Chemistry · Hydrocarbons

Alkanes — Preparation, Properties, Reactions

Alkanes are the saturated hydrocarbons of general formula $\ce{C_{n}H_{2n+2}}$, the structural backbone of organic chemistry on which every more complex molecule is built. This note follows NCERT Class 11 Chemistry §9.2, supplemented by NIOS §24.1, to cover their nomenclature and isomerism, every standard method of preparation, their physical trends, and the full slate of reactions — culminating in the free-radical halogenation mechanism that the NTA tests almost every year.

What Are Alkanes

Alkanes are saturated hydrocarbons in which every carbon atom is $sp^3$ hybridised and joined to its neighbours by single covalent bonds alone. Successive members differ by a $\ce{-CH2-}$ unit, and constructing the family member by member shows that the molecular formula always satisfies the relation $\ce{C_{n}H_{2n+2}}$. Because each carbon already carries the maximum number of hydrogen atoms permitted by its four bonds, alkanes are described as saturated.

The tetrahedral geometry around each carbon dictates a $\ce{C-C-C}$ bond angle close to $109.5^\circ$, so an extended alkane chain is not straight but zig-zag. The C–C and C–H bonds are nearly non-polar because the electronegativity difference between carbon and hydrogen is small, and this electronic character is the root of almost every property discussed below — from low solubility in water to chemical inertness. Alkanes are historically called paraffins, from the Latin parum affinis, meaning of little affinity.

PropertyDescriptionUnderlying reason
General formulaCnH2n+2Each C is fully saturated with single bonds
Hybridisation$sp^3$ at every carbonFour sigma bonds per carbon
Bond angle$\approx 109.5^\circ$Tetrahedral arrangement
PolarityAlmost non-polarSmall C–H electronegativity gap
Intermolecular forceWeak van der WaalsNon-polar covalent molecules

Nomenclature and Isomerism

The first three alkanes — methane, ethane and propane — have only one possible structure each. From butane onwards, the same molecular formula can be arranged into more than one skeleton, giving rise to chain (skeletal) isomerism. Butane, $\ce{C4H10}$, exists as the straight-chain n-butane and the branched 2-methylpropane (isobutane); pentane, $\ce{C5H12}$, has three isomers: n-pentane, 2-methylbutane and 2,2-dimethylpropane (neopentane).

In IUPAC nomenclature the longest continuous carbon chain is selected as the parent and numbered so that substituents receive the lowest set of locants. Alkyl groups, obtained by removing one hydrogen from an alkane ($\ce{-CH3}$ methyl, $\ce{-C2H5}$ ethyl), are cited as prefixes with their position numbers. Hydrogen atoms themselves are classified by the carbon they sit on — a point that becomes decisive when we reach halogenation selectivity.

Hydrogen typeCarbon it is attached toExample carbon
Primary (1°)Carbon bonded to one other carbon$\ce{-CH3}$ of propane
Secondary (2°)Carbon bonded to two other carbonsMiddle $\ce{-CH2-}$ of propane
Tertiary (3°)Carbon bonded to three other carbonsCentral C of 2-methylpropane

Preparation of Alkanes

Petroleum and natural gas are the principal natural sources of alkanes, but several laboratory routes allow targeted synthesis. NCERT §9.2.2 groups them by the starting material: unsaturated hydrocarbons, alkyl halides, and carboxylic acids. Each method has a characteristic constraint that examiners exploit, so the carbon-counting logic matters as much as the equation itself.

1. From unsaturated hydrocarbons (hydrogenation)

Dihydrogen adds across the multiple bonds of alkenes and alkynes over finely divided platinum, palladium or nickel — the Sabatier–Senderens reaction. The metal adsorbs $\ce{H2}$ on its surface and activates the H–H bond. Platinum and palladium operate at room temperature, while nickel needs higher temperature and pressure.

$$\ce{CH2=CH2 + H2 ->[Pt/Pd/Ni] CH3-CH3}$$ $$\ce{CH3-C#C-H + 2H2 ->[Pt/Pd/Ni] CH3-CH2-CH3}$$

2. From alkyl halides

Alkyl halides (except fluorides) on reduction with zinc and dilute hydrochloric acid give alkanes with the same carbon count:

$$\ce{CH3-Cl + H2 ->[Zn, H+] CH4 + HCl}$$

The more important route is the Wurtz reaction: two molecules of an alkyl halide treated with sodium metal in dry ether couple to give a higher alkane containing twice the carbon atoms of the halide. The product therefore always has an even number of carbons and a symmetrical skeleton.

$$\ce{2CH3Br + 2Na ->[dry\ ether] CH3-CH3 + 2NaBr}$$ $$\ce{2C2H5Br + 2Na ->[dry\ ether] CH3-CH2-CH2-CH3 + 2NaBr}$$

Build the chapter

The hydrogenation route runs backwards from alkenes — see how addition of $\ce{H2}$ across the double bond fits the wider unsaturation story.

3. From carboxylic acids

Heating the sodium salt of a carboxylic acid with soda lime (sodium hydroxide and calcium oxide, typically in a 3:1 ratio) eliminates carbon dioxide and yields an alkane with one carbon atom fewer than the acid. This loss of $\ce{CO2}$ is called decarboxylation.

$$\ce{CH3COONa + NaOH ->[CaO][\Delta] CH4 + Na2CO3}$$

Kolbe's electrolytic method takes the aqueous sodium or potassium salt of a carboxylic acid and electrolyses it. At the anode the carboxylate loses an electron and then $\ce{CO2}$ to give an alkyl free radical; two such radicals combine, so the alkane formed has an even number of carbon atoms.

$$\ce{2CH3COONa + 2H2O ->[electrolysis] C2H6 + 2CO2 + H2 + 2NaOH}$$

MethodReagent / conditionCarbon-count rule
Hydrogenation (Sabatier–Senderens)$\ce{H2}$, Pt / Pd / NiSame as parent alkene/alkyne
Reduction of alkyl halideZn + dilute HClSame as halide
Wurtz reactionNa, dry etherDouble the halide; even, symmetrical
Decarboxylation (soda lime)NaOH + CaO, $\Delta$One carbon less than acid
Kolbe electrolysisElectrolysis of carboxylate saltEven; couples two alkyl radicals
NEET Trap

Wurtz works only for symmetrical, even-carbon alkanes

Because Wurtz couples two alkyl halides, the product is restricted to an even number of carbons and a symmetrical chain. Attempting an unsymmetrical (often odd-carbon) alkane from two different halides yields a mixture of all three coupling products plus side-reaction alkenes and alkanes, so the target is impure and low-yielding.

Memory hook: n-heptane (7 C, odd) cannot be made cleanly by Wurtz — this exact framing was asked in NEET 2020.

Physical Properties

Alkanes are colourless, odourless and almost non-polar; they dissolve in non-polar solvents but are essentially insoluble in water, following the rule like dissolves like. Their molecules attract one another only through weak van der Waals forces, which is why the lightest members are gases. The first four members $\ce{C1}$ to $\ce{C4}$ are gases, $\ce{C5}$ to $\ce{C17}$ are liquids, and those with 18 or more carbons are solids at 298 K.

Boiling point rises steadily with molecular mass because the van der Waals forces grow with molecular size and surface area. Among isomers, branching lowers the boiling point: as a molecule becomes more branched it approaches a spherical shape, reducing the surface area available for contact and weakening the intermolecular attraction. Hence the three isomeric pentanes boil in the order $n\text{-pentane} > 2\text{-methylbutane} > 2{,}2\text{-dimethylpropane}$.

Figure 1 · Branching vs surface contact

Why neopentane boils lower than n-pentane

n-pentane (extended) Large contact area → strong van der Waals → higher b.p. neopentane (spherical) Point contact → weak van der Waals → lower b.p.

The same five carbons pack very differently: the extended chain offers a long line of contact, the spherical isomer only a near-point of contact.

Chemical Properties

Alkanes are generally inert towards acids, bases, oxidising agents and reducing agents — the very inertness implied by the name paraffin. Under forcing conditions, however, they undergo a defined set of reactions. The most heavily examined of these is the substitution reaction, in which one or more hydrogen atoms are replaced by halogen, nitro or sulphonic acid groups.

Halogenation proceeds in diffused sunlight or ultraviolet light, or at high temperature ($573$–$773$ K). Chlorination of methane does not stop cleanly at one stage but proceeds through successive substitution to a mixture of products:

$$\ce{CH4 + Cl2 ->[h\nu] CH3Cl + HCl}$$ $$\ce{CH3Cl + Cl2 ->[h\nu] CH2Cl2 + HCl}$$ $$\ce{CH2Cl2 + Cl2 ->[h\nu] CHCl3 + HCl}$$ $$\ce{CHCl3 + Cl2 ->[h\nu] CCl4 + HCl}$$

The reactivity of the halogens themselves follows $\ce{F2 > Cl2 > Br2 > I2}$. Fluorination is too violent to control, while iodination is slow and reversible — it must be driven by an oxidising agent such as $\ce{HIO3}$ or $\ce{HNO3}$ to remove the $\ce{HI}$ produced. The ease with which different hydrogens are replaced follows the order $3^\circ > 2^\circ > 1^\circ$, which governs the major product when an alkane has more than one type of hydrogen.

NEET Trap

Halogenation selectivity: tertiary beats secondary beats primary

When an alkane carries different kinds of hydrogen, the radical preferentially abstracts the one that gives the most stable carbon radical. Stability order $3^\circ > 2^\circ > 1^\circ$ therefore sets the order of replacement $3^\circ > 2^\circ > 1^\circ$. Do not confuse this with halogen reactivity, which runs $\ce{F2 > Cl2 > Br2 > I2}$.

Two separate orders: which halogen reacts fastest ($\ce{F2>Cl2>Br2>I2}$) vs which H is replaced fastest ($3^\circ>2^\circ>1^\circ$).

Free-Radical Halogenation Mechanism

Halogenation proceeds by a free-radical chain mechanism in three phases: initiation, propagation and termination. The chain is set off because the weak $\ce{Cl-Cl}$ bond breaks more easily than the stronger $\ce{C-C}$ and $\ce{C-H}$ bonds when energised by light or heat.

Initiation — homolysis of chlorine produces two chlorine radicals: $$\ce{Cl2 ->[h\nu] 2Cl^.}$$ Propagation — a chlorine radical abstracts a hydrogen from methane to give a methyl radical; this radical then attacks a fresh chlorine molecule, forming the product and regenerating a chlorine radical, so the chain continues: $$\ce{CH4 + Cl^. -> CH3^. + HCl}$$ $$\ce{CH3^. + Cl2 -> CH3Cl + Cl^.}$$ Termination — the chain stops when two radicals combine, consuming radicals without making new ones. Combination of two methyl radicals explains why ethane appears as a by-product of methane chlorination: $$\ce{Cl^. + Cl^. -> Cl2}$$ $$\ce{CH3^. + CH3^. -> CH3-CH3}$$ $$\ce{CH3^. + Cl^. -> CH3Cl}$$

Figure 2 · Chain mechanism schematic

Initiation → Propagation → Termination flow for $\ce{CH4 + Cl2}$

INITIATION Cl–Cl → 2 Cl• (hν) PROPAGATION (chain cycle) CH₄ + Cl• → CH₃• + HCl CH₃• + Cl₂ → CH₃Cl + Cl• Cl• fed back TERMINATION Cl• + Cl• → Cl₂ CH₃• + CH₃• → C₂H₆ CH₃• + Cl• → CH₃Cl Radicals are regenerated in propagation; the chain ends only when two radicals pair off.

The propagation step is self-sustaining: each turn consumes one radical and makes another, so a single initiation event can convert many molecules before termination breaks the chain.

Combustion, Oxidation and Cracking

Beyond substitution, alkanes undergo combustion, controlled oxidation, isomerisation, aromatisation, reaction with steam and pyrolysis. These define their industrial value as fuels and feedstocks.

Combustion — in excess air or oxygen, alkanes are completely oxidised to $\ce{CO2}$ and water with the release of a large amount of heat, which is why they are used as fuels. The general equation is: $$\ce{C_{n}H_{2n+2} + ( \frac{3n+1}{2} )O2 -> nCO2 + (n+1)H2O}$$ With insufficient air, incomplete combustion gives carbon black, used in inks and pigments: $$\ce{CH4 + O2 -> C + 2H2O}$$

ReactionConditionsRepresentative equation
Controlled oxidation to methanolCu, 523 K, 100 atm$\ce{2CH4 + O2 -> 2CH3OH}$
Oxidation to methanal$\ce{Mo2O3}$, $\Delta$$\ce{CH4 + O2 -> HCHO + H2O}$
3° H to alcohol$\ce{KMnO4}$$\ce{(CH3)3CH -> (CH3)3COH}$
IsomerisationAnhydrous $\ce{AlCl3}$/HCl$n$-hexane → 2- & 3-methylpentane
Aromatisation (reforming)773 K, 10–20 atm, oxide catalyst$\ce{C6H14 -> C6H6 + 4H2}$
Reaction with steamNi, 1273 K$\ce{CH4 + H2O -> CO + 3H2}$
Pyrolysis (cracking)High temperature$\ce{C12H26 -> C7H16 + C5H10}$

Controlled oxidation with a regulated supply of oxygen and the right catalyst halts at a useful intermediate such as methanol, methanal or acetic acid. Isomerisation over anhydrous aluminium chloride converts straight-chain into branched alkanes (raising the octane number of fuel), while aromatisation dehydrogenates and cyclises $\ce{C6+}$ alkanes into benzene and its homologues over oxides of vanadium, molybdenum or chromium on alumina. Finally, pyrolysis (cracking) breaks higher alkanes into smaller alkanes and alkenes by heat, the basis of converting kerosene into oil gas and lighter fuels.

Quick Recap

Alkanes in one screen

  • General formula $\ce{C_{n}H_{2n+2}}$; all $sp^3$, non-polar, weak van der Waals forces, hence inert "paraffins".
  • Preparation: hydrogenation of alkenes/alkynes (Pt/Pd/Ni), reduction of alkyl halides (Zn/HCl), Wurtz (Na, dry ether → even, symmetrical), decarboxylation with soda lime (one C less), Kolbe electrolysis (even C).
  • Boiling point rises with mass; branching lowers it: $n$-pentane > 2-methylbutane > neopentane.
  • Halogenation = free-radical chain (initiation, propagation, termination); halogen reactivity $\ce{F2>Cl2>Br2>I2}$; H replacement $3^\circ>2^\circ>1^\circ$.
  • Other reactions: combustion (fuel), controlled oxidation, isomerisation, aromatisation, steam reforming, pyrolysis (cracking).

NEET PYQ Snapshot — Alkanes — Preparation, Properties, Reactions

Real NEET questions on alkane preparation, the Wurtz reaction, halogenation and physical-property trends.

NEET 2020

Which of the following alkanes cannot be made in good yield by the Wurtz reaction?

  • (1) 2,3-Dimethylbutane
  • (2) n-Heptane
  • (3) n-Butane
  • (4) n-Hexane
Answer: (2) n-Heptane

The Wurtz reaction couples two alkyl halides, giving only even-carbon, symmetrical alkanes. n-Heptane has 7 carbons (odd, unsymmetrical) and so cannot be obtained in good yield by this method.

NEET 2018

Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to a gaseous hydrocarbon containing less than four carbon atoms. (A) is:

  • (1) $\ce{CH#CH}$
  • (2) $\ce{CH2=CH2}$
  • (3) $\ce{CH3-CH3}$
  • (4) $\ce{CH4}$
Answer: (4) CH₄

$\ce{CH4 ->[Br2] CH3Br ->[Na,\ dry\ ether] CH3-CH3}$. Ethane has 2 carbons (less than four) and is gaseous, satisfying the condition; the alkyl bromide came from a substitution on methane.

NEET 2021

$\ce{CH3CH2COO^- Na^+ ->[NaOH, +Heat] CH3CH3 + Na2CO3}$. Identify the missing reagent/chemical.

  • (1) DIBAL-H
  • (2) $\ce{B2H6}$
  • (3) Red Phosphorus
  • (4) CaO
Answer: (4) CaO

The alkane is produced by heating the sodium salt of the carboxylic acid with soda lime — NaOH and CaO in a 3:1 ratio. The missing component is calcium oxide (CaO); this is decarboxylation giving an alkane with one carbon less.

NEET 2024

Statement I: The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane. Statement II: When branching increases, the molecule attains the shape of a sphere; this results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. Choose the most appropriate answer.

  • (1) Both Statement I and Statement II are correct.
  • (2) Both Statement I and Statement II are incorrect.
  • (3) Statement I is correct but Statement II is incorrect.
  • (4) Statement I is incorrect but Statement II is correct.
Answer: (1) Both correct

Branching makes the molecule more spherical, reducing surface contact and weakening van der Waals forces; Statement II correctly explains the boiling-point order in Statement I.

FAQs — Alkanes — Preparation, Properties, Reactions

The recurring conceptual doubts on alkane preparation, selectivity and physical trends.

Why does the Wurtz reaction fail to give a good yield of an alkane with an odd number of carbon atoms?
The Wurtz reaction couples two alkyl halides through sodium in dry ether, so the product always has an even number of carbon atoms and a symmetrical skeleton. To make an odd-carbon or unsymmetrical alkane you must mix two different alkyl halides, but this gives all three possible coupling products (two symmetrical and one cross-coupled) along with alkenes and alkanes from side reactions. The desired unsymmetrical alkane is therefore obtained only as a minor, impure fraction, so the Wurtz reaction is not used for odd-carbon alkanes such as n-heptane.
Which hydrogen is replaced fastest during free-radical halogenation of an alkane?
The ease of replacement of hydrogen atoms follows the order tertiary greater than secondary greater than primary. A tertiary C–H bond gives the most stable carbon radical on abstraction, so it is broken most readily. This selectivity means that in an alkane containing different types of hydrogen, the substitution product arising from the tertiary position is favoured.
Why can methane not be prepared by Kolbe's electrolytic method?
Kolbe's electrolysis couples two alkyl free radicals produced at the anode, so the alkane formed always contains an even number of carbon atoms. Methane has only one carbon, which cannot arise from the coupling of two radicals; the smallest possible product is ethane from two methyl radicals. Hence methane cannot be obtained by this method.
Why does the boiling point of pentane exceed that of its branched isomer neopentane?
Boiling point depends on the strength of intermolecular van der Waals forces, which increase with molecular surface area. n-Pentane has an extended chain with a large contact area, while 2,2-dimethylpropane (neopentane) is nearly spherical and has the smallest surface area among the three pentane isomers. The spherical shape reduces contact between molecules, weakens van der Waals forces, and lowers the boiling point, giving the order n-pentane greater than 2-methylbutane greater than neopentane.
What is the role of soda lime in the decarboxylation of carboxylic acid salts?
Soda lime is a mixture of sodium hydroxide and calcium oxide, usually taken in a 3:1 ratio. When the sodium salt of a carboxylic acid is heated with soda lime, carbon dioxide is eliminated and an alkane containing one carbon atom less than the acid is formed. The calcium oxide keeps the sodium hydroxide dry and prevents it from absorbing moisture and carbon dioxide, allowing the decarboxylation to proceed cleanly.
Why are alkanes called paraffins and why are they chemically inert?
Alkanes are saturated hydrocarbons with the general formula CnH2n+2 containing only strong C–C and C–H sigma bonds. The small electronegativity difference between carbon and hydrogen makes these bonds almost non-polar, so alkanes are not attacked easily by acids, bases, oxidising agents or reducing agents under ordinary conditions. This low reactivity earned them the name paraffins, from the Latin for little affinity.