Alkanes: Preparation and Properties

Saturation and shape. Every carbon in an alkane is $sp^3$-hybridised, bonded to four other atoms (C or H) by single $\sigma$ bonds only, with bond angles close to $109.5^\circ$. There are no lone pairs, no $\pi$ bonds, and no polar functional groups — which is why alkanes are called "paraffins" (Latin, "little affinity") and stay inert toward acids, bases, and most ionic reagents.

Nomenclature and isomerism. IUPAC naming picks the longest continuous carbon chain as the parent, numbers it to give substituents the lowest possible locants, and names branches as substituent prefixes (methyl, ethyl, …) in alphabetical order. Alkanes with four or more carbons show chain (structural) isomerism: the same molecular formula, $\text{C}_n\text{H}_{2n+2}$, can be arranged as a straight chain or with one or more branches, and each arrangement is a distinct compound with its own physical properties.

Free-radical halogenation. Chlorination of an alkane (e.g., methane) needs light or heat and proceeds in three stages. Initiation homolytically splits $\text{Cl}_2$: $\text{Cl}_2 \xrightarrow{h\nu} 2\text{Cl}^{\bullet}$. Propagation is a repeating two-step chain,

where: the dot denotes an unpaired electron (a free radical), and the regenerated $\text{Cl}^{\bullet}$ re-enters the first propagation step, sustaining the chain until radicals are used up. Termination consumes radicals by combining any two of them, e.g. $2\text{Cl}^{\bullet}\rightarrow \text{Cl}_2$ or $2\text{CH}_3^{\bullet}\rightarrow \text{C}_2\text{H}_6$.

Combustion is complete oxidation to $\text{CO}_2$ and $\text{H}_2\text{O}$, strongly exothermic:

where: $\Delta H < 0$ marks the reaction as exothermic; with limited $\text{O}_2$, incomplete combustion gives soot (carbon) or $\text{CO}$ instead of $\text{CO}_2$.

Pyrolysis/cracking heats a higher alkane without air, homolytically breaking C–C bonds into a mixture of smaller alkanes and alkenes — the same free-radical logic as halogenation, applied to C–C instead of C–H bonds, and the basis of refinery "cracking" of heavy petroleum fractions into lighter, more useful ones.

Physical property trend. Boiling point rises steadily with chain length, because more carbons give more surface area for van der Waals (London dispersion) contact between neighbouring molecules. For a fixed carbon count, branching lowers the boiling point, because a more compact, spherical shape reduces that contact area — the reasoning developed quantitatively below.

Q. Write the IUPAC name of $\text{CH}_3\text{-CH(CH}_3)\text{-CH}_2\text{-CH}_2\text{-CH}_3$, confirm its molecular formula matches $\text{C}_n\text{H}_{2n+2}$, and explain why it boils lower than its chain isomer, n-hexane.

Step 1. Identify the longest continuous chain: counting through the backbone gives five carbons in a row, so the parent chain is pentane.

Step 2. Locate the substituent and number for the lowest locant: a methyl group sits on the second carbon, counting from the end nearer the branch.

Step 3. Assemble the name: 2-methylpentane.

Step 4. Check the formula against $\text{C}_n\text{H}_{2n+2}$: the molecule has $n=6$ carbons (five in the chain plus one methyl branch), so it should have $2(6)+2=14$ hydrogens, giving $\text{C}_6\text{H}_{14}$ — the same formula as n-hexane, confirming the two are chain (structural) isomers, not different compounds.

Step 5. Compare boiling points: 2-methylpentane (b.p. ≈ 60 °C) boils lower than n-hexane (b.p. ≈ 69 °C) because branching makes the molecule more compact and nearly spherical, shrinking the surface area available for van der Waals contact between neighbouring molecules and so weakening the net intermolecular attraction that must be overcome to boil. Final answer: 2-methylpentane, formula C6H14 (matching CnH2n+2 with n=6), boiling about 9 degrees lower than n-hexane due to reduced van der Waals contact from branching.

Competitive stretch

Q. Outline the free-radical mechanism for monochlorination of ethane, $\text{C}_2\text{H}_6 + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{HCl}$, naming the type of bond fission in initiation.

Step 1. Initiation: UV light homolytically cleaves the weak Cl–Cl bond, $\text{Cl}_2 \xrightarrow{h\nu} 2\text{Cl}^{\bullet}$ — homolytic fission means each chlorine keeps one electron from the shared pair, giving two chlorine free radicals.

Step 2. Propagation 1: a chlorine radical abstracts a hydrogen from ethane, $\text{C}_2\text{H}_6 + \text{Cl}^{\bullet} \rightarrow \text{C}_2\text{H}_5^{\bullet} + \text{HCl}$.

Step 3. Propagation 2: the ethyl radical reacts with another $\text{Cl}_2$ molecule, $\text{C}_2\text{H}_5^{\bullet} + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}^{\bullet}$, regenerating $\text{Cl}^{\bullet}$ to sustain the chain. Answer: initiation is homolytic fission; the chain alternates the two propagation steps until termination (e.g. $2\text{Cl}^{\bullet}\rightarrow \text{Cl}_2$) uses up the radicals, and in practice further-substituted chloroethanes also form since the reaction does not stop cleanly at mono-substitution.

• Writing "cracking"/pyrolysis as if industrial-only — at this level it is simply thermal C–C bond homolysis, applicable to any alkane in principle.
• Assuming "bigger or more branched always boils higher" — branching lowers boiling point for a fixed carbon count; only increasing chain length raises it.
• Confusing homolytic (radical) and heterolytic (ionic) fission — alkane halogenation is strictly homolytic/free-radical; ionic/electrophilic mechanisms belong to alkenes and benzene, not alkanes.
• Getting combustion completeness wrong — complete combustion needs excess $\text{O}_2$ and gives only $\text{CO}_2$ and $\text{H}_2\text{O}$; incomplete combustion (limited $\text{O}_2$) gives $\text{CO}$ or soot instead, a separate question from the exothermicity sign.

Ask the tutor: Want to check your own free-radical mechanism arrows or predict the boiling-point order for a set of C7H16 isomers — ask the AI tutor to walk through it with you.