Chapter 11: Organic Chemistry & Alkanes | Complete Guide with Solved Exercises

Chapter 11: Organic Chemistry & Alkanes

Complete Guide with IUPAC Nomenclature, Preparation Methods & Solved Exercises
Hydrocarbons | Saturated Compounds | Methane | Ethane | Propane | Butane
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Introduction to Organic Chemistry

Organic chemistry is the study of carbon compounds. Carbon has unique properties that allow it to form millions of compounds, more than all other elements combined.

KEY NOTES: Fundamentals of Organic Chemistry

DEFINITION & SCOPE

• Organic chemistry = Chemistry of carbon compounds

• Exceptions: Carbonates, cyanides, carbides, oxides of carbon → inorganic

• Most organic compounds contain C, H, O

• Millions of compounds exist naturally or synthetically

UNIQUE PROPERTIES OF CARBON

• Forms 4 covalent bonds (tetravalency)

Catenation: Self-linking ability

• Forms chains, rings, branched structures

• Strong C-C and C-H bonds

• Small size → short, strong bonds

EXAMPLES OF ORGANIC COMPOUNDS

• Biomolecules: Proteins, enzymes, carbohydrates

• Lipids, vitamins, nucleic acids

• Pharmaceuticals, synthetic fibers, plastics

• Fuels: Natural gas, petrol, diesel

Memory Trick:

“Carbon is King” – Forms millions of compounds, tetravalent, catenation power.

Remember: Organic = Carbon compounds (except carbonates, oxides).

11.1 Hydrocarbons

Hydrocarbons are organic compounds containing only carbon and hydrogen. They are the simplest organic compounds and serve as fuels and feedstocks.

Common Hydrocarbon Fuels

Fuel Type Main Hydrocarbons Uses
Natural Gas Gas Methane (CH₄) Heating, cooking, electricity
LPG Liquid/Gas Propane, Butane Cooking, heating, vehicles
CNG Compressed Gas Methane Vehicle fuel
Petrol Liquid C₅–C₁₂ alkanes Vehicle fuel
Diesel Liquid C₁₂–C₂₀ alkanes Trucks, generators
Kerosene Liquid C₁₂–C₁₅ alkanes Heating, jet fuel
CH₄

METHANE

Simplest alkane

Natural gas

C₂H₆

ETHANE

2 carbon atoms

Petrochemical feedstock

C₃H₈

PROPANE

LPG component

Heating fuel

C₄H₁₀

BUTANE

Lighter fuel

n-Butane & iso-Butane

KEY NOTES: Classification of Hydrocarbons

ALKANES (SATURATED)

• Only C-C and C-H single bonds

• General formula: CₙH₂ₙ₊₂

• Also called paraffins

• Examples: Methane, Ethane, Propane

• Low reactivity (inert)

ALKENES (UNSATURATED)

• Contain C=C double bonds

• General formula: CₙH₂ₙ

• Also called olefins

• Examples: Ethene, Propene

• More reactive

ALKYNES (UNSATURATED)

• Contain C≡C triple bonds

• General formula: CₙH₂ₙ₋₂

• Examples: Ethyne (acetylene)

• Used in welding

AROMATIC HYDROCARBONS

• Contain benzene rings

• Examples: Benzene, Toluene

• Planar cyclic structures

• Delocalized π electrons

11.2 Alkanes & IUPAC Nomenclature

Alkanes are saturated hydrocarbons with only single bonds. The IUPAC system provides systematic names for organic compounds.

KEY NOTES: IUPAC Naming System

THREE PARTS OF IUPAC NAME

1. Root: Number of carbons in longest chain

2. Suffix: Class of compound (-ane for alkanes)

3. Prefix: Substituents (branches) with positions

ROOTS FOR CARBON CHAINS

• C1: Meth- • C2: Eth- • C3: Prop-

• C4: But- • C5: Pent- • C6: Hex-

• C7: Hept- • C8: Oct- • C9: Non- • C10: Dec-

NAMING STEPS

1. Find longest continuous carbon chain

2. Number chain from end nearest branch

3. Identify substituents (methyl, ethyl, etc.)

4. Write name: Prefix + Root + Suffix

5. Use hyphens to separate numbers from words

Solved Example: Naming Alkanes

Compound: CH₃–CH(CH₃)–CH₂–CH₃

Step 1: Longest chain = 4 carbons → Butane

Step 2: Numbering from left: CH₃–CH(CH₃)–CH₂–CH₃

• Branch at carbon 2

Step 3: Branch = methyl group

Step 4: Name = 2-Methylbutane (iso-Butane)

CH₃–CH–CH₂–CH₃
   │
   CH₃

Table 11.1: Common Alkanes

Name Molecular Formula Structural Formula Condensed Formula
Methane CH₄ H–C–H (tetrahedral) CH₄
Ethane C₂H₆ CH₃–CH₃ C₂H₆
Propane C₃H₈ CH₃–CH₂–CH₃ C₃H₈
n-Butane C₄H₁₀ CH₃–CH₂–CH₂–CH₃ C₄H₁₀
iso-Butane C₄H₁₀ CH₃–CH(CH₃)–CH₃ (CH₃)₃CH

Memory Trick:

“MEPB” – Methane, Ethane, Propane, Butane (first 4 alkanes)

Roots: “Monkeys Eat Peanut Butter” → Meth, Eth, Prop, But

11.3 Preparation of Alkanes

Alkanes can be prepared by several methods including cracking, hydrogenation, and reduction of alkyl halides.

KEY NOTES: Three Main Methods

1. CRACKING OF HIGHER HYDROCARBONS

• Breaking larger hydrocarbons into smaller ones

• Catalyst: Zeolite

• Temperature: ~500°C

• Source: Naphtha from petroleum

• Product: C₅–C₁₀ hydrocarbons

• Balances supply and demand of fuels

2. REDUCTION OF ALKENES/ALKYNES

• Also called hydrogenation

• Catalyst: Nickel (Ni)

• Temperature: ~200°C

• Addition of H₂ to double/triple bonds

• Example: CH₂=CH₂ + H₂ → CH₃–CH₃

• Used to make margarine, banaspati ghee

3. REDUCTION OF ALKYL HALIDES

• Reducing agent: Zn/HCl

• Produces atomic hydrogen [H]

• R–X + 2[H] → R–H + HX

• Example: CH₃Cl + 2[H] → CH₄ + HCl

• Magnesium (Mg) can also be used instead of Zn

Interesting Information!

Cracking Economics: Cracking helps balance petroleum supply with demand. When large hydrocarbons are cracked into smaller ones, fuel supply increases to meet demand.

Hydrogenation in Food: Adding hydrogen to vegetable oils makes them solid at room temperature (margarine, banaspati ghee). This is an industrial application of alkene reduction.

Reaction Equations

1. Cracking:

C₁₆H₃₄ (Naphtha) → C₈H₁₈ + C₈H₁₆

2. Hydrogenation of Ethene:

CH₂=CH₂ + H₂ → CH₃–CH₃ (Ni, 200°C)

3. Reduction of Chloromethane:

CH₃Cl + 2[H] → CH₄ + HCl (Zn/HCl)

Solved Exercises with Explanations

1 Multiple Choice Questions (Tick the correct answer):

(i) Which other atom is almost always present along with carbon atom in all organic compounds?

(a) Oxygen
(b) Hydrogen
(c) Nitrogen
(d) Halogen
Show Explanation

Explanation: Almost all organic compounds contain hydrogen along with carbon. While many contain oxygen, nitrogen, or halogens, hydrogen is nearly universal.

• Hydrocarbons: C + H only

• Most biomolecules: C, H, O, N

• Organic definition: Compounds of carbon, usually with hydrogen

(ii) Which other metal can be used to reduce alkyl halides?

(a) Al
(b) Mg
(c) Ni
(d) Co
Show Explanation

Explanation: Magnesium (Mg) can be used instead of zinc (Zn) to reduce alkyl halides with hydrochloric acid.

• Reaction: R–X + 2[H] → R–H + HX

• Source of [H]: Zn + 2HCl → ZnCl₂ + 2[H]

• Mg works similarly: Mg + 2HCl → MgCl₂ + 2[H]

• Ni is used for hydrogenation, not reduction with HCl

(iv) Why does a mixture of zinc and hydrochloric acid act as a reducing agent?

(a) Because zinc acts as a reducing agent.
(b) Because atomic hydrogen is produced with Zn/HCl which acts as a reducing agent.
(c) Because molecular hydrogen is produced with Zn/HCl which acts as a reducing agent.
(d) Because chloride ions are produced with Zn/HCl which act as a reducing agent.
Show Explanation

Explanation: Zn/HCl produces atomic hydrogen [H], which is a powerful reducing agent.

• Reaction: Zn + 2HCl → ZnCl₂ + 2[H]

• Atomic hydrogen [H] is highly reactive

• It reduces alkyl halides: R–X + 2[H] → R–H + HX

• Molecular H₂ is less reactive than atomic hydrogen

(v) Which alkane will evolve the most amount of heat when it is burnt with oxygen?

(a) Ethane
(b) Propane
(c) n-Butane
(d) iso-Butane
Show Explanation

Explanation: n-Butane (C₄H₁₀) has the highest heat of combustion among the options because it has the most carbon and hydrogen atoms per molecule.

• Heat of combustion increases with molecular size

• Order: Methane < Ethane < Propane < Butane

• n-Butane and iso-Butane have same formula but n-Butane has slightly higher heat of combustion due to structure

• Complete combustion: CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → n CO₂ + (n+1) H₂O + heat

(vii) Which hydrocarbon is responsible for explosions in coal mines?

(a) Butane
(b) Pentane
(c) Methane
(d) Ethene
Show Explanation

Explanation: Methane (CH₄) is the main component of “firedamp” in coal mines.

• Coal mines contain trapped methane gas

• Methane-air mixtures are explosive (5-15% methane in air)

• Ignition sources: Sparks, flames, electrical equipment

• Methane explosions have caused many mining disasters

• Proper ventilation and methane detection are crucial for mine safety

(viii) Which product will be formed when ethyl bromide (CH₃CH₂Br) is treated with Zn/HCl?

(a) CH₄
(b) CH₃–CH₃
(c) CH₃–CH₂–CH₂–CH₃
(d) CH₃–CH₂–CH₃
Show Explanation

Explanation: Ethyl bromide (CH₃CH₂Br) reduces to ethane (CH₃–CH₃) with Zn/HCl.

• Reaction: CH₃CH₂Br + 2[H] → CH₃CH₃ + HBr

• Source of [H]: Zn + 2HCl → ZnCl₂ + 2[H]

• The alkyl halide loses Br and gains H

• General: R–X + 2[H] → R–H + HX

• Here R = CH₃CH₂–, so product is CH₃CH₃ (ethane)

2 Questions for Short Answers:

i. Differentiate between an organic and an inorganic compound.

Answer:
Organic Compounds Inorganic Compounds
• Contain carbon (except carbonates, cyanides, oxides) • May or may not contain carbon
• Usually contain C-H bonds • Usually lack C-H bonds
• Covalent bonding predominates • Ionic bonding common
• Millions of compounds known • Fewer compounds
• Examples: Methane, glucose, proteins • Examples: NaCl, H₂SO₄, CaCO₃

ii. Why are organic compounds found in large numbers?

Answer:

Organic compounds exist in large numbers due to unique properties of carbon:

1. Tetravalency: Carbon forms 4 covalent bonds

2. Catenation: Carbon atoms link to each other forming chains, rings, branches

3. Isomerism: Same molecular formula can give different structures

4. Multiple bonding: Can form single, double, triple bonds

5. Bonding with many elements: Forms stable bonds with H, O, N, halogens

6. Small size: Strong C-C bonds due to small atomic radius

Result: More carbon compounds than all other elements combined.

iv. How naphtha fraction is decomposed to give lower hydrocarbons?

Answer:

Naphtha is decomposed by cracking:

1. Source: Naphtha from fractional distillation of petroleum

2. Process: Thermal cracking with catalyst

3. Catalyst: Zeolite

4. Temperature: ~500°C

5. Reaction: Larger hydrocarbons → smaller hydrocarbons

6. Example: C₁₆H₃₄ → C₈H₁₈ + C₈H₁₆

7. Product: Hydrocarbons with 5-10 carbon atoms (gasoline range)

8. Purpose: Convert less useful heavy fractions to more valuable lighter fuels

v. Write down the molecular formula, structural formula and the condensed structural formula for iso-butane.

Answer:

iso-Butane (2-Methylpropane):

1. Molecular Formula: C₄H₁₀

2. Structural Formula:

  CH₃
  │
CH₃–C–CH₃
  │
  H

3. Condensed Structural Formula: (CH₃)₃CH or CH₃CH(CH₃)CH₃

4. IUPAC Name: 2-Methylpropane

5. Common Name: iso-Butane

vi. How are organic compounds useful for us?

Answer:

Organic compounds are essential for life and modern society:

1. Biological Importance:

• Proteins, enzymes, DNA (genetic material)

• Carbohydrates (energy source)

• Lipids (cell membranes, energy storage)

• Vitamins, hormones (metabolic regulation)

2. Materials & Industry:

• Plastics, synthetic fibers (nylon, polyester)

• Rubber, paints, varnishes

• Detergents, soaps, cosmetics

3. Energy:

• Fuels: Natural gas, petrol, diesel, coal

• LPG, CNG for cooking and vehicles

4. Medicine:

• Pharmaceuticals: Antibiotics, painkillers, vaccines

• Anesthetics, antiseptics

5. Agriculture:

• Fertilizers, pesticides, herbicides

• Plant growth regulators

vii. Write down the names of five such organic compounds which exist naturally.

Answer:

Five naturally occurring organic compounds:

1. Glucose (C₆H₁₂O₆) – Sugar in fruits, honey

2. Methane (CH₄) – Natural gas, swamp gas

3. Cellulose ((C₆H₁₀O₅)ₙ) – Plant cell walls

4. Ethanol (C₂H₅OH) – Produced by fermentation

5. Urea (CH₄N₂O) – Mammalian urine

Other examples: Proteins, fats, vitamins, DNA, chlorophyll, caffeine, penicillin.

3 Constructed Response Questions:

i. Why do alkanes show little reactivity towards the other reagents?

Answer:

Alkanes show little reactivity due to:

1. Strong Covalent Bonds:

• C–C and C–H bonds are strong (high bond energies)

• C–H bond: ~413 kJ/mol, C–C bond: ~347 kJ/mol

2. Non-polar Nature:

• C–C bonds are non-polar (same electronegativity)

• C–H bonds are nearly non-polar (small electronegativity difference: C=2.5, H=2.1)

• No charge separation → no sites for attack

3. Saturation:

• All carbon valencies satisfied with single bonds

• No π-bonds (double/triple bonds) for addition reactions

4. Lack of Functional Groups:

• No reactive sites like –OH, –COOH, –NH₂

• Only C–C and C–H bonds present

5. High Ionization Energy:

• Difficult to remove electrons from alkanes

Result: Alkanes are relatively inert, hence called “paraffins” (little affinity). They mainly undergo substitution reactions under drastic conditions (UV light, high temperature).

ii. Why does a mixture of natural gas and air explode?

Answer:

A mixture of natural gas (mainly methane) and air explodes due to rapid combustion reaction:

1. Composition:

• Natural gas: ~90% methane (CH₄)

• Air: ~21% oxygen (O₂)

2. Explosive Range:

• Methane-air mixtures explode at 5-15% methane concentration

• Below 5%: Too little fuel – no explosion

• Above 15%: Too little oxygen – incomplete combustion

3. Rapid Reaction:

• CH₄ + 2O₂ → CO₂ + 2H₂O + heat

• Highly exothermic (releases large energy quickly)

• Rapid gas expansion creates pressure wave (explosion)

4. Ignition Source:

• Spark, flame, or heat initiates reaction

• Once started, reaction propagates through mixture

5. Applications:

• Controlled explosion in internal combustion engines

• Uncontrolled explosion in mine disasters, gas leaks

Safety: Proper ventilation, gas detectors, and avoiding ignition sources prevent explosions.

iii. How do you compare the melting and boiling points of inorganic and organic compounds?

Answer:
Property Inorganic Compounds Organic Compounds
Bonding Type Mainly ionic (metal + non-metal) Mainly covalent (non-metal + non-metal)
Melting Points Generally HIGH (NaCl: 801°C, CaO: 2572°C) Generally LOW (Methane: -182°C, Ethanol: -114°C)
Boiling Points Generally HIGH (H₂O: 100°C, NaCl: 1413°C) Generally LOW (Methane: -162°C, Benzene: 80°C)
Reason Strong ionic bonds require much energy to break Weak intermolecular forces (van der Waals) easy to overcome
State at Room Temp Often solids (NaCl, CaCO₃) Often liquids or gases (CH₄ gas, C₂H₅OH liquid)
Exceptions Some covalent inorganic compounds have low MP/BP (CO₂: sublimes at -78°C) Some large organic compounds have high MP/BP (waxes, polymers)

Key Point: Ionic bonds (inorganic) are stronger than van der Waals forces (organic), leading to higher melting/boiling points for inorganic compounds.