Alkanes

Alkanes are SATURATED HYDROCARBONS — molecules containing only carbon and hydrogen, with only single covalent bonds.

General formula: $C_n H_{2n+2}$.

Substituting n = 1, 2, 3, 4 gives the first four alkanes:

n	Name	Molecular formula	Boiling point	State at room T
1	Methane	CH₄	−161 °C	Gas
2	Ethane	C₂H₆	−89 °C	Gas
3	Propane	C₃H₈	−42 °C	Gas
4	Butane	C₄H₁₀	−1 °C	Gas

(Pentane C₅H₁₂ bp +36 °C is the first liquid alkane at room T.)

Displayed structural formulae (all bonds shown).

Methane (CH₄) — tetrahedral with 4 C–H bonds:

    H
    |
H - C - H
    |
    H

Ethane (C₂H₆) — two tetrahedral C atoms connected by one C–C bond:

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

Propane (C₃H₈) — three C in a chain:

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

Butane (C₄H₁₀) — four C in a chain (n-butane / straight-chain isomer):

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

Geometry of alkanes.

Each carbon in an alkane is sp³ hybridised:

• 4 single bonds.
• TETRAHEDRAL geometry — bonds point to corners of a tetrahedron.
• Bond angle ~ 109.5° (the tetrahedral angle).

These are very strong, stable structures — no strain, no weak spots. The bonds are all SINGLE (no π bond), so there's no exposed electron-rich region for reagents to attack.

Bond energies in alkanes:

• C–H ~ 412 kJ/mol (strong).
• C–C ~ 348 kJ/mol (strong).

These are MUCH stronger than typical intermolecular forces (~ 10-20 kJ/mol). On melting or boiling, only the intermolecular forces are overcome; the covalent bonds stay intact.

Physical-property trends.

As chain length increases (more carbons in the alkane):

• Boiling point INCREASES (e.g. CH₄ −161 °C → C₅H₁₂ +36 °C → C₁₀H₂₂ +174 °C).
• Viscosity INCREASES — longer chains tangle more.
• Flammability DECREASES — longer alkanes vaporise less easily.
• State transitions: C1-C4 gases; C5-C16 liquids; C17+ solids (waxes).

The reason is stronger intermolecular forces (van der Waals / London dispersion) in longer molecules — they have more electrons + larger surface area for intermolecular contact.

NOT 'stronger covalent bonds' — those don't change. The number of bonds increases, but each bond is the same type and strength.

Why so many possible alkanes.

For molecules with 4+ carbons, multiple structural isomers exist. The number of isomers grows quickly:

Carbon count	Number of structural isomers
C4 (butane)	2
C5 (pentane)	3
C6 (hexane)	5
C7	9
C8 (octane)	18
C10	75
C20	366,319

Despite this complexity, all these isomers belong to the same homologous series (alkanes) and have similar chemistry — they all undergo combustion and substitution with halogens.

The two isomers of butane (C₄H₁₀):

1. n-Butane: straight 4-C chain. CH₃CH₂CH₂CH₃. bp −1 °C.
2. 2-Methylpropane: 3-C propane chain with a methyl branch on C2. (CH₃)₃CH. bp −12 °C.

Same molecular formula; different structures. The branched isomer has a lower bp because branching reduces surface contact between molecules → weaker intermolecular forces.

The most important reaction of alkanes is COMBUSTION — burning in oxygen. This releases LARGE amounts of heat → alkanes are the principal FUELS of modern society:

• Methane (natural gas) — domestic heating, cooking, power stations.
• Propane and butane (LPG) — bottled gas for cooking, heating, vehicles.
• Pentane → octane (petrol) — cars.
• Decane → hexadecane (kerosene, diesel) — aircraft, trucks.

Combustion is HIGHLY exothermic. For methane: ΔH ≈ −890 kJ/mol. This is much more energy than is released by most other chemical reactions per mole.

Complete combustion.

With PLENTY of oxygen (excess O₂), every carbon atom is oxidised to CO₂ and every hydrogen to H₂O:

$\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O(l)}$

$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$

$2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$

Features of complete combustion:

• BLUE NON-LUMINOUS flame (Bunsen with air hole open).
• High temperature (~ 1500 °C).
• Maximum heat released.
• Only CO₂ + H₂O as products (no soot, no CO).
• Most efficient — full energy of fuel released.

How to balance combustion equations.

For C_x H_y + ? O₂ → x CO₂ + (y/2) H₂O:

1. Carbons: x atoms in alkane → x molecules of CO₂.
2. Hydrogens: y atoms in alkane → y/2 molecules of H₂O.
3. Oxygen atoms in products: 2x (from CO₂) + y/2 (from H₂O) = 2x + y/2.
4. So number of O₂ molecules = (2x + y/2) / 2 = x + y/4.

Example: C₃H₈ + ? O₂ → 3 CO₂ + 4 H₂O.

• O in products = 3 × 2 + 4 × 1 = 10. O₂ needed = 10/2 = 5.
• So: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. ✓

If the answer isn't a whole number, MULTIPLY THROUGH by 2. e.g. C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O → multiply by 2 → 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.

Incomplete combustion.

With LIMITED oxygen, carbon is only partially oxidised. Products are a MIXTURE of:

• CO (carbon monoxide) — partial oxidation.
• C (soot / particulates) — no oxidation.
• CO₂ — still some complete oxidation.
• H₂O — H always becomes H₂O (it has nowhere else to go).

Example equations:

$2\text{CH}_4 + 3\text{O}_2 \rightarrow 2\text{CO} + 4\text{H}_2\text{O}$ (CO product)

$\text{CH}_4 + \text{O}_2 \rightarrow \text{C(s)} + 2\text{H}_2\text{O}$ (soot product)

Features of incomplete combustion:

• YELLOW LUMINOUS flame (Bunsen with air hole closed). Glowing soot is incandescent → yellow colour.
• Lower temperature (less energy released).
• Deposits black SOOT on cool surfaces (e.g. blackens a porcelain dish held in the flame).
• Releases TOXIC CO.
• LESS EFFICIENT — wasted fuel.

Where it happens:

• Faulty gas boilers / heaters in poorly-ventilated rooms.
• Old or badly-tuned engines.
• A Bunsen with the air hole closed (deliberately for some experiments — but the flame is not hot enough for chemistry).

Toxicity of carbon monoxide.

CO is the MAIN health hazard from incomplete combustion. It is:

• Colourless and odourless — undetectable without a CO alarm.
• Binds to HAEMOGLOBIN in red blood cells ~200× more strongly than O₂.
• Prevents the blood from carrying O₂ to tissues → oxygen starvation.
• Symptoms: headache, dizziness, drowsiness, unconsciousness, DEATH.
• 'Silent killer' — kills hundreds of people each year worldwide from faulty gas heaters.

Safety:

• Install CO detectors in homes with gas heaters / boilers.
• Have gas appliances serviced regularly.
• Ensure good ventilation when using gas (open windows, working flues).
• Never run a car engine in an enclosed garage.

Why combustion is exothermic.

Combustion forms strong O–H and C=O bonds (in H₂O and CO₂) which release MORE energy than is absorbed breaking the C–H and O=O bonds in the reactants. The 'extra' energy is released as heat.

For methane: bonds broken = 4(C–H) + 2(O=O) = 1648 + 992 = 2640 kJ; bonds made = 2(C=O) + 4(O–H) = 1486 + 1852 = 3338 kJ; ΔH = 2640 − 3338 = −698 kJ/mol. (Calculated value differs slightly from the experimental −890 kJ/mol due to mean-bond-energy approximations.)

Apart from combustion, alkanes also react with HALOGENS (Cl₂, Br₂) via SUBSTITUTION — a H atom is replaced by a halogen atom.

Conditions: UV LIGHT is essential.

Without UV light (or strong sunlight), no reaction occurs at room T. With UV, the reaction is moderate-paced.

General equation:

$\text{alkane} + X_2 \xrightarrow{\text{UV}} \text{haloalkane} + HX$

Examples:

Methane + chlorine: $\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV}} \text{CH}_3\text{Cl} + \text{HCl}$

Methane + bromine: $\text{CH}_4 + \text{Br}_2 \xrightarrow{\text{UV}} \text{CH}_3\text{Br} + \text{HBr}$

Ethane + chlorine: $\text{C}_2\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{UV}} \text{C}_2\text{H}_5\text{Cl} + \text{HCl}$

Why UV light is essential.

UV photons carry enough energy (~ 250-400 kJ/mol) to break the relatively WEAK halogen-halogen bond:

• Cl–Cl bond energy = 242 kJ/mol — easily broken by UV.
• Br–Br = 193 kJ/mol — even more easily broken.

UV breaks the halogen molecule into FREE RADICALS (atoms with unpaired electrons):

$\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}\bullet$

These reactive Cl atoms attack the alkane, abstracting a H atom:

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

The methyl radical then reacts with another Cl₂ molecule:

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

This is a free-radical CHAIN REACTION — the Cl• regenerated at the end of one cycle starts the next. One UV photon can produce many CH₃Cl molecules.

(The full mechanism with initiation/propagation/termination is A-level. At IGCSE, just say 'UV breaks Cl₂ into reactive atoms / radicals'.)

Why visible light doesn't work.

Visible light photons carry ~ 150-300 kJ/mol — usually not enough to break Cl–Cl. (Some UV-visible boundary photons can; that's why sunlight works.) Heat (very high T) can also break Cl–Cl thermally, but that's a different mechanism.

Multiple substitutions — products are a mixture.

Once CH₃Cl forms, it STILL has 3 H atoms. Each can be substituted in turn:

$\text{CH}_3\text{Cl} + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}_2 + \text{HCl}$ (dichloromethane) $\text{CH}_2\text{Cl}_2 + \text{Cl}_2 \rightarrow \text{CHCl}_3 + \text{HCl}$ (trichloromethane / chloroform) $\text{CHCl}_3 + \text{Cl}_2 \rightarrow \text{CCl}_4 + \text{HCl}$ (tetrachloromethane)

In practice, a MIXTURE of all four products forms (plus unreacted starting materials). The ratio depends on the relative amounts of CH₄ and Cl₂:

• Excess CH₄ + little Cl₂ → mostly CH₃Cl + unreacted CH₄.
• Excess Cl₂ + little CH₄ → mostly CCl₄ + intermediates.

So substitution gives a mixture — not great for making a single pure compound. To get one specific haloalkane, chemists use different methods (e.g. alcohol + HX → haloalkane).

Important contrast: substitution (alkanes) vs addition (alkenes).

Same reagent (Br₂) gives DIFFERENT reactions:

• Alkane + Br₂ →UV→ substitution: 1 Br atom replaces 1 H atom. HBr formed. Slow, needs UV.
• Alkene + Br₂ → addition: 2 Br atoms add across C=C double bond. No HBr formed. Fast, no UV needed.

Bromine water test exploits this difference: at room T without UV, only alkenes react (decolourise the orange Br₂). Alkanes don't react (Br₂ water stays orange).

Why alkanes do undergo substitution but not addition.

• Alkanes have no C=C double bond → no electron-rich site for Br₂ to add across.
• C–H bonds can be broken (with UV-activated Cl/Br radicals) → substitution possible.
• Substitution is SLOWER than addition because (a) bond-breaking is needed first (Cl–Cl, then C–H); (b) only one of the H atoms is replaced per cycle.

Industrial relevance.

The mass-produced chloromethanes (CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄) have various uses:

• CH₃Cl — refrigerant (formerly), methylating agent in chemistry.
• CH₂Cl₂ (dichloromethane) — paint stripper, degreaser.
• CHCl₃ (chloroform) — historical anaesthetic (now replaced); laboratory solvent.
• CCl₄ — was used as dry-cleaning solvent; banned due to environmental ozone damage.

Modern environmental concerns: chlorinated solvents are 'organic pollutants' and many are restricted under the Montreal Protocol (ozone-depleting chemicals).

Alkanes are described as 'relatively unreactive' compared with most other organic compound classes (alkenes, alcohols, carboxylic acids). Here's why:

Reason 1: Strong covalent bonds.

The bonds in alkanes are STRONG single covalent bonds:

• C–H bond ≈ 412 kJ/mol.
• C–C bond ≈ 348 kJ/mol.

These are typical bond strengths for non-polar organic single bonds — much stronger than weak intermolecular forces (~ 10-20 kJ/mol). At room temperature, few reactant molecules have enough kinetic energy to break a C–H or C–C bond — so direct attack is very slow.

Reason 2: No functional group / no electron-rich site.

Reactive organic compounds typically have a 'functional group' — a region that's chemically different from the rest of the molecule and serves as a site for reactions:

• Alkenes have C=C → electron-rich → attacked by electrophiles (Br₂, HBr).
• Alcohols have −OH → polar → reacts with acids, bases, oxidisers.
• Carboxylic acids have −COOH → polar, acidic → reacts with metals, bases.

Alkanes have NONE of these. They are just chains of C–C and C–H. There's no 'attack point' for polar reagents.

Reason 3: Non-polar molecules.

C and H have similar electronegativities (C = 2.55, H = 2.20 on the Pauling scale — small difference). The C–H bond is essentially NON-POLAR. The molecule overall has no permanent dipole.

Polar reagents (acids, bases, water, salts) don't 'see' anything to attack in a non-polar alkane molecule. The two layers (water + alkane) don't even mix.

What alkanes DO react with — only two main reactions.

(1) Combustion with O₂. Why this works despite the strong bonds:

• High temperature (flame) → many molecules have energy > Ea.
• Reaction releases enormous energy (forming strong O–H and C=O bonds) → exothermic → self-sustaining once started.
• A spark or flame provides the initial Ea.

(2) Substitution with halogens in UV light. Why this works:

• UV provides energy to break the WEAK halogen-halogen bond first.
• The resulting halogen ATOM (free radical) is highly reactive.
• It can then attack the C–H bond.

Both require an external energy input (heat / UV) to overcome the inherent unreactivity of the alkane.

What alkanes do NOT react with under normal conditions:

• Bromine water (Br₂(aq)) at room T — no UV in this test → no substitution → no reaction → bromine water stays orange.
• Acids (HCl, H₂SO₄) — no functional group to react with.
• Alkalis (NaOH) — no functional group; no H atoms acidic enough to deprotonate.
• Oxidising agents (KMnO₄, K₂Cr₂O₇) at room T — alkanes don't oxidise easily without combustion conditions.
• Reducing agents at room T — already in a reduced state; nothing further to reduce.
• Polar solvents like water — immiscible because of polarity mismatch.

This UNREACTIVITY is actually USEFUL: alkanes can be safely stored, transported, and used as fuels without worrying about them reacting with everything around them. They only react when we DELIBERATELY ignite them (combustion) or expose them to UV (substitution).

Why alkanes are different from alkenes.

Both alkanes and alkenes are hydrocarbons. But alkenes have a C=C double bond:

• The π bond of the C=C is electron-rich → ATTRACTS electrophilic reagents like Br₂.
• The π bond is weaker (~ 264 kJ/mol) than a single bond → breaks more easily.
• Addition reactions can occur fast at room T with no catalyst, no UV.

Alkanes don't have this π bond → don't undergo addition → can't be polymerised → can't be hydrogenated (already saturated) → can't decolourise bromine water.

Summary: alkanes are 'inert' in the sense that they don't react with most common reagents. They're useful precisely because of this stability — they don't decompose in air, don't react with metal containers, can be stored long-term.