Solved Examples

Comprehensive Solved Examples for Chapter 4

30+ board-focused examples covering all topics — from covalent bonding to soaps.

Example 1: NCERT — Why Carbon Forms Covalent Bonds

Why does carbon form covalent bonds rather than ionic bonds?

Solution:

Carbon's electronic configuration: 2, 4 (4 valence electrons).

Three options for octet:

Option 1: Lose 4 electrons → C⁴⁺

• Requires very high energy.
• Too unstable.

Option 2: Gain 4 electrons → C⁴⁻

• Carbon nucleus (only 6 protons) cannot hold 10 electrons stably.
• Highly unstable.

Option 3: Share electrons (Covalent bonding)

• Both atoms achieve octet without transfer.
• Energetically favourable.
• Stable.

Hence carbon prefers covalent bonding — sharing 4 electrons with other atoms.

Examples:

• $CH_4$: 4 C-H single bonds.
• $CO_2$: 2 C=O double bonds.
• $C_2H_4$: 1 C=C + 4 C-H.

[NCERT — fundamental]

Example 2: NCERT — Catenation and Tetravalency

Define catenation. Why is carbon the king of catenation?

Solution:

Catenation

Property of an element to form long chains by bonding with its own atoms. From Latin: 'catena' = chain.

Why Carbon is the King

1. Strong C-C bonds:

• C-C bond energy: 348 kJ/mol (very high).
• Stable in long chains.

2. Small atomic size:

• Atomic radius: ~77 pm.
• Strong nucleus pull on bonding electrons.
• Stable bonds.

3. Tetravalency:

• C has 4 valence electrons → forms 4 bonds.
• Allows linear chains, branched chains, rings.

Examples

Methane (1 C), ethane (2 C), ethane (3 C), butane (4 C), …, long polymers (thousands of C atoms).

Compared to Other Elements

Si: bond too weak (222 kJ/mol). Chains break after 7-8 atoms. N, O: lone pairs cause repulsion. Limited catenation.

Carbon — uniquely versatile in catenation.

[NCERT — important]

Example 3: NCERT — Allotropes of Carbon

Compare diamond and graphite — structure, properties, uses.

Solution:

Comparison Table

Property	Diamond	Graphite
Bonds per C	4	3
Free electrons	0	1 per C
Structure	3D tetrahedral	2D hexagonal layers
Hardness	Hardest substance	Soft, slippery
Conductivity	None (insulator)	Good conductor
Density	3.5 g/cm³	2.2 g/cm³
Appearance	Transparent, sparkling	Black, opaque

Reason for Difference

Both are pure carbon — same atoms, but different arrangement.

Diamond: Each C bonded to 4 others in 3D rigid network. All electrons in bonds → no free electrons → no conduction; rigid → hardest.

Graphite: Each C bonded to only 3 others in 2D layers. 4th electron is free → conducts. Layers held by weak forces → slippery, soft.

Uses

Diamond: jewellery, cutting tools, drilling.

Graphite: pencils, lubricants, electrodes, brushes.

A beautiful example of how structure determines function.

[NCERT — every year]

Example 4: NCERT — Saturated and Unsaturated

State differences between saturated and unsaturated hydrocarbons. Give examples.

Solution:

Saturated Hydrocarbons

All bonds are single (C-C and C-H). 'Saturated' = filled with H atoms; can't accept more. Series: Alkanes. General formula: $C_nH_{2n+2}$. Examples: methane (CH₄), ethane (C₂H₆), propane (C₃H₈).

Unsaturated Hydrocarbons

Have double or triple bonds between C atoms. 'Unsaturated' = can accept more H.

A. Alkenes (1 double bond): $C_nH_{2n}$. Examples: ethene, propene. B. Alkynes (1 triple bond): $C_nH_{2n-2}$. Examples: ethyne, propyne.

Differences

Property	Saturated	Unsaturated
Bonds	All single	Double/triple
Reactivity	Less	More
Bromine water	No reaction	Decolourises
Flame	Clean blue	Yellow, sooty

Bromine Water Test

Reddish-brown bromine water:

• Saturated: unchanged.
• Unsaturated: becomes colourless.

Reaction: $CH_2=CH_2 + Br_2 \rightarrow CH_2Br-CH_2Br$

[NCERT — important]

Example 5: NCERT — Structural Isomerism

What is structural isomerism? Draw 3 isomers of pentane (C₅H₁₂).

Solution:

Definition

'Structural isomerism' = phenomenon where compounds have same molecular formula but different structural arrangements.

3 Isomers of Pentane (C₅H₁₂)

1. n-Pentane (straight chain):

CH₃-CH₂-CH₂-CH₂-CH₃

B.P.: 36°C.

2. Iso-Pentane (1 branch):

CH₃-CH(CH₃)-CH₂-CH₃

Or:

         CH₃
          |
CH₃ — CH — CH₂ — CH₃

B.P.: 28°C.

3. Neo-Pentane (2 branches on one C):

         CH₃
          |
CH₃ — C — CH₃
          |
         CH₃

B.P.: 9°C.

Verification

All three: 5 C + 12 H = C₅H₁₂. Different structures, different boiling points.

Why Different B.P.?

Branching → more compact → less surface contact → weaker intermolecular forces → lower b.p.

Pattern: more branching → lower b.p.

[NCERT — every year]

Example 6: NCERT — Functional Groups

List 5 important functional groups with structures and examples.

Solution:

5 Important Functional Groups

1. Halogen (-X) X = F, Cl, Br, I. Example: CH₃Cl (chloromethane).

2. Hydroxyl/Alcohol (-OH) Structure: R-O-H Example: CH₃OH (methanol), C₂H₅OH (ethanol). Suffix: -ol.

3. Aldehyde (-CHO) Structure: R-C(=O)-H Example: HCHO (methanal), CH₃CHO (ethanal). Suffix: -al.

4. Ketone (>C=O) Structure: R-C(=O)-R' Example: (CH₃)₂CO (propanone, acetone). Suffix: -one.

5. Carboxylic Acid (-COOH) Structure: R-COOH Example: HCOOH (methanoic acid), CH₃COOH (ethanoic acid). Suffix: -oic acid.

Why Functional Groups Matter

They determine the chemical properties of the compound.

Same C-H backbone with different functional groups → very different chemistry.

Comparison

Compound	Functional Group	Type
C₂H₆	None	Alkane
C₂H₅OH	-OH	Alcohol
CH₃CHO	-CHO	Aldehyde
CH₃COOH	-COOH	Acid

[NCERT — every year]

Example 7: NCERT — Homologous Series

What is a homologous series? State 4 properties.

Solution:

Definition

'Homologous series' = a 'family' of organic compounds with:

• Same functional group.
• Same general formula.
• Members differ by CH₂ (14 amu).
• Similar chemistry, gradually changing physical properties.

4 Properties

1. Same general formula. Example: alkanes — $C_nH_{2n+2}$ (methane CH₄, ethane C₂H₆, etc.).

2. Successive members differ by CH₂.

• C₂H₆ - CH₄ = CH₂ ✓
• C₃H₈ - C₂H₆ = CH₂ ✓ All consistent.

3. Similar chemical properties. All alkanes burn similarly. All alcohols react with Na. All acids react with bases.

4. Gradual change in physical properties. Boiling points of alkanes:

• CH₄: -161°C
• C₂H₆: -89°C
• C₃H₈: -42°C
• C₄H₁₀: -1°C Steady increase.

Examples of Common Homologous Series

Alkanes, alkenes, alkynes, alcohols, aldehydes, ketones, acids.

Each forms a 'family' with predictable properties.

[NCERT — important]

Example 8: NCERT — IUPAC Naming

Give IUPAC names for: (a) CH₃CH₂OH (b) HCHO (c) CH₃COOH (d) CH₃-CO-CH₃ (e) C₂H₅Cl

Solution:

(a) CH₃CH₂OH

2 C + -OH = ethanol

(b) HCHO

1 C + -CHO = methanal (Common name: formaldehyde)

(c) CH₃COOH

2 C + -COOH = ethanoic acid (Common name: acetic acid)

(d) CH₃-CO-CH₃

3 C + >C=O = propanone (Common name: acetone)

(e) C₂H₅Cl

2 C + -Cl = chloroethane (Common name: ethyl chloride)

Summary Table

Formula	IUPAC Name	Common Name
CH₃CH₂OH	Ethanol	Ethyl alcohol
HCHO	Methanal	Formaldehyde
CH₃COOH	Ethanoic acid	Acetic acid
CH₃COCH₃	Propanone	Acetone
C₂H₅Cl	Chloroethane	Ethyl chloride

Tips for Naming

1. Count C atoms → root (meth, eth, prop, but…). 2. Identify functional group → suffix (-ol, -al, -one, -oic acid). 3. Add prefix for substituents (chloro, bromo, methyl…).

[NCERT — important]

Example 9: NCERT — Combustion

Write balanced equations for combustion of: (a) ethanol (b) butane

Solution:

(a) Combustion of Ethanol

$$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O + \text{heat}$$

Heat: ~1370 kJ/mol.

Verification:

• C: 2 = 2 ✓
• H: 6 = 6 ✓
• O: 7 = 7 ✓

(b) Combustion of Butane

$$2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O + \text{heat}$$

Heat: ~2880 kJ/mol of butane.

Verification:

• C: 8 = 8 ✓
• H: 20 = 20 ✓
• O: 26 = 26 ✓

Used in LPG cooking gas.

Comparison

Fuel	Heat (kJ/mol)	Heat (kJ/g)
Ethanol	1370	30
Butane	2880	50

Both burn cleanly with blue flame. Used as fuels — butane in LPG, ethanol as petrol additive.

Why Important?

Combustion releases heat → cooking, heating, transport. Burning hydrocarbons = main energy source globally.

[NCERT — important]

Example 10: NCERT — Esterification

Write the esterification reaction. What are esters used for?

Solution:

Esterification Reaction

Carboxylic acid + alcohol → ester + water (in the presence of conc. H₂SO₄).

General reaction: $$\text{R-COOH} + \text{R'-OH} \rightleftharpoons \text{R-COO-R'} + H_2O$$

Specifically: $$CH_3COOH + C_2H_5OH \xrightarrow{\text{conc. } H_2SO_4} CH_3COOC_2H_5 + H_2O$$

Acetic acid + ethanol → ethyl ethanoate + water.

Conditions

• Catalyst: concentrated $H_2SO_4$.
• Heat: gentle warming.
• Reaction is reversible (note ⇌).

Uses of Esters

1. Fragrance and Flavour Industry:

• Esters have sweet fruity smells.
• Used in:
• Perfumes (rose, jasmine).
• Artificial fruit flavours (banana, apple).
• Air fresheners.

2. Solvents:

• Ethyl acetate — common laboratory solvent.
• Used in nail polish removers, glues.

3. Pharmaceuticals:

• Aspirin (acetylsalicylic acid) — an ester.
• Many other drugs contain ester groups.

4. Polymers:

• Polyester fabrics.
• Plastic bottles (PET).

5. Food Industry:

• Artificial flavours in candies, soft drinks.

Esters — small molecules, big economic impact.

Saponification (Reverse Reaction)

Ester + NaOH → salt + alcohol Used in soap making.

[NCERT — important]

Example 11: NCERT — Soap and Detergent

Explain the cleaning action of soap. Why doesn't soap work in hard water?

Solution:

Cleaning Action of Soap

Soap molecule has two parts:

• Hydrophilic head (-COO⁻Na⁺): water-loving.
• Hydrophobic tail (long C chain): oil-loving.

Steps:

1. Soap molecules surround oily dirt. Tails embed in oil; heads point outward.

2. Micelle forms. Spherical structure with dirt inside, water outside.

3. Mechanical washing loosens dirt. Many micelles disperse in water.

4. Rinse with water. Water carries away micelles; dirt removed.

Result: clean surface.

Hard Water Problem

Hard water = water with $Ca^{2+}$, $Mg^{2+}$ ions.

Soap reacts with these ions: $$2C_{17}H_{35}COO^-Na^+ + Ca^{2+} \rightarrow (C_{17}H_{35}COO)_2Ca\downarrow + 2Na^+$$

Forms insoluble calcium soap (scum) which:

• Wastes soap.
• Stains clothes.
• Doesn't lather.

Solution: Detergents

Detergents have -SO₃⁻Na⁺ instead of -COO⁻Na⁺. Don't form scum with Ca²⁺/Mg²⁺. Lather well in hard water. Used in modern washing powders.

Comparison

Property	Soap	Detergent
Head	-COO⁻Na⁺	-SO₃⁻Na⁺
Hard water	Forms scum	Lathers well
Biodegradable	Yes	Some

[NCERT — every year, important]

Example 12: NCERT — Properties of Ethanol

List 5 properties of ethanol and 3 main reactions.

Solution:

5 Properties of Ethanol (C₂H₅OH)

1. Physical state: colourless liquid. 2. Smell: mild, sweet, alcoholic. 3. Boiling point: 78°C. 4. Solubility: highly soluble in water (any ratio). 5. Volatile: evaporates easily.

3 Main Reactions

1. Combustion: $$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O + \text{heat}$$ Burns with blue flame, used as fuel.

2. Reaction with sodium: $$2C_2H_5OH + 2Na \rightarrow 2C_2H_5O^-Na^+ + H_2\uparrow$$ Test for alcohols (slow bubbling).

3. Oxidation to acetic acid: $$C_2H_5OH \xrightarrow{KMnO_4} CH_3COOH$$ Industrial vinegar production.

Why Ethanol is Important

1. Alcoholic beverages. 2. Solvent (in cosmetics, pharmaceuticals). 3. Fuel (E10, E20). 4. Disinfectant (in sanitisers). *5. Industrial chemical for esters, ethers.

Common Names

• Drinking alcohol.
• Ethyl alcohol.
• Grain alcohol (when from grains).

[NCERT — important]

Example 13: NCERT — Properties of Acetic Acid

List properties of acetic acid and its 3 main reactions.

Solution:

Properties of Acetic Acid (CH₃COOH)

1. Physical state: colourless liquid. 2. Smell: pungent, sharp (vinegar smell). 3. Taste: sour. 4. Boiling point: 118°C (high due to H-bonding). 5. Melting point: 16.6°C — pure (glacial) form solidifies in cold weather. 6. Highly soluble in water.

3 Main Reactions

1. With sodium hydroxide (neutralisation): $$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$$ Acid + base → salt + water.

2. With sodium carbonate/bicarbonate: $$CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2\uparrow$$ Releases CO₂ — fizz with baking soda!

3. Esterification with ethanol: $$CH_3COOH + C_2H_5OH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O$$ Forms ethyl acetate (ester) with sweet smell.

Why Acidic?

-COOH releases H⁺ in water: $$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$ pKa = 4.76 (weakly acidic).

Glacial Acetic Acid

Pure acetic acid (no water). Freezes at 16.6°C → forms ice-like crystals → 'glacial'.

Uses

Industrial: vinegar, plastics, dyes, drugs. Food: vinegar (5-8%), pickling. Lab: buffers, reagents.

Found in vinegar (5-8% solution).

[NCERT — every year]

Example 14: NCERT — Hydrogenation

What is hydrogenation? Why is it industrially important?

Solution:

Hydrogenation

Reaction where H₂ is added across a multiple bond (C=C or C≡C). Catalyst: Ni, Pt, or Pd.

General Reaction

$$\text{C=C} + H_2 \xrightarrow{Ni} \text{C-C}$$

Examples

Ethene + H₂

$$CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3-CH_3$$

Ethyne + 2H₂

$$HC≡CH + 2H_2 \xrightarrow{Ni} CH_3-CH_3$$

Industrial Importance

Vanaspati Ghee Production (Most Important)

Vegetable oils (with C=C, unsaturated) + H₂ → Vanaspati (saturated, solid).

Why:

• Solid fat easier to handle, store, transport.
• Longer shelf life.
• Cheaper than animal ghee.

Margarine

Similar process — used as butter substitute.

Other Uses

Drug manufacturing. Industrial chemicals from unsaturated precursors.

Health Concerns

Hydrogenation produces some trans fats (unhealthy):

• Raise bad cholesterol (LDL).
• Lower good cholesterol (HDL).
• Linked to heart disease.

Modern recommendation: limit vanaspati, prefer natural oils (mustard, sunflower, olive).

India Context

Many Indian foods use vanaspati — historically important for cooking (cheaper than ghee). Now FSSAI regulates trans fats in food.

[NCERT — important]

Example 15: A Concluding Question

(a) Explain why ethanol's b.p. (78°C) is much higher than ethane's (-89°C). (b) How do you distinguish acetic acid from ethanol? (c) Why is benzene important? (d) Define micelle.

Solution:

(a) Ethanol vs Ethane B.P.

Ethanol: C₂H₅OH — has -OH group → forms hydrogen bonds with other ethanol molecules. Ethane: C₂H₆ — only weak van der Waals forces.

Hydrogen bonds are much stronger → much higher b.p. for ethanol. Difference: 167°C!

(b) Distinguishing Acetic Acid from Ethanol

Test 1: Litmus paper.

• Acetic acid: turns blue litmus red.
• Ethanol: no change.

Test 2: With Na₂CO₃ or NaHCO₃.

• Acetic acid: effervescence (CO₂).
• Ethanol: no change.

Test 3: Smell.

• Acetic acid: pungent, vinegar smell.
• Ethanol: mild, sweet smell.

(c) Why Benzene is Important

Benzene (C₆H₆) is a special aromatic compound:

1. Industrial raw material:

• For plastics (polystyrene).
• Dyes, drugs, perfumes.
• Solvent.

2. Found in petroleum (5-7%).

3. Many medicines contain benzene rings:

• Aspirin, paracetamol.

Caution: Benzene is carcinogenic (cancer-causing). Limited in petrol.

(d) Micelle

'Micelle' = a spherical structure formed by soap molecules in water:

• Hydrophobic tails inside (in oily dirt).
• Hydrophilic heads outside (in water).

Function: traps oily dirt and washes it away.

This is how soap cleans — by forming micelles.

Summary

Carbon chemistry covers:

• Bonding (covalent).
• Allotropes (diamond, graphite, fullerene).
• Hydrocarbons (alkanes, alkenes, alkynes).
• Functional groups (-OH, -COOH, etc.).
• Reactions (combustion, oxidation, addition, substitution).
• Important compounds (ethanol, acetic acid).
• Soaps and detergents.

All — based on the unique properties of carbon.

[Board: 5-mark mixed]

Example 16: Numerical — Combustion

If 23 g of ethanol is completely burnt, calculate: (a) O₂ required (mass). (b) CO₂ produced (mass). (c) Heat released (kJ). [Heat of combustion of ethanol = 1370 kJ/mol] (C=12, H=1, O=16)

Solution:

Reaction

$$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$$

Molar Masses

• $C_2H_5OH$ = 24 + 6 + 16 = 46 g/mol
• $O_2$ = 32 g/mol
• $CO_2$ = 44 g/mol

Moles of Ethanol

23 g ÷ 46 g/mol = 0.5 mol

From Stoichiometry

1 mol ethanol → 3 mol O₂ → 2 mol CO₂

For 0.5 mol ethanol:

• O₂ used: 0.5 × 3 = 1.5 mol = 1.5 × 32 = 48 g
• CO₂ produced: 0.5 × 2 = 1 mol = 1 × 44 = 44 g
• Heat released: 0.5 × 1370 = 685 kJ

Final Answers

(a) O₂ required: 48 g (b) CO₂ produced: 44 g (c) Heat released: 685 kJ

Real-world Significance

This calculation is useful for:

• Sizing fuel tanks.
• Calculating CO₂ emissions.
• Designing engines.
• Power plants.

Carbon compounds are major energy sources — but also produce CO₂ (greenhouse gas).

[Board: 3-5 mark numerical]