NCERT Solutions Class 12 Chemistry Chapter 8 - Aldehydes, Ketones and Carboxylic Acids

NCERT Solutions for Class 12 Chemistry Chapter 8 (Aldehydes, Ketones and Carboxylic Acids) play a crucial role in helping students build a strong conceptual foundation for both CBSE board exams and competitive exams like JEE Main, JEE Advanced, and NEET. This chapter serves as the backbone of carbonyl chemistry, introducing the mathematical and chemical foundations for understanding the Carbonyl Group. Understanding these concepts is essential for students as they lead to many of the systems that utilize this technology, including the synthesis of fragrances, plastics, and essential metabolic intermediates like pyruvic acid.

The NCERT Solutions for Class 12 Chemistry Chapter 8 will help students learn how to use nucleophilic addition mechanisms and alpha-hydrogen acidity to predict complex organic transformations. Well-explained NCERT solutions enable students to strengthen fundamentals, practice important multi-step conversions, and perform better in term exams, board exams, and national-level competitive tests.

1.0Class 12 Chemistry Chapter 8: Key Concepts

Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids covers the structure, nomenclature, preparation, and reactions of carbonyl compounds and carboxylic acids. It includes important reactions, chemical tests, and factors affecting acidity.

Key Concepts

Nomenclature and Structure: Learn the structure and IUPAC nomenclature of aliphatic and aromatic aldehydes, ketones, and carboxylic acids.
Preparation of Aldehydes and Ketones: Study important preparation methods such as Rosenmund Reduction, Stephen Reaction, and hydration of alkynes.
Nucleophilic Addition Reactions: Understand the reactions of aldehydes and ketones with HCN, NaHSO₃, Grignard reagents, and alcohols, including the formation of acetals and ketals.
Reactions Due to α-Hydrogen: Learn the importance of α-hydrogen in carbonyl compounds and study Aldol Condensation and Cross-Aldol Condensation.
Cannizzaro Reaction: Understand the disproportionation reaction shown by aldehydes that do not contain α-hydrogen.
Chemical Tests: Study Tollens’ Test, Fehling’s Test, and Iodoform Test and learn how these tests help identify specific carbonyl compounds.
Acidity of Carboxylic Acids: Understand why carboxylic acids are acidic and how different substituents affect their acid strength.
Hell-Volhard-Zelinsky Reaction: Learn the HVZ Reaction and its role in the α-halogenation of carboxylic acids.

2.0NCERT Solutions Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids : Detailed Solutions

INTEXT QUESTIONS

Write the structures of products of the following reactions :

Ans.

Arrange the following compounds in increasing order of their boiling points. $CH_{3} CHO, CH_{3} CH_{2} OH, CH_{3} OCH_{3}, CH_{3} CH_{2} CH_{3}$ Ans.

Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i) Enthanal, Propanal, Propanone, Butanone. (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone. Hint : Consider steric effect and electronic effect.

Ans. Reactivity of carbonyl compound $\propto (+)$ ve charge on $sp^{2} C \propto \frac{1}{steric hindrence}$

(i)

ques-3-(i)-sol-chap-8-class-12-chemistry

The +I effect of the alkyl group increases in the order :

Ethanal < Propanal < Propanone < Butanone If +I effect increases then (+)ve charge on sp2 C decreases.

So reactivity order is : Butanone < Propanone < Propanal < Ethanal

(ii)

chemistry-ques-3-(ii)-sol-class-12-chap-8

Predict the product of the following reactions :

Ans.

5. Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene)

Ans.

Which acid of each pair shown here would you expect to be stronger?

Ans. Acidic strength

7. Give names of the reagents to bring about the following transformations: (i) Hexan-1-ol to hexanal (ii) Cyclohexanol to cyclohexanone (iii) p-Fluorotoluene to p-fluorobenzaldehyde (iv) Ethanenitrile to ethanal (v) Allyl alcohol to propenal (vi) But-2-ene to ethanal

Ans.

(i) $C_{5} H_{5} NH^{+} CrO_{3} Cl^{-} (PCC)$ (ii) $K_{2} Cr_{2} O_{7}$ in acidic medium (iii) $CrO_{3}$ in the presence of acetic anhydride/1. $CrO_{2} Cl_{2}$ 2. HOH (iv) (Diisobutyl) aluminium hydride (DIBAL-H) (v) PCC (vi) $O_{3} / H_{2} O - Zn$ dust

Arrange the following compounds in the increasing order of their boiling points: $CH_{3} CH_{2} CH_{2} CHO, CH_{3} CH_{2} CH_{2} CH_{2} OH, H_{5} C_{2}$ $- O - C_{2} H_{5}, CH_{3} CH_{2} CH_{2} CH_{2} CH_{3}$

Ans. The molecular masses of these compounds are in the range of 72 to 74. Since only butan-1-ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan-1-ol would be the highest. Butanal is more polar than ethoxyethane. Therefore, the intermolecular dipole-dipole attraction is stronger in the former. n-Pentane molecules have only weak van der Waals forces. Hence increasing order of boiling points of the given compounds is as follows : $CH_{3} CH_{2} CH_{2} CH_{2} CH_{3} < H_{5} C_{2} - O - C_{2} H_{5} <$ $CH_{3} CH_{2} CH_{2} CHO < CH_{3} CH_{2} CH_{2} CH_{2} OH$

Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal ? Explain your answer. Ans. The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.
An organic compound (A) with molecular formula $C_{8} H_{8} O$ forms an orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens' or Fehlings' reagent, nor does it decolourise bromine water or Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula $C_{7} H_{6} O_{2}$ . Identify the compounds (A) and (B) and explain the reactions involved. Ans. (A) forms 2,4-DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens' or Fehling reagent, (A) must be a ketone. (A) responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyer's reagent. This indicates the presence of unsaturation due to an aromatic ring. Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows :

Write chemical reactions to affect the following transformations: (i) Butan-1-ol to butanoic acid (ii) Benzyl alcohol to phenylethanoic acid (iii) 3-Nitrobromobenzene to 3-nitrobenzoic acid (iv) 4-Methylacetophenone to benzene-1, 4- dicarboxylic acid (v) Cyclohexene to hexane-1,6-dioic acid (vi) Butanal to butanoic acid.

Ans.

NCERT EXERCISE

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

Ans.

(i) Cyanohydrin : Organic compound which have -CN and -OH group in structure.

(ii) Acetal : Acetals are gem-dialkoxy alkanes in which two alkoxy groups are present on the terminal carbon atom.

(iii) Semicarbazone : Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

(iv) Aldol : A $β$ -hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones with $α H$ in the presence of a base.

(v) Hemiacetal : Hemiacetal are $α$ -alkoxyalcohols Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas. (vi) Oxime : Oximes are a class of organic compounds having the general formula .

On treatment with hydroxylamine in a weakly acidic medium, aldehydes or ketones form oximes.

(vii)Ketal : Ketals are gem-dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain.

(viii) Imine : Imines are $> C = N - R$ or $> C = NH$ . Imines are produced when aldehydes and ketones react with $R - NH_{2}$

class-12-chap-8-ncert-exer-ques-1-chemistry

(ix)2, 4-DNP-derivative : 2, 4-dinitrophenylhydrazones are 2, 4-DNP-derivatives, which are produced when aldehydes or ketones react with 2, 4- dinitrophenylhydrazine in a weakly acidic medium.

To identify and characterize aldehydes and ketones, 2, 4-DNP derivatives are used. (x) Schiff's base : is $> C = NR$ (R is not Hydrogen) (R may be -Ph). Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of trace of an acid yields a Schiff's base.

ncert-exe-sol-1-class-12-chemistry-chap-8

Draw structures of the following derivatives. (i) The 2, 4-dinitrophenylhydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehydedimethylacetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde

Ans.

ncert-exe-ques-2-sol-class-12-chap-8-che

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i) PhMgBr and then $H_{3} O^{+}$ (ii) Tollen's reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid

Ans.

Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither ? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2, 2-Dimethylbutanal

Ans. Aldehydes and ketones having at least one $α$ hydrogen undergo aldol condensation (ii), (v), (vi), (vii) gives aldol condensation.

Aldehydes (only) having no $α$ -hydrogen undergo Cannizzaro reactions. (i), (iii), (ix) gives cannizzaro reaction.

Compound (iv) is a ketone having no $α$ hydrogen atom and compound (viii) Butan-1ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or cannizzaro reactions.

Aldol condensation

chemistry-ncert-exer-4-sol-class-12-chap-8

Cannizzaro reaction :

How will you convert ethanal into the following compounds ? (i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid

Ans.

ncert-exer-ques-5-sol-che-class-12-chap-8

Write structural formulas and names of four possible aldol from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

Ans.

(i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.

(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.

(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.

che-class-12-ncert-exer-ques-6-sol-chap-8

(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

An organic compound (A) (molecular formula $C_{8} H_{16} O_{2}$ ) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Ans. Hydrolysis of ester gives acid and alcohol

class-12-chap-8-ncert-exer-7-sol-chemistry

Arrange the following compounds in increasing order of their property as indicated : (a) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone [Reactivity towards HCN] (b) $CH_{3} CH_{2} CH (Br) COOH$ , $CH_{3} CH (Br) CH_{2} COOH, (CH_{3})_{2} CHCOOH$ , $CH_{3} CH_{2} CH_{2} COOH$ [Acid strength] (c) Benzoic acid, 4-Nitrobenzoic acid, 3,4- Dinitrobenzoic acid, 4-Methoxybenzoic acid [Acid strength] (d) $CH_{3} COCl, CH_{3} CONH_{2}, CH_{3} COOCH_{3}$ , $(CH_{3} CO)_{2} O$ [Reactivity in hydrolysis]

Ans.

(a) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde (b) $(CH_{3})_{2} CHCOOH, < CH_{3} CH_{2} CH_{2} COOH$ $< CH_{3} CH (Br) CH_{2} COOH, < CH_{3} CH_{2} CH (Br)$ COOH (c) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3, 4-Dinitrobenzoic acid (d) $CH_{3} CONH_{2} < CH_{3} COOCH_{3}$ $< (CH_{3} CO)_{2} O < CH_{3} COCl$

Give simple chemical tests to distinguish between the following pairs of compounds. (i) Propanal and Propanone (ii) Acetophenone and Benzophenone (iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate (v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone (vii) Ethanal and Propanal

Ans.

(i) By Tollen's test (a) Propanal reduces Tollen's reagent, fehling solution, Benedict solution but propanone (ketone) does not reduces. (b) Iodoform test :

class-12-ncert-exer-ques-9-sol-chap-8-che

(ii) Acetophenone and Benzophenone can be distinguished using the iodoform test. Acetophenone gives iodoform test but benzophenone does not. (iii) Phenol and benzoic acid can be distinguished by ferric chloride test and $NaHCO_{3}$ test. (a) Phenol gives voilet colour with $FeCl_{3}$ . (b) Benzoic acid gives $CO_{2}$ with $NaHCO_{3}$ while phenol does not. (iv) Benzoic acid and Ethyl benzoate can be distinguished by sodium bicarbonate test. Sodium bicarbonate test : Benzoic acid react with $NaHCO_{3}$ to produce brisk effervescence due to the evolution of $CO_{2}$ gas but ethylbenzoate does not. (v) Pentan-2-one and pentan-3- one can be distinguished by iodoform test. Pentan-2-one gives iodoform test but pentan-3-one does not. (vi) Benzaldehyde and acetophenone can be distinguished by the (a) Tollen's test (b) Iodoform test. Benzaldehyde gives Tollen's test where as acetophenone gives iodoform test. (vii) Ethanal and propanal can be distinguished by iodoform test. Ethanal gives iodoform test but propanal does not.

How will you prepare the following compounds from benzene ? You may use any inorganic reagent and any organic reagent having not more than one carbon atom (i) Methylbenzoate (ii) m - Nitrobenzoic acid (iii) p - Nitrobenzoic acid (iv) Phenylacetic acid (v) p - Nitrobenzaldehyde

Ans.

How will you bring about the following conversions in not more than two steps ? (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to $m$ -Nitroacetophenone (v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benazaldehyde to $α$ -Hydroxyphenylacetic acid Ans.

chap-8-ncert-exer-ques-11-sol-class-12-chemistry

Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation

Ans.

(i) Acetylation :

ncert-exe-12-sol-class-12-chap-8-chemistry

(ii) Cannizzaro reaction : The self oxidationreduction (disproportionation) reaction of aldehydes having no $α$ -hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction.In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

class-12-ncert-exer-ques-12-sol-chap-8-chemistry

(iii) Cross-aldol condensation : Reaction between two different aldehydes, or two different ketones, or an aldehyde and a ketone (having $α - H$ ) in presence of base is called a cross-aldol condensation.

(iv) Decarboxylation : Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

ncert-exer-ques-12-sol-chap-8-che-class-12

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

Ans.

ncert-exer-13-sol-class-12-chap-8-chemistry

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- $NH_{2}$ group in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii) During the preparation of esters a carboxylic acid and an alcohol in the pressence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Ans.

(i) Cyclohexanones form cyanohydrins according to the following equation.

In 2, 2, 6 - trimethyl cyclohexanone, methyl groups at $α$ -positions offer steric hindrance and as a result, $CN^{-}$ cannot attack effectively.

2, 2, 6 - Trimethyl cyclohexanone For the reason, it does not form a cyanohydrin. (ii)

chem-ncert-exer-14-sol-chapter-8-class-12

The electron density on $- NH_{2}$ group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other - $NH_{2}$ group is not involved in resonance; it can act as a nucleophile and can attack carbonyl - carbon atoms of aldehydes and ketones to produce semicarbazones. (iii)

class-12-che-ncert-exer-ques-14-sol-chapter-8

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible.

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86 . It does not reduce Tollen's reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound. Ans. % of carbon $= 69.77%$ % of hydrogen = 11.63% $%$ of oxygen $= {100 - (69.77 + 11.63)} %$ $= 18.6%$ Thus, the ratio of the number of carbon, hydrogen and oxygen atoms in the organic compound can be given as: $C : H : O = \frac{69.77}{12} : \frac{11.63}{1} : \frac{18.6}{16}$ $= 5.81 : 11.63 : 1.16 = 5 : 10 : 1$ Therefore, the empirical formula of the compound is $C_{5} H_{10} O$ . Now, the empirical formula mass of the compound can be given as $5 \times 12 + 10 \times 1 + 1 \times 16 = 86$ Molecular mass of the compound $= 86$ (given) Therefore, the molecular formula of the compound is given by $C_{5} H_{10} O$ . Since the given compound does not reduce Tollen's reagent, it is not an aldehyde. Again the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.

The given compound also gives a mixture of ethanoic acid and propanoic acid. Hence, the given compound is pentan-2-one.

class-12-ncert-exer-ques-15-sol-chap-8-chemistry

The given reactions can be explained by the following equations :

Although phenoxide ion has more number or resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why ?

Ans. Resonance structures of phenoxide ion are :

class-12-chap-8-chemsitry-ncert-exer-16-sol

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

Explore chapter-wise NCERT Solutions for Class 12 Chemistry with simple explanations, important concepts, key formulae, and step-by-step solutions to numerical problems.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Solutions
Chapter 2	Electrochemistry
Chapter 3	Chemical Kinetics
Chapter 4	d- and f-Block Elements
Chapter 5	Coordination Compounds
Chapter 6	Haloalkanes and Haloarenes
Chapter 7	Alcohols, Phenols and Ethers
Chapter 9	Amines
Chapter 10	Biomolecules

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 8

Reaction Mechanisms: Step-by-step explanations of Aldol and Cannizzaro reactions.
Concept-Based Answers: Clear explanations of why aldehydes are more reactive than ketones toward nucleophiles.
Acidity Comparison: Easy comparison of carboxylic acid strength using pKₐ values.
Chemical Tests: Simple explanations of important tests used to identify functional groups.

NCERT Solutions Class 12 Chemistry Chapter 8 - Aldehydes, Ketones and Carboxylic Acids

1.0Class 12 Chemistry Chapter 8: Key Concepts

Key Concepts

Nomenclature and Structure: Learn the structure and IUPAC nomenclature of aliphatic and aromatic aldehydes, ketones, and carboxylic acids.
Preparation of Aldehydes and Ketones: Study important preparation methods such as Rosenmund Reduction, Stephen Reaction, and hydration of alkynes.
Nucleophilic Addition Reactions: Understand the reactions of aldehydes and ketones with HCN, NaHSO₃, Grignard reagents, and alcohols, including the formation of acetals and ketals.
Reactions Due to α-Hydrogen: Learn the importance of α-hydrogen in carbonyl compounds and study Aldol Condensation and Cross-Aldol Condensation.
Cannizzaro Reaction: Understand the disproportionation reaction shown by aldehydes that do not contain α-hydrogen.
Chemical Tests: Study Tollens’ Test, Fehling’s Test, and Iodoform Test and learn how these tests help identify specific carbonyl compounds.
Acidity of Carboxylic Acids: Understand why carboxylic acids are acidic and how different substituents affect their acid strength.
Hell-Volhard-Zelinsky Reaction: Learn the HVZ Reaction and its role in the α-halogenation of carboxylic acids.

2.0NCERT Solutions Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids : Detailed Solutions

INTEXT QUESTIONS

Write the structures of products of the following reactions :

Ans.

Arrange the following compounds in increasing order of their boiling points. $CH_{3} CHO, CH_{3} CH_{2} OH, CH_{3} OCH_{3}, CH_{3} CH_{2} CH_{3}$ Ans.

Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i) Enthanal, Propanal, Propanone, Butanone. (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone. Hint : Consider steric effect and electronic effect.

Ans. Reactivity of carbonyl compound $\propto (+)$ ve charge on $sp^{2} C \propto \frac{1}{steric hindrence}$

(i)

The +I effect of the alkyl group increases in the order :

Ethanal < Propanal < Propanone < Butanone If +I effect increases then (+)ve charge on sp2 C decreases.

So reactivity order is : Butanone < Propanone < Propanal < Ethanal

(ii)

Predict the product of the following reactions :

Ans.

5. Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene)

Ans.

Which acid of each pair shown here would you expect to be stronger?

Ans. Acidic strength

Ans.

Arrange the following compounds in the increasing order of their boiling points: $CH_{3} CH_{2} CH_{2} CHO, CH_{3} CH_{2} CH_{2} CH_{2} OH, H_{5} C_{2}$ $- O - C_{2} H_{5}, CH_{3} CH_{2} CH_{2} CH_{2} CH_{3}$

Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal ? Explain your answer. Ans. The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.
An organic compound (A) with molecular formula $C_{8} H_{8} O$ forms an orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens' or Fehlings' reagent, nor does it decolourise bromine water or Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula $C_{7} H_{6} O_{2}$ . Identify the compounds (A) and (B) and explain the reactions involved. Ans. (A) forms 2,4-DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens' or Fehling reagent, (A) must be a ketone. (A) responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyer's reagent. This indicates the presence of unsaturation due to an aromatic ring. Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows :

Write chemical reactions to affect the following transformations: (i) Butan-1-ol to butanoic acid (ii) Benzyl alcohol to phenylethanoic acid (iii) 3-Nitrobromobenzene to 3-nitrobenzoic acid (iv) 4-Methylacetophenone to benzene-1, 4- dicarboxylic acid (v) Cyclohexene to hexane-1,6-dioic acid (vi) Butanal to butanoic acid.

Ans.

NCERT EXERCISE

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

Ans.

(i) Cyanohydrin : Organic compound which have -CN and -OH group in structure.

(ii) Acetal : Acetals are gem-dialkoxy alkanes in which two alkoxy groups are present on the terminal carbon atom.

(iii) Semicarbazone : Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

On treatment with hydroxylamine in a weakly acidic medium, aldehydes or ketones form oximes.

(vii)Ketal : Ketals are gem-dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain.

(viii) Imine : Imines are $> C = N - R$ or $> C = NH$ . Imines are produced when aldehydes and ketones react with $R - NH_{2}$

(ix)2, 4-DNP-derivative : 2, 4-dinitrophenylhydrazones are 2, 4-DNP-derivatives, which are produced when aldehydes or ketones react with 2, 4- dinitrophenylhydrazine in a weakly acidic medium.

Draw structures of the following derivatives. (i) The 2, 4-dinitrophenylhydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehydedimethylacetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde

Ans.

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i) PhMgBr and then $H_{3} O^{+}$ (ii) Tollen's reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid

Ans.

Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither ? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2, 2-Dimethylbutanal

Ans. Aldehydes and ketones having at least one $α$ hydrogen undergo aldol condensation (ii), (v), (vi), (vii) gives aldol condensation.

Aldehydes (only) having no $α$ -hydrogen undergo Cannizzaro reactions. (i), (iii), (ix) gives cannizzaro reaction.

Compound (iv) is a ketone having no $α$ hydrogen atom and compound (viii) Butan-1ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or cannizzaro reactions.

Aldol condensation

Cannizzaro reaction :

How will you convert ethanal into the following compounds ? (i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid

Ans.

Write structural formulas and names of four possible aldol from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

Ans.

(i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.

(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.

(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.

(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

An organic compound (A) (molecular formula $C_{8} H_{16} O_{2}$ ) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Ans. Hydrolysis of ester gives acid and alcohol

Arrange the following compounds in increasing order of their property as indicated : (a) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone [Reactivity towards HCN] (b) $CH_{3} CH_{2} CH (Br) COOH$ , $CH_{3} CH (Br) CH_{2} COOH, (CH_{3})_{2} CHCOOH$ , $CH_{3} CH_{2} CH_{2} COOH$ [Acid strength] (c) Benzoic acid, 4-Nitrobenzoic acid, 3,4- Dinitrobenzoic acid, 4-Methoxybenzoic acid [Acid strength] (d) $CH_{3} COCl, CH_{3} CONH_{2}, CH_{3} COOCH_{3}$ , $(CH_{3} CO)_{2} O$ [Reactivity in hydrolysis]

Ans.

Give simple chemical tests to distinguish between the following pairs of compounds. (i) Propanal and Propanone (ii) Acetophenone and Benzophenone (iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate (v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone (vii) Ethanal and Propanal

Ans.

(i) By Tollen's test (a) Propanal reduces Tollen's reagent, fehling solution, Benedict solution but propanone (ketone) does not reduces. (b) Iodoform test :

How will you prepare the following compounds from benzene ? You may use any inorganic reagent and any organic reagent having not more than one carbon atom (i) Methylbenzoate (ii) m - Nitrobenzoic acid (iii) p - Nitrobenzoic acid (iv) Phenylacetic acid (v) p - Nitrobenzaldehyde

Ans.

How will you bring about the following conversions in not more than two steps ? (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to $m$ -Nitroacetophenone (v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benazaldehyde to $α$ -Hydroxyphenylacetic acid Ans.

Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation

Ans.

(i) Acetylation :

(iv) Decarboxylation : Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

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

Ans.

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- $NH_{2}$ group in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii) During the preparation of esters a carboxylic acid and an alcohol in the pressence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Ans.

(i) Cyclohexanones form cyanohydrins according to the following equation.

In 2, 2, 6 - trimethyl cyclohexanone, methyl groups at $α$ -positions offer steric hindrance and as a result, $CN^{-}$ cannot attack effectively.

2, 2, 6 - Trimethyl cyclohexanone For the reason, it does not form a cyanohydrin. (ii)

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible.

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86 . It does not reduce Tollen's reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound. Ans. % of carbon $= 69.77%$ % of hydrogen = 11.63% $%$ of oxygen $= {100 - (69.77 + 11.63)} %$ $= 18.6%$ Thus, the ratio of the number of carbon, hydrogen and oxygen atoms in the organic compound can be given as: $C : H : O = \frac{69.77}{12} : \frac{11.63}{1} : \frac{18.6}{16}$ $= 5.81 : 11.63 : 1.16 = 5 : 10 : 1$ Therefore, the empirical formula of the compound is $C_{5} H_{10} O$ . Now, the empirical formula mass of the compound can be given as $5 \times 12 + 10 \times 1 + 1 \times 16 = 86$ Molecular mass of the compound $= 86$ (given) Therefore, the molecular formula of the compound is given by $C_{5} H_{10} O$ . Since the given compound does not reduce Tollen's reagent, it is not an aldehyde. Again the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.

The given compound also gives a mixture of ethanoic acid and propanoic acid. Hence, the given compound is pentan-2-one.

The given reactions can be explained by the following equations :

Although phenoxide ion has more number or resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why ?

Ans. Resonance structures of phenoxide ion are :

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

Explore chapter-wise NCERT Solutions for Class 12 Chemistry with simple explanations, important concepts, key formulae, and step-by-step solutions to numerical problems.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Solutions
Chapter 2	Electrochemistry
Chapter 3	Chemical Kinetics
Chapter 4	d- and f-Block Elements
Chapter 5	Coordination Compounds
Chapter 6	Haloalkanes and Haloarenes
Chapter 7	Alcohols, Phenols and Ethers
Chapter 9	Amines
Chapter 10	Biomolecules

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 8

Reaction Mechanisms: Step-by-step explanations of Aldol and Cannizzaro reactions.
Concept-Based Answers: Clear explanations of why aldehydes are more reactive than ketones toward nucleophiles.
Acidity Comparison: Easy comparison of carboxylic acid strength using pKₐ values.
Chemical Tests: Simple explanations of important tests used to identify functional groups.

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NCERT Solutions Class 12 Chemistry Chapter 8 - Aldehydes, Ketones and Carboxylic Acids

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2.0NCERT Solutions Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids : Detailed Solutions

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3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

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Frequently Asked Questions

Why are NCERT Solutions for Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids important?

Why are NCERT Solutions for Class 12 Chemistry Chapter 8 helpful in board and other competitive exams?

Are these NCERT Solutions for Class 9 Chapter 8 updated as per the latest syllabus? How these help?

Which reactions should students practise from the Aldehydes, Ketones and Carboxylic Acids chapter?

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NCERT Solutions Class 12 Chemistry Chapter 8 - Aldehydes, Ketones and Carboxylic Acids

1.0Class 12 Chemistry Chapter 8: Key Concepts

Key Concepts

2.0NCERT Solutions Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids : Detailed Solutions

INTEXT QUESTIONS

NCERT EXERCISE

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 8

Table of Contents