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Aldehydes, Ketones and Carboxylic Acids – Class 12 Chemistry notes, important questions and MCQs

NCERT Class 12 Chemistry, chapter 8. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Aldehydes and ketones contain the carbonyl group (C=O) and undergo nucleophilic addition; carboxylic acids (–COOH) are acidic and form derivatives.

Aldehydes and ketones

Nucleophilic addition: HCN (cyanohydrin), NaHSO₃, Grignard reagents; aldehydes are more reactive than ketones.
Tests: Tollens' reagent (silver mirror) and Fehling's solution (red precipitate) for aldehydes; iodoform test for CH₃CO– groups.
Named reactions: aldol condensation (α-H present), Cannizzaro reaction (no α-H, e.g. HCHO, C₆H₅CHO), Clemmensen and Wolff–Kishner reductions (C=O → CH₂).

Carboxylic acids

Acidity increases with electron-withdrawing groups (Cl, NO₂): ClCH₂COOH > CH₃COOH.
HVZ reaction gives α-halo acids; decarboxylation with soda lime gives alkanes.
Esterification: acid + alcohol ⇌ ester + water (acid catalyst).

Key terms

Carbonyl group: C=O group in aldehydes and ketones
Aldol condensation: reaction of aldehydes/ketones with α-H in dilute alkali to give β-hydroxy carbonyl compounds
Cannizzaro reaction: disproportionation of aldehydes without α-hydrogen in concentrated alkali into an alcohol and a carboxylate salt

Formulas

RCOOH + NaOH + CaO (soda lime, heat) → RH + Na₂CO₃

Important questions with answers

Why are aldehydes more reactive than ketones towards nucleophilic addition?2 marks

Ketones have two alkyl groups that push electrons (+I) and crowd the carbonyl carbon (steric hindrance), making it less positive and less accessible than in aldehydes.

Write the aldol condensation of ethanal with the mechanism outline and product.3 marks

In dilute NaOH, OH⁻ removes an α-hydrogen to form an enolate (carbanion), which attacks the carbonyl carbon of another ethanal molecule. The product 3-hydroxybutanal (aldol) on heating loses water to give but-2-enal (crotonaldehyde).

Write equations for (i) Rosenmund reduction, (ii) Clemmensen reduction, (iii) HVZ reaction, and give two tests to distinguish ethanal from propanone.5 marks

(i) CH₃COCl + H₂ → (Pd/BaSO₄) → CH₃CHO + HCl. (ii) CH₃COCH₃ + 4[H] → (Zn-Hg, conc. HCl) → CH₃CH₂CH₃ + H₂O. (iii) CH₃COOH + Cl₂ → (red P) → ClCH₂COOH + HCl. Tests: ethanal gives a silver mirror with Tollens' reagent and a red precipitate with Fehling's solution; propanone does not. (Both give iodoform, so that test does not distinguish them.)

Quick revision

Cannizzaro needs no α-hydrogen.
Iodoform test shows CH₃CO– or CH₃CH(OH)–.
Electron-withdrawing groups increase acid strength.

Exam tips

Only aldehydes reduce Tollens' and Fehling's reagents (ketones do not).
Tollens' and Fehling's tests distinguish aldehydes from ketones (aromatic aldehydes do not give Fehling's test).

Practice MCQs

Tollens' reagent gives a silver mirror with
1. aldehydes
2. ketones
3. alcohols
4. ethers
Show answer
Answer: (a) aldehydes. Aldehydes reduce Ag⁺.
The functional group of carboxylic acids is
1. –CHO
2. >C=O
3. –COOH
4. –OH
Show answer
Answer: (c) –COOH. Carboxyl.
Which gives the Cannizzaro reaction?
1. CH₃CHO
2. CH₃CH₂CHO
3. CH₃COCH₃
4. HCHO
Show answer
Answer: (d) HCHO. No α-hydrogen.
Which is the stronger acid?
1. ClCH₂COOH
2. CH₃COOH
3. CH₃CH₂OH
4. C₆H₅OH
Show answer
Answer: (a) ClCH₂COOH. –I effect of Cl.
Which gives the iodoform test
1. acetone
2. methanol
3. benzaldehyde
4. formaldehyde
Show answer
Answer: (a) acetone. CH₃CO– group.
Clemmensen reduction converts a ketone into
1. alcohol
2. alkane
3. alkene
4. acid
Show answer
Answer: (b) alkane. Zn–Hg/HCl, C=O → CH₂.
Which compound gives a positive iodoform test?
1. propanone
2. methanol
3. benzaldehyde
4. formaldehyde
Show answer
Answer: (a) propanone. Has CH₃CO– group.
The strongest acid among these is
1. CH₃COOH
2. ClCH₂COOH
3. Cl₃CCOOH
4. Cl₂CHCOOH
Show answer
Answer: (c) Cl₃CCOOH. Most –I groups.

All chapters of Class 12 Chemistry

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.