Answers are written in a clear, exam-focused style. Use the bullet points to pick mark-earning phrases during tests.

1. Question 1: Explain why carbon forms a large number of compounds. Mention the structural features of carbon responsible for this behaviour.

   Answer:

   • Tetravalency: Carbon has four valence electrons and can form four covalent bonds allowing complex structures.
   • Catenation: Carbon atoms form strong C–C bonds enabling the formation of chains and rings of varying lengths.
   • Small atomic size: Leads to strong overlapping of orbitals and stable covalent bonds.
   • Multiple bonding: Carbon forms single, double and triple bonds, increasing diversity (e.g., alkanes, alkenes, alkynes).
   • Variety of functional groups: Combining with different atoms (O, N, halogens) creates diverse chemical behaviour.
2. Question 2: Define homologous series. Describe two characteristic features of a homologous series and give an example.

   Answer:

   • Definition: A homologous series is a group of organic compounds having the same functional group and similar chemical properties, with successive members differing by a —CH₂— unit.
   • Feature 1 — Same functional group: Members display similar chemical reactions (e.g., alkanes undergo substitution).
   • Feature 2 — Gradation in physical properties: Physical properties like boiling point change gradually with molecular mass.
   • Example: Alkane series — methane (CH₄), ethane (C₂H₆), propane (C₃H₈), etc. (general formula CₙH₂ₙ₊₂).
3. Question 3: Explain isomerism with an example. How does isomerism affect physical properties?

   Answer:

   • Isomerism: Occurs when compounds have same molecular formula but different arrangements of atoms.
   • Example: C₄H₁₀ has two structural isomers — n‑butane (straight chain) and isobutane (branched).
   • Effect on physical properties: Isomers differ in boiling points, melting points and densities — branching usually lowers boiling point by reducing surface area and intermolecular forces.
   • Note: Chemical properties may be similar if functional groups are same, but reactivity can vary due to steric effects.
4. Question 4: Describe the general properties of alcohols and explain why their boiling points are higher than those of corresponding hydrocarbons.

   Answer:

   • General properties: Contain —OH group; may be liquids at room temperature; show hydrogen bonding; soluble in water (short chains).
   • Boiling point: Alcohols form intermolecular hydrogen bonds which are stronger than the van der Waals forces in hydrocarbons; more energy is required to break these bonds, resulting in higher boiling points.
   • Example: Ethanol (bp 78°C) vs ethane (bp −89°C).
5. Question 5: Explain the formation of ethyl ethanoate (an ester) from ethanol and ethanoic acid. Include the type of catalyst used and why the reaction is reversible.

   Answer:

   • Reaction: Ethanol + Ethanoic acid ⇌ Ethyl ethanoate + Water.
   • Catalyst: Concentrated sulfuric acid (H₂SO₄) acts as an acid catalyst and a dehydrating agent.
   • Mechanism overview: Protonation of acid increases electrophilicity; nucleophilic attack by alcohol on carbonyl carbon, followed by elimination of water to form ester.
   • Reversibility: The reaction is an equilibrium; removal of water (or using excess alcohol) shifts equilibrium toward ester formation (Le Chatelier's principle).
6. Question 6: What is saponification? Write a balanced equation for the saponification of a simple triglyceride and explain products formed.

   Answer:

   • Definition: Saponification is the alkaline hydrolysis of esters (fats/oils) to produce glycerol and soap (salts of fatty acids).
   • Equation (simplified): Triglyceride + 3 NaOH → Glycerol + 3 RCOONa (soap).
   • Products: Glycerol (a triol) and sodium salts of fatty acids which act as soap molecules (hydrophilic head, hydrophobic tail).
   • Significance: Soap molecules help emulsify oily dirt and allow it to be washed away with water.
7. Question 7: Describe the difference between soaps and detergents. Give two advantages of detergents over soaps.

   Answer:

   • Soaps: Sodium or potassium salts of fatty acids; prepared by saponification of fats with alkali.
   • Detergents: Synthetic surfactants (often sulphonates) designed for cleaning; not necessarily derived from fats.
   • Advantages of detergents:
     • Work well in hard water — do not form insoluble scum with Ca²⁺/Mg²⁺ ions.
     • Can be formulated for specific purposes (heavy‑duty washing, dishwashing, industrial cleaning).
8. Question 8: Discuss the oxidation of primary alcohols. Include examples and products formed under mild and strong oxidation conditions.

   Answer:

   • Primary alcohol oxidation: Primary alcohols (R–CH₂–OH) are oxidised first to aldehydes (R–CHO) and further to carboxylic acids (R–COOH) under stronger or prolonged oxidation.
   • Example (ethanol):
     • Ethanal (acetaldehyde) formation (mild, controlled oxidation): CH₃CH₂OH → CH₃CHO + H₂.
     • Ethanoic acid formation (strong oxidation): CH₃CH₂OH → CH₃COOH.
   • Common oxidising agents: Acidified potassium dichromate (K₂Cr₂O₇/H₂SO₄), potassium permanganate (KMnO₄).
9. Question 9: Write a detailed answer explaining why ethanol is miscible with water but hexane is not. Include intermolecular forces in your explanation.

   Answer:

   • Ethanol (CH₃CH₂OH): Contains a polar —OH group capable of hydrogen bonding with water molecules; the small non‑polar hydrocarbon part is not large enough to prevent solubility, hence ethanol is miscible with water.
   • Hexane (C₆H₁₄): Non‑polar hydrocarbon with only London dispersion forces; it cannot form hydrogen bonds with water and therefore is immiscible (separates into a distinct layer).
   • Conclusion: Like dissolves like — polar solvents dissolve polar solutes via dipole–dipole or hydrogen bonding; non‑polar solvents dissolve non‑polar solutes via dispersion forces.
10. Question 10: Explain the terms 'saturated' and 'unsaturated' hydrocarbons with examples and highlight one chemical test to distinguish them.

    Answer:

    • Saturated hydrocarbons: Contain only single C–C bonds (alkanes); general formula CₙH₂ₙ₊₂ (e.g., methane, ethane). They are less reactive towards addition reactions.
    • Unsaturated hydrocarbons: Contain one or more double or triple bonds (alkenes, alkynes) and can undergo addition reactions (e.g., ethene, acetylene).
    • Chemical test: Bromine water test — unsaturated compounds decolourise bromine water due to addition across multiple bonds; saturated hydrocarbons do not react under mild conditions.
11. Question 11: Give an account of the importance and uses of esters, with two practical examples from daily life.

    Answer:

    • Importance: Esters are widely used because of pleasant fragrances and volatility; they are also used as solvents and plasticizers.
    • Daily-life examples:
      • Ethyl acetate used as a solvent in nail polish removers and glues.
      • Isoamyl acetate used to impart banana flavour in confectionery and as a fragrance in perfumes.
    • Note: Esters are often prepared by esterification reactions in industries and laboratories.
12. Question 12: Describe an experimental method to prepare ethanoic acid from ethanol in the laboratory. Mention reagents, conditions and the principle involved.

    Answer (brief experimental outline):

    • Principle: Oxidation of primary alcohol (ethanol) to carboxylic acid (ethanoic acid).
    • Reagents: Acidified potassium dichromate (K₂Cr₂O₇ + H₂SO₄) or potassium permanganate (KMnO₄).
    • Conditions: Refluxing the alcohol with oxidising agent ensures complete oxidation to ethanoic acid.
    • Observation: Colour change depending on oxidant (e.g., orange dichromate → green chromium ions) and formation of carboxylic acid (characteristic smell).
    • Workup: Distillation and purification of the acid may be carried out if needed.
13. Question 13: Explain why functional groups are important in organic chemistry and list three functional groups discussed in this chapter with one characteristic reaction each.

    Answer:

    • Importance: Functional groups determine chemical reactivity and properties of organic molecules; they allow classification and predictability of reactions.
    • Three functional groups & reactions:
      • —OH (alcohol): Undergoes oxidation (primary alcohol → aldehyde → acid).
      • —COOH (carboxylic acid): Reacts with alcohols to form esters (esterification).
      • —COOR (ester): Undergoes hydrolysis (saponification) to give alcohol and carboxylate salt.
14. Question 14: Write balanced chemical equations for the following and provide short explanations: (a) Combustion of methane, (b) Reaction of ethanol with sodium.

    Answer:

    • (a) Combustion of methane (complete):
      • CH₄ + 2O₂ → CO₂ + 2H₂O
      • Explanation: In excess oxygen, carbon oxidises to CO₂ and hydrogen to H₂O releasing energy.
    • (b) Reaction of ethanol with sodium:
      • 2CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂↑
      • Explanation: Ethanol reacts with active metals like sodium to form alkoxide salts and liberate hydrogen gas.
15. Question 15: Discuss the environmental and health hazards of incomplete combustion of carbon compounds. Include preventive measures.

    Answer:

    • Hazards:
      • Carbon monoxide (CO) is produced — a colourless, odourless toxic gas that binds to haemoglobin reducing oxygen transport, causing asphyxiation.
      • Soot and particulate matter cause respiratory problems and contribute to air pollution and climate effects.
    • Prevention:
      • Ensure adequate oxygen supply and proper ventilation during combustion.
      • Use catalytic converters in vehicles and cleaner fuels to reduce incomplete combustion products.
16. Question 16: Explain the role of concentrated sulfuric acid in esterification reactions. Why is it described as both catalyst and dehydrating agent?

    Answer:

    • Catalytic role: H₂SO₄ protonates the carbonyl oxygen of the carboxylic acid, increasing electrophilicity of the carbonyl carbon and facilitating nucleophilic attack by the alcohol.
    • Dehydrating role: H₂SO₄ helps remove water produced in the reaction, shifting the equilibrium toward ester formation (Le Chatelier's principle).
    • Practical note: Using excess alcohol or removing water by distillation also favours ester formation.
17. Question 17: Describe a simple laboratory test to distinguish between an alcohol and a carboxylic acid.

    Answer:

    • Test with sodium carbonate (Na₂CO₃):
      • Add dilute sodium carbonate solution to the unknown liquid.
      • If effervescence (CO₂ bubbles) occurs, the compound is likely a carboxylic acid (acid reacts with carbonate to produce CO₂).
      • No effervescence indicates it is not a carboxylic acid (could be an alcohol).
    • Safety note: Conduct tests in small amounts and with appropriate safety precautions.
18. Question 18: How would you prepare soap in the laboratory? Provide a stepwise outline and precautions.

    Answer (outline):

    • Step 1: Heat a mixture of vegetable oil (triglyceride) and ethanolic NaOH solution under reflux.
    • Step 2: Maintain gentle boiling until saponification is complete (glycerol separates).
    • Step 3: Cool the mixture and add salt (NaCl) to precipitate soap which can be filtered and washed.
    • Precautions: Use gloves and goggles; NaOH is caustic; work in well‑ventilated area; avoid skin contact and inhalation.
19. Question 19: Explain with reactions how an aldehyde differs from a ketone in terms of oxidation and typical reactions.

    Answer:

    • Structural difference: Aldehyde has —CHO group at end of chain; ketone has carbonyl inside chain (R–CO–R').
    • Oxidation behaviour: Aldehydes oxidise easily to carboxylic acids (e.g., R–CHO → R–COOH) while ketones resist oxidation under mild conditions.
    • Example reaction: Oxidation of ethanal: CH₃CHO + [O] → CH₃COOH.
    • Other reactions: Both can undergo nucleophilic addition; aldehydes give positive Tollens' or Fehling's tests (silver mirror or red precipitate) whereas ketones do not (except alpha‑hydroxy ketones).
20. Question 20: Give a structured answer on how boiling points of organic compounds depend on molecular mass, branching and intermolecular forces. Use examples where appropriate.

    Answer:

    • Molecular mass: Boiling point generally increases with molecular mass due to stronger London dispersion forces (e.g., pentane bp > methane bp).
    • Branching: More branching reduces surface area, decreases van der Waals interactions and thus lowers boiling point (isobutane bp < n‑butane bp).
    • Intermolecular forces:
      • Hydrogen bonding (—OH, —COOH) greatly raises boiling point (e.g., ethanol vs ethane).
      • Dipole–dipole interactions increase bp compared to non‑polar molecules of similar mass.
21. Question 21: Explain the mechanism and industrial importance of the oxidation of ethanol to ethanoic acid.

    Answer:

    • Mechanism (overview): Ethanol undergoes dehydrogenation to ethanal (acetaldehyde) which is further oxidised to ethanoic acid in presence of oxidants or catalysts.
    • Industrial importance: Ethanoic acid (acetic acid) is a key chemical used to produce vinyl acetate monomer, esters, acetic anhydride and in food industry (vinegar).
    • Process note: Controlled catalytic oxidation (e.g., using air over catalysts) is used industrially for high yield.
22. Question 22: Discuss the role of soaps in cleaning action. Use bullet points to explain the mechanism of micelle formation and dirt removal.

    Answer:

    • Soap molecule: Has a hydrophilic (ionic) head and hydrophobic (hydrocarbon) tail.
    • Micelle formation:
      • In water, soap molecules arrange with hydrophobic tails inward and hydrophilic heads outward forming micelles.
      • Oil and grease (non‑polar dirt) dissolve in the micelle core.
    • Dirt removal: Agitation and water wash away micelles containing entrapped oil, effectively cleaning the surface.
    • Limitation: In hard water, soap reacts with Ca²⁺/Mg²⁺ to form scum reducing cleaning efficiency.
23. Question 23: Provide a comprehensive answer on how naming of simple organic compounds (IUPAC) is done — include steps and a short example.

    Answer (steps):

    • Step 1: Identify the longest continuous carbon chain — this gives the parent name (meth-, eth-, prop-, but- etc.).
    • Step 2: Number the chain from the end nearest a substituent or functional group to give the lowest possible locants.
    • Step 3: Name substituents (alkyl groups) and list them alphabetically with their position numbers.
    • Step 4: Indicate functional groups using appropriate suffixes (e.g., -ol for alcohols, -oic acid for carboxylic acids) and include locant if required.
    • Example: CH₃–CH(CH₃)–CH₂–OH → Longest chain = 3 carbons → propan‑1‑ol (numbering gives OH at carbon 1; methyl substituent at C‑2 → 2‑methylpropan‑1‑ol).
24. Question 24: Explain how you would carry out a qualitative test to distinguish between an aldehyde and a ketone using Tollens' reagent. State observations.

    Answer:

    • Procedure: Add Tollens' reagent (ammoniacal silver nitrate) to the unknown compound and gently warm the mixture.
    • Observation: Aldehydes reduce Ag⁺ to metallic silver forming a silver mirror on the inner walls of the test tube; ketones generally do not give this test.
    • Conclusion: Presence of silver mirror indicates aldehyde; absence suggests ketone (or other non‑reducing compound).
    • Safety: Prepare and handle reagents carefully; dispose of silver wastes responsibly.
25. Question 25: Summarise the chapter in five key points that a student must remember for board exams.

    Answer:

    • Tetravalency & catenation: Core reasons for carbon's extensive chemistry.
    • Functional groups: Recognise —OH, —COOH, —COOR and their characteristic reactions.
    • Important reactions: Esterification (acid + alcohol → ester), saponification (ester + NaOH → salt + alcohol), oxidation of alcohols.
    • Physical properties: Understand H‑bonding, solubility patterns (polar vs non‑polar), and boiling point trends.
    • Practical/Applications: Soaps vs detergents, uses of esters and industrial importance of ethanoic acid.