Alkane reactions

Alkanes are nicknamed "paraffins" — Latin for little affinity — because their strong, non-polar C–H and C–C bonds make them resist attack by most acids, bases, and ordinary reagents. Yet under the right push (UV light, heat, or a flame) alkanes do react, and ISC Class 11 asks you to master exactly three such reactions: free-radical halogenation, combustion, and pyrolysis (cracking). Halogenation replaces an H atom with a halogen through a three-stage radical chain; combustion burns alkanes in oxygen to release heat, the basis of their use as fuels; pyrolysis breaks bigger alkanes into smaller alkanes and alkenes when heated without air. This article works through the mechanism, the balanced equations, and a full stoichiometry calculation you can reuse on any alkane.

The core insight

Alkanes are largely unreactive but undergo three characteristic reactions — free-radical halogenation (initiation, propagation, termination with $\text{Cl}_2$ or $\text{Br}_2$ ), complete combustion to $\text{CO}_2$ and $\text{H}_2\text{O}$ following $C_nH_{2n+2} + \left(\frac{3n+1}{2}\right)O_2 \rightarrow nCO_2 + (n+1)H_2O$ , and pyrolysis/cracking into smaller alkanes and alkenes on strong heating without air.

The core idea

Free-radical halogenation. Alkanes react with $\text{Cl}_2$ or $\text{Br}_2$ in sunlight or heat via a radical chain mechanism with three stages. Initiation absorbs energy and splits the halogen molecule homolytically:

Cl_2 \xrightarrow{h\nu} 2\,Cl^{\bullet}

where: $h\nu$ is the photon energy (UV light or heat) supplying the bond-dissociation energy, and $Cl^{\bullet}$ is a chlorine free radical (one unpaired electron). Propagation is two repeating steps that consume methane and chlorine while regenerating the chain carrier:

Cl^{\bullet} + CH_4 \longrightarrow CH_3^{\bullet} + HCl, \qquad CH_3^{\bullet} + Cl_2 \longrightarrow CH_3Cl + Cl^{\bullet}

The regenerated $Cl^{\bullet}$ re-enters the first step, so the cycle repeats thousands of times before termination, when two radicals combine and destroy the chain carriers, e.g. $Cl^{\bullet}+Cl^{\bullet}\rightarrow Cl_2$ or $CH_3^{\bullet}+CH_3^{\bullet}\rightarrow C_2H_6$ . Because chloromethane still has C–H bonds, further substitution can occur, giving a mixture of $\text{CH}_2\text{Cl}_2$ , $\text{CHCl}_3$ , and $\text{CCl}_4$ unless the halogen is limited.

Combustion. Alkanes burn in excess oxygen to give $\text{CO}_2$ and $\text{H}_2\text{O}$ with a large release of heat — the reason they are used as fuels:

C_nH_{2n+2} + \left(\frac{3n+1}{2}\right)O_2 \longrightarrow n\,CO_2 + (n+1)\,H_2O

where: $n$ is the number of carbon atoms in the alkane, and $\frac{3n+1}{2}$ is the number of moles of $\text{O}_2$ needed per mole of alkane for complete combustion. With a limited oxygen supply, incomplete combustion occurs, producing carbon monoxide or soot instead of (or along with) $\text{CO}_2$ .

Pyrolysis / cracking. Higher alkanes, strongly heated without air, break down into smaller alkanes and alkenes through C–C bond homolysis and radical fragmentation:

C_{10}H_{22} \xrightarrow{\Delta} C_5H_{12} + C_5H_{10}

This is industrially important for converting heavy petroleum fractions into lighter, more useful ones such as petrol, since crude oil naturally yields far more heavy alkanes than the light fractions that fuel vehicles and heat homes. Cracking proceeds through a free-radical mechanism, mirroring the same C–C bond homolysis idea used to explain radical chains in halogenation, so once you understand one you can reason through the other.

Figure

Three boxes showing free radical halogenation combustion and pyrolysis of an alkane with their reagents and products

Derivation ABalancing the general alkane combustion equation

The goal is to derive the coefficient of oxygen, $\frac{3n+1}{2}$ , needed to completely burn one mole of any alkane $C_nH_{2n+2}$ into carbon dioxide and water, using nothing more than atom-counting balance, so that the same formula can be applied instantly to methane, propane, or any higher alkane without re-deriving it each time.

Step 1. Write the unbalanced skeleton equation. One mole of alkane $C_nH_{2n+2}$ reacts with oxygen to give carbon dioxide and water: $C_nH_{2n+2} + O_2 \rightarrow CO_2 + H_2O$ . All the carbon in the alkane must end up in $\text{CO}_2$ , and all the hydrogen must end up in $\text{H}_2\text{O}$ .

Step 2. Balance carbon and hydrogen first, since oxygen appears in two different product molecules and is easiest to balance last. There are $n$ carbon atoms per alkane molecule, so the $\text{CO}_2$ coefficient must be $n$ . There are $2n+2$ hydrogen atoms per alkane molecule, and each water molecule holds 2 hydrogens, so the $\text{H}_2\text{O}$ coefficient must be $\frac{2n+2}{2} = n+1$ .

Step 3. Count total oxygen atoms needed on the product side. The $n$ molecules of $\text{CO}_2$ contribute $2n$ oxygen atoms, and the $(n+1)$ molecules of $\text{H}_2\text{O}$ contribute $(n+1)$ oxygen atoms, giving a total of $2n + (n+1) = 3n+1$ oxygen atoms.

Step 4. Convert oxygen atoms to $\text{O}_2$ molecules by dividing by 2, since each $\text{O}_2$ molecule supplies two oxygen atoms: coefficient of $\text{O}_2 = \frac{3n+1}{2}$ . The fully balanced equation is therefore $C_nH_{2n+2} + \left(\frac{3n+1}{2}\right)O_2 \rightarrow n\,CO_2 + (n+1)\,H_2O$ , and it stays balanced for every value of $n$ — check $n=1$ (methane): $\frac{3(1)+1}{2}=2$ , giving the familiar $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ .

Worked example

Problem: Write the balanced equation for complete combustion of propane ( $\text{C}_3\text{H}_8$ ), and calculate (a) the volume of $\text{O}_2$ at STP needed to burn 22 g of propane completely, and (b) the mass of $\text{CO}_2$ produced. (Molar mass of propane $= 44\ \text{g mol}^{-1}$ ; molar mass of $\text{CO}_2 = 44\ \text{g mol}^{-1}$ ; molar volume at STP $= 22.4\ \text{L mol}^{-1}$ .)

Step 1. Balance the equation using the general formula with $n=3$ : $C_3H_8 + 5\,O_2 \longrightarrow 3\,CO_2 + 4\,H_2O$ . Check: $\frac{3n+1}{2} = \frac{10}{2} = 5$ , matching the coefficient of $\text{O}_2$ derived above.

Step 2. Find moles of propane reacting: $\dfrac{22\ \text{g}}{44\ \text{g mol}^{-1}} = 0.5\ \text{mol}$ .

Step 3. Use the mole ratio from the balanced equation to find moles of $\text{O}_2$ . Since 1 mol propane requires 5 mol $\text{O}_2$ , 0.5 mol propane requires $0.5 \times 5 = 2.5\ \text{mol}\ O_2$ . Volume at STP $= 2.5 \times 22.4 = 56\ \text{L}$ .

Step 4. Find the mass of $\text{CO}_2$ produced. Since 1 mol propane gives 3 mol $\text{CO}_2$ , 0.5 mol propane gives $0.5 \times 3 = 1.5\ \text{mol}\ CO_2$ , so mass $= 1.5 \times 44 = 66\ \text{g}$ . Answer: 56 L of $\text{O}_2$ at STP is required, and 66 g of $\text{CO}_2$ is produced.

Common traps

Calling free-radical halogenation an electrophilic or nucleophilic reaction — it proceeds via neutral radicals, not charged intermediates.
Assuming halogenation stops cleanly at the mono-substituted product; without a large excess of methane, further substitution gives a mixture of chloromethanes.
Confusing homolytic fission (each atom keeps one electron, giving radicals) with heterolytic fission (both electrons go to one atom, giving ions) — halogenation initiation is homolytic.
Mixing up combustion (reaction with excess $\text{O}_2$ , gives $\text{CO}_2/\text{H}_2\text{O}$ , exothermic) with pyrolysis/cracking (thermal decomposition in the absence of air, breaking C–C bonds to give smaller alkanes/alkenes).
Forgetting to balance moles before converting to mass or volume in a stoichiometry problem — always find moles of the given substance first, apply the mole ratio from the balanced equation, then convert the answer to whatever unit the question asks for.

Ask the tutor: if the radical-chain mechanism steps or a combustion stoichiometry calculation trips you up, ask the AI tutor to work through a similar problem with you interactively.

Why does methane keep reacting even after forming chloromethane?

Because chloromethane, $\text{CH}_3\text{Cl}$ , still has three C–H bonds that a chlorine radical can attack just as easily as those in methane. Once some $\text{CH}_3\text{Cl}$ has formed, the same propagation cycle can abstract one of its hydrogens, producing $\text{CH}_2\text{Cl}_2$ , and then $\text{CHCl}_3$ , and eventually $\text{CCl}_4$ . To get mostly mono-chlorinated product, chemists use a large excess of methane so that a chlorine radical is far more likely to collide with an unreacted methane molecule than with a partly chlorinated one.

Is combustion the same reaction whether oxygen is plentiful or limited?

No. With excess oxygen (complete combustion), every carbon atom is fully oxidised to $\text{CO}_2$ and every hydrogen ends up as $\text{H}_2\text{O}$ , releasing the maximum possible heat. With a restricted oxygen supply (incomplete combustion), carbon cannot be fully oxidised, so you get carbon monoxide (CO), or even unburnt carbon as soot, mixed in with or instead of $\text{CO}_2$ . Incomplete combustion also releases less heat and is dangerous because CO is toxic, which is why appliances burning alkanes need good ventilation.

Why does cracking need heat but no oxygen, while combustion needs oxygen?

Combustion is an oxidation reaction — the alkane's carbon and hydrogen atoms bond to oxygen atoms from $\text{O}_2$ , so oxygen must be present as a reactant. Pyrolysis (cracking) is not an oxidation at all; it is simply the thermal breaking of C–C bonds by strong heating, with no new element added. If air were present during cracking, the alkane would combust instead of cracking cleanly into smaller alkanes and alkenes, which is why industrial cracking units keep air out and rely purely on heat (and often a catalyst) to snap the carbon chain.

Alkane reactions

The core idea

Worked example

Common traps

Before this

Where this leads