehydes

   Identify the major product obtained in the following reaction: A benzene ring with an ortho-acetyl group (COCH3) and an ortho-aldehyde group (CHO) is treated with 2[Ag(NH3)2]+ + 3OH- (Tollens' reagent) and heat (Δ). Identify the major product.

   • A. The aldehyde reduced to a -CH(OH)- group while the ketone remains unchanged (COCH3 intact)
   • B. The aldehyde oxidized to a carboxylate (-COO-) group while the ketone (COCH3) remains unchanged
   • C. Both the aldehyde and the ketone converted to -CH(OH)- and -COO- respectively
   • D. Both carbonyl groups reduced to -CH(OH)- groups
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   Answer: A — The aldehyde reduced to a -CH(OH)- group while the ketone remains unchanged (COCH3 intact)

   Tollens' reagent ([Ag(NH3)2]+/OH-) selectively oxidizes aldehydes to carboxylate ions (via the silver mirror test) but does not oxidize ketones. In this substrate, the -CHO group is oxidized to -COO-, while the -COCH3 (ketone) group remains unchanged, giving the major product with an intact acetyl group and a carboxylate group on the ring.

10. Q10NEET 2022Nucleophilic Addition Reactions

    Match List-I with List-II. List-I (Products formed): (a) Cynohydrin (b) Acetal (c) Schiff's base (d) Oxime List-II (Reaction of carbonyl Compound with): (i) NH2OH (ii) RNH2 (iii) alcohol (iv) HCN Choose the correct answer from the options given below:

    • A. a-(ii), b-(iii), c-(iv), d-(i)
    • B. a-(i), b-(iii), c-(ii), d-(iv)
    • C. a-(iv), b-(iii), c-(ii), d-(i)
    • D. a-(iii), b-(iv), c-(ii), d-(i)
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    Answer: B — a-(i), b-(iii), c-(ii), d-(iv)

    Cyanohydrin is formed from carbonyl compound + HCN (iv); Acetal from carbonyl compound + alcohol (iii); Schiff's base from carbonyl compound + RNH2 (ii); Oxime from carbonyl compound + NH2OH (i).