Alkanes: Saturated Hydrocarbons

Complete guide to alkanes - preparation methods, Newman projections, conformational analysis, and halogenation mechanisms for JEE

11 min read

The Hook: From Petroleum to Your Car

Connect: Real Life → Chemistry

Ever wondered what happens when you fill petrol in your car or use LPG for cooking? The petrol that powers vehicles is primarily a mixture of alkanes (C₅ to C₈), while LPG contains propane and butane. Natural gas, used in millions of homes, is almost pure methane - the simplest alkane!

Here’s the JEE question: Why are alkanes called “saturated” hydrocarbons? And why can’t they undergo addition reactions like alkenes?

The Core Concept

What are Alkanes?

Alkanes are saturated hydrocarbons containing only single bonds between carbon atoms. The general formula is:

$$\boxed{\text{C}_n\text{H}_{2n+2}}$$

In simple terms: “Saturated” means the carbon atoms are bonded to the maximum number of hydrogen atoms possible - they’re “saturated” with hydrogen, just like a sponge saturated with water can’t hold more!

Why Alkanes are Unreactive

  • All bonds are σ (sigma) bonds - strong and stable
  • No π (pi) bonds to break
  • C-H bonds are non-polar (electronegativity difference: C = 2.5, H = 2.1)
  • Called “paraffins” (Latin: parum affinis = little affinity)
JEE Weightage
Alkanes: 2-3 questions in JEE Main, 1-2 in JEE Advanced Focus areas: Conformational analysis, free radical halogenation mechanisms

Preparation of Alkanes

Method 1: Wurtz Reaction

Best for: Symmetrical alkanes with even number of carbons

$$\boxed{2\text{R-X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R-R} + 2\text{NaX}}$$

Example:

$$\text{2CH}_3\text{Br} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3\text{-CH}_3 + 2\text{NaBr}$$
Trap: Mixed Products

Common JEE Mistake: Using two different alkyl halides gives a mixture!

If you react CH₃Br + C₂H₅Br, you get:

  • CH₃-CH₃ (ethane)
  • C₂H₅-C₂H₅ (butane)
  • CH₃-C₂H₅ (propane) ← desired product

Yield of desired product = only 25%!

JEE Tip: Wurtz reaction is practical only for symmetrical alkanes.

Method 2: Kolbe’s Electrolysis

Best for: Symmetrical alkanes from carboxylic acid salts

$$\boxed{2\text{RCOO}^-\text{Na}^+ \xrightarrow{\text{electrolysis}} \text{R-R} + 2\text{CO}_2 + 2\text{Na}^+ + 2e^-}$$

Mechanism at anode:

  1. Discharge: RCOO⁻ → RCOO• + e⁻
  2. Decarboxylation: RCOO• → R• + CO₂
  3. Coupling: R• + R• → R-R

Example:

$$2\text{CH}_3\text{COO}^-\text{Na}^+ \xrightarrow{\text{electrolysis}} \text{CH}_3\text{-CH}_3 + 2\text{CO}_2$$

Method 3: Reduction of Alkyl Halides

Method 3a: Using Zinc + HCl

$$\boxed{\text{R-X} + 2[\text{H}] \xrightarrow{\text{Zn + HCl}} \text{R-H} + \text{HX}}$$

Method 3b: Using LiAlH₄ (Milder reducing agent)

$$\text{R-X} + \text{LiAlH}_4 \rightarrow \text{R-H}$$

Method 4: Reduction of Carbonyl Compounds

Clemmensen Reduction (For aldehydes/ketones)

$$\boxed{\text{R-CO-R'} \xrightarrow{\text{Zn-Hg/HCl}} \text{R-CH}_2\text{-R'}}$$

Wolff-Kishner Reduction (Alternative method)

$$\boxed{\text{R-CO-R'} \xrightarrow{\text{NH}_2\text{NH}_2, \text{KOH}} \text{R-CH}_2\text{-R'}}$$
Memory Trick: Which Reduction When?

“Clemmensen is Acidic, Wolff-Kishner is Basic”

  • Clemmensen → Zn-Hg + HCl (acidic medium)
  • Wolff-Kishner → NH₂NH₂ + KOH (basic medium)

JEE Strategy: If substrate has acid-sensitive groups → use Wolff-Kishner If substrate has base-sensitive groups → use Clemmensen

Method 5: Decarboxylation

Sodalime Decarboxylation

$$\boxed{\text{RCOONa} \xrightarrow{\text{CaO + NaOH, } \Delta} \text{R-H} + \text{Na}_2\text{CO}_3}$$

Example:

$$\text{CH}_3\text{COONa} \xrightarrow{\text{sodalime, } \Delta} \text{CH}_4 + \text{Na}_2\text{CO}_3$$

Conformational Analysis: Newman Projections

Why This Matters for JEE Advanced

Conformational analysis questions appear regularly in JEE Advanced. Understanding Newman projections helps you:

  • Predict stability of conformers
  • Calculate dihedral angles
  • Understand rotation barriers

What are Conformations?

Conformations = Different 3D arrangements of atoms produced by rotation around single σ bonds

Key Difference

Conformers vs Configurational Isomers:

  • Conformers: Interconvertible by rotation (no bond breaking)
  • Configurational isomers: Require bond breaking (like cis-trans isomers)

Related concept: Stereoisomerism

Newman Projection: Ethane (C₂H₆)

How to draw Newman Projections:

  1. Front carbon = Center point (•)
  2. Back carbon = Circle (○)
  3. View along the C-C bond axis

Two extreme conformations:

Staggered Conformation (Most Stable)

        H
        |
    H—•—H
      / \
     /   \
    H     H
     \   /
      \ /
       H
  • Dihedral angle: 60° between H atoms
  • Minimum torsional strain
  • Minimum electron repulsion
  • Most stable (0 kJ/mol)

Eclipsed Conformation (Least Stable)

        H
        |
    H—•—H
       |
       H
       |
       H
  • Dihedral angle: 0° between H atoms
  • Maximum torsional strain
  • Maximum electron repulsion
  • Least stable (+12.5 kJ/mol)

Energy Barrier: 12.5 kJ/mol (rotation barrier between conformations)

Newman Projection: Butane (CH₃-CH₂-CH₂-CH₃)

For butane, we look at rotation around the C₂-C₃ bond:

Four Important Conformations:

1. Anti (Most Stable)

      CH₃
       |
    H—•—H
      / \
     /   \
    H     H
     \   /
      \ /
      CH₃
  • CH₃ groups 180° apart
  • Zero steric strain
  • Energy: 0 kJ/mol (reference)

2. Gauche (Moderately Stable)

      CH₃
       |
    H—•—H
      / \
     /   \
   CH₃   H
     \   /
      \ /
       H
  • CH₃ groups 60° apart
  • Some steric repulsion
  • Energy: +3.8 kJ/mol

3. Eclipsed (CH₃ and H eclipsed)

  • Energy: +16 kJ/mol

4. Fully Eclipsed (CH₃ groups eclipsed)

      CH₃
       |
    H—•—H
       |
      CH₃
       |
       H
  • CH₃ groups directly overlap (0° angle)
  • Maximum steric strain
  • Energy: +19 kJ/mol (least stable)

Energy Diagram for Butane Rotation

Energy
(kJ/mol)
   20│                    Fully Eclipsed
     │                    •
   16│         Eclipsed  / \  Eclipsed
     │            •     /   \    •
   12│           / \   /     \  / \
     │          /   \ /       \/   \
    4│  Gauche /     X         X    \ Gauche
     │    •   /     / \       / \    •
    0│___/_________/___\_____/___\________
       0°   60°   120°  180°  240°  300°  360°
           Anti         Anti

X = Gauche conformations
Memory Trick for Stability

“Anti is the Answer for Stability”

Stability order: Anti > Gauche > Eclipsed > Fully Eclipsed

Mnemonic: “Always Go Early for Front seats”

  • Anti (most stable)
  • Gauche
  • Eclipsed
  • Fully eclipsed (least stable)

Reactions of Alkanes

1. Combustion

$$\boxed{\text{C}_n\text{H}_{2n+2} + \frac{3n+1}{2}\text{O}_2 \rightarrow n\text{CO}_2 + (n+1)\text{H}_2\text{O} + \text{Heat}}$$

Example:

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \quad \Delta H = -890 \text{ kJ/mol}$$

Uses: Fuel source (LPG, CNG, petroleum)

Related: Thermochemistry

2. Halogenation (Most Important for JEE)

Free Radical Substitution Mechanism

Overall Reaction:

$$\boxed{\text{R-H} + \text{X}_2 \xrightarrow{h\nu \text{ or } \Delta} \text{R-X} + \text{HX}}$$

Three Steps:

Step 1: Chain Initiation

$$\text{Cl}_2 \xrightarrow{h\nu} 2\text{Cl}^\bullet$$

Homolytic cleavage of Cl-Cl bond (242 kJ/mol)

Step 2: Chain Propagation

$$\text{Cl}^\bullet + \text{CH}_4 \rightarrow \text{CH}_3^\bullet + \text{HCl}$$ $$\text{CH}_3^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet$$

Step 3: Chain Termination

$$\text{CH}_3^\bullet + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl}$$ $$\text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6$$ $$\text{Cl}^\bullet + \text{Cl}^\bullet \rightarrow \text{Cl}_2$$
JEE Trap: Reactivity Order

Question Type: “Which alkane reacts fastest with Cl₂?”

Reactivity of Hydrogen Atoms:

$$\boxed{3° > 2° > 1° > \text{CH}_4}$$

Why?

  • Carbocation stability of the intermediate
  • 3° radicals are most stable (hyperconjugation + inductive effect)

Bond Dissociation Energies:

  • CH₃-H: 435 kJ/mol (1°)
  • (CH₃)₂CH-H: 410 kJ/mol (2°)
  • (CH₃)₃C-H: 390 kJ/mol (3°)

Lower BDE = Easier to break = More reactive

Reactivity of Halogens

Reactivity order:

$$\boxed{\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2}$$
  • F₂: Too reactive, explosive, non-selective
  • Cl₂: Moderately reactive, some selectivity
  • Br₂: Slow, highly selective
  • I₂: Reaction is endothermic, doesn’t proceed
Memory Trick

“Fluorine is Fast and Furious, Iodine is Idle”

For JEE:

  • Cl₂ questions → selectivity based on radical stability
  • Br₂ questions → only most stable radical forms

Example: Chlorination of Propane

$$\text{CH}_3\text{-CH}_2\text{-CH}_3 + \text{Cl}_2 \xrightarrow{h\nu} ?$$

Two possible products:

  1. 1-chloropropane: CH₃-CH₂-CH₂Cl (attack at 1° carbon)
  2. 2-chloropropane: CH₃-CHCl-CH₃ (attack at 2° carbon)

Statistical factor:

  • 6 hydrogen atoms at 1° positions (two CH₃ groups)
  • 2 hydrogen atoms at 2° position (one CH₂ group)

Relative reactivity (per H atom):

  • 1° : 2° = 1 : 4

Product ratio:

$$\frac{\text{2-chloropropane}}{\text{1-chloropropane}} = \frac{2 \times 4}{6 \times 1} = \frac{8}{6} = \frac{4}{3}$$
JEE Advanced Type Calculation

Problem: What is the percentage of 2-chloro-2-methylbutane in the monochlorination of 2-methylbutane?

Solution: Structure: (CH₃)₂CH-CH₂-CH₃

Hydrogen types:

  • 1° (9H): 6H from two CH₃ on central carbon + 3H from terminal CH₃
  • 2° (2H): CH₂ group
  • 3° (1H): Central CH

Relative reactivity: 1° : 2° : 3° = 1 : 4 : 5.5

For 3° product:

$$\text{Fraction} = \frac{1 \times 5.5}{9(1) + 2(4) + 1(5.5)} = \frac{5.5}{22.5} = 24.4\%$$

Common Mistakes to Avoid

Mistake #1: Confusing Conformers with Isomers

Wrong: “Staggered and eclipsed are two different isomers”

Correct: They are conformers (same connectivity, different rotation)

  • Interconvertible at room temperature
  • Cannot be isolated separately

JEE Tip: Conformers have same molecular formula AND same connectivity. Only spatial arrangement differs.

Mistake #2: Wrong Product in Wurtz Reaction

Wrong: CH₃Br + C₂H₅Br + Na → CH₃-C₂H₅ (only)

Correct: You get a mixture of THREE alkanes

  • Statistical mixture based on coupling probabilities
  • Use Wurtz only for symmetrical alkanes

JEE Trap: Questions ask “major product” - it’s the symmetrical ones!

Mistake #3: Ignoring Statistical Factors in Halogenation

Wrong: Equal amounts of all monochloro products

Correct: Product ratio = (Number of H) × (Relative reactivity)

Always count:

  1. How many H atoms at each position
  2. Type of H (1°, 2°, 3°)
  3. Multiply and calculate ratio

Practice Problems

Level 1: Foundation (NCERT)

Problem 1: Nomenclature and Formula

Q: Write the molecular formula and structure of 2,2,3-trimethylpentane.

Solution:

  • Parent chain: Pentane (5 carbons)
  • Substituents: 3 methyl groups (2 at C-2, 1 at C-3)
      CH₃  CH₃
       |    |
CH₃—C—CH—CH₂—CH₃
       |
      CH₃

Molecular formula: C₈H₁₈ Check: n = 8, so 2n+2 = 18 ✓

Problem 2: Preparation

Q: How will you convert chloroethane to ethane?

Solution: Two methods:

Method 1: Reduction with Zn/HCl

$$\text{CH}_3\text{CH}_2\text{Cl} + 2[\text{H}] \xrightarrow{\text{Zn/HCl}} \text{CH}_3\text{CH}_3 + \text{HCl}$$

Method 2: Wurtz reaction

$$2\text{CH}_3\text{CH}_2\text{Cl} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3\text{CH}_2\text{-CH}_2\text{CH}_3 + 2\text{NaCl}$$

Wait! Method 2 gives butane, not ethane!

Correct answer: Method 1 only (reduction)

Level 2: JEE Main

Problem 3: Conformation Energy

Q: The energy difference between staggered and eclipsed conformations of ethane is: (A) 2.5 kJ/mol (B) 12.5 kJ/mol (C) 25 kJ/mol (D) 50 kJ/mol

Solution: (B) 12.5 kJ/mol

This is the torsional strain energy.

  • Staggered: 0 kJ/mol (reference)
  • Eclipsed: +12.5 kJ/mol

JEE Tip: Remember this exact value - it’s frequently asked!

Problem 4: Product Ratio

Q: When 2-methylbutane reacts with Cl₂ in the presence of light, how many monochloro products are possible?

Solution: Structure: CH₃-CH(CH₃)-CH₂-CH₃

Possible positions for Cl:

  1. CH₃-CHCl-CH(CH₃)-CH₂-CH₃ (C-1)
  2. CH₃-CH(CH₃)-CHCl-CH₂-CH₃ (C-2) → 3° carbon
  3. CH₃-CH(CH₃)-CH₂-CHCl-CH₃ (C-3)
  4. Two equivalent CH₃ groups attached to C-2

Wait, need to check for duplicates!

Position 1 and terminal CH₃ on C-2 give different products.

Answer: 4 different monochloro products

JEE Trick: Draw the structure and mark each unique carbon!

Level 3: JEE Advanced

Problem 5: Mechanism-based

Q: In the free radical chlorination of 2,3-dimethylbutane, what is the percentage of 2-chloro-2,3-dimethylbutane in the monochloro products?

Solution:

Structure: (CH₃)₂CH-CH(CH₃)₂

Hydrogen inventory:

  • 1° hydrogens: 12 H (four CH₃ groups)
  • 2° hydrogens: 2 H (two CH groups)
  • 3° hydrogens: 0 H

Relative reactivity per H:

  • 1° : 2° : 3° = 1 : 4 : 5.5

Product from 2° carbon (chlorination at CH):

  • This gives 2-chloro-2,3-dimethylbutane
$$\text{Fraction} = \frac{2 \times 4}{12(1) + 2(4)} = \frac{8}{20} = 0.4 = 40\%$$

Answer: 40%

Advanced concept: Even though there are only 2 secondary H atoms, their high reactivity (4×) makes them contribute significantly!

Problem 6: Multi-step Synthesis

Q: How will you prepare 2,3-dimethylbutane from acetone?

Solution: This requires multiple steps!

Step 1: Convert acetone to pinacol

$$2(\text{CH}_3)_2\text{CO} \xrightarrow{\text{Mg-Hg/H}^+} (\text{CH}_3)_2\text{C(OH)}-\text{C(OH)}(\text{CH}_3)_2$$

Step 2: Pinacol-pinacolone rearrangement

$$(\text{CH}_3)_2\text{C(OH)}-\text{C(OH)}(\text{CH}_3)_2 \xrightarrow{\text{H}^+} (\text{CH}_3)_3\text{C-CO-CH}_3$$

Step 3: Reduction using Clemmensen or Wolff-Kishner

$$(\text{CH}_3)_3\text{C-CO-CH}_3 \xrightarrow{\text{Zn-Hg/HCl}} (\text{CH}_3)_3\text{C-CH}_2\text{-CH}_3$$

This gives 2,2-dimethylbutane (not 2,3!)

Alternative approach:

Start with 2-butanone → reduce to 2-butanol → dehydrate → add HBr → Wurtz reaction

JEE Advanced loves: Multi-step synthesis with retrosynthetic analysis!

Related: Retrosynthetic Analysis

Quick Revision Box

TopicKey PointsJEE Formula/Rule
General FormulaSaturated hydrocarbons$\text{C}_n\text{H}_{2n+2}$
Wurtz Reaction2R-X + 2Na → R-ROnly for symmetrical alkanes
Kolbe Electrolysis2RCOO⁻ → R-R + 2CO₂Decarboxylation at anode
ReductionR-X → R-HZn/HCl or LiAlH₄
Carbonyl ReductionR-CO-R’ → R-CH₂-R'Clemmensen (acidic), WK (basic)
ConformationsRotation around C-CAnti > Gauche > Eclipsed
HalogenationR-H + X₂ → R-XFree radical mechanism
H Reactivity3° > 2° > 1°Based on radical stability
Halogen ReactivityF₂ > Cl₂ > Br₂ > I₂F₂ too reactive, I₂ unreactive

When to Use What: Decision Tree

Need to prepare an alkane?
│
├─ From alkyl halide?
│  ├─ Same alkyl groups → Wurtz reaction
│  ├─ Single product → Reduction (Zn/HCl)
│  └─ From carboxylate → Kolbe electrolysis
│
├─ From carbonyl compound?
│  ├─ Acid-sensitive groups → Wolff-Kishner
│  └─ Base-sensitive groups → Clemmensen
│
└─ From carboxylic acid?
   └─ Decarboxylation (sodalime)

Connection to Other Topics

Prerequisites:

Related Topics:

Applications:

Teacher’s Summary

Key Takeaways

1. Alkanes are saturated (C-C single bonds only) - formula: CₙH₂ₙ₊₂

2. Preparation Methods:

  • Wurtz: Only for symmetrical alkanes (R-X + Na)
  • Reduction: Alkyl halides → Alkanes (Zn/HCl)
  • Clemmensen/WK: Carbonyl → Alkanes (choose based on medium)

3. Conformational Analysis (High-yield for JEE Advanced):

  • Newman projections show rotation around C-C bonds
  • Anti > Gauche > Eclipsed > Fully Eclipsed
  • Energy barrier for ethane: 12.5 kJ/mol

4. Halogenation - Free Radical Mechanism:

  • Reactivity: 3° > 2° > 1° (radical stability)
  • Product ratio = (No. of H) × (Relative reactivity)
  • Cl₂ moderately selective, Br₂ highly selective

5. JEE Strategy:

  • Conformational analysis → Always draw Newman projection
  • Halogenation → Count H atoms at each position
  • Synthesis → Consider acid/base sensitivity for reduction method

“In alkanes, stability is all about minimizing strain - whether it’s conformational strain or radical stability!”

For advanced organic synthesis involving alkanes, next study alkenes to understand how to introduce unsaturation and functional groups!

Interactive Demo: Visualize Molecular Structure

Explore 3D molecular structures of alkanes and understand conformational changes in real-time.