BrainIt

Aldehydes, Ketones and Carboxylic Acids: NEET previous year questions

Every Aldehydes, Ketones and Carboxylic Acids question from the official NEET papers on this site, with the answer and a worked solution.

The questions come from NEET 2026, Re-NEET 2026 and NEET 2020. Each shows its answer and solution when you open it, so you can attempt it first. For what the chapter covers and how NEET tests it, see the Aldehydes, Ketones and Carboxylic Acids guide.

NEET 2026 · the full paper

  1. NEET 2026 · Q58

    Compound P (C8H8OC_8H_8O) gives a red-orange precipitate with 2,4-DNP reagent but does not reduce Fehling's reagent or give a positive Tollens' test. On drastic oxidation with chromic acid, P gives an aromatic product Q that produces effervescence with aqueous NaHCO3NaHCO_3. Compounds P and Q, respectively, are:

    1. (1)
      opt A
    2. (2)
      opt B
    3. (3)
      opt C
    4. (4)
      opt D
    Answer and solution

    Answer: (4)

    P is acetophenone, C8H8O\mathrm{C}_{8}\mathrm{H}_{8}\mathrm{O}. Its aryl-ketone carbonyl gives the red-orange 2,4-DNP precipitate, while as a ketone it is negative to BOTH Fehling's and Tollens'; a positive Tollens' test would mean an aldehyde, so an aromatic aldehyde such as p-tolualdehyde is excluded. Drastic oxidation with chromic acid cleaves the methyl ketone to benzoic acid: C6H5COCH3→CrO3/H+C6H5COOH\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COCH}_{3} \xrightarrow{\text{CrO3/H+}} \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH}. Q is benzoic acid, whose −COOH liberates CO2\mathrm{CO}_{2} with aqueous NaHCO3\mathrm{NaHCO}_{3}: C6H5COOH+NaHCO3→C6H5COONa+CO2+H2O\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH} + \mathrm{NaHCO}_{3} \rightarrow \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COONa} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}. Option (1) gives the wrong oxidation product for acetophenone; options (2) and (3) take P to be p-tolualdehyde, which the Tollens' clue rules out.

    figure

Re-NEET 2026 · the full paper

  1. Re-NEET 2026 · Q46

    For the following reaction sequence, choose the correct option.

    figure
    1. (1)

      If P\mathbf{P} is the sodium salt of a carboxylic acid, Q\mathbf{Q} is a primary alcohol

    2. (2)

      P\mathbf{P} and Q\mathbf{Q} are aromatic compounds

    3. (3)

      If P\mathbf{P} gives a carboxylic acid on acidification, Q\mathbf{Q} gives a poisonous gas on exposure to air and light

    4. (4)

      Both P\mathbf{P} and Q\mathbf{Q} are carbonyl compounds

    Answer and solution

    Answer: (3)

    Friedel-Crafts acylation converts benzene to acetophenone, and sodium hypochlorite then carries out the haloform reaction on the methyl ketone, giving sodium benzoate (P) and chloroform (Q). Acidifying sodium benzoate liberates benzoic acid, and chloroform slowly oxidises in air and light to phosgene, COCl2\mathrm{COCl}_{2}, which is highly poisonous - the reason chloroform is stored in dark bottles with a little ethanol. NCERT Class 12, Aldehydes, Ketones and Carboxylic Acids (haloform reaction); chloroform's oxidation is in Haloalkanes and Haloarenes.

    Why the other options are wrong:

    - (1) identifies P correctly but calls chloroform a primary alcohol, confusing the haloform product with a reduction product
    - (2) chloroform contains no ring and is not aromatic, so both products cannot be aromatic
    - (4) neither the carboxylate salt nor chloroform retains a C=O available as a carbonyl compound in the sense intended

  2. Re-NEET 2026 · Q82

    The compound that CANNOT be obtained from the aldol condensation reaction shown below, is

    figure
    1. (1)
      opt A
    2. (2)
      opt B
    3. (3)
      opt C
    4. (4)
      opt D
    Answer and solution

    Answer: (1)

    The ketone has α\alpha-hydrogens only at C-5, because C-2 already carries two methyl groups and no hydrogen. Cross-aldol with benzaldehyde followed by dehydration therefore gives the benzylidene product at C-5, available as both geometrical isomers (options (2) and (4)). Self-condensation between two ketone molecules places the new double bond BETWEEN the α\alpha-carbon of one ring and the former carbonyl carbon of the other, which is the exocyclic alkene of option (3). Option (1) shows the rings joined by a single bond with the alkene sitting inside the second ring - no addition-elimination sequence produces that connectivity. NCERT Class 12, Aldehydes, Ketones and Carboxylic Acids.

    Why the other options are wrong:

    - (2) is obtainable - the standard cross-aldol condensation product with the non-enolisable aldehyde
    - (3) is obtainable - the self-aldol condensation product, since the ketone does possess one enolisable position
    - (4) is obtainable - simply the other geometrical isomer of the same benzylidene product

  3. Re-NEET 2026 · Q86

    Consider the following reaction, and choose the correct option.

    figure
    1. (1)

      On treating compound P\mathbf{P} with saturated NaHCO3\mathrm{NaHCO}_{3} solution, brisk effervescence is observed

    2. (2)

      Compound P\mathbf{P} can be prepared by treating benzene with anhydrous AlCl3\mathrm{AlCl}_{3} and CH3COCl\mathrm{CH}_{3}\mathrm{COCl}

    3. (3)

      On treatment with bromine water, compound P\mathbf{P} gives a white precipitate

    4. (4)

      Compound P\mathbf{P} is obtained by the hydrogenation of benzoyl chloride with Pd on BaSO4\mathrm{BaSO}_{4}

    Answer and solution

    Answer: (4)

    Chromyl chloride in carbon disulphide oxidises the methyl group of toluene only as far as the aldehyde, and hydrolysis of the chromium complex releases benzaldehyde - the Etard reaction. Benzaldehyde is exactly the product of Rosenmund reduction, in which benzoyl chloride is hydrogenated over palladium poisoned with barium sulphate, so option (4) is a genuine alternative preparation. NCERT Class 12, Aldehydes, Ketones and Carboxylic Acids (preparation of aldehydes).

    Why the other options are wrong:

    - (1) effervescence with bicarbonate is a test for carboxylic ACIDS; an aldehyde is not acidic enough to release CO2\mathrm{CO}_{2}
    - (2) acylation of benzene with acetyl chloride gives acetophenone, a ketone, not benzaldehyde
    - (3) a white precipitate with bromine water is characteristic of activated rings such as phenol or aniline, not of benzaldehyde

NEET 2020 · the full paper

  1. NEET 2020 · Q104

    Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give :

    1. (1)

      Sec. butyl alcohol

    2. (2)

      Tert. butyl alcohol

    3. (3)

      Isobutyl alcohol

    4. (4)

      Isopropyl alcohol

    Answer and solution

    Answer: (2)

    The reaction of acetone with methylmagnesium chloride forms a tertiary alkoxide, which upon hydrolysis yields tert-butyl alcohol. This follows the Grignard reagent mechanism where the nucleophilic attack occurs at the carbonyl carbon, leading to the formation of a tertiary alcohol. Chapter: Aldehydes, Ketones and Carboxylic Acids.

  2. NEET 2020 · Q118

    Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as

    1. (1)

      Cannizzaro's reaction

    2. (2)

      Cross Cannizzaro's reaction

    3. (3)

      Cross Aldol condensation

    4. (4)

      Aldol condensation

    Answer and solution

    Answer: (3)

    The reaction between benzaldehyde and acetophenone in the presence of dilute NaOH is a cross aldol condensation. This reaction involves the condensation of an aldehyde with a ketone to form a β\beta-hydroxy ketone. Chapter: Aldehydes, Ketones and Carboxylic Acids.

Practise Aldehydes, Ketones and Carboxylic Acids in the app: it brings back the questions you get wrong, so the chapter improves, not just your memory of these. Or try the free CBT mock. All chapters: NEET chapter-wise PYQs.

Questions

Are these the official NEET questions on Aldehydes, Ketones and Carboxylic Acids?

Yes. They are the questions NTA set in each paper, in its wording. The solutions are written by BrainIt.

Something here wrong or out of date? Email hello@brainit.pro and a person will read it.