Chapter 14: Biomolecules

Updated NCERT Solutions for Class 12 Chemistry Chapter 14: Biomolecules + Important Board Exam Questions 2026

Scoring full marks in organic chemistry can sometimes feel tricky, but Biomolecules is a highly theoretical, direct, and high-scoring chapter that can easily boost your percentage . This complete guide provides the ultimate Updated NCERT Solutions, fundamental concept breakdowns, and curated Board Exam Questions 2026 . Whether you are preparing for your CBSE board exams or gearing up for competitive tests like NEET and JEE, these student-friendly explanations are designed to help you secure top marks .

Introduction

Unlock the secrets of life's chemistry ! In this chapter, you will explore how complex organic compounds like carbohydrates, proteins, enzymes, and nucleic acids drive living systems . It serves as an essential bridge between biology and chemistry, making it highly valuable for both CBSE Board Exam Questions 2026 and competitive medical exams like NEET . Let's master these concepts together and maximize your final chemistry score .

Chapter Overview

Here is a quick structural snapshot of the chapter to help you plan your study sessions :

Feature Details
Chapter Name Biomolecules (Chapter 14)
Subject Chemistry
Class Class 12
Board CBSE & State Boards (Latest NCERT Syllabus)
Important Topics Glucose structure, D/L configurations, Peptide bonds, Protein denaturation, DNA vs RNA
Difficulty Level Easy to Moderate (Requires memorizing structures and direct classifications)
Exam Weightage Approximately 7 Marks

Learning Objectives

After completing this chapter, students will be able to :

  • Classify carbohydrates into monosaccharides, oligosaccharides, and polysaccharides .
  • Understand the chemical reactions that elucidate the open-chain and cyclic structures of Glucose .
  • Explain the properties of amino acids, the formation of peptide bonds, and the structural levels of proteins .
  • Define denaturation of proteins and its everyday examples .
  • Distinguish between the structural frameworks and functions of DNA and RNA .

Key Concepts, Definitions & Formulas

Before tackling the core textbook questions, let us review the vital terms that examiners frequently target :

  • Carbohydrates: Optically active polyhydroxy aldehydes or ketones, or compounds which produce such units on hydrolysis .
  • Anomers: Diastereomers of cyclic carbohydrates that differ in configuration specifically at the first carbon ($C-1$ for aldoses, or $C-2$ for ketoses), which is known as the anomeric carbon .
  • Zwitterion: An internal neutral salt form of an amino acid where the carboxylic group loses a proton (becoming $-COO^-$) and the amino group gains that proton (becoming $-NH_3^+$) .
  • Peptide Linkage: An amide linkage ($-CO-NH-$) formed between the carboxyl group of one amino acid and the amino group of another, releasing a water molecule .
  • Inversion of Cane Sugar: Hydrolysis of sucrose (dextrorotatory) yields an equimolar mixture of glucose (dextrorotatory) and fructose (strongly levorotatory), resulting in a net change of optical rotation from dextro to levo .

Full NCERT Solutions for Class 12 Chemistry Chapter 14

Here are the complete, step-by-step updated NCERT solutions for all textbook exercise questions .

Question 1: What are monosaccharides?

Answer: Monosaccharides are the simplest units of carbohydrates that cannot be hydrolyzed further into smaller polyhydroxy aldehyde or ketone units . They serve as the structural building blocks for more complex carbohydrates .

General Formula: $(CH_2O)_n$ where $n=3$ to $7$ .
Examples: Glucose ($C_6H_{12}O_6$), Fructose ($C_6H_{12}O_6$), and Ribose ($C_5H_{10}O_5$) .

Question 2: What are reducing sugars?

Answer: Carbohydrates that are capable of reducing Fehling's solution and Tollen's reagent are classified as reducing sugars .
  • Key Condition: They must contain a free, unbonded aldehyde ($-CHO$) or ketone ($-CO-$) group capable of forming an open chain .
  • Examples: All monosaccharides (like glucose and galactose) and several disaccharides like maltose and lactose . Note that sucrose is a non-reducing sugar because its reducing groups are locked in the glycosidic bond .

Question 3: Write two functions of carbohydrates in plants.

Answer: Carbohydrates are essential to plant survival and structure in the following ways :
  1. Structural Framework: Cellulose is a complex carbohydrate that forms the cell wall of plant cells, providing mechanical strength, rigidity, and shape .
  2. Energy Storage: Starch serves as the primary storage polysaccharide in plants . It acts as an energy reserve utilized when the plant requires nutrition .

Question 4: Classify the following into monosaccharides and disaccharides: Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.

Answer: Based on their behavior upon chemical hydrolysis, we classify them as :
  • Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose . (Cannot be broken down into simpler sugars ).
  • Disaccharides: Maltose, lactose . (Yield two monosaccharide units upon hydrolysis ).

Question 5: What do you understand by the term glycosidic linkage?

Answer: A glycosidic linkage is an oxide linkage that joins two monosaccharide units together through an oxygen atom, accompanied by the loss of a water molecule . It typically forms between the anomeric carbon of one sugar molecule and a hydroxyl group of another sugar molecule . This linkage is the core structural bond found in all oligosaccharides and polysaccharides .

Question 6: What is glycogen? How is it different from starch?

Answer: Glycogen is a highly branched polysaccharide composed of glucose units that serves as the primary energy storage molecule in animals and fungi . It is frequently called "animal starch" .

Key Differences from Starch:
  • Source: Glycogen is found in animals (stored mainly in the liver and muscles), whereas starch is exclusively produced and stored in plants .
  • Structure: Glycogen is structurally similar to amylopectin (a component of starch) but is much more highly branched, making it compact and readily broken down by enzymes when animals need a rapid energy burst .

Question 7: What are the hydrolysis products of (i) sucrose and (ii) lactose?

Answer: Upon enzymatic or acidic hydrolysis, these disaccharides split into their constituent monosaccharide building blocks :
  • (i) Sucrose: Yields an equimolar mixture of $\alpha$-D-glucose and $\beta$-D-fructose . $$\text{Sucrose} + H_2O \rightarrow \text{D-Glucose} + \text{D-Fructose}$$
  • (ii) Lactose: Yields an equimolar mixture of $\beta$-D-galactose and $\beta$-D-glucose . $$\text{Lactose} + H_2O \rightarrow \text{D-Galactose} + \text{D-Glucose}$$

Question 8: What is the basic structural difference between starch and cellulose?

Answer: While both are polymers of glucose, their configurations and arrangements create completely different properties :
  • Starch: Made of two components—Amylose (linear polymer of $\alpha$-D-glucose with $\alpha$-1,4-glycosidic links) and Amylopectin (branched polymer with additional $\alpha$-1,6 branches) .
  • Cellulose: A strictly linear, straight-chain polymer consisting exclusively of $\beta$-D-glucose units linked together via $\beta$-1,4-glycosidic bonds . This linear geometry allows the chains to pack closely via strong hydrogen bonding, forming rigid structural fibers .

Question 9: What happens when D-glucose is treated with the following reagents? (i) $HI$ (ii) $HNO_3$ (iii) Bromine water

Answer: These chemical tests are vital for proving the structural makeup of glucose :
  • (i) With $HI$ (Prolonged heating): D-glucose undergoes complete reduction to form n-hexane . This reaction proves that all six carbon atoms are linked in a straight, unbranched chain . $$C_6H_{12}O_6 + 6HI \xrightarrow{\Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3 + 3I_2 + 3H_2O$$
  • (ii) With $HNO_3$ (Strong Oxidation): Both the terminal aldehyde group and the primary alcohol group ($-CH_2OH$) are oxidized, forming a dicarboxylic acid called saccharic acid . $$C_6H_{12}O_6 \xrightarrow{HNO_3} HOOC-(CHOH)_4-COOH$$
  • (iii) With Bromine water (Mild Oxidation): Only the sensitive aldehyde group gets oxidized to a carboxylic acid group, yielding gluconic acid . This proves the presence of an aldehyde group rather than a ketone . $$C_6H_{12}O_6 \xrightarrow{Br_2/H_2O} HOOC-(CHOH)_4-CH_2OH$$

Question 10: Enumerate the reactions of D-glucose which cannot be explained by its open-chain structure.

Answer: The open-chain model fails to account for several experimental facts, which led to the discovery of its cyclic structures :
  1. Schiff's and Bisulfite Test: Despite having a free aldehyde group, glucose does not form a hydrogen sulfite addition product with $NaHSO_3$ and fails to give a positive Schiff's test .
  2. Acetylation: The pentaacetate derivative of glucose does not react with hydroxylamine ($NH_2OH$) . This indicates the absence of a free open-chain $-CHO$ group in the acetylated form .
  3. Existence of Mutarotation ($\alpha$ and $\beta$ forms): Glucose crystallizes in two distinct forms with different melting points and optical rotations ($\alpha$-glucose and $\beta$-glucose), which slowly convert into an equilibrium mixture in solution .

Question 11: What are essential and non-essential amino acids? Give two examples of each type.

Answer:
  • Essential Amino Acids: Amino acids that cannot be synthesized by the human body and must be supplied regularly through our daily dietary intake .
    Examples: Valine, Leucine, Lysine .
  • Non-Essential Amino Acids: Amino acids that our body is fully capable of synthesizing internally from basic metabolic intermediates; hence, they do not need to be specifically tracked in our diet .
    Examples: Glycine, Alanine, Aspartic acid .

Question 12: Define the following as related to proteins: (i) Peptide linkage (ii) Primary structure (iii) Denaturation

Answer:
  • (i) Peptide Linkage: An amide bond ($-CO-NH-$) created between the carboxylic acid group of one amino acid and the amino group of another, driving protein synthesis .
  • (ii) Primary Structure: The specific linear sequence in which various amino acids are linked together by peptide bonds within a polypeptide chain . Any change in this sequence alters the entire protein function .
  • (iii) Denaturation: A process where a native protein is exposed to physical or chemical changes (like heat, radiation, or extreme pH shifts), causing its complex secondary, tertiary, and quaternary structures to unfold and disrupt without cleaving the strong primary peptide bonds .

Question 13: What are the common types of secondary structure of proteins?

Answer: The secondary structure describes the local conformation of the polypeptide backbone, stabilized by hydrogen bonds between the carbonyl oxygen and amide hydrogen atoms . The two primary types are :
  1. $\alpha$-Helix Structure: The polypeptide chain coils into a right-handed screw configuration where each amide group forms a hydrogen bond with the fourth amide group ahead of it down the chain .
  2. $\beta$-Pleated Sheet Structure: Polypeptide chains are stretched out side-by-side in an extended conformation and run parallel or anti-parallel, bound together by intermolecular hydrogen bonds to resemble pleated drapery .

Question 14: What type of bonding helps in stabilizing the $\alpha$-helix structure of proteins?

Answer: The $\alpha$-helix structure is stabilized exclusively by intramolecular hydrogen bonding . These hydrogen bonds form between the carbonyl oxygen atom ( $>C=O$) of one amino acid residue and the amide hydrogen atom ($-NH-$) of an amino acid located four residues further along the chain .

Question 15: Differentiate between globular and fibrous proteins.

Answer: Based on their overall molecular shape, proteins are grouped as follows :
Fibrous Proteins Globular Proteins
Polypeptide chains run parallel to each other, forming long, wire-like fibers . Polypeptide chains fold upon themselves into compact, spherical shapes .
Generally insoluble in water . Highly soluble in water .
Main functions are structural and protective . Primarily involved in biological regulation and maintenance .
Examples: Keratin (hair, nails), Myosin (muscles) . Examples: Insulin, Egg Albumin, Hemoglobin .

Question 16: How do you explain the amphoteric behaviour of amino acids?

Answer: In aqueous solution, an amino acid undergoes an internal acid-base neutralization . The carboxylic acid group ($-COOH$) acts as an acid and loses a proton, while the basic amino group ($-NH_2$) accepts that proton . This creates a dipolar ion known as a Zwitterion : $$R-CH(NH_2)-COOH \rightleftharpoons R-CH(NH_3^+)-COO^-$$ Because the Zwitterion contains both a basic negatively charged carboxylate group ($-COO^-$) and an acidic positively charged ammonium group ($-NH_3^+$), it can react with both acids and bases, demonstrating classic amphoteric behavior .

Question 17: What are enzymes?

Answer: Enzymes are specialized globular proteins that function as highly efficient, sophisticated biological catalysts in living organisms . They speed up the rates of complex biochemical reactions by lowering the required activation energy pathway . They are exceptionally specific, meaning a single enzyme typically catalyzes only one specific chemical reaction for a unique substrate molecule .

Question 18: What is the effect of denaturation on the structure of proteins?

Answer: During denaturation, the weak hydrogen bonds, hydrophobic interactions, and disulfide linkages stabilizing the secondary, tertiary, and quaternary structures break open . As a result, the globular coils unfold and the fibers stretch out .

Crucially, the covalent primary structure remains completely intact because peptide bonds are strong enough to resist mild thermal or chemical denaturation . However, because the specific three-dimensional conformation is lost, the protein completely loses its biological activity .

Question 19: How are vitamins classified? Name the vitamin responsible for the coagulation of blood.

Answer: Vitamins are classified into two broad categories based on their solubility behavior :
  1. Fat-Soluble Vitamins: Vitamins that dissolve in fats and organic solvents but are insoluble in water . They are stored in the liver and adipose tissues .
    • Examples: Vitamins A, D, E, and K .
  2. Water-Soluble Vitamins: Vitamins that dissolve easily in water . Because they are excreted regularly in urine, they cannot be stored easily in the body and must be consumed daily .
    • Examples: Vitamin C and Vitamin B-complex .
The specific vitamin responsible for blood coagulation (clotting) is Vitamin K .

Question 20: Why are vitamin A and vitamin C essential to us? Give their important sources.

Answer:
  • Vitamin A: Essential for maintaining healthy night vision, cell growth, and robust immune defenses .
    • Deficiency disease: Xerophthalmia and night blindness .
    • Sources: Carrots, fish liver oil, milk, and butter .
  • Vitamin C (Ascorbic Acid): Essential for collagen synthesis, wound healing, bone development, and acting as a powerful antioxidant .
    • Deficiency disease: Scurvy (bleeding gums) .
    • Sources: Citrus fruits (lemons, oranges), amla, and green leafy vegetables .

Question 21: What are nucleic acids? Mention their two important functions.

Answer: Nucleic acids are long-chain biopolymers (macromolecules) made of repeating units called nucleotides . They serve as the chemical repositories of genetic specifications within living cells .

Two Primary Functions:
  1. Heredity: They are responsible for storing and transferring genetic information from parents to offspring over successive generations (primarily handled by DNA) .
  2. Protein Synthesis: They direct and manage the cellular assembly of amino acids into specific proteins needed for cellular function and growth (primarily handled by RNA) .

Question 22: What is the difference between a nucleoside and a nucleotide?

Answer:
  • Nucleoside: A molecular unit formed when a nitrogenous base (purine or pyrimidine) attaches to the $C-1$ position of a pentose sugar molecule . $$\text{Nucleoside} = 1 \text{ Sugar} + 1 \text{ Base}$$
  • Nucleotide: A complete molecular unit formed when the hydroxyl group at the $C-5$ position of the pentose sugar in a nucleoside binds to a phosphoric acid molecule . It is the building block of nucleic acids . $$\text{Nucleotide} = 1 \text{ Sugar} + 1 \text{ Base} + 1 \text{ Phosphoric Acid}$$

Question 23: The two strands in DNA are not identical but are complementary. Explain.

Answer: The two strands of a DNA double helix are held together by specific hydrogen bonds formed between pairs of nitrogenous bases . Due to structural constraints and hydrogen-bonding geometries :
  • Adenine ($A$) always pairs exclusively with Thymine ($T$) through two hydrogen bonds ($A=T$) .
  • Guanine ($G$) always pairs exclusively with Cytosine ($C$) through three hydrogen bonds ($G \equiv C$) .
Because of this obligatory base-pairing rule, if we know the exact linear sequence of bases along one strand, the sequence of the opposite strand is automatically determined . Hence, the strands are not identical copies, but are complementary matchings of each other .

Question 24: Write the important structural and functional differences between DNA and RNA.

Answer: The key distinctions between these two nucleic acids include :
Feature DNA (Deoxyribonucleic Acid) RNA (Ribonucleic Acid)
Pentose Sugar Contains $\beta$-D-2-deoxyribose . Contains $\beta$-D-ribose .
Nitrogenous Bases Contains Adenine, Guanine, Cytosine, and Thymine . Contains Adenine, Guanine, Cytosine, and Uracil .
Physical Geometry Exists as a stable, double-stranded helix . Exists primarily as a single-stranded chain .
Primary Location Found mainly inside the cell nucleus . Found mainly in the cytoplasm .
Core Function Controls genetic inheritance and replication . Directs protein synthesis inside ribosomes .

Question 25: What are the different types of RNA found in the cell?

Answer: Cells utilize three major functional types of RNA to orchestrate protein synthesis :
  1. Messenger RNA (m-RNA): Carries the transcribed genetic code from the DNA inside the nucleus to the ribosomes in the cytoplasm .
  2. Transfer RNA (t-RNA): Acts as an adapter molecule that identifies and carries specific amino acids to the ribosome during protein assembly .
  3. Ribosomal RNA (r-RNA): Forms the structural component of the ribosome and provides the catalytic site where peptide bonds are synthesized .

Extra Important Questions (Board Style 2026)

Test your understanding with these highly anticipated practice questions designed based on current CBSE evaluation trends .

Section A: Multiple Choice Questions (MCQs)

Q1. Which of the following is a non-reducing sugar?

A) Glucose
B) Fructose
C) Sucrose
D) Maltose

Answer: C
Explanation: In sucrose, the reducing functional groups of both glucose ($C-1$) and fructose ($C-2$) are involved in forming the glycosidic linkage, leaving no free aldehyde or ketone group to reduce Tollen's or Fehling's reagents .
Difficulty Level: Easy

Q2. Which base is present in RNA but completely absent in DNA?

A) Uracil
B) Thymine
C) Adenine
D) Cytosine

Answer: A
Explanation: RNA incorporates Uracil ($U$) to pair with Adenine, whereas DNA utilizes Thymine ($T$) instead .
Difficulty Level: Easy

Section B: Assertion-Reason Questions

Directions: Choose the correct answer option:
Option A: Both (A) and (R) are true and (R) is the correct explanation of (A) .
Option B: Both (A) and (R) are true but (R) is NOT the correct explanation of (A) .
Option C: (A) is true but (R) is false .
Option D: (A) is false but (R) is true .

Q3. Assertion (A): Boiling an egg causes the soluble globular protein to harden into an insoluble mass.
Reason (R): High temperatures break the primary peptide bonds, causing coagulation.

Answer: Option C
Explanation: While the assertion is correct, the reason is completely false . High temperatures break the weak hydrogen bonds and secondary/tertiary structures during denaturation, but the covalent primary peptide bonds remain completely unbroken .
Difficulty Level: Medium

Section C: Short Answer Questions

Q4. Define what an anomeric carbon is and give an example using glucose.

Answer: The anomeric carbon is the stereocenter generated at the carbonyl carbon atom ($C-1$ for aldoses like glucose) during the intramolecular cyclization process . It determines whether the cyclic carbohydrate is in the $\alpha$ or $\beta$ configuration based on the orientation of its hydroxyl group .
Difficulty Level: Medium

Q5. What is a peptide bond? Show the chemical configuration of a dipeptide linkage.

Answer: A peptide bond is a covalent amide linkage formed by condensation between the carboxyl group of one amino acid and the amino group of another .
Visual Guide: To draw this, write out $R_1-CH(NH_2)-COOH$ next to $R_2-CH(NH_2)-COOH$, remove a molecule of water ($H_2O$), and link them via a structural $-CO-NH-$ bridge .
Difficulty Level: Medium

Section D: Case-Based Questions

Q6. Read the passage and answer the questions below:
Living cells perform biochemical tasks using large biocatalysts known as enzymes . Unlike traditional chemical catalysts, enzymes show extreme specificity toward their target molecules and operate within precise homeostatic boundaries . If body temperature spikes during a severe fever or tissue pH levels shift sharply, these delicate protein structures can undergo denaturation, immediately rendering the enzymes inactive .

(i) What structural type of protein forms almost all enzymes?
(ii) Why does a severe spike in human body temperature pose a risk to physiological enzyme function?

Answer (i): Globular proteins .
Answer (ii): High temperatures alter the structural conformation of the enzyme's active site via thermal denaturation, preventing substrate molecules from binding correctly .

Section E: Long Answer Question

Q7. (a) Provide three distinct chemical reactions that help confirm the open-chain molecular structure of D-glucose.
(b) Explain the chemical concept behind the "Inversion of Cane Sugar".

Answer (a): Refer to the detailed step-by-step chemical proofs using $HI$, $HNO_3$, and Bromine water detailed in NCERT Exercise Question 9 above .
Answer (b): Sucrose is naturally dextrorotatory (+66.5°) . When hydrolyzed, it forms a mixture of equal parts D-glucose (+52.7°) and D-fructose (-92.4°) . Because the negative levorotation value of fructose is significantly stronger than the positive dextrorotation of glucose, the total resulting solution becomes levorotatory . Since the direction of optical rotation flips ("inverts") from dextro (+) to levo (-), this process is called the inversion of cane sugar .
Difficulty Level: Hard

Common Mistakes Students Make

  • Calling Fructose an Aldose: Many students categorize all reducing sugars as aldoses . Remember that Fructose is a ketose sugar, but it is still a reducing sugar because it easily isomerizes to an aldose under alkaline test conditions .
  • Miscalculating the Unbroken Bonds in Denaturation: In exams, students often incorrectly write that denaturation destroys the entire protein structure . Always emphasize that the primary structure remains completely intact .
  • Confusing Nucleosides with Nucleotides: Remember this memory trick: Nucleoside contains Sugar + Base (2 components), while Nucleotide contains Three components: Sugar + Base + Phosphate .
  • Flipping the base pairs: Do not mismatch DNA bases . Always map $A$ to $T$ and $G$ to $C$ .

Exam Preparation Tips

  • Focus on Chemical Equations: Make sure you can write out the chemical proofs for glucose structures from memory, as these are highly tested board topics .
  • Understand Cyclic Forms: Practice drawing the basic ring representations (Haworth structures) of $\alpha$-D-glucose and $\beta$-D-glucose .
  • Bullet Your Answers: When writing answers for questions about fibrous vs globular proteins or DNA vs RNA, use clean, side-by-side tabular formats to maximize presentation marks .
  • Time Tracking: This chapter is largely conceptual and requires minimal mathematical calculation . Aim to complete questions from this section efficiently to save time for calculation-heavy chapters like Electrochemistry or Chemical Kinetics .

FAQ Section

Q1. Is the Biomolecules chapter important for CBSE Class 12 Chemistry?
Ans: Absolutely! It typically contributes around 7 marks to the theory exam . Because it is direct and theoretical, it offers a great opportunity to pick up quick marks .
Q2. Why is sucrose a non-reducing sugar?
Ans: Sucrose is non-reducing because the reducing functional groups (the aldehyde group on $C-1$ of glucose and the ketone group on $C-2$ of fructose) are bound up in the glycosidic linkage, leaving them unable to react with test reagents .
Q3. What exactly happens during the denaturation of a protein?
Ans: Denaturation causes the secondary and tertiary structures of a protein to unfold due to changes in temperature or pH, leading to a loss of biological activity while leaving the primary structure untouched .
Q4. What is a Zwitterion?
Ans: A Zwitterion is a dipolar ion formed when an amino acid transfers its internal proton from the acidic carboxyl group to the basic amino group, creating a molecule with balanced positive and negative charges .
Q5. Where can I find the latest official NCERT Class 12 Chemistry PDF?
Ans: You can access and download the most up-to-date, revised textbooks for free from the official NCERT portal (ncert.nic.in) .

Conclusion

Mastering Chapter 14: Biomolecules comes down to understanding structural trends, remembering key definitions like Zwitterions and glycosidic links, and reviewing past exam patterns . Make sure to practice writing out the reactions of glucose and summarizing comparisons in tables . Use this guide to streamline your review, work through previous years' questions, and approach your 2026 board exams with clear confidence !

More Class 12 Chemistry Chapters

Ch 1: The Solid State Ch 2: Solutions Ch 3: Electrochemistry Ch 4: Chemical Kinetics Ch 5: Surface Chemistry Ch 6: General Principles Ch 7: The p-Block Elements Ch 8: The d- & f-Block Elements Ch 9: Coordination Compounds Ch 10: Haloalkanes Ch 11: Alcohols & Phenols Ch 12: Aldehydes & Ketones Ch 13: Amines Ch 14: Biomolecules All Subjects Dashboard