Organic Chemistry – Chapter 16 Hydrocarbons | Interactive Revision

16.1 Hydrocarbons – Classification

⚛️ Definition & Broad Classification

  • Hydrocarbons: Organic compounds containing only C and H elements
  • Aliphatic Hydrocarbons: From Greek “fatty” – open chain or cyclic compounds resembling open chains
  • Aromatic Hydrocarbons: From Greek “fragrant” – contain benzene ring or resemble benzene
Hydrocarbons
→
Aliphatic
(Open/Closed chain)
→
Aromatic
(Benzene ring)

🔗 Open Chain Hydrocarbons

Characteristics: No cycles/rings, terminal carbon atoms present
TypeBondingHybridizationExamples
Saturated (Alkanes) Only single bonds sp³ Methane (CH₄), Ethane (C₂H₆)
Unsaturated – Alkenes At least one C=C sp² Ethene (C₂H₄), Propene (C₃H₆)
Unsaturated – Alkynes At least one C≡C sp Ethyne (C₂H₂), Propyne (C₃H₄)

🌀 Closed Chain Hydrocarbons

Alicyclic: Cycloalkanes (CₙH₂ₙ)
Aromatic: Benzene (C₆H₆)
Key Difference: Alicyclic hydrocarbons have 2H less than corresponding open-chain alkanes

16.2 Reactivity of Alkanes

⚡ Why Alkanes are Less Reactive?

  • Non-polar Bonds: C (2.5) and H (2.1) electronegativity difference negligible
  • Strong Sigma (σ) Bonds: Electrons close to nuclei, strongly held
  • No Pi (π) Bonds: π bonds are weaker and more reactive
  • Do not react with acids, bases, oxidizing/reducing agents under normal conditions

🔥 Free Radical Substitution of Ethane

Three Step Mechanism

1. Initiation: Cl₂ → 2Cl• (UV light)

2. Propagation: Chain reaction producing free radicals

3. Termination: Free radicals combine to end reaction

Propagation Step Examples:
C₂H₆ + Cl• → C₂H₅• + HCl
C₂H₅• + Cl₂ → C₂H₅Cl + Cl•

🔬 Shapes of Molecules

Ethane (C₂H₆) Cyclopropane (C₃H₆) • Tetrahedral around each C • sp³ hybridized carbons • Bond angles: 109.5° • C-C-C angle: 60° • Not planar • Not planar (3D shape)

16.4 Preparation of Alkenes

⚗️ 1. Dehydration of Alcohols

CH₃-CH₂-OH → CH₂=CH₂ + H₂O

Catalysts: Al₂O₃ (alumina), H₂SO₄, H₃PO₄

🧪 2. Dehydrohalogenation of Halogenoalkanes

CH₃-CH₂-Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O

Conditions: Hot ethanolic NaOH/KOH

But-1-ene Preparation:
CH₃-CH₂-CH₂-CH₂-OH → CH₃-CH₂-CH=CH₂ + H₂O (Dehydration)
CH₃-CH₂-CH₂-CH₂-Br → CH₃-CH₂-CH=CH₂ + HBr (Dehydrohalogenation)

16.5 Ethene Structure & Reactivity

🔬 Molecular Structure

  • Formula: C₂H₄
  • Bonding: One σ bond + one π bond between C atoms
  • Hybridization: sp² (trigonal planar)
  • Geometry: Planar molecule, bond angles ~120°
  • π bond: Formed by parallel p-orbital overlap, weaker than σ bond
H H
\ /
C=C
/ \
H H

⚡ Reactivity Order

Most Reactive
Alkenes
→
Moderate
Alkynes
→
Least Reactive
Alkanes
Why Alkenes > Alkynes? π bond in ethene more exposed/diffused than in ethyne

16.6 Chemical Reactions of Alkenes

🔥 1. Hydrogenation

CH₂=CH₂ + H₂ → CH₃-CH₃ (Ni/Pt catalyst, 200°C)

Application: Margarine production

🧪 2. Hydrohalogenation

Markovnikov’s Rule

“Hydrogen adds to carbon with more hydrogens”

Propene + HBr → 2-Bromopropane (major) + 1-Bromopropane (minor)

Carbocation Stability:
Tertiary > Secondary > Primary > Methyl

💧 3. Hydration

CH₂=CH₂ + H₂O → CH₃-CH₂-OH (Acid catalyst)

🧬 Other Important Reactions

ReactionReagentProduct
HalogenationBr₂/CCl₄1,2-Dibromoethane
HalohydrationHOBrBromohydrin
EpoxidationPeracidsEpoxyethane
OzonolysisO₃ then ZnAldehydes/Ketones
PolymerizationHigh P/TPolyethene
Bromine Water Test: Brown color disappears with alkenes (test for unsaturation)

16.9 Stereoisomerism

👓 Optical Isomerism

  • Chiral Carbon: Carbon with 4 different groups
  • Enantiomers: Mirror images, non-superimposable
  • Optical Activity: Rotates plane-polarized light
  • Racemic Mixture: Equal amounts of enantiomers, optically inactive

🔄 Geometric (Cis-Trans) Isomerism

Cis Isomer Trans Isomer Same groups on same side Same groups on opposite sides Latin: “on this side” Latin: “across”
Requirements for Geometric Isomerism:
1. Restricted rotation (C=C or ring)
2. Different groups on each carbon of double bond
3. Not in propene, but-1-ene (same H on one C)

Exercise Questions – Complete Solutions

Multiple Choice Questions:

1. The most likely reaction alkanes undergo is:

a. electrophilic addition
b. free radical substitution
c. nucleophilic addition
d. electrophilic substitution

Explanation: Alkanes undergo free radical substitution due to non-polar C-H and C-C bonds. Electrophilic/nucleophilic reactions require polar bonds.

2. The condensed structural formula of the major product obtained by hydrobromination of 2-methylpent-2-ene is:

a. CH₃CH(Br)CH(CH₃)CH₂CH₃
b. (CH₃)₂C(Br)(CH₂)₂CH₃
c. (CH₃)₂CHCH(Br)CH₂CH₃
d. CH₂(Br)CH(CH₃)(CH₂)₂CH₃

Explanation: Markovnikov’s rule: H adds to C with more H’s. Tertiary carbocation forms → 2-bromo-2-methylpentane.

3. Which one is a possible propagation step in chlorination of ethane?

a. C₂H₆ + Cl → C₂H₅Cl + H⁺
b. C₂H₅Cl + Cl → C₂H₄Cl• + HCl
c. C₂H₅• + Cl → C₂H₅Cl•
d. C₂H₄Cl₂ + Cl• → C₂H₄Cl• + Cl₂

Correct Answer: None exactly match, but closest is B with correction
Propagation steps: C₂H₆ + Cl• → C₂H₅• + HCl and C₂H₅• + Cl₂ → C₂H₅Cl + Cl•

4. Which is the most stable (least reactive) hydrocarbon?

a. hex-1-ene
b. propene
c. penta-1,2-diene
d. hepta-1,3,5-triene

Explanation: Conjugated systems (alternate double bonds) are more stable due to delocalization of π electrons.

Short Answer Questions:

i. Justify why the second step in halogenation of alkanes is named propagation?

Answer: The second step is called propagation because it propagates (continues) the chain reaction. Each step generates a new free radical that can continue the reaction, creating a chain of reactions until termination.

ii. Draw the structure of ethane molecule.

Answer:

H H
| |
H-C-C-H
| |
H H

3D Structure: Tetrahedral around each carbon (109.5° bond angles)

iii. Alkenes show addition reactions while alkanes cannot. Why?

Answer: Alkenes have π bonds which are weaker and more reactive. π electrons are exposed and can be attacked by electrophiles. Alkanes have only strong σ bonds with no exposed electron cloud for addition.

iv. Halogen is a non-polar molecule. How can it act as electrophile in halogenation of alkenes?

Answer: When Br₂ approaches alkene π cloud, electron repulsion polarizes Br₂ making one Br partially positive (electrophile) and the other partially negative.

Exercise from Textbook:

Differentiate between saturated and unsaturated compounds. Give examples.

Saturated Compounds Unsaturated Compounds • Only single bonds
• sp³ hybridization
• Cannot add more atoms
• Less reactive
• Examples: Alkanes (CH₄, C₂H₆) • Contain double/triple bonds
• sp² or sp hybridization
• Can add more atoms
• More reactive
• Examples: Alkenes (C₂H₄), Alkynes (C₂H₂)

Define aliphatic and aromatic compounds. Give two examples of each.

Aliphatic Compounds: Open chain or cyclic compounds resembling open chains. Examples: Methane (CH₄), Cyclohexane (C₆H₁₂)

Aromatic Compounds: Contain benzene ring or resemble benzene. Examples: Benzene (C₆H₆), Toluene (C₆H₅CH₃)

What type of hybridization is shown by each carbon of hexa-1,2-diene?

Answer: CH₂=C=CH-CH₂-CH₂-CH₃

• C1 (CH₂=): sp² (double bond)
• C2 (=C=): sp (two double bonds)
• C3 (-CH-): sp² (double bond)
• C4, C5, C6: sp³ (single bonds)

Prepare but-1-ene by: i) dehydration of alcohols ii) dehydrohalogenation of halogenoalkanes.

i) Dehydration of Alcohol:

CH₃-CH₂-CH₂-CH₂-OH → CH₃-CH₂-CH=CH₂ + H₂O

Catalyst: Al₂O₃ or H₂SO₄ at 180°C

ii) Dehydrohalogenation:

CH₃-CH₂-CH₂-CH₂-Br + KOH(alc) → CH₃-CH₂-CH=CH₂ + KBr + H₂O

Constructed Response Questions:

1. Generally, hydrocarbons are less reactive. Arrange alkanes, alkenes and alkynes in their reactivity order and justify.

Reactivity Order: Alkenes > Alkynes > Alkanes

Justification:

Alkenes (Most Reactive): Have π bonds with exposed electron cloud, susceptible to electrophilic attack

Alkynes (Moderate): Have π bonds but electrons are more tightly held (sp hybridization)

Alkanes (Least Reactive): Only strong σ bonds, non-polar, require harsh conditions for reaction

2. Define primary, secondary and tertiary carbocations. Justify their order of reactivity.

Definitions:

Primary (1°): Carbon attached to one alkyl group (e.g., CH₃-CH₂⁺)

Secondary (2°): Carbon attached to two alkyl groups (e.g., (CH₃)₂CH⁺)

Tertiary (3°): Carbon attached to three alkyl groups (e.g., (CH₃)₃C⁺)

Stability Order: 3° > 2° > 1° > Methyl⁺

Reason: Alkyl groups donate electrons via inductive effect, stabilizing positive charge

3. Define Markovnikov’s rule. Give mechanism of hydrobromination of 2-methylhex-2-ene.

Markovnikov’s Rule: “In addition of unsymmetrical reagent to unsymmetrical alkene, hydrogen adds to carbon with more hydrogens.”

Mechanism for 2-methylhex-2-ene + HBr:

Step 1: Protonation at C=C

(CH₃)₂C=CH-CH₂-CH₂-CH₃ + H⁺ → (CH₃)₃C⁺-CH-CH₂-CH₂-CH₃ (Tertiary carbocation)

Step 2: Bromide attack

(CH₃)₃C⁺-CH-CH₂-CH₂-CH₃ + Br⁻ → (CH₃)₃C-CH(Br)-CH₂-CH₂-CH₃

Product: 2-Bromo-2-methylhexane (Major product)

4. Define geometric isomerism. Apply this concept on open chain alkenes and cycloalkanes.

Geometric Isomerism: Stereoisomerism due to restricted rotation around double bond or in rings, with different spatial arrangement of groups.

Open Chain Alkenes (e.g., But-2-ene):

Cis: CH₃ on same side
Trans: CH₃ on opposite sides

Cycloalkanes (e.g., 1,2-dimethylcyclopropane):

Cis: Both CH₃ on same side of ring
Trans: CH₃ on opposite sides of ring

Investigative Question:

Design a synthetic route for: a) Bromoethane b) 1,2-dibromoethane

a) Bromoethane (C₂H₅Br):

Route 1:
Ethane + Br₂ (UV light)
Free radical substitution
→
Route 2:
Ethene + HBr
Electrophilic addition

b) 1,2-Dibromoethane (C₂H₄Br₂):

CH₂=CH₂ + Br₂ → Br-CH₂-CH₂-Br

Direct bromination of ethene at room temperature