NCERT Solutions for Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids (English Medium)
🧪 Get NCERT Solutions for Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids with accurate answers and step-by-step explanations in a clean, exam-ready format. This chapter covers nomenclature & structures, methods of preparation, physical properties, nucleophilic addition reactions, key oxidation & reduction, aldol condensation, Cannizzaro reaction, haloform reaction, plus acidity of carboxylic acids and important tests (Tollens’, Fehling’s, iodoform). Ideal for CBSE boards and NEET/JEE revision. ✅
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Q.1:
What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin
(ii) Acetal
(iii) Semicarbazone
(iv) Aldol
(v) Hemiacetal
(vi) Oxime
(vii) Ketal
(viii) Imine
(ix) 2,4-DNP-derivative
(x) Schiff’s base
(i) Cyanohydrin
(ii) Acetal
(iii) Semicarbazone
(iv) Aldol
(v) Hemiacetal
(vi) Oxime
(vii) Ketal
(viii) Imine
(ix) 2,4-DNP-derivative
(x) Schiff’s base
Answer ✅
Meanings + one typical reaction example for each term
(i) Cyanohydrin
✅ Meaning: Compound having –OH and –CN on the same carbon.
👉 General: R–CH(OH)–CN (from aldehyde) or R2C(OH)–CN (from ketone)
✅ Example reaction:
👉 CH3CHO + HCN → CH3CH(OH)CN (acetaldehyde cyanohydrin)
✅ Meaning: Compound having –OH and –CN on the same carbon.
👉 General: R–CH(OH)–CN (from aldehyde) or R2C(OH)–CN (from ketone)
✅ Example reaction:
👉 CH3CHO + HCN → CH3CH(OH)CN (acetaldehyde cyanohydrin)
(ii) Acetal
✅ Meaning: Product formed when an aldehyde reacts with 2 alcohol molecules (acid catalysed), giving two –OR groups on the same carbon.
👉 General: R–CH(OR)2
✅ Example reaction:
👉 CH3CHO + 2 C2H5OH (H+) → CH3CH(OC2H5)2 + H2O
✅ Meaning: Product formed when an aldehyde reacts with 2 alcohol molecules (acid catalysed), giving two –OR groups on the same carbon.
👉 General: R–CH(OR)2
✅ Example reaction:
👉 CH3CHO + 2 C2H5OH (H+) → CH3CH(OC2H5)2 + H2O
(iii) Semicarbazone
✅ Meaning: Derivative formed when aldehyde/ketone reacts with semicarbazide (NH2NHCONH2).
✅ Example reaction:
👉 CH3COCH3 + NH2NHCONH2 → (CH3)2C=NNHCONH2 + H2O
✅ Meaning: Derivative formed when aldehyde/ketone reacts with semicarbazide (NH2NHCONH2).
✅ Example reaction:
👉 CH3COCH3 + NH2NHCONH2 → (CH3)2C=NNHCONH2 + H2O
(iv) Aldol
✅ Meaning: β-hydroxy aldehyde/ketone formed by aldol addition of aldehydes/ketones having α-H.
✅ Example reaction:
👉 2 CH3CHO (dil. NaOH) → CH3CH(OH)CH2CHO (3-hydroxybutanal, aldol)
✅ Meaning: β-hydroxy aldehyde/ketone formed by aldol addition of aldehydes/ketones having α-H.
✅ Example reaction:
👉 2 CH3CHO (dil. NaOH) → CH3CH(OH)CH2CHO (3-hydroxybutanal, aldol)
(v) Hemiacetal
✅ Meaning: Product with one –OR and one –OH on the same carbon, formed by addition of one alcohol to an aldehyde.
👉 General: R–CH(OH)–OR
✅ Example reaction:
👉 CH3CHO + CH3OH (H+) ⇌ CH3CH(OH)OCH3
✅ Meaning: Product with one –OR and one –OH on the same carbon, formed by addition of one alcohol to an aldehyde.
👉 General: R–CH(OH)–OR
✅ Example reaction:
👉 CH3CHO + CH3OH (H+) ⇌ CH3CH(OH)OCH3
(vi) Oxime
✅ Meaning: Product formed when aldehyde/ketone reacts with hydroxylamine (NH2OH).
✅ Example reaction:
👉 CH3CHO + NH2OH → CH3CH=NOH + H2O
✅ Meaning: Product formed when aldehyde/ketone reacts with hydroxylamine (NH2OH).
✅ Example reaction:
👉 CH3CHO + NH2OH → CH3CH=NOH + H2O
(vii) Ketal
✅ Meaning: Product formed when a ketone reacts with 2 alcohol molecules (acid catalysed), giving two –OR groups on the same carbon.
👉 General: R2C(OR)2
✅ Example reaction:
👉 (CH3)2CO + 2 C2H5OH (H+) → (CH3)2C(OC2H5)2 + H2O
✅ Meaning: Product formed when a ketone reacts with 2 alcohol molecules (acid catalysed), giving two –OR groups on the same carbon.
👉 General: R2C(OR)2
✅ Example reaction:
👉 (CH3)2CO + 2 C2H5OH (H+) → (CH3)2C(OC2H5)2 + H2O
(viii) Imine
✅ Meaning: Product containing C=N formed when aldehyde/ketone reacts with primary amine.
👉 General: R–CH=NR' or R2C=NR'
✅ Example reaction:
👉 CH3CHO + CH3NH2 → CH3CH=NCH3 + H2O
✅ Meaning: Product containing C=N formed when aldehyde/ketone reacts with primary amine.
👉 General: R–CH=NR' or R2C=NR'
✅ Example reaction:
👉 CH3CHO + CH3NH2 → CH3CH=NCH3 + H2O
(ix) 2,4-DNP derivative
✅ Meaning: Derivative formed when aldehyde/ketone reacts with 2,4-dinitrophenylhydrazine (2,4-DNP) giving hydrazone (orange/yellow ppt).
✅ Example reaction:
👉 CH3COCH3 + H2NNH–C6H3(NO2)2 → (CH3)2C=NNH–C6H3(NO2)2 + H2O
✅ Meaning: Derivative formed when aldehyde/ketone reacts with 2,4-dinitrophenylhydrazine (2,4-DNP) giving hydrazone (orange/yellow ppt).
✅ Example reaction:
👉 CH3COCH3 + H2NNH–C6H3(NO2)2 → (CH3)2C=NNH–C6H3(NO2)2 + H2O
(x) Schiff’s base
✅ Meaning: An imine specifically formed by reaction of an aldehyde/ketone with a primary aromatic amine (commonly), giving C=N.
✅ Example reaction:
👉 C6H5CHO + C6H5NH2 → C6H5CH=NC6H5 + H2O
✅ Meaning: An imine specifically formed by reaction of an aldehyde/ketone with a primary aromatic amine (commonly), giving C=N.
✅ Example reaction:
👉 C6H5CHO + C6H5NH2 → C6H5CH=NC6H5 + H2O
Explanation (Step by Step) 👉
✅ Most of these are
addition/condensation derivatives
of aldehydes and ketones:• Addition products: cyanohydrin, hemiacetal
• Condensation (water removed) products: oxime, imine, Schiff’s base, semicarbazone, 2,4-DNP derivative
• With alcohols: aldehyde → hemiacetal → acetal, ketone → hemiketal → ketal
• Base-catalysed coupling: aldol reaction gives β-hydroxy carbonyl compound
Did You Know? 🤔📌
✅
2,4-DNP test
is a classic lab test: aldehydes/ketones give
yellow/orange precipitate,
so it helps detect the carbonyl
(C=O)
group quickly.✅ Acetals and ketals are often used as protecting groups for aldehydes/ketones in organic synthesis.
Q.2:
Name the following compounds according to
IUPAC system of nomenclature:
(i) CH3CH(CH3)CH2CH2CHO
(ii) CH3CH2COCH(C2H5)CH2CH2Cl
(iii) CH3CH=CHCHO
(iv) CH3COCH2COCH3
(v) CH3CH(CH3)CH2C(CH3)2COCH3
(vi) (CH3)3CCH2COOH
(vii) OHCC6H4CHO-p
(i) CH3CH(CH3)CH2CH2CHO
(ii) CH3CH2COCH(C2H5)CH2CH2Cl
(iii) CH3CH=CHCHO
(iv) CH3COCH2COCH3
(v) CH3CH(CH3)CH2C(CH3)2COCH3
(vi) (CH3)3CCH2COOH
(vii) OHCC6H4CHO-p
Answer ✅
(i)
👉 4-methylpentanal
(ii) 👉 6-chloro-4-ethylhexan-3-one
(iii) 👉 but-2-enal
(iv) 👉 pentane-2,4-dione
(v) 👉 3,3,5-trimethylhexan-2-one
(vi) 👉 3,3-dimethylbutanoic acid
(vii) 👉 benzene-1,4-dicarbaldehyde (common name: terephthalaldehyde)
(ii) 👉 6-chloro-4-ethylhexan-3-one
(iii) 👉 but-2-enal
(iv) 👉 pentane-2,4-dione
(v) 👉 3,3,5-trimethylhexan-2-one
(vi) 👉 3,3-dimethylbutanoic acid
(vii) 👉 benzene-1,4-dicarbaldehyde (common name: terephthalaldehyde)
Explanation (Step by Step) 👉
✅ Quick rules used here (same for all):• Choose the longest chain containing the functional group.
• Give the functional group the lowest number (aldehyde/ketone/acid priority).
• Name and number substituents, then arrange substituents in alphabetical order.
(i)
Aldehyde carbon is C-1
→ methyl at C-4
→ 4-methylpentanal
(ii) Longest chain = 6C, C=O at C-3, ethyl at C-4, chloro at C-6 → 6-chloro-4-ethylhexan-3-one
(iii) Aldehyde is C-1, double bond starts at C-2 → but-2-enal
(iv) 5-carbon chain with ketone at C-2 and C-4 → pentane-2,4-dione
(v) 6-carbon chain, ketone at C-2, methyl at C-3 (two) and C-5 (one) → 3,3,5-trimethylhexan-2-one
(vi) Acid carbon is C-1, two methyl at C-3 → 3,3-dimethylbutanoic acid
(vii) Benzene with –CHO at 1 and 4 → benzene-1,4-dicarbaldehyde
(ii) Longest chain = 6C, C=O at C-3, ethyl at C-4, chloro at C-6 → 6-chloro-4-ethylhexan-3-one
(iii) Aldehyde is C-1, double bond starts at C-2 → but-2-enal
(iv) 5-carbon chain with ketone at C-2 and C-4 → pentane-2,4-dione
(v) 6-carbon chain, ketone at C-2, methyl at C-3 (two) and C-5 (one) → 3,3,5-trimethylhexan-2-one
(vi) Acid carbon is C-1, two methyl at C-3 → 3,3-dimethylbutanoic acid
(vii) Benzene with –CHO at 1 and 4 → benzene-1,4-dicarbaldehyde
Did You Know? 🤔📌
✅ The suffix
-al
is used for
aldehydes,
-one
for
ketones,
and
-oic acid
for
carboxylic acids.✅ Compound (vii) is the para-dialdehyde of benzene, commonly called terephthalaldehyde (para arrangement = 1,4).
Q.4:
Write the IUPAC names
of the following ketones
and aldehydes.
Wherever possible, give also common names.
(i) CH3CO(CH2)4CH3
(ii) CH3CH2CHBrCH2CH(CH3)CHO
(iii) CH3(CH2)5CHO
(iv) Ph–CH=CH–CHO
(i) CH3CO(CH2)4CH3
(ii) CH3CH2CHBrCH2CH(CH3)CHO
(iii) CH3(CH2)5CHO
(iv) Ph–CH=CH–CHO
Answer ✅
(i)
CH3CO(CH2)4CH3
👉 IUPAC: heptan-2-one
👉 Common name: methyl n-amyl ketone (also called methyl pentyl ketone)
👉 IUPAC: heptan-2-one
👉 Common name: methyl n-amyl ketone (also called methyl pentyl ketone)
(ii)
CH3CH2CHBrCH2CH(CH3)CHO
👉 IUPAC: 5-bromo-2-methylhexanal
👉 Common name: (not commonly used in NCERT; usually written by IUPAC)
👉 IUPAC: 5-bromo-2-methylhexanal
👉 Common name: (not commonly used in NCERT; usually written by IUPAC)
(iii)
CH3(CH2)5CHO
👉 IUPAC: heptanal
👉 Common name: enanthaldehyde (also written as oenanthaldehyde)
👉 IUPAC: heptanal
👉 Common name: enanthaldehyde (also written as oenanthaldehyde)
(iv)
Ph–CH=CH–CHO
👉 IUPAC: 3-phenylprop-2-enal
👉 Common name: cinnamaldehyde
👉 IUPAC: 3-phenylprop-2-enal
👉 Common name: cinnamaldehyde
Explanation (Step by Step) 👉
✅ Rule for ketones:
choose the longest chain containing
C=O
and give it the
lowest number
→ suffix -one.✅ Rule for aldehydes: –CHO carbon is always C-1 → suffix -al.
✅ Substituents (like bromo, methyl, phenyl) get numbering accordingly and are written in alphabetical order.
(i)
7-carbon chain, C=O at C-2 → heptan-2-one
(ii) Aldehyde carbon is C-1, methyl at C-2, bromo at C-5 → 5-bromo-2-methylhexanal
(iii) Straight 7-carbon aldehyde → heptanal
(iv) Aldehyde chain is 3 carbons with double bond at 2 and phenyl at 3 → 3-phenylprop-2-enal
(ii) Aldehyde carbon is C-1, methyl at C-2, bromo at C-5 → 5-bromo-2-methylhexanal
(iii) Straight 7-carbon aldehyde → heptanal
(iv) Aldehyde chain is 3 carbons with double bond at 2 and phenyl at 3 → 3-phenylprop-2-enal
Did You Know? 🤔📌
✅ Many common names
of aldehydes come from old fatty-acid names:👉 heptanal = enanthaldehyde (from enanthic acid).
✅ Cinnamaldehyde (compound iv) is a key compound responsible for the smell of cinnamon.
Q.18:
Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Answer ✅
(i)
Cyclohexanone
gives cyanohydrin easily because the
carbonyl carbon
is accessible, but
2,2,6-trimethylcyclohexanone
is heavily crowded, so
CN− addition
is hindered.
(ii) In semicarbazide, only the terminal –NH2 acts as the nucleophile; the other –NH2 attached to the carbonyl is resonance-stabilized and much less reactive.
(iii) Esterification is a reversible equilibrium reaction; removing water (or the ester) shifts equilibrium to the product side and increases yield.
(ii) In semicarbazide, only the terminal –NH2 acts as the nucleophile; the other –NH2 attached to the carbonyl is resonance-stabilized and much less reactive.
(iii) Esterification is a reversible equilibrium reaction; removing water (or the ester) shifts equilibrium to the product side and increases yield.
Explanation (Step by Step) 👉
(i) Cyclohexanone vs 2,2,6-trimethylcyclohexanone (cyanohydrin formation)
✅ Cyanohydrin formation is a nucleophilic addition:
👉 ketone + HCN ⇌ cyanohydrin
• Cyclohexanone: carbonyl carbon is relatively open, so CN− attacks easily → good yield.
• 2,2,6-trimethylcyclohexanone: three methyl groups at 2,2,6-positions create strong steric hindrance (crowding) near the C=O group.
✅ So CN− cannot approach properly → very low / no cyanohydrin formation.
✅ Cyanohydrin formation is a nucleophilic addition:
👉 ketone + HCN ⇌ cyanohydrin
• Cyclohexanone: carbonyl carbon is relatively open, so CN− attacks easily → good yield.
• 2,2,6-trimethylcyclohexanone: three methyl groups at 2,2,6-positions create strong steric hindrance (crowding) near the C=O group.
✅ So CN− cannot approach properly → very low / no cyanohydrin formation.
(ii) Two –NH2 in semicarbazide, but only one reacts
Semicarbazide: NH2–NH–CO–NH2
✅ Semicarbazone formation needs a nucleophilic –NH2 to attack the carbonyl carbon of aldehyde/ketone.
• The terminal –NH2 (left side) is free and more nucleophilic, so it reacts.
• The –NH2 attached to –CO– (right side) has its lone pair involved in resonance with the carbonyl (–CONH2), so it is less available for attack.
👉 इसलिए only one –NH2 participates in semicarbazone formation.
Semicarbazide: NH2–NH–CO–NH2
✅ Semicarbazone formation needs a nucleophilic –NH2 to attack the carbonyl carbon of aldehyde/ketone.
• The terminal –NH2 (left side) is free and more nucleophilic, so it reacts.
• The –NH2 attached to –CO– (right side) has its lone pair involved in resonance with the carbonyl (–CONH2), so it is less available for attack.
👉 इसलिए only one –NH2 participates in semicarbazone formation.
(iii) Why remove water/ester during esterification?
Esterification (Fischer esterification) is reversible:
👉 RCOOH + R'OH ⇌ RCOOR' + H2O (H+)
✅ Because it is an equilibrium reaction, product yield can be improved by shifting equilibrium to the right.
According to Le Chatelier’s principle:
• Removing H2O as it forms drives reaction forward.
• Or removing ester as it forms also drives reaction forward.
✅ Result: higher ester yield.
Esterification (Fischer esterification) is reversible:
👉 RCOOH + R'OH ⇌ RCOOR' + H2O (H+)
✅ Because it is an equilibrium reaction, product yield can be improved by shifting equilibrium to the right.
According to Le Chatelier’s principle:
• Removing H2O as it forms drives reaction forward.
• Or removing ester as it forms also drives reaction forward.
✅ Result: higher ester yield.
Did You Know? 🤔📌
✅
Steric hindrance
is one of the biggest reasons why bulky ketones react slowly in nucleophilic addition reactions.✅ Semicarbazones and 2,4-DNP derivatives are widely used because they are usually solid with sharp melting points, helping in identification of carbonyl compounds.
✅ In labs, water is often removed using techniques like Dean–Stark apparatus (or by using excess alcohol) to push ester formation forward.