CBSE Class 12 Chemistry Chapter 14 Biomolecules NCERT Solutions
This chapter delves into the essential molecules of life, covering carbohydrates, proteins, nucleic acids, and vitamins. The NCERT Solutions for Class 12 Chemistry, Chapter 14: Biomolecules, provide clear and concise explanations for intext questions. These solutions help students understand the structure, properties, and functions of these vital organic compounds. Key topics include monosaccharides, disaccharides, polysaccharides, amino acids, protein structure, DNA, RNA, and the role of vitamins. The detailed step-by-step solutions are designed to clarify complex concepts, aiding students in their preparation for board examinations and competitive entrance tests by reinforcing their understanding of fundamental biochemical principles.
Quick info
| Board | CBSE |
|---|---|
| Class | Class 12 |
| Subject | Chemiry |
| Session | 2026 |
| Language | English |
| Type | NCERT Solutions |
| Chapter | Chapter 14: Biomolecules - Intext Questions Solutions |
Chapter summary
Chapter 14 of the NCERT Class 12 Chemistry textbook focuses on Biomolecules. This section provides solutions to intext questions, covering the classification and properties of carbohydrates, the structure and function of proteins, and an introduction to nucleic acids and vitamins. The solutions explain concepts like the solubility of glucose and sucrose, the hydrolysis of lactose, and the structural implications for the reactivity of glucose derivatives like pentaacetate.
Learning outcomes
- Understand the reasons for solubility differences between polar and non-polar organic compounds in water.
- Identify the products formed from the hydrolysis of disaccharides like lactose.
- Explain the chemical behavior of glucose, particularly the absence of an aldehyde group in its pentaacetate derivative.
- Relate molecular structure to chemical reactivity in carbohydrates.
Topics covered
Paper topics
- Carbohydrates
- Monosaccharides
- Disaccharides
- Hydrolysis
- Glucose
- Sucrose
- Lactose
- Functional groups
- Hydrogen bonding
- Solubility
- Chemical reactions of glucose
- Pentaacetate of glucose
Important topics
- Solubility of biomolecules
- Hydrolysis of disaccharides
- Reactivity of glucose and its derivatives
- Role of functional groups in biomolecules
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Questions and Solutions
Question 14.1
The solubility of a compound in water depends on its ability to form hydrogen bonds with water molecules. Glucose possesses five hydroxyl (-OH) groups, and sucrose has eight hydroxyl (-OH) groups. These numerous -OH groups allow both glucose and sucrose to form extensive hydrogen bonds with water molecules. This strong interaction facilitates their dissolution in water.
In contrast, cyclohexane and benzene are non-polar organic compounds that lack hydroxyl or other polar functional groups capable of forming hydrogen bonds. Consequently, they cannot interact favorably with polar water molecules and remain insoluble.
Question 14.2
Lactose is a disaccharide. Upon complete hydrolysis in the presence of an acid or enzyme, it breaks down into its constituent monosaccharides. Lactose is specifically composed of one molecule of β-D-galactose and one molecule of β-D-glucose, linked by a glycosidic bond.
The hydrolysis reaction can be represented as:
Therefore, the expected products of lactose hydrolysis are D-glucose and D-galactose.
Question 14.3
The presence of an aldehyde group in D-glucose is confirmed by its reaction with hydroxylamine (NH2OH) to form an oxime. This reaction implies that glucose can exist in an open-chain form with an aldehyde group (-CHO) at one end, in equilibrium with its cyclic forms in aqueous solution.
However, when glucose is treated with acetic anhydride ((CH3CO)2O), it forms D-glucose pentaacetate. In this derivative, all five hydroxyl groups of glucose, including the anomeric hydroxyl group (which is part of the hemiacetal in the cyclic form), are esterified with acetyl groups (-COCH3). The acetylation of the anomeric hydroxyl group stabilizes the cyclic structure and prevents the opening of the ring to form the free aldehyde group.
Since the pentaacetate derivative does not possess a free aldehyde group, it does not react with hydroxylamine to form an oxime. This observation supports the cyclic structure of glucose and the fact that the anomeric position is blocked in the pentaacetate.
The reaction sequence is:
Glucose (in equilibrium with open-chain aldehyde form) + NH2OH → Glucose oxime
Glucose + Acetic anhydride → Glucose pentaacetate
Glucose pentaacetate + NH2OH → No reaction (No oxime formation)
Common mistakes
- Confusing the open-chain and cyclic structures of glucose and their implications.
- Incorrectly predicting the products of disaccharide hydrolysis.
- Overlooking the role of hydrogen bonding in solubility.
Revision tips
- Focus on the role of functional groups, especially hydroxyl (-OH) and carbonyl (aldehyde/ketone), in determining molecular properties.
- Draw out the structures of monosaccharides and disaccharides to visualize hydrolysis and bonding.
- Pay close attention to how acetylation (forming pentaacetate) affects the reactivity of glucose.
- Review the concept of hydrogen bonding and its impact on solubility.
Practice MCQs
Q1. Why are glucose and sucrose soluble in water, while cyclohexane and benzene are not?
Explanation: Glucose and sucrose possess multiple hydroxyl (-OH) groups that can form extensive hydrogen bonds with water molecules, leading to their solubility. Cyclohexane and benzene lack these polar groups and thus cannot form hydrogen bonds with water.
Q2. What are the monosaccharide units obtained from the complete hydrolysis of lactose?
Explanation: Lactose is a disaccharide composed of one molecule of β-D-galactose and one molecule of β-D-glucose, linked by a glycosidic bond. Hydrolysis breaks this bond, yielding these two monosaccharides.
Q3. The reaction of glucose with hydroxylamine to form an oxime indicates the presence of which functional group?
Explanation: Glucose reacts with hydroxylamine to form an oxime, which is a characteristic reaction of aldehydes. This reaction occurs when the cyclic form of glucose opens up to its open-chain aldehyde form in aqueous solution.
Q4. Why does the pentaacetate of D-glucose not react with hydroxylamine?
Explanation: In glucose pentaacetate, all the hydroxyl groups, including the one involved in the hemiacetal linkage (which would open to form the aldehyde), are esterified with acetate groups. This prevents the formation of the open-chain aldehyde and thus the reaction with hydroxylamine.
Frequently asked questions
What is the main reason for the solubility of glucose in water?
Glucose is soluble in water primarily because its multiple hydroxyl (-OH) groups can form hydrogen bonds with water molecules, making it hydrophilic.
What products are formed when lactose is hydrolyzed?
Hydrolysis of lactose yields one molecule of β-D-galactose and one molecule of β-D-glucose.
Why is the aldehyde group of glucose not detected in its pentaacetate form?
In glucose pentaacetate, the anomeric hydroxyl group (which would open to form the aldehyde) is acetylated, preventing the formation of the free aldehyde group and its characteristic reactions.
How does the structure of cyclohexane differ from glucose in terms of water solubility?
Cyclohexane is a non-polar hydrocarbon and cannot form hydrogen bonds with water, making it insoluble. Glucose, with its polar hydroxyl groups, readily forms hydrogen bonds and is soluble.
What is the significance of the reaction of glucose with hydroxylamine?
The reaction of glucose with hydroxylamine to form an oxime demonstrates the presence of a carbonyl group, specifically an aldehyde group, in the open-chain form of glucose.
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