Aldehydes and Ketones — NEET UG Questions

18 NEET UG practice questions on Aldehydes and Ketones, part of Chemistry. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. Which of the following compounds has the most acidic alpha-hydrogen?

  1. Acetone
  2. Acetaldehyde
  3. Diethyl ketone (3-pentanone)
  4. Ethyl acetoacetate

Answer: Ethyl acetoacetate

Alpha-hydrogens are acidic due to the resonance stabilization of the enolate ion formed after deprotonation. The acidity increases with the number of carbonyl groups that can stabilize the negative charge. Ethyl acetoacetate is a β-ketoester with two carbonyl groups (ketone and ester) that can delocalize the negative charge, making its alpha-hydrogen significantly more acidic (pKa ~11) than those of simple ketones or aldehydes (pKa ~17-20). Acetone, acetaldehyde, and diethyl ketone have only one carbonyl group and thus less stabilized enolates, making their alpha-hydrogens much less acidic.

2. Which of the following compounds will NOT give a positive iodoform test?

  1. Butanone (CH3COCH2CH3)
  2. Ethanol (CH3CH2OH)
  3. 2-Propanol (CH3CHOHCH3)
  4. 1-Propanol (CH3CH2CH2OH)

Answer: 1-Propanol (CH3CH2CH2OH)

The iodoform test is positive for compounds containing a CH3CO- group (methyl ketones) or compounds that can be oxidized to such a group (e.g., ethanol gives acetaldehyde, 2-propanol gives acetone). Butanone is a methyl ketone (CH3COCH2CH3) and gives a positive test. Ethanol is oxidized to acetaldehyde (CH3CHO), which has the CH3CO- group. 2-Propanol is oxidized to acetone (CH3COCH3), a methyl ketone. 1-Propanol is oxidized to propanal (CH3CH2CHO), which does not have a methyl group directly attached to the carbonyl carbon, so it does not give the iodoform test.

3. Aldehydes are generally more reactive than ketones towards nucleophilic addition reactions. Which of the following is the most important reason for this difference?

  1. Aldehydes have a greater +I effect from alkyl groups
  2. Aldehydes have less steric hindrance at the carbonyl carbon
  3. Ketones have a more polar C=O bond
  4. Ketones form more stable tetrahedral intermediates

Answer: Aldehydes have less steric hindrance at the carbonyl carbon

In ketones, two alkyl groups flank the carbonyl carbon, creating greater steric hindrance that makes nucleophilic attack more difficult. Additionally, the electron-donating alkyl groups in ketones slightly reduce the electrophilicity of the carbonyl carbon. Aldehydes have only one alkyl group (or H), so both steric and electronic factors favour easier nucleophilic addition.

4. Rosenmund reduction of benzoyl chloride (C₆H₅COCl) yields:

  1. benzoic acid
  2. benzaldehyde
  3. benzyl alcohol
  4. benzene

Answer: benzaldehyde

Rosenmund reduction uses H₂ gas with Pd/BaSO₄ poisoned with sulfur to selectively reduce an acyl chloride to an aldehyde without over-reduction to the alcohol. Benzoyl chloride thus gives benzaldehyde (C₆H₅CHO). Benzoic acid would require hydrolysis, benzyl alcohol would need a stronger reducing agent, and benzene would require decarboxylation.

5. The Gattermann–Koch reaction directly introduces which functional group onto a benzene ring?

  1. –CHO (formyl)
  2. –COOH (carboxyl)
  3. –COCH₃ (acetyl)
  4. –OH (hydroxyl)

Answer: –CHO (formyl)

The Gattermann–Koch reaction treats benzene with carbon monoxide and hydrogen chloride in the presence of AlCl₃/CuCl to give benzaldehyde, thus introducing a formyl group (–CHO). Carboxylation (–COOH) requires other methods; acetylation is done by Friedel–Crafts acylation; hydroxylation needs electrophilic substitution with specific reagents.

6. When ethanol vapour is passed over heated copper at 573 K, the organic product formed is:

  1. ethene
  2. ethanoic acid
  3. ethanal
  4. ethoxyethane

Answer: ethanal

Heated copper (573 K) acts as a dehydrogenation catalyst. Primary alcohols like ethanol lose two hydrogen atoms (from –CH₂OH) to form the corresponding aldehyde (ethanal). Ethene requires dehydration with conc. H₂SO₄, ethanoic acid requires further oxidation, and ethoxyethane is formed by dehydration with conc. H₂SO₄ at lower temperature.

7. In the nucleophilic addition reaction of an aldehyde with a nucleophile, the intermediate formed before protonation is:

  1. A carbocation
  2. A carbanion
  3. An alkoxide ion
  4. A free radical

Answer: An alkoxide ion

The nucleophile attacks the electrophilic carbonyl carbon, breaking the C=O π-bond and forming a tetrahedral intermediate with a negative charge on the oxygen – an alkoxide ion. This intermediate is then protonated to give the alcohol product. Carbocations, carbanions, and free radicals are not intermediates in this typical mechanism.

8. Which of the following aldehydes will undergo the Cannizzaro reaction when treated with concentrated NaOH?

  1. Acetaldehyde (CH₃CHO)
  2. Propanal (CH₃CH₂CHO)
  3. Benzaldehyde (C₆H₅CHO)
  4. Butanal (CH₃CH₂CH₂CHO)

Answer: Benzaldehyde (C₆H₅CHO)

The Cannizzaro reaction occurs only for aldehydes that lack an α‑hydrogen atom. Benzaldehyde has no α‑hydrogen (the α‑carbon is part of the aromatic ring), so it undergoes disproportionation into benzyl alcohol and benzoate salt. Acetaldehyde, propanal, and butanal all have α‑hydrogens and undergo aldol condensation instead.

9. Which of the following reactions converts an acyl chloride into a ketone?

  1. Rosenmund reduction
  2. Friedel–Crafts acylation
  3. Stephen reduction
  4. Cannizzaro reaction

Answer: Friedel–Crafts acylation

Friedel–Crafts acylation of an aromatic compound with an acyl chloride (RCOCl) in the presence of AlCl₃ yields a ketone (RCO–Ar). Rosenmund reduction gives an aldehyde from an acyl chloride; Stephen reduction converts nitriles to aldehydes; the Cannizzaro reaction is a disproportionation of aldehydes without α-hydrogens.

10. Stephen reduction is a method to convert a nitrile (R–C≡N) into:

  1. a primary amine
  2. a secondary amine
  3. an aldehyde
  4. a carboxylic acid

Answer: an aldehyde

Stephen reduction uses SnCl₂ in HCl followed by hydrolysis to reduce a nitrile to an imine intermediate, which on hydrolysis gives an aldehyde. Primary amines are obtained by LiAlH₄ or catalytic hydrogenation of nitriles; secondary amines require other routes; carboxylic acids come from complete hydrolysis of nitriles.

11. Two molecules of acetaldehyde undergo aldol condensation in the presence of dilute aqueous NaOH. The initial organic product formed is:

  1. But-2-enal
  2. 3-Hydroxybutanal
  3. Butan-1-ol
  4. Butanoic acid

Answer: 3-Hydroxybutanal

In the aldol reaction, the α-carbon of one aldehyde adds to the carbonyl group of another, forming a β‑hydroxy aldehyde (aldol). For acetaldehyde, this product is 3‑hydroxybutanal (CH₃–CHOH–CH₂–CHO). Upon heating, it can dehydrate to but‑2‑enal, but the immediate product is the β‑hydroxy compound.

12. When hydrogen cyanide (HCN) is added to an aldehyde in the presence of a trace of base, the organic product formed is:

  1. A cyanohydrin
  2. An oxime
  3. A hydrazone
  4. An acetal

Answer: A cyanohydrin

HCN adds across the carbonyl group to give a cyanohydrin (α‑hydroxynitrile). The base generates CN⁻, which is the nucleophile that attacks the carbonyl carbon, followed by protonation. Oximes come from hydroxylamine, hydrazones from hydrazine, and acetals from alcohols – all different reactions.

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