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Alkanes (Saturated Hydrocarbons)

CBSE Class 11 & JEE Mains • Module 05 of 20 • The Simplest but Most Fundamental Carbon Compounds

📍 Chapter Overview

Alkanes — Mind Map

Topics Covered: General Formula CₙH₂ₙ₊₂ · sp³ Hybridization · Nomenclature · Conformational Isomers (Newman Projections) · Preparation Methods · Physical Properties · Chemical Reactions (Free Radical Halogenation Mechanism, Combustion, Cracking, Isomerisation)

🤖 AI Prompt — Chapter Mind Map: Warm amber-brown themed mind map on dark background. Central node: "ALKANES — CₙH₂ₙ₊₂" in white. Six radiating branches in golden and orange tones: (1) "Structure" — shows methane CH₄ tetrahedral structure with 109.5° bond angles, sp3 hybridization label; (2) "Nomenclature" — branches showing meth, eth, prop, but, pent, hex as chain lengths; (3) "Conformations" — Newman projection circles for eclipsed vs staggered ethane; (4) "Preparation" — shows 3 routes: Wurtz reaction (2RX + 2Na), Kolbe electrolysis, reduction of alkyl halide; (5) "Physical Properties" — BP increases with molecular mass, branching lowers BP, all non-polar; (6) "Chemical Reactions" — main reaction FREE RADICAL HALOGENATION: shows 3 colored boxes: Initiation (Cl₂ → 2Cl•, hν), Propagation (Cl• + CH₄ → CH₃• + HCl, then CH₃• + Cl₂ → CH₃Cl + Cl•), Termination (radical + radical combinations). High-resolution educational illustration, clean bold labels.

1. Structure and Hybridization

General formula: CₙH₂ₙ₊₂ (for acyclic/open-chain alkanes)

Hybridization: All carbons are sp³ hybridized — 4 equivalent sp³ orbitals pointing to corners of a tetrahedron.

Bond angle: 109°28' (tetrahedral angle)

C–C bond length: 1.54 Å (longest C–C single bond — only sigma bonds)

C–H bond length: 1.09 Å

Methane Tetrahedral Structure with sp³ Hybridization Orbital Diagram
Draw two diagrams side by side on white background. Left diagram: 3D tetrahedral structure of methane (CH₄) drawn in perspective. Central carbon atom with 4 hydrogen atoms at the corners of a tetrahedron. Show one C-H bond as a solid wedge (coming toward viewer), one as a dashed wedge (going away from viewer), and two in the plane of the page as regular bonds. Mark the H-C-H bond angle as 109° 28' with a curved arc. Label: "Methane — tetrahedral geometry". Right diagram: sp³ hybridization orbital diagram of carbon. Show the energy level diagram: 2s (one orbital) + 2p (three orbitals) → hybridize → 4 equal sp³ orbitals. Each sp³ orbital shown as a dumbell with one lobe much larger (the bonding lobe). Show all four sp³ orbitals pointing toward tetrahedral corners at 109.5°. Label: "sp³ hybridization — 4 equivalent orbitals". White background, chemistry textbook quality.

2. Conformational Isomers

Alkanes can rotate freely around C–C single bonds. Different spatial arrangements due to rotation are called conformations (not true isomers — they interconvert rapidly).

Newman Projections of Ethane — Staggered vs Eclipsed Conformations
Draw two Newman projections of ethane (CH₃CH₃) on white background, clearly labeled. Left projection: Newman projection looking down the C1–C2 bond axis. Front carbon shown as a circle with a dot in center, three C-H bonds of front carbon radiating outward at 120° intervals. Back carbon shown as the outer circle, three C-H bonds of back carbon visible between the front bonds — alternating, 60° offset from front bonds. This is the STAGGERED conformation. Label: "Staggered conformation — most stable (lowest energy)". Show anti (180°) and gauche (60°) arrangements labeled. Right projection: Same ethane in ECLIPSED conformation — back carbon H atoms exactly behind front carbon H atoms (0° dihedral angle). Label: "Eclipsed conformation — least stable (highest torsional strain, energy difference = 12 kJ/mol)". Between the two projections, draw an energy level bar showing staggered is lower. Show the dihedral angle (torsional angle) measurement in each. Clean, standard chemistry Newman projection notation, white background, textbook quality.

Conformations of Butane (more complex):

Stability Order: Anti > Gauche > Eclipsed > Fully Eclipsed

3. Preparation of Alkanes

3.1 Wurtz Reaction

Reaction: 2 R–X + 2Na → R–R + 2NaX  (in dry ether)

Best used for: Preparing symmetrical alkanes with even carbon numbers. For unsymmetrical alkanes (different RX + R'X + 2Na), you get a mixture of R–R, R–R', and R'–R' — not useful preparatively.

Example: 2 CH₃Br + 2Na → CH₃–CH₃ + 2NaBr  (ethane from methyl bromide)

Mechanism: SN2 attack by R–Na carbanion on R–X → R–R. Or two single-electron transfers → radical coupling.

3.2 Kolbe's Electrolysis

Reaction: 2RCOONa →(electrolysis, anode)→ R–R + 2CO₂ + 2e⁻

At anode: RCOO⁻ → RCOO• → R• + CO₂ → R• + R• → R–R (coupling of radicals)

Best for: Symmetrical alkanes. Important: decarboxylation removes 1 carbon from each acid → product alkane has 2n−2 carbons if starting from n-carbon acid.

Example: CH₃COONa (sodium acetate, 2C) → CH₃–CH₃ (ethane, 2C) + 2CO₂

Wait — from two 2-carbon acids → R = CH₃, R–R = C₂H₆ (ethane, still 2C since each contributes CH₃).

3.3 Reduction of Alkyl Halides

Method 1 — Zn/HCl (nascent hydrogen): R–X + 2[H] → R–H + HX

Method 2 — LiAlH₄ (in dry ether): R–X + LiAlH₄ → R–H (powerful reducing agent)

Method 3 — Hydrogenation of alkenes: R–CH=CH₂ + H₂ →(Ni or Pt, heat)→ R–CH₂–CH₃ (Sabatier-Senderens reduction)

3.4 From Carboxylic Acids — Decarboxylation

RCOONa + NaOH →(CaO, Δ)→ R–H + Na₂CO₃ (soda lime fusion)

Product alkane has ONE FEWER carbon than the carboxylic acid starting material.

Example: CH₃CH₂COONa (sodium propanoate, 3C) + NaOH/CaO → CH₃CH₃ (ethane, 2C) + Na₂CO₃

4. Physical Properties

AlkaneFormulaBoiling PointState at room T
MethaneCH₄−161.5°CGas
EthaneC₂H₆−88.6°CGas
PropaneC₃H₈−42.1°CGas
ButaneC₄H₁₀−0.5°CGas
PentaneC₅H₁₂36.1°CLiquid
HexadecaneC₁₆H₃₄286°CLiquid

Key trends:

5. Chemical Reactions

5.1 Free Radical Halogenation — Most Important!

CH₄ + Cl₂ →(UV light/hν)→ CH₃Cl + HCl  → further: CH₂Cl₂, CHCl₃, CCl₄ (substitution products)

This is a free radical chain reaction involving three stages.

Free Radical Chlorination of Methane — 3-Stage Mechanism
Draw a detailed, step-by-step free radical halogenation mechanism diagram for chlorination of methane (CH₄ + Cl₂ → CH₃Cl + HCl) on a white background. Divide into three clearly labeled colored sections: SECTION 1 — "INITIATION (start the chain)" in red box: Show Cl-Cl bond breaking homolytically under UV light (hν). Use fish-hook (half-headed) arrows showing one electron from each Cl going to each Cl atom. Equation: Cl₂ + hν → 2 Cl• (two chlorine radicals). Show the full/empty bullet point (•) notation on Cl atoms. SECTION 2 — "PROPAGATION (the chain grows)" in blue box: Step 2a: Cl• + CH₄ → •CH₃ + HCl. Show fish-hook arrow from C-H bond to H going to Cl, leaving •CH₃. Step 2b: •CH₃ + Cl₂ → CH₃Cl + Cl•. Show •CH₃ attacking Cl-Cl, fish-hook arrow taking Cl, leaving Cl•. Explain: "Cl• is regenerated — this is the CHAIN — each cycle makes one CH₃Cl". SECTION 3 — "TERMINATION (chain ends)" in green box: Show 3 termination possibilities: (i) Cl• + Cl• → Cl₂, (ii) •CH₃ + Cl• → CH₃Cl, (iii) •CH₃ + •CH₃ → C₂H₆. Explain: "Radicals combine with each other — chain ends, no new radical formed". Add a note: "Overall: CH₄ + Cl₂ → CH₃Cl + HCl. Products of further halogenation: CH₂Cl₂, CHCl₃, CCl₄". White background, clear bold text, educational chemistry textbook quality, high resolution.

Reactivity of halogens: F₂ > Cl₂ > Br₂ >> I₂

Selectivity order (for H abstraction): Br > Cl. Bromine radical attacks 3° H preferentially (more selective), chlorine is much less selective.

3° H reacts faster than 2° H reacts faster than 1° H in free radical halogenation (stability of transition state mirrors stability of radical formed)

5.2 Combustion

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

Alkanes are excellent fuels (high heat of combustion). CH₄ (natural gas), C₄H₁₀ (LPG butane), C₈H₁₈ (octane in petrol).

Incomplete combustion: CO and soot (carbon) formed — toxic CO is the danger in poorly ventilated rooms with gas appliances.

Octane number: Measures anti-knock quality of petrol. Isooctane (2,2,4-trimethylpentane) = octane number 100 (reference). n-Heptane = octane number 0. Adding branched alkanes or aromatic hydrocarbons raises octane number.

5.3 Cracking (Pyrolysis)

Higher alkanes (from petroleum) → heated strongly (450–750°C) → break into smaller alkanes + alkenes.

Thermal cracking: Free radical mechanism at high T.

Catalytic cracking: Uses zeolite catalysts, lower T → produces more branched alkanes (better octane number) → carbocation mechanism.

Example: C₁₆H₃₄ → C₈H₁₈ + C₈H₁₆ (octane + octene)

5.4 Isomerisation

Straight-chain alkanes → converted to branched isomers using AlCl₃ catalyst + HCl at 25–50°C.

Example: n-butane →(AlCl₃/HCl)→ 2-methylpropane (isobutane)

Useful industrially: branched alkanes have higher octane numbers and are better fuels.

Worked Examples

Ex 1 M: Which gives a higher boiling point — n-pentane or 2,2-dimethylpropane (neopentane)?

Solution: n-Pentane has a more extended shape → larger surface area → stronger London dispersion forces → higher BP (36°C vs 9.5°C for neopentane). Branching always lowers BP.


Ex 2 M: In free radical chlorination of propane (CH₃CH₂CH₃), predict the major product.

Solution: Propane has 2 types of H: six 1° H (on C1 and C3) and two 2° H (on C2). Cl• is not very selective but 2° radical is more stable → C2 is preferentially attacked. However, since 1° H outnumber 2° H (6:2 = 3:1 ratio), the 1-chloropropane and 2-chloropropane form in approximately (6×1):(2×3.8) statistical/selectivity product ratio. For Cl halogenation, relative reactivity 1°:2°:3° ≈ 1:3.8:5 (approx). 1-chloro: 6×1=6; 2-chloro: 2×3.8=7.6. So 2-chloropropane is the major product (≈56%).

Practice Problems

E Q1. Write the Wurtz reaction for the preparation of propane from ethyl bromide.

E Q2. Arrange in increasing BP: neopentane, n-pentane, isopentane (2-methylbutane).

M Q3. Write the 3 stages of free radical bromination of ethane with Br₂/hν. Mark each fish-hook arrow and intermediate.

M Q4. Predict the major product of free radical bromination of propane. Is it different from the chlorination major product? Why?

H Q5. In Kolbe's electrolysis, a mixture of sodium acetate (CH₃COONa) and sodium propanoate (CH₃CH₂COONa) is electrolysed. List ALL the alkane products possible and explain why this method gives a mixture.

H Q6. Draw Newman projections for all conformations of butane as C2–C3 bond is rotated from 0° to 360°. Rank them in order of stability.

  1. Writing double-headed curved arrows for free radical steps — WRONG! Free radical steps use half-headed fish-hook arrows (one electron). Full arrow = ionic, half = radical.
  2. Thinking Kolbe products: from CH₃COONa you do NOT get methane — you get ethane (CH₃• + CH₃• → C₂H₆).
  3. Saying branched alkanes have HIGHER BP than straight-chain — WRONG! Branching always DECREASES BP.
  4. Saying I₂/alkane reaction works easily — it does NOT (thermodynamically unfavorable, endothermic).
  5. Forgetting that decarboxylation removes one C — sodium propanoate (3C) gives ethane (2C), NOT propane.
ReactionReagent/ConditionProduct
Wurtz2RX + 2Na, dry etherR–R (higher alkane)
Kolbe electrolysisRCOONa, electrolysis, anodeR–R + CO₂
Reduction (RX)Zn/HCl or LiAlH₄R–H (same C, no C loss)
DecarboxylationRCOONa + NaOH, CaO, ΔR–H (n−1 C alkane)
Free radical halogenationX₂, hνR–X (alkyl halide)
CombustionO₂, sparkCO₂ + H₂O
CrackingHigh T or catalystSmaller alkanes + alkenes