Biomolecules

Introduction to Biomolecules

Biomolecules are naturally occurring organic compounds that are essential constituents of living organisms. They include carbohydrates, proteins, nucleic acids, vitamins, and hormones, each playing vital roles in biological processes.

Carbohydrates

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that yield such units upon hydrolysis. They serve as primary energy sources and structural components in organisms.

Proteins

Proteins are complex polyamides formed from amino acids linked by peptide bonds. They are crucial for growth, repair, enzymatic activities, and various biological functions.

Nucleic Acids

Nucleic acids, including DNA and RNA, are polymers of nucleotides responsible for storing and transferring genetic information and protein synthesis.

Vitamins

Vitamins are organic micronutrients required in small quantities for normal metabolism and physiological functions. They are classified based on solubility into fat-soluble and water-soluble vitamins.

Hormones

Hormones are chemical messengers produced by endocrine glands that regulate physiological activities and maintain homeostasis.

Solved Examples

Example 1: Find the amount of O₂ required for complete combustion of 11.2 L of CH₄ at STP.

• Balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O
• At the same temperature and pressure, gas volumes are proportional to mole ratios.
• According to the equation, 1 volume of CH₄ requires 2 volumes of O₂.
• So, 11.2 L of CH₄ requires 22.4 L of O₂ at STP.

Answer: 22.4 L

Example 2: Calculate the percentage by mass of oxygen in H₂SO₄.

• Molar mass of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 98
• Mass of oxygen in H₂SO₄ = 4 × 16 = 64
• Percentage of oxygen = (64/98) × 100
• Percentage of oxygen ≈ 65.3%

Answer: 65.3%

Example 3: How many moles are present in 9 g of water?

• Molar mass of H₂O = 18 g mol⁻¹
• Number of moles = mass/molar mass = 9/18
• Number of moles = 0.5 mol

Answer: 0.5 mol

Practice Set

Answer Key

• 1. The functional group in alcohols is hydroxyl (–OH).
• 2. Ethanol and dimethyl ether are functional isomers because both have the molecular formula C₂H₆O but different functional groups.
• 3. Tertiary alcohols are oxidized least readily because they do not have an α-hydrogen on the carbon bearing the –OH group.
• 4. The IUPAC name of CH₃CH₂CH₂OH is propan-1-ol.
• 5. Phenols are more acidic than alcohols because the phenoxide ion is stabilized by resonance.
• 6. Ethanol is prepared by fermentation of sugars using yeast at about 303 K.
• 7. The Lucas test distinguishes primary, secondary and tertiary alcohols by the rate of formation of insoluble alkyl chloride.
• 8. Dehydration of ethanol with concentrated H₂SO₄ at 443 K gives ethene.

Carbohydrates Classification and Importance

Definition and Classification

Carbohydrates are optically active polyhydroxy aldehydes or ketones. They are classified based on molecular size into monosaccharides, oligosaccharides, and polysaccharides.

Monosaccharides

Monosaccharides are simple sugars that cannot be hydrolyzed further. They are soluble in water and classified as aldoses (with aldehyde group) or ketoses (with ketone group). Examples include glucose (aldohexose) and fructose (ketohexose).

Oligosaccharides

Oligosaccharides yield two to ten monosaccharide units upon hydrolysis. Disaccharides, such as sucrose, lactose, and maltose, are common examples.

Polysaccharides

Polysaccharides are complex carbohydrates yielding many monosaccharide units on hydrolysis. Examples include starch, cellulose, and glycogen. They are generally non-sweet and serve as storage or structural materials.

Reducing and Non-Reducing Sugars

Reducing sugars contain free aldehyde or ketone groups and can reduce Fehling's or Tollen's reagents (e.g., glucose, maltose). Non-reducing sugars lack free aldehyde or ketone groups and do not reduce these reagents (e.g., sucrose).

Importance of Carbohydrates

• Primary energy source for the body.
• Fibre aids digestion and regulates cholesterol.
• Structural components in plants and bacteria.
• Storage molecules as starch in plants and glycogen in animals.

Solved Examples

No solved examples are provided in this section.

Practice Set

• 1. A 5.0 g sample of CaCO₃ is heated strongly. Calculate the mass of CaO formed and the volume of CO₂ liberated at STP. Solution: CaCO₃ → CaO + CO₂. Molar mass of CaCO₃ = 100 g mol⁻¹, CaO = 56 g mol⁻¹, CO₂ = 44 g mol⁻¹. 100 g CaCO₃ gives 56 g CaO and 22.4 L CO₂ at STP. Therefore, 5.0 g CaCO₃ gives 2.8 g CaO and 1.12 L CO₂.
• 2. Write the balanced chemical equations for the following reactions: (a) baking soda on heating, (b) lime stone on heating, (c) thermal decomposition of potassium chlorate in the presence of MnO₂. Solution: (a) 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂, (b) CaCO₃ → CaO + CO₂, (c) 2KClO₃ → 2KCl + 3O₂.
• 3. Define thermal decomposition. Give one example each of a thermal decomposition reaction of a carbonate, a hydrogen carbonate and a chlorate. Solution: Thermal decomposition is the breakdown of a compound into simpler substances on heating. Examples are CaCO₃ → CaO + CO₂, 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂ and 2KClO₃ → 2KCl + 3O₂.
• 4. Explain why many metal carbonates and metal hydrogen carbonates decompose on heating, and state one use of this property. Solution: On heating, these compounds break down to give a more stable oxide, carbon dioxide and/or water. This property is used in the extraction of quicklime from limestone and in the decomposition of baking soda during cooking.
• 5. If 0.2 mol of potassium chlorate is completely decomposed, calculate the number of moles of oxygen produced. Solution: From 2KClO₃ → 2KCl + 3O₂, 2 mol KClO₃ give 3 mol O₂. Therefore, 0.2 mol KClO₃ gives 0.3 mol O₂.

Answer Key

• 1. The key factor affecting the rate of a reaction is the concentration of the reactants, because a higher concentration increases the number of effective collisions per unit time.
• 2. Rate of reaction can be expressed as the change in concentration of a reactant or product per unit time.
• 3. A catalyst increases the rate of reaction by lowering the activation energy and is not consumed in the reaction.
• 4. For a zero-order reaction, the rate is independent of concentration and the plot of concentration versus time is a straight line.
• 5. For a first-order reaction, the rate is directly proportional to concentration, and the half-life is constant.
• 6. The unit of rate constant depends on the order of the reaction.

Glucose and Fructose Structure and Reactions

Glucose Structure

Glucose is a six-carbon aldose sugar with an aldehyde group and multiple hydroxyl groups. It exists in open-chain and cyclic forms (α and β anomers) that interconvert via mutarotation. In aqueous solution, glucose mainly exists in cyclic hemiacetal forms, with the pyranose ring being the major form.

Reactions of Glucose

• Reaction with HI converts glucose to n-hexane, indicating a straight chain of six carbons.
• Reactions with hydroxylamine, HCN, and bromine water confirm the presence of the aldehyde group.
• Reaction with acetic anhydride forms glucose pentaacetate, showing five hydroxyl groups.
• Oxidation with nitric acid forms saccharic acid, indicating primary alcohol groups.

Fructose Structure

Fructose is a ketohexose sugar obtained from sucrose hydrolysis. It exists in open-chain and cyclic furanose forms with α and β anomers differing in hydroxyl group orientation. In aqueous solution, fructose mainly forms a five-membered hemiketal ring, though a small amount of the open-chain form is also present.

Solved Examples

• Example 1: Convert 5 g of O₂ to moles. Solution: Molar mass of O₂ = 32 g mol^-1. Number of moles = mass/molar mass = 5/32 = 0.15625 mol ≈ 0.156 mol.
• Example 2: Find the mass of 0.25 mol of NaCl. Solution: Molar mass of NaCl = 58.5 g mol^-1. Mass = moles × molar mass = 0.25 × 58.5 = 14.625 g ≈ 14.6 g.
• Example 3: How many molecules are present in 0.5 mol of H₂O? Solution: Number of molecules = moles × Avogadro constant = 0.5 × 6.02 × 10²³ = 3.01 × 10²³ molecules.
• Example 4: Calculate the number of moles in 11 g of CO₂. Solution: Molar mass of CO₂ = 44 g mol^-1. Number of moles = 11/44 = 0.25 mol.
• Example 5: Find the volume occupied by 2 mol of a gas at STP. Solution: At STP, 1 mol of any gas occupies 22.4 L. Therefore, volume = 2 × 22.4 = 44.8 L.

Practice Set

• Q1. State the IUPAC name of the coordination compound [Co(NH₃)₆]Cl₃.
• Q2. Write the oxidation state of cobalt in [Co(NH₃)₆]Cl₃.
• Q3. How many ions are produced by one formula unit of [Co(NH₃)₆]Cl₃ in water?
• Q4. Name the ligand present in [Co(NH₃)₆]Cl₃ and state whether it is monodentate or polydentate.
• Q5. Identify the coordination entity and the counter ions in [Co(NH₃)₆]Cl₃.
• Q6. Write the coordination number of cobalt in [Co(NH₃)₆]Cl₃.
• Q7. Give the IUPAC name of the coordination compound K₄[Fe(CN)₆].
• Q8. Determine the oxidation state of iron in K₄[Fe(CN)₆].
• Q9. State the coordination number of iron in K₄[Fe(CN)₆].
• Q10. Name the ligand and mention its charge in K₄[Fe(CN)₆].

Answer Key

• 1. Water soluble salts are those salts which dissolve in water to give ions in solution.
• 2. A salt is formed by the neutralization reaction between an acid and a base.
• 3. Common examples of salts include sodium chloride (NaCl), potassium nitrate (KNO₃) and copper sulfate (CuSO₄).
• 4. A neutral salt is formed when the strong acid and strong base both get completely neutralized.
• 5. Acidic salts and basic salts can also be formed depending on whether the parent acid or base is only partially neutralized.

Polysaccharides: Starch, Cellulose, Glycogen

Starch

Starch is the main plant storage polysaccharide composed of amylose (linear chains of α-D-glucose linked by C1–C4 glycosidic bonds) and amylopectin (branched chains with C1–C4 and C1–C6 linkages).

Cellulose

Cellulose is a structural polysaccharide in plants, consisting of β-D-glucose units linked by C1–C4 glycosidic bonds forming straight chains, providing rigidity to cell walls.

Glycogen

Glycogen is the animal storage polysaccharide found in liver and muscles, similar to amylopectin but more highly branched, serving as a glucose reserve.

Starch

Starch is the main plant storage polysaccharide composed of amylose (linear chains of α-D-glucose linked by C1–C4 glycosidic bonds) and amylopectin (branched chains with C1–C4 and C1–C6 linkages).

Cellulose

Cellulose is a structural polysaccharide in plants, consisting of β-D-glucose units linked by C1–C4 glycosidic bonds forming straight chains, providing rigidity to cell walls.

Glycogen

Glycogen is the animal storage polysaccharide found in liver and muscles, similar to amylopectin but more highly branched, serving as a glucose reserve.

Solved Examples

• Example 1: A hydrocarbon contains 85.7% carbon and 14.3% hydrogen. Find its empirical formula.
  Solution: Assume 100 g of compound. Moles of C = 85.7/12 = 7.14, moles of H = 14.3/1 = 14.3. Divide by the smaller value: C : H = 7.14 : 14.3 = 1 : 2. So the empirical formula is CH₂.
• Example 2: The molecular mass of a compound is 56 and its empirical formula is CH₂. Find the molecular formula.
  Solution: Empirical formula mass of CH₂ = 12 + 2 = 14. Multiplier n = 56/14 = 4. Molecular formula = (CH₂)₄ = C₄H₈.
• Example 3: A compound has the percentage composition: C = 40%, H = 6.67%, O = 53.33%. Determine the empirical formula.
  Solution: Assume 100 g compound. Moles of C = 40/12 = 3.33, H = 6.67/1 = 6.67, O = 53.33/16 = 3.33. Divide by 3.33: C : H : O = 1 : 2 : 1. Therefore, the empirical formula is CH₂O.
• Example 4: The empirical formula of a compound is NO₂ and its molecular mass is 92. Find the molecular formula.
  Solution: Empirical formula mass of NO₂ = 14 + 32 = 46. Multiplier n = 92/46 = 2. Molecular formula = (NO₂)₂ = N₂O₄.

Practice Set

• 1. A sample of gas occupies 2.0 L at 1 atm pressure. What will be its volume at 2 atm pressure at constant temperature?
• 2. State Boyle’s law and write its mathematical expression.
• 3. A gas has a volume of 500 mL at 27 °C. What will be its volume at 127 °C at constant pressure?
• 4. State Charles’s law.
• 5. A gas occupies 4.0 L at 300 K and 1 atm. What will be its volume at 600 K if pressure remains constant?
• 6. Define absolute zero. Why is it considered the zero point of the Kelvin scale?
• 7. Explain why temperature must be taken in Kelvin while applying gas laws.
• 8. A gas occupies 10 L at 1 atm and 273 K. What will be its volume at 2 atm and 546 K?

Answer Key

• No practice questions are included in this section. An answer key is only needed when a matching set of questions appears in the same topic.
• If this heading is meant to support omitted questions, the corresponding questions should be restored first so that each answer can be matched properly.

Proteins, Nucleic Acids, Vitamins and Hormones

Proteins

Proteins are polymers of amino acids linked by peptide bonds. They have primary, secondary (α-helix and β-pleated sheet), tertiary, and quaternary structures. Proteins perform structural, enzymatic, transport, motor, hormonal, and storage functions.

Denaturation of Proteins

Denaturation involves the disruption of hydrogen bonds due to changes in temperature or pH, leading to loss of protein structure and biological activity, e.g., coagulation of egg white.

Enzymes

Enzymes are protein biocatalysts that accelerate specific biochemical reactions. Examples include amylase, maltase, lipase, and trypsin.

Nucleic Acids

Nucleic acids are polymers of nucleotides composed of a pentose sugar, phosphate group, and nitrogenous base. DNA contains deoxyribose and thymine, forming a double helix, while RNA contains ribose and uracil, usually single-stranded.

Vitamins

Vitamins are essential organic micronutrients classified as fat-soluble (A, D, E, K) and water-soluble (B-complex, C). They perform specific biological functions and must be obtained from diet or synthesized by gut bacteria.

Hormones

Hormones are intercellular messengers produced by endocrine glands, including steroids (e.g., testosterone), polypeptides (e.g., insulin), and amino acid derivatives (e.g., epinephrine). They regulate physiological processes and maintain homeostasis.

Solved Examples

1. Example: Calculate the pH of a 0.01 M HCl solution.

Solution: HCl is a strong acid and dissociates completely in water.

So, [H⁺] = 0.01 M = 10^-2 M

pH = -log[H⁺] = -log(10^-2) = 2

Answer: pH = 2

2. Example: Find the pOH of a solution having [OH^-] = 10^-3 M.

Solution: pOH = -log[OH^-] = -log(10^-3) = 3

Answer: pOH = 3

3. Example: A solution has pH = 5. Find [H⁺].

Solution: pH = -log[H⁺]

5 = -log[H⁺]

[H⁺] = 10^-5 M

Answer: [H⁺] = 10^-5 M

4. Example: If the pOH of a solution is 4, calculate its pH.

Solution: For aqueous solutions, pH + pOH = 14

pH = 14 - 4 = 10

Answer: pH = 10

Practice Set

• 1. What is the IUPAC name of the compound CH₃CH₂OH?
  Answer: Ethanol.
• 2. What is the common name of methanal?
  Answer: Formaldehyde.
• 3. Write the molecular formula of ethanoic acid.
  Answer: CH₃COOH.
• 4. Which functional group is present in aldehydes?
  Answer: The aldehydic or formyl group, —CHO.
• 5. What is the general formula of alkanes?
  Answer: C_nH_2n+2.
• 6. Name the compound having formula CH₃COCH₃.
  Answer: Propanone.
• 7. What is the oxidation state of carbon in CO₂?
  Answer: +4.
• 8. State one use of ethanol.
  Answer: It is used as a solvent and as a fuel.

Answer Key

• 1. The cathode is the electrode where reduction occurs.
• 2. The anode is the electrode where oxidation occurs.
• 3. In a galvanic cell, electrons flow from anode to cathode through the external circuit.
• 4. The salt bridge maintains electrical neutrality by allowing ion migration between the two half-cells.
• 5. The standard electrode potential of the standard hydrogen electrode is taken as 0 V.
• 6. For a spontaneous cell reaction, E°cell is positive.
• 7. Cell notation is written as anode | anode solution || cathode solution | cathode.
• 8. In the Daniell cell, zinc acts as the anode and copper acts as the cathode.

Quick Reference Table

• Biomolecules: Naturally occurring organic compounds essential to living organisms, including carbohydrates, proteins, nucleic acids, vitamins and hormones.
• Carbohydrates: Optically active polyhydroxy aldehydes or ketones, or compounds that yield such units on hydrolysis; act as major energy sources and structural materials.
• Proteins: Polyamides formed from amino acids linked by peptide bonds; important for growth, repair, enzymes and other biological functions.
• Nucleic acids: Polymers of nucleotides that store and transfer genetic information; DNA and RNA are the two main types.
• Vitamins: Organic micronutrients needed in small amounts for normal metabolism and physiological functions; classified as fat-soluble or water-soluble.
• Hormones: Chemical messengers produced by endocrine glands that regulate body activities and maintain homeostasis.
• Monosaccharides: Simple sugars that cannot be hydrolyzed further; examples include glucose and fructose.
• Oligosaccharides: Carbohydrates that yield two to ten monosaccharide units on hydrolysis; disaccharides such as sucrose and lactose are common examples.
• Polysaccharides: Complex carbohydrates that yield many monosaccharide units on hydrolysis; examples include starch, cellulose and glycogen.
• Reducing sugars: Sugars with a free aldehyde or ketone group that can reduce Fehling's or Tollen's reagents.
• Glucose: An aldohexose that shows mutarotation and gives reactions such as formation of glucose pentaacetate and saccharic acid.
• Fructose: A ketohexose obtained from sucrose hydrolysis and existing in cyclic furanose forms.
• Starch: Plant storage polysaccharide made of amylose and amylopectin.
• Cellulose: Structural polysaccharide of plants made of β-D-glucose units linked by C1–C4 bonds.
• Glycogen: Highly branched animal storage polysaccharide found in liver and muscles.
• Denaturation of proteins: Loss of native structure and biological activity due to heat or pH change.
• Enzymes: Protein biocatalysts that accelerate specific biochemical reactions.

Common Mistakes and Misconceptions

• Confusing carbohydrates with only sugars, instead of remembering that polysaccharides like starch, cellulose, and glycogen are also carbohydrates.
• Thinking all carbohydrates are sweet; polysaccharides are generally non-sweet.
• Mixing up monosaccharides, oligosaccharides, and polysaccharides by number of sugar units, especially disaccharides under oligosaccharides.
• Assuming every carbohydrate is a reducing sugar; sucrose is a common non-reducing sugar.
• Forgetting that glucose is an aldose and fructose is a ketose, even though both have the same molecular formula.
• Confusing the open-chain and cyclic forms of glucose and fructose, especially the α and β anomers.
• Misidentifying starch as a single uniform polymer and ignoring its two components, amylose and amylopectin.
• Thinking cellulose is a storage polysaccharide; it is actually a structural polysaccharide in plant cell walls.
• Confusing glycogen with starch; glycogen is the storage polysaccharide in animals and is more highly branched.
• Believing denaturation breaks peptide bonds; it mainly disrupts the higher-order structure of proteins and causes loss of biological activity.

Glossary

• Biomolecules: Naturally occurring organic compounds essential for living organisms.
• Carbohydrates: Optically active polyhydroxy aldehydes or ketones, or compounds that yield such units on hydrolysis.
• Monosaccharides: Simple sugars that cannot be hydrolyzed further, such as glucose and fructose.
• Disaccharides: Oligosaccharides that yield two monosaccharide units on hydrolysis, such as sucrose, lactose and maltose.
• Polysaccharides: Complex carbohydrates that yield many monosaccharide units on hydrolysis, such as starch, cellulose and glycogen.
• Reducing sugar: A sugar that contains a free aldehyde or ketone group and reduces Fehling's or Tollen's reagents.
• Non-reducing sugar: A sugar that does not reduce Fehling's or Tollen's reagents because it lacks a free aldehyde or ketone group.
• Protein: A complex polyamide formed from amino acids linked by peptide bonds.
• Peptide bond: The linkage formed between amino acids in proteins.
• Denaturation: The loss of native protein structure and biological activity due to heat, pH change or other factors.
• Enzyme: A protein biocatalyst that speeds up a specific biochemical reaction.
• Nucleic acid: A polymer of nucleotides responsible for storage and transfer of genetic information.
• Nucleotide: The basic unit of nucleic acids, made of a pentose sugar, phosphate group and nitrogenous base.
• DNA: Deoxyribonucleic acid; the genetic material in most organisms, usually double-stranded.
• RNA: Ribonucleic acid; usually single-stranded and involved in protein synthesis.
• Vitamin: An organic micronutrient required in small amounts for normal metabolism and physiological functions.
• Hormone: A chemical messenger produced by endocrine glands that regulates physiological activities.