Aldehydes, Ketones and Carboxylic Acids — Class 12 MCQs with Answers

Aldehydes, Ketones and Carboxylic Acids — important questions & MCQs with answers (Class 12 Chemistry)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 82 are timed and scored when you sign in.

1. Q1Easy

   Aldehyde functional group:

   A.−COOH

   B.−CHO✓ Correct

   C.−OH

   D.C=O within chain

   Answer: B. −CHO

   Explanation: The −CHO (formyl) group defines aldehydes — the carbonyl carbon bonded to one H and one R group. −COOH is the acid group, and >C=O with two carbon substituents is a ketone, not an aldehyde.

2. Q2Easy

   Carboxylic acid functional group:

   A.−OH

   B.−CHO

   C.−COOH✓ Correct

   D.−NH₂

   Answer: C. −COOH

   Explanation: The −COOH group (a carbonyl fused with a hydroxyl) makes a compound a carboxylic acid; −CHO is an aldehyde and −OH alone is an alcohol, both far less acidic since neither forms a resonance-stabilised carboxylate.

3. Q3Easy

   Which functional group is present in aldehydes?

   A.-OH

   B.-COOH

   C.>C=O (ketone)

   D.-CHO✓ Correct

   Answer: D. -CHO

   Explanation: Aldehydes are defined by the −CHO group — a carbonyl carbon bearing exactly one H; a bare >C=O with two carbon substituents (as shown) is the ketone carbonyl, not an aldehyde.

4. Q4Easy

   Carboxylic acids contain which functional group?

   A.-CHO

   B.-COOH✓ Correct

   C.-OH

   D.-CO-

   Answer: B. -COOH

   Explanation: Carboxylic acids are defined by −COOH, formally a −C(=O)OH group; −CHO is the aldehyde group and −CO− without the OH is a ketone linkage embedded in a chain.

5. Q5Easy

   What is the hybridisation of the carbonyl carbon in HCHO?

   A.sp²✓ Correct

   B.sp³

   C.sp

   D.sp³d

   Answer: A. sp²

   Explanation: The carbonyl carbon in HCHO uses three sp² orbitals for its two C−H σ-bonds and one C−O σ-bond, leaving an unhybridised p-orbital for the π-bond to O — trigonal planar geometry, not sp³ or sp.

6. Q6Easy

   The IUPAC name of CH₃COOH is:

   A.Methanoic acid

   B.Butanoic acid

   C.Propanoic acid

   D.Ethanoic acid✓ Correct

   Answer: D. Ethanoic acid

   Explanation: CH₃COOH has two carbons, so its IUPAC name takes the two-carbon acid stem: ethanoic acid (common name acetic acid); methanoic acid is instead the one-carbon acid, HCOOH.

7. Q7Medium

   Tollens' reagent gives silver mirror with:

   A.Ketones

   B.Aldehydes✓ Correct

   C.Acids

   D.Alkanes

   Answer: B. Aldehydes

   Explanation: Tollens' reagent [Ag(NH₃)₂]⁺ oxidises the aldehyde's C−H to −COOH while Ag⁺ is reduced to metallic Ag, depositing the silver mirror; ketones have no such H on the carbonyl carbon and give no reaction.

8. Q8Medium

   Carboxylic acids are more acidic than alcohols because:

   A.Lower mass

   B.Smaller

   C.Carboxylate ion is resonance-stabilised✓ Correct

   D.More polar only

   Answer: C. Carboxylate ion is resonance-stabilised

   Explanation: Losing H⁺ from −COOH gives RCOO⁻, whose negative charge is delocalised equally over two O atoms by resonance; RO⁻ from an alcohol has no such stabilisation, so carboxylic acids ionise far more readily.

9. Q9Medium

   Aldol condensation of two molecules of CH₃CHO followed by dehydration gives:

   A.Butan-1-ol

   B.But-2-enal✓ Correct

   C.Butan-2-one

   D.Butanoic acid

   Answer: B. But-2-enal

   Explanation: Two CH₃CHO molecules give the aldol CH₃CH(OH)CH₂CHO, and base-catalysed dehydration removes H₂O across the α,β-carbons to form the conjugated enal CH₃CH=CHCHO — but-2-enal, not an alcohol, ketone or acid.

10. Q10Medium

    Correct acidity order:

    A.All equal

    B.C₂H₅COOH > CH₃COOH > HCOOH

    C.CH₃COOH > HCOOH > C₂H₅COOH

    D.HCOOH > CH₃COOH > C₂H₅COOH✓ Correct

    Answer: D. HCOOH > CH₃COOH > C₂H₅COOH

    Explanation: Each added alkyl carbon pushes more electron density onto the carboxylate via +I, destabilising the negative charge and weakening the acid; HCOOH has no alkyl group at all, so it sits at the top of the order.

11. Q11Medium

    Cannizzaro reaction occurs with aldehydes that have:

    A.α-H atoms

    B.Aromatic ring only

    C.No α-H atom✓ Correct

    D.Branched α-carbon

    Answer: C. No α-H atom

    Explanation: Cannizzaro disproportionation needs an aldehyde with no α-hydrogen to enolise, so it can only act as a hydride donor/acceptor — HCHO and C₆H₅CHO qualify; any aldehyde with α-H undergoes aldol condensation instead.

12. Q12Hard

    Aldol condensation needs:

    A.No hydrogen

    B.α-hydrogen on the carbonyl compound✓ Correct

    C.Strong oxidiser

    D.Water only

    Answer: B. α-hydrogen on the carbonyl compound

    Explanation: Aldol condensation starts with base pulling off an α-hydrogen to generate a resonance-stabilised enolate, which then attacks a second carbonyl; without α-H there is no enolate, so the aldol step can't begin.

13. Q13Hard

    α-bromo acid from HVZ can be converted to α-amino acid by:

    A.Treatment with HCl

    B.Treatment with NaOH

    C.Treatment with excess NH₃✓ Correct

    D.Esterification

    Answer: C. Treatment with excess NH₃

    Explanation: Excess ammonia performs an SN2 substitution at the α-carbon bearing the good leaving group Br, displacing it with −NH₂ to give an α-amino acid; a shortage of NH₃ would instead favour over-alkylation of the product.

14. Q14Hard

    Cannizzaro reaction occurs with:

    A.Aldehydes lacking α-H (e.g., HCHO, PhCHO)✓ Correct

    B.Any aldehyde

    C.Ketones only

    D.Acids only

    Answer: A. Aldehydes lacking α-H (e.g., HCHO, PhCHO)

    Explanation: Cannizzaro disproportionation needs an aldehyde with no α-hydrogen — HCHO and C₆H₅CHO qualify — so hydride transfer in conc. NaOH is the only path: one molecule is oxidised to RCOO⁻, the other reduced to RCH₂OH.

Aldehydes, Ketones and Carboxylic Acids — FAQs

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• 02Electrochemistry
• 03Chemical Kinetics
• 04The d- and f-Block Elements
• 05Coordination Compounds
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