Alkanes

Light a gas stove and you are burning the simplest family of organic compounds in front of you: the alkanes. They are hydrocarbons built entirely from single carbon–carbon and carbon–hydrogen bonds — no double bonds, no triple bonds, nothing left "unsaturated." That plain-vanilla bonding is exactly why alkanes behave the way they do: chemically calm towards most reagents, yet happy to burn and, under sunlight, to swap a hydrogen for a halogen one atom at a time. This article builds the ICSE Class 10 picture of alkanes from the ground up — where the general formula $C_nH_{2n+2}$ comes from, how methane is made in the laboratory, why substitution (not addition) is their signature reaction, and how to turn the combustion equation into an exam-ready mole calculation.

The core insight

Alkanes are saturated hydrocarbons of general formula $C_nH_{2n+2}$ , made only of single bonds, which makes them largely unreactive except for combustion and sunlight-driven substitution with halogens.

The core idea

Saturation is the whole story. Every carbon atom in an alkane forms four single bonds — to other carbons or to hydrogens — and there is no spare bonding capacity left over for anything to "add" across. This single structural fact explains almost every property discussed below: why alkanes are chemically dull compared to alkenes, and why their one characteristic reaction with halogens has to be substitution rather than addition.

The general formula. Counting carbons and hydrogens in methane, ethane, propane and butane reveals a pattern:

C_nH_{2n+2}

where: $n$ is the number of carbon atoms in the chain (so $n = 1$ gives $CH_4$ , $n = 2$ gives $C_2H_6$ , and so on), and $2n+2$ is the number of hydrogen atoms needed to saturate every remaining bond on those carbons. The formula is derived from simple valency counting below.

Laboratory preparation of methane. Sodium ethanoate is heated with soda lime (a mixture of NaOH and CaO that absorbs moisture and moderates the reaction), which strips away the $-COONa$ part of the molecule as sodium carbonate and releases methane:

CH_3COONa + NaOH \xrightarrow{\text{soda lime, heat}} CH_4\uparrow + Na_2CO_3

This is a decarboxylation — a carboxylate group is removed from the parent compound, leaving a shorter hydrocarbon behind.

Building longer chains: the Wurtz reaction. Two alkyl halide molecules can be joined using sodium metal in dry ether, doubling the carbon count in one step:

2C_2H_5Br + 2Na \xrightarrow{\text{dry ether}} C_4H_{10} + 2NaBr

This is only an outline — the mechanism is not required at this level — but it shows how a four-carbon alkane can be assembled from two two-carbon alkyl halides.

Physical properties. The lower alkanes (methane to butane) are colourless, odourless gases at room temperature. Being non-polar molecules, they dissolve in organic solvents but not in water, and both their density and boiling point rise steadily as the chain gets longer — more carbons mean stronger intermolecular attractions, even though every bond inside the molecule is still a plain single bond.

Figure

Bar chart of boiling point in degrees Celsius for methane ethane propane and butane rising steadily as the carbon chain lengthens

Combustion. Alkanes burn in excess oxygen to give carbon dioxide, water and heat — the reaction that makes methane and propane/butane useful fuels:

CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{heat}

Substitution with halogens. In sunlight, alkanes react with halogens by replacing hydrogen atoms one at a time; because there is no double or triple bond for the halogen to "add" across, this must be a substitution:

CH_4 + Cl_2 \xrightarrow{\text{sunlight}} CH_3Cl + HCl

Alkanes are otherwise unreactive towards dilute acids, alkalis and common oxidising agents at room temperature — a direct consequence of being saturated.

Uses. Methane is the chief component of natural gas and biogas; ethane and propane/butane (as LPG) are widely used fuels; higher alkanes serve as solvents and as starting materials for other organic compounds.

Derivation AWhere the general formula CnH2n+2 comes from

The goal is to show, using nothing more than the rule that carbon always forms four bonds, that a straight-chain alkane of $n$ carbon atoms must carry exactly $2n + 2$ hydrogen atoms — turning the memorised formula $C_nH_{2n+2}$ into something you can rebuild from scratch if you forget it in an exam.

Step 1. Count the total bonding capacity of the carbon skeleton. Every carbon atom forms exactly four single covalent bonds (carbon's valency is 4). With $n$ carbon atoms in the chain, the total number of bonds available across all of them is $4n$ .

Step 2. Subtract the bonds already used to link the carbons together. A straight chain of $n$ carbons has $(n-1)$ carbon-to-carbon bonds joining consecutive atoms. Each such C–C bond uses up one bonding slot on each of the two carbons it connects, so it removes $2$ slots from the total pool of $4n$ . With $(n-1)$ such bonds, the slots used up are $2(n-1)$ .

Step 3. What remains must go to hydrogen. The leftover bonding capacity is $4n - 2(n-1) = 4n - 2n + 2 = 2n + 2$ . Every one of these remaining slots is filled by a single covalent bond to one hydrogen atom (hydrogen's valency is 1), so the number of hydrogen atoms is exactly $2n + 2$ .

Step 4. Check against the known members. For $n = 1$ (methane), $2(1)+2 = 4$ , matching $CH_4$ . For $n = 4$ (butane), $2(4)+2 = 10$ , matching $C_4H_{10}$ . The formula $C_nH_{2n+2}$ therefore is not an arbitrary pattern to memorise — it falls directly out of carbon's fixed valency of four and the fact that a saturated chain has no bonds left over for anything but hydrogen.

Worked example

Question: 4.0 g of methane is burnt completely in excess oxygen. Calculate (a) the mass of carbon dioxide produced and (b) the volume of oxygen (at STP) consumed. (Molar mass: $CH_4 = 16\ \text{g/mol}$ , $CO_2 = 44\ \text{g/mol}$ ; 1 mole of any gas occupies 22.4 L at STP.)

Step 1. Write the balanced combustion equation and read off the mole ratio: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ . One mole of methane needs two moles of oxygen and produces one mole of carbon dioxide.

Step 2. Convert the given mass of methane to moles: $\text{moles of } CH_4 = \dfrac{4.0\ \text{g}}{16\ \text{g/mol}} = 0.25\ \text{mol}$ .

Step 3. Use the 1:1 ratio to find $CO_2$ produced: $0.25\ \text{mol}$ of $CH_4$ gives $0.25\ \text{mol}$ of $CO_2$ , which is a mass of $0.25 \times 44 = 11.0\ \text{g}$ .

Step 4. Use the 1:2 ratio to find the oxygen consumed: $0.25\ \text{mol}$ of $CH_4$ needs $0.25 \times 2 = 0.50\ \text{mol}$ of $O_2$ . At STP this is a volume of $0.50 \times 22.4 = 11.2\ \text{L}$ . Answer: 11.0 g of $CO_2$ is produced, and 11.2 L of $O_2$ (at STP) is consumed.

Common traps

Calling the alkane–halogen reaction an "addition" reaction — alkanes are saturated, so halogens can only substitute an existing hydrogen, never add across a bond that isn't there.
Forgetting the exact reagent (soda lime = NaOH + CaO) for methane's laboratory preparation, or writing plain NaOH alone.
Assuming alkanes are highly reactive because they burn readily — combustion needs an ignition source and oxygen; otherwise alkanes are quite unreactive towards common laboratory reagents at room temperature.
Using the wrong mole ratio from the combustion equation — one mole of $CH_4$ needs two moles of $O_2$ , not one, and produces one mole of $CO_2$ but two moles of $H_2O$ .

Ask the tutor: Want to practise more mole calculations on alkane combustion, or work through the Wurtz reaction with a different alkyl halide? Ask the AI tutor for extra examples.

Why do alkanes need sunlight to react with chlorine, but no special conditions to burn?

Combustion is driven by heat once a flame ignites the mixture, and the reaction is highly exothermic, so it runs on its own once started. Substitution with chlorine is different: it needs sunlight (or UV light) to break the Cl-Cl bond into reactive chlorine atoms in the first place, because alkanes have no double bond to attract the halogen the way alkenes do. Without that initial energy input, methane and chlorine can simply sit together in the dark with no visible reaction.

If alkanes are saturated and unreactive, why are they used as fuels at all?

"Unreactive" here means unreactive towards ordinary reagents like acids, alkalis and oxidising agents at room temperature — it does not mean alkanes cannot release energy. Combustion is a special case: given enough activation energy (a spark or flame) and oxygen, the strong C-H and C-C bonds break and reform into the much more stable bonds of $CO_2$ and $H_2O$ , releasing a large amount of heat. That is exactly the property that makes methane, propane and butane good fuels.

Why is soda lime used instead of just heating sodium ethanoate with NaOH?

Pure NaOH alone is corrosive to the glass and can also char the organic compound at the temperatures needed, giving a messy reaction. Soda lime — a mixture of NaOH and CaO — supplies the NaOH needed for decarboxylation while the CaO absorbs moisture and moderates the reaction, keeping it controlled and giving a cleaner yield of methane gas.

Alkanes

The core idea

Worked example

Common traps

Before this

Where this leads