JEE Mains Chemistry · Hydrocarbons

Alkanes: Preparation, Structure and Conformations

Alkanes (CₙH₂ₙ₊₂) are made by adding hydrogen to a C=C or C≡C, by removing a halogen or a carboxyl group, or by joining two alkyl groups; their carbons are classed 1°, 2°, 3° or 4°, and rotation about each C–C bond gives staggered and eclipsed conformations.

All Hydrocarbons notes JEE Mains Chemistry strategy

Why this matters

Nineteen PYQs, fourteen of them multiple choice, and seven from 2026. Seven ask which route gives which alkane: Kolbe electrolysis, the Wurtz reaction, soda-lime decarboxylation or a Grignard reagent with water. Eight work from the formula, count primary or secondary carbons, or order the conformations of ethane and butane. Four are about isomerisation, aromatisation and oxidation of alkanes. Five of the nineteen ask for a number.

Concept 1 of 3: Preparing alkanes and what each route does to the carbon count

Each route to an alkane does one of three things to the carbon chain. It keeps it (adding H₂ to an alkene, reducing an alkyl halide, a Grignard reagent with water), it doubles it (Wurtz, Kolbe) or it shortens it by one carbon (soda-lime decarboxylation). So before you choose a route, ask how many carbons the product needs. A route that doubles the chain can never give methane, because methane has only one carbon.

Definition

Hydrogenation: $\mathrm{R{-}CH{=}CH_2 + H_2 \xrightarrow{Pt/Pd/Ni} R{-}CH_2{-}CH_3}$ . Same carbon count; needs at least two carbons, so no methane.
Reduction of alkyl halides: $\mathrm{R{-}X + H_2 \xrightarrow{Zn,\ H^+} R{-}H + HX}$ . Same carbon count; $\mathrm{CH_3Cl}$ gives $\mathrm{CH_4}$ .
Wurtz reaction: $\mathrm{2R{-}X + 2Na \xrightarrow{dry\ ether} R{-}R + 2NaX}$ . Doubles the chain; two different halides give three alkanes.
Kolbe electrolysis: $\mathrm{2RCOO^-Na^+ + 2H_2O \rightarrow R{-}R + 2CO_2 + H_2 + 2NaOH}$ . Doubles the alkyl group; $\mathrm{R{-}R}$ and $\mathrm{CO_2}$ form at the anode.
Soda-lime decarboxylation: $\mathrm{RCOONa + NaOH \xrightarrow{CaO,\ \Delta} R{-}H + Na_2CO_3}$ . One carbon fewer.
Grignard reagent + any compound with an acidic H (water, an alcohol, an amine): $\mathrm{R{-}MgX + H_2O \rightarrow R{-}H + Mg(OH)X}$ . One mole of gas per mole of $\mathrm{RMgX}$ ; with $\mathrm{D_2O}$ the product is $\mathrm{R{-}D}$ .

Route	Reagents	Carbon count of product	Example
Hydrogenation	$\mathrm{H_2}$ with Pt, Pd or Ni	Same as the alkene or alkyne	Propene gives propane
Reduction of R–X	Zn and dilute HCl	Same as the halide	$\mathrm{CH_3CH_2Br}$ gives ethane
Wurtz	Na in dry ether	Twice the alkyl group	$\mathrm{CH_3CH_2Br}$ gives butane
Kolbe electrolysis	Electrolysis of the aqueous sodium salt	Twice the alkyl group	Sodium propanoate gives butane
Soda-lime decarboxylation	NaOH with CaO, heat	One carbon fewer than the salt	Sodium propanoate gives ethane
Grignard + acidic H	$\mathrm{H_2O}$ , ROH or $\mathrm{RNH_2}$	Same as the alkyl group	$\mathrm{C_2H_5MgBr}$ gives ethane
Clemmensen reduction	Zn–Hg and conc. HCl	Same; C=O becomes CH₂	Propanone gives propane

Wurtz and Kolbe double the chain, so neither can make methane; decarboxylation removes one carbon.

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The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q40Moderate

Example 1 · Hydrocarbons · Alkanes: Preparation, Structure and Conformations

Given below are two statements : Statement-I : Methane can be prepared by decarboxylation of sodium ethanoate, Kolbe's electrolysis of sodium acetate and reaction of

{CH}_{3}MgBr

with water. Statement-II : Methane cannot be prepared from unsaturated hydrocarbons and by Wurtz reaction. In the light of the above statements, choose the correct answer from the options given below.

Kolbe electrolysis of sodium ethanoate gives ethane

Two methyl radicals join at the anode, so the product is

\mathrm{CH_3{-}CH_3}

. Methane from sodium ethanoate needs soda lime, not electrolysis.

A mixture of two salts or two halides gives three alkanes

Kolbe electrolysis of

\mathrm{CH_3COONa}

with

\mathrm{C_2H_5COONa}

couples methyl–methyl, methyl–ethyl and ethyl–ethyl: ethane, propane and butane. The Wurtz reaction on two different halides does the same.

Every acidic H counts for a Grignard reagent

Water, alcohols, amines and terminal alkynes all protonate RMgX. One mole of RMgX gives one mole of RH gas whatever the proton source, so the gas volume gives the moles.

Drill 6 more on preparing alkanes and what each route does to the carbon count

Concept 2 of 3: Alkane formula, carbon classes and conformations

An open-chain alkane is CₙH₂ₙ₊₂, so its molar mass is 14n + 2. That one fact turns a molar mass or an oxygen demand into the number of carbons. Then draw the isomers and class each carbon by how many carbons it is bonded to: one (primary), two (secondary), three (tertiary) or four (quaternary). The hydrogens take the class of the carbon they sit on. Rotation about a C–C single bond is free, but not equally easy: eclipsed forms put bonds (or groups) face to face and cost energy.

Definition

Molar mass $= 14n + 2$ ; complete combustion needs $\dfrac{3n+1}{2}$ mol $\mathrm{O_2}$ per mole of alkane.
A 1° carbon is bonded to one carbon, 2° to two, 3° to three, 4° to four. 1° H sits on a 1° carbon, and so on; a 4° carbon has no H.
Ethane: staggered (dihedral angle 60°) is the most stable, eclipsed (0°) the least; they differ by about 12.5 kJ mol⁻¹ (torsional strain). Between them lie infinitely many conformations, and they interconvert at room temperature, so they cannot be separated.
n-Butane (along C2–C3), increasing energy: anti (CH₃ groups 180° apart) < gauche (60°) < eclipsed with CH₃ over H < fully eclipsed with CH₃ over CH₃ (0°).

Open-chain alkane

\mathrm{C_nH_{2n+2}}:\quad M = 14n + 2,\qquad \mathrm{C_nH_{2n+2}} + \tfrac{3n+1}{2}\,\mathrm{O_2} \rightarrow n\,\mathrm{CO_2} + (n+1)\,\mathrm{H_2O}

nnumber of carbon atoms
Mmolar mass in g mol⁻¹ (C = 12, H = 1)

Worked example

An alkane has molar mass 86 g mol⁻¹ and exactly two tertiary carbons. Name it and count its primary carbons.

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The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q39Moderate

Example 2 · Hydrocarbons · Alkanes: Preparation, Structure and Conformations

Identify the correct IUPAC name of hydrocarbon (x) containing three primary carbon atoms and with molar mass

72\text{ }g{\text{ }mol}^{- 1}

A ring has two hydrogens fewer

A cycloalkane is CₙH₂ₙ, so its molar mass is 14n. A ring with the same carbons as an alkane is 2 g mol⁻¹ lighter, which rules it out when the molar mass fits CₙH₂ₙ₊₂.

The fully eclipsed form of butane is the highest in energy

Two eclipsed conformations exist. The one with CH₃ over CH₃ (0°) is higher than the one with CH₃ over H (120°). Gauche is higher than anti but lower than both eclipsed forms.

Conformers are not isolable

The barrier in ethane is only about 12.5 kJ mol⁻¹, so the conformations change into one another at room temperature. A statement that they can be separated is wrong.

Drill 7 more on alkane formula, carbon classes and conformations

Concept 3 of 3: Isomerisation, aromatisation and oxidation of alkanes

Alkanes are unreactive, so each of their reactions needs a special reagent or harsh conditions, and each one is named by what it does to the chain. Isomerisation keeps the formula but branches the chain. Aromatisation closes a six-carbon ring and removes hydrogen, so the carbon count stays the same. Oxidation with KMnO₄ attacks only a tertiary C–H.

Definition

Isomerisation: anhydrous $\mathrm{AlCl_3}$ and HCl gas, heat. n-Hexane gives 2-methylpentane and 3-methylpentane (same formula).
Aromatisation: $\mathrm{Cr_2O_3}$ , $\mathrm{V_2O_5}$ or $\mathrm{Mo_2O_3}$ on alumina, 773 K, 10–20 atm. n-Hexane gives benzene; n-heptane gives toluene.
Oxidation by KMnO₄: a tertiary C–H becomes C–OH: $\mathrm{(CH_3)_3CH \rightarrow (CH_3)_3COH}$ .
Controlled oxidation: $\mathrm{2CH_4 + O_2 \xrightarrow{Cu,\ 523\,K,\ 100\,atm} 2CH_3OH}$ ; $\mathrm{CH_4 + O_2 \xrightarrow{Mo_2O_3} HCHO + H_2O}$ .
Pyrolysis (cracking): heating without air breaks a large alkane into smaller alkanes and alkenes.

Reaction	Conditions	What changes	Example
Isomerisation	Anhydrous $\mathrm{AlCl_3}$ , HCl gas, heat	Chain branches; formula unchanged	n-Hexane → 2-methylpentane and 3-methylpentane
Aromatisation	$\mathrm{Cr_2O_3}$ or $\mathrm{V_2O_5}$ on alumina, 773 K, 10–20 atm	Six-carbon ring closes; H₂ is lost	n-Hexane → benzene
KMnO₄ oxidation	$\mathrm{KMnO_4}$	Tertiary C–H becomes C–OH	2-Methylpropane → 2-methylpropan-2-ol
Controlled oxidation	Cu at 523 K and 100 atm, or $\mathrm{Mo_2O_3}$	Methane becomes methanol or methanal	$\mathrm{CH_4 \rightarrow CH_3OH}$
Steam reforming	$\mathrm{H_2O}$ , Ni, 1273 K	Methane becomes CO and H₂	$\mathrm{CH_4 + H_2O \rightarrow CO + 3H_2}$
Pyrolysis	Strong heat, no air	Chain breaks into smaller alkanes and alkenes	Hexane → butene + ethane, among others

Aromatisation keeps the carbon count: count the carbons of the arene to find the alkane.

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The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q39Moderate

Example 3 · Hydrocarbons · Alkanes: Preparation, Structure and Conformations

The compound (X) on (i) on heating in the presence of anhydrous

{AlCl}_{3}

and HCl gas gives 2, 4-dimethyl pentane (ii) aromatization gives toluene and (iii) cyclisation gives methyl cyclohexane The correct name of compound

(X)

is :

Isomerisation does not change the formula

An alkane and its isomerised product have the same molecular formula. If a product has fewer hydrogens, the reaction was aromatisation or cyclisation, not isomerisation.

KMnO₄ needs a tertiary hydrogen

n-Alkanes have no tertiary C–H, so KMnO₄ does not give an alcohol from them. 2-Methylbutane has one, at C-2, and gives 2-methylbutan-2-ol.

Drill 3 more on isomerisation, aromatisation and oxidation of alkanes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Alkane formula, carbon classes and conformations
Open-chain alkane
$\mathrm{C_nH_{2n+2}}:\quad M = 14n + 2,\qquad \mathrm{C_nH_{2n+2}} + \tfrac{3n+1}{2}\,\mathrm{O_2} \rightarrow n\,\mathrm{CO_2} + (n+1)\,\mathrm{H_2O}$

Reference tables (2)

Preparing alkanes and what each route does to the carbon count7 rows

Route	Reagents	Carbon count of product	Example
Hydrogenation	$\mathrm{H_2}$ with Pt, Pd or Ni	Same as the alkene or alkyne	Propene gives propane
Reduction of R–X	Zn and dilute HCl	Same as the halide	$\mathrm{CH_3CH_2Br}$ gives ethane
Wurtz	Na in dry ether	Twice the alkyl group	$\mathrm{CH_3CH_2Br}$ gives butane
Kolbe electrolysis	Electrolysis of the aqueous sodium salt	Twice the alkyl group	Sodium propanoate gives butane
Soda-lime decarboxylation	NaOH with CaO, heat	One carbon fewer than the salt	Sodium propanoate gives ethane
Grignard + acidic H	$\mathrm{H_2O}$ , ROH or $\mathrm{RNH_2}$	Same as the alkyl group	$\mathrm{C_2H_5MgBr}$ gives ethane
Clemmensen reduction	Zn–Hg and conc. HCl	Same; C=O becomes CH₂	Propanone gives propane

Wurtz and Kolbe double the chain, so neither can make methane; decarboxylation removes one carbon.

Isomerisation, aromatisation and oxidation of alkanes6 rows

Reaction	Conditions	What changes	Example
Isomerisation	Anhydrous $\mathrm{AlCl_3}$ , HCl gas, heat	Chain branches; formula unchanged	n-Hexane → 2-methylpentane and 3-methylpentane
Aromatisation	$\mathrm{Cr_2O_3}$ or $\mathrm{V_2O_5}$ on alumina, 773 K, 10–20 atm	Six-carbon ring closes; H₂ is lost	n-Hexane → benzene
KMnO₄ oxidation	$\mathrm{KMnO_4}$	Tertiary C–H becomes C–OH	2-Methylpropane → 2-methylpropan-2-ol
Controlled oxidation	Cu at 523 K and 100 atm, or $\mathrm{Mo_2O_3}$	Methane becomes methanol or methanal	$\mathrm{CH_4 \rightarrow CH_3OH}$
Steam reforming	$\mathrm{H_2O}$ , Ni, 1273 K	Methane becomes CO and H₂	$\mathrm{CH_4 + H_2O \rightarrow CO + 3H_2}$
Pyrolysis	Strong heat, no air	Chain breaks into smaller alkanes and alkenes	Hexane → butene + ethane, among others

Aromatisation keeps the carbon count: count the carbons of the arene to find the alkane.

Watch out for (8)

Kolbe electrolysis of sodium ethanoate gives ethane
→ Preparing alkanes and what each route does to the carbon count
A mixture of two salts or two halides gives three alkanes
→ Preparing alkanes and what each route does to the carbon count
Every acidic H counts for a Grignard reagent
→ Preparing alkanes and what each route does to the carbon count
A ring has two hydrogens fewer
→ Alkane formula, carbon classes and conformations
The fully eclipsed form of butane is the highest in energy
→ Alkane formula, carbon classes and conformations
Conformers are not isolable
→ Alkane formula, carbon classes and conformations
Isomerisation does not change the formula
→ Isomerisation, aromatisation and oxidation of alkanes
KMnO₄ needs a tertiary hydrogen
→ Isomerisation, aromatisation and oxidation of alkanes

Test yourself on Hydrocarbons

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