Chapter 12: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 12

Aldehydes, Ketones and Carboxylic Acids – Exercise Questions with Answers

Question 12.1: Write the structures of the following compounds:

Answer:
(i) α-Methoxypropionaldehyde: CH₃–CH(OCH₃)–CHO
(ii) 3-Hydroxybutanal: CH₃–CH(OH)–CH₂–CHO
(iii) 2-Hydroxycyclopentane carbaldehyde: Hydroxy group on C-2 and –CHO attached to the cyclopentane ring
(iv) 4-Oxopentanal: CH₃–CO–CH₂–CH₂–CHO
(v) Di-sec-butyl ketone: (CH₃)₂CH–CO–CH(CH₃)₂
(vi) 4-Fluoroacetophenone: Fluorine substituted at para-position of acetophenone ring

Question 12.2: Write the structures of the products of the following reactions.

Answer:
The products involve:
– Oxidation of primary alcohols to aldehydes
– Clemmensen reduction of ketones to alkanes
– Reaction of Grignard reagents with formaldehyde to give primary alcohols
(Refer to NCERT textbook for detailed structures.)

Question 12.3: Arrange the following compounds in increasing order of their boiling points:
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃

Answer:
Order of increasing boiling point:
CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH

Explanation:
– Ethanol forms hydrogen bonds → highest boiling point
– Acetaldehyde shows dipole-dipole interactions
– Dimethyl ether has weaker dipole forces
– Propane shows only weak van der Waals forces

Question 12.4: Arrange the following carbonyl compounds in increasing order of their reactivity in nucleophilic addition reactions:

Answer:
(a) Butanone < Propanone < Propanal < Ethanal
(b) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde

Explanation:
– Aldehydes are more reactive than ketones.
– Electron-withdrawing groups (like –NO₂) increase reactivity.
– Electron-donating groups (like –CH₃) decrease reactivity.

Question 12.5: Predict the products of the following reactions.

Answer:
General outcomes include:
– Reduction of aldehydes/ketones to alcohols
– Aldol condensation to form β-hydroxy aldehydes/ketones
– Cannizzaro reaction for aldehydes without α-hydrogen
(Refer NCERT textbook for complete reaction schemes.)

Question 12.6: Give the IUPAC names of the following compounds:

Answer:
(i) 3-Phenylpropanoic acid
(ii) 3-Methylbut-2-enoic acid
(iii) 2-Methylcyclohexanecarboxylic acid (IUPAC name already given)
(iv) 2,4,6-Trinitrobenzoic acid (IUPAC name already given)

Question 12.7: Show how each of the following compounds can be converted into benzoic acid:

Answer:
(i) Ethylbenzene → (Oxidation with KMnO₄) → Benzoic acid
(ii) Acetophenone → (Oxidation with acidified KMnO₄) → Benzoic acid
(iii) Bromobenzene → (Grignard reagent formation, then CO₂, then acidification) → Benzoic acid
(iv) Phenylethene (Styrene) → (Oxidation with hot alkaline KMnO₄) → Benzoic acid

Question 12.8: Which acid from each of the following pairs would you expect to be a stronger acid?

Answer:
(i) CH₂FCOOH is stronger than CH₃COOH (due to strong –I effect of fluorine)
(ii) CH₂FCOOH is stronger than CH₂ClCOOH (fluorine is more electronegative than chlorine)
(iii) CH₃CHFCH₂COOH is stronger than CH₂FCH₂CH₂COOH (fluorine is nearer to –COOH group in CH₃CHFCH₂COOH)

NCERT Solutions for Class 12 Chemistry Chapter 12

Aldehydes, Ketones and Carboxylic Acids – Exercise Questions with Answers

Question 12.1: What is meant by the following terms? Give an example of the reaction in each case.

Answer:
(i) Cyanohydrin: Organic compounds having both –OH and –CN groups attached to the same carbon atom.
Example: CH₃CHO + HCN → CH₃CH(OH)CN

(ii) Acetal: Geminal-dialkoxy compounds formed when aldehydes react with alcohols in dry HCl.
Example: CH₃CHO + 2CH₃OH → CH₃CH(OCH₃)₂ + H₂O

(iii) Semicarbazone: Derivatives formed by the reaction of aldehydes or ketones with semicarbazide.
Example: CH₃CHO + H₂NNHCONH₂ → CH₃CH=NNHCONH₂ + H₂O

(iv) Aldol: β-Hydroxy aldehydes or ketones formed by base-catalyzed condensation.
Example: 2CH₃CHO → CH₃CH(OH)CH₂CHO

(v) Hemiacetal: Compounds where one alkoxy and one hydroxyl group are attached to the same carbon.
Example: CH₃CHO + CH₃OH → CH₃CH(OH)OCH₃

(vi) Oxime: Formed by the reaction of aldehydes/ketones with hydroxylamine (NH₂OH).
Example: CH₃CHO + NH₂OH → CH₃CH=NOH + H₂O

(vii) Ketal: Formed when ketones react with dihydric alcohols.
Example: CH₃COCH₃ + HOCH₂CH₂OH → Ketal + H₂O

(viii) Imine: Compounds with a C=N bond formed by reaction with primary amines.
Example: CH₃CHO + NH₂CH₃ → CH₃CH=NHCH₃ + H₂O

(ix) 2,4-DNP Derivative: Formed by reaction with 2,4-dinitrophenylhydrazine.
Example: CH₃COCH₃ + H₂NNHC₆H₃(NO₂)₂ → DNP derivative

(x) Schiff’s Base: Imines formed by aldehydes or ketones reacting with primary amines.
Example: CH₃CHO + C₆H₅NH₂ → CH₃CH=N–C₆H₅ + H₂O

Question 12.2: Name the following compounds according to the IUPAC system of nomenclature:

Answer:
(i) CH₃CH(CH₃)CH₂CH₂CHO – 4-Methylpentanal
(ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl – 6-Chloro-4-ethylhexan-3-one
(iii) CH₃CH=CHCHO – But-2-enal
(iv) CH₃COCH₂COCH₃ – Pentane-2,4-dione
(v) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃ – 3,3,5-Trimethylhexan-2-one
(vi) (CH₃)₃CCH₂COOH – 3,3-Dimethylbutanoic acid
(vii) OHCC₆H₄CHO (para position) – Benzene-1,4-dicarbaldehyde

Question 12.3: Draw the structures of the following compounds:

Answer:
(i) 3-Methylbutanal: CH₃–CH(CH₃)–CH₂–CHO
(ii) p-Methylbenzaldehyde: Benzene ring with –CHO at position 1 and –CH₃ at position 4
(iii) 4-Chloropentan-2-one: CH₃–CO–CH₂–CHCl–CH₃
(iv) p, p’-Dihydroxybenzophenone: Two benzene rings linked by C=O group with –OH at para positions
(v) p-Nitropropiophenone: C₆H₄(NO₂)–CO–CH₂CH₃
(vi) 4-Methylpent-3-en-2-one: CH₃–CO–CH=CH–CH₃
(vii) 3-Bromo-4-phenylpentanoic acid: Br–CH₂–CH(C₆H₅)–CH₂–COOH
(viii) Hex-2-en-4-ynoic acid: HOOC–CH=CH–C≡C–CH₂–CH₃

Question 12.4: Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also the common names.

Answer:
(i) CH₃CO(CH₂)₄CH₃ – Heptan-2-one (Methyl n-pentyl ketone)
(ii) CH₃CH₂CHBrCH₂CH(CH₃)CHO – 6-Bromo-2-methylhexanal
(iii) CH₃(CH₂)₅CHO – Heptanal (n-Heptaldehyde)
(iv) Ph–CH=CH–CHO – 3-Phenylprop-2-enal (Cinnamaldehyde)

Question 12.5: Draw structures of the following derivatives:

Answer:
(i) 2,4-Dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehyde dimethyl acetal
(iv) Semicarbazone of cyclobutanone
(v) Ethylene ketal of hexan-3-one
(vi) Methyl hemiacetal of formaldehyde

Question 12.6: Predict the product when cyclohexanecarbaldehyde reacts with the following reagents:

Answer:
(i) C₆H₅MgBr followed by H₃O⁺ – Secondary alcohol with phenyl group
(ii) Tollen’s reagent – Cyclohexanecarboxylic acid
(iii) Semicarbazide – Semicarbazone derivative
(iv) Excess ethanol + Acid – Diethyl acetal
(v) Zinc amalgam + HCl – Cyclohexylmethane

Question 12.7: Which of the following compounds would undergo aldol condensation, Cannizzaro reaction, or neither? Write the expected product structures.

Answer:
(i) Methanal – Cannizzaro reaction (no α-hydrogen)
(ii) 2-Methylpentanal – Aldol condensation (has α-hydrogen)
(iii) Benzaldehyde – Cannizzaro reaction (no α-hydrogen)
(iv) Benzophenone – Neither (no α-hydrogen and not an aldehyde)
(v) Cyclohexanone – Aldol condensation (has α-hydrogen)
(vi) 1-Phenylpropanone – Aldol condensation (has α-hydrogen)
(vii) Phenylacetaldehyde – Aldol condensation (has α-hydrogen)
(viii) Butan-1-ol – Neither (not a carbonyl compound)
(ix) 2,2-Dimethylbutanal – Cannizzaro reaction (no α-hydrogen)

Question 12.8: How will you convert ethanal into the following compounds?

Answer:
(i) Butane-1,3-diol: Ethanal → 3-Hydroxybutanal (aldol product) → Hydrogenation → Butane-1,3-diol
(ii) But-2-enal: Ethanal → 3-Hydroxybutanal → Dehydration → But-2-enal
(iii) But-2-enoic acid: Ethanal → But-2-enal → Oxidation → But-2-enoic acid

Question 12.9: Write structural formulas and names of four possible aldol condensation products from propanal and butanal.

Answer:
(i) Propanal + Propanal → Pent-2-enal
(ii) Butanal + Butanal → Hex-2-enal
(iii) Propanal + Butanal → Hex-2-enal (isomeric form)
(iv) Butanal + Propanal → Hex-2-enal
Products are α,β-unsaturated aldehydes formed after dehydration.

Question 12.10: An organic compound (C₉H₁₀O) forms a 2,4-DNP derivative, reduces Tollen’s reagent, undergoes Cannizzaro reaction, and gives phthalic acid on oxidation. Identify the compound.

Answer:
The compound is o-Ethylbenzaldehyde (C₆H₄(CHO)(C₂H₅)), with ethyl and formyl groups ortho to each other.

Question 12.11: An organic compound (C₈H₁₆O₂) was hydrolyzed to give a carboxylic acid (B) and an alcohol (C). Identify A, B, and C.

Answer:
Compound A: Butyl butanoate
Compound B: Butanoic acid
Compound C: Butan-1-ol
Reactions:
Butyl butanoate + H₂O → Butanoic acid + Butan-1-ol
Butan-1-ol + [O] → Butanoic acid

Question 12.12: Arrange the following in increasing order of the property indicated:

Answer:
(i) Reactivity towards HCN: Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde
(ii) Acid strength: (CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH
(iii) Acid strength: 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,5-Dinitrobenzoic acid

Question 12.13: Give simple chemical tests to distinguish between the following pairs of compounds:

Answer:
(i) Propanal vs Propanone: Tollen’s/Fehling’s test positive for propanal.
(ii) Acetophenone vs Benzophenone: Iodoform test positive for acetophenone.
(iii) Phenol vs Benzoic acid: NaHCO₃ test positive for benzoic acid.
(iv) Benzoic acid vs Ethyl benzoate: NaHCO₃ test positive for benzoic acid.
(v) Pentan-2-one vs Pentan-3-one: Iodoform test positive for Pentan-2-one.
(vi) Benzaldehyde vs Acetophenone: Tollen’s test positive for benzaldehyde.
(vii) Ethanal vs Propanal: Iodoform test positive for ethanal.

Question 12.14: How will you prepare the following compounds from benzene?

Answer:
(i) Methyl benzoate: Benzene → Benzoic acid → Methyl benzoate (esterification)
(ii) m-Nitrobenzoic acid: Benzene → Nitrobenzene → m-Nitrotoluene → Oxidation
(iii) p-Nitrobenzoic acid: Benzene → Toluene → p-Nitrotoluene → Oxidation
(iv) Phenylacetic acid: Benzene → Benzyl chloride → Phenylacetonitrile → Hydrolysis
(v) p-Nitrobenzaldehyde: Benzene → p-Nitrotoluene → p-Nitrobenzaldehyde (oxidation)

Question 12.15: How will you bring about the following conversions in not more than two steps?

Answer:
(i) Propanone → Propan-2-ol → Propene
(ii) Benzoic acid → Benzoyl chloride → Benzaldehyde (Rosenmund reduction)
(iii) Ethanol → Acetaldehyde → 3-Hydroxybutanal
(iv) Benzene → m-Nitrobenzene → m-Nitroacetophenone
(v) Benzaldehyde → Benzoin → Benzophenone
(vi) Bromobenzene → Phenylethyl bromide → 1-Phenylethanol
(vii) Benzaldehyde → Cinnamaldehyde → 3-Phenylpropan-1-ol
(viii) Benzaldehyde → α-Hydroxyphenylacetic acid
(ix) Benzoic acid → m-Nitrobenzoic acid → m-Nitrobenzyl alcohol

Question 12.16: Describe the following terms:

Answer:
(i) Acetylation: Introduction of an acetyl group (CH₃CO–) using acetyl chloride or acetic anhydride.
(ii) Cannizzaro Reaction: Disproportionation of aldehydes without α-hydrogen under strong base.
(iii) Cross Aldol Condensation: Aldol condensation involving two different carbonyl compounds.
(iv) Decarboxylation: Removal of CO₂ from carboxylic acids using soda lime.

Question 12.17: Complete each synthesis by giving missing starting material, reagent or products.

Answer:
Typical reagents:
– LiAlH₄ for reduction
– KMnO₄ or CrO₃ for oxidation
– CH₃MgBr for Grignard reactions
– NaOH for aldol condensations
– Acetic anhydride for acetylation
(Refer NCERT reaction schemes.)

Question 12.18: Give plausible explanations for the following:

Answer:
(i) 2,2,6-Trimethylcyclohexanone does not form cyanohydrin due to steric hindrance.
(ii) Only one –NH₂ group of semicarbazide reacts as the other is resonance-stabilized.
(iii) Water/ester is removed during esterification to shift equilibrium toward product.

Question 12.19: Identify the compound with formula C₅H₁₀O giving positive iodoform test and no Tollen’s test.

Answer:
The compound is Pentan-2-one.

Question 12.20: Although phenoxide ion has more resonance structures than carboxylate ion, carboxylic acid is a stronger acid. Why?

Answer:
In phenoxide ion, some structures place negative charge on less electronegative carbon. In carboxylate ion, negative charge is delocalized equally over two oxygens, making it more stable, thus carboxylic acids are stronger acids.