Fundamentholfundamenthol

Properties of Alkanes

ChemistryHydrocarbonsFor NEET aspirants

The properties of alkanes follow from one fact: the and bonds are strong and almost non-polar. That makes alkanes the least reactive class of organic compounds, which is why they are called paraffins. Physically their properties are set by London dispersion forces alone, so boiling point rises with chain length and falls with branching. Chemically the properties of alkanes are dominated by free-radical substitution, plus oxidation and a few high-temperature industrial reactions.

State at 298 K- gas- liquid, + solid
Stable conformerStaggeredanti for butane, dihedral
Key reactionFree-radical substitutionneeds UV light or heat
Radical stabilitysets every product ratio
Key Formulas - Quick Reference
  1. Combustion:
  2. Relative amount of a product (number of equivalent H) (relative reactivity)
  3. Percentage of a product
  4. Relative reactivity for chlorination,
  5. Halogen reactivity:
  6. Ease of H abstraction:
  7. Torsional barrier in ethane

1Physical Properties

The only intermolecular force available to an alkane is the London dispersion force, because the molecules carry no dipole and no hydrogen-bonding site. Everything in this section follows from that.

Physical state of alkanes at room temperature against chain length Alkanes with one to four carbons are gases at 298 kelvin, those with five to seventeen carbons are liquids, and those with eighteen or more carbons are solids. The trend arises because the only forces between alkane molecules are London dispersion forces, which grow stronger as the chain gets longer and the contact surface increases. Gases C₁ to C₄ Liquids C₅ to C₁₇ Solids C₁₈ and above increasing number of carbon atoms at 298 K Why the trend exists The only forces between alkane molecules are London dispersion forces A longer chain has more surface contact, so the forces add up and the boiling point rises Physical state of the straight-chain alkanes at room temperature
Fig 1State at 298 K tracks chain length, because dispersion forces grow with molecular surface area.
  • All alkanes are colourless and odourless
  • They are insoluble in water and soluble in non-polar solvents such as benzene and , following the like-dissolves-like rule
  • Density increases with chain length but stays below , so alkanes float on water
  • Boiling point rises steadily down the homologous series as the chain, and therefore the contact surface, gets longer
Boiling point and melting point of straight chain alkanes against carbon number Graph of boiling point and melting point in degrees Celsius for the first ten straight chain alkanes. Boiling point rises smoothly from minus 162 for methane to 174 for decane, crossing room temperature between butane and pentane, which is why the first four alkanes are gases and the next ones are liquids. Melting point also rises but not smoothly, because even numbered chains pack better in the solid and melt a little higher than the trend predicts. Boiling and melting points against chain length boiling point melting point 2001000 -100-200 GASES at 298 K LIQUIDS at 298 K 0 °C C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 number of carbon atoms in the straight chain temperature / °C Boiling point climbs smoothly. Melting point climbs in steps. An even number of carbons lets the zigzag chain pack more tightly in the crystal.
Fig 2Boiling point rises steadily with chain length because dispersion forces grow with molecular size, and it crosses between and . Melting point rises too, but unevenly: an even number of carbons lets the zigzag chain pack more tightly, so those members melt higher than the trend line suggests.

Branching always lowers the boiling point

Effect of branching on the boiling point of the pentane isomers The three isomers of C5H12 have boiling points of 309.1 kelvin for n-pentane with no branching, 300.9 kelvin for 2-methylbutane with one branch, and 282.5 kelvin for 2,2-dimethylpropane with two branches. Branching makes the molecule more compact and spherical, reducing surface contact, weakening dispersion forces and lowering the boiling point. n-Pentane no branching b.p. 309.1 K 2-Methylbutane one branch b.p. 300.9 K 2,2-Dimethylpropane two branches, nearly spherical b.p. 282.5 K All three are C₅H₁₂, yet the boiling point falls by 27 K More branching means a more compact, more spherical molecule Less surface contact means weaker dispersion forces and a lower boiling point
Fig 3Same formula , three boiling points. Branching shrinks the contact surface, so it always lowers the boiling point.
Melting point does not follow the same simple rule. A highly symmetric branched alkane such as neopentane packs efficiently into a crystal lattice, so its melting point is unusually high even though its boiling point is the lowest of the three isomers.

2Conformations of Alkanes

Rotation about a single bond is easy but not completely free. The different arrangements produced by that rotation are conformations, and they are drawn most clearly using Newman projections, looking straight down the bond.

Newman projections of staggered and eclipsed ethane Two Newman projections of ethane viewed along the carbon carbon bond. In the staggered form the three front hydrogens sit at sixty degrees from the three back hydrogens, bonds are as far apart as possible and the energy is at a minimum. In the eclipsed form the front and back hydrogens line up, the dihedral angle is zero and the energy is at a maximum. Staggered H H H H H H more stable, energy minimum dihedral 60°, bonds maximally apart Eclipsed H H H H H H less stable, energy maximum dihedral 0°, bonds line up red bonds belong to the front carbon blue bonds belong to the back carbon, the circle is that back carbon
Fig 4The two limiting conformations of ethane. Staggered is the energy minimum, eclipsed the maximum, and rotation interconverts them constantly at room temperature.
Torsional energy profile of ethane through 360 degrees of rotation A smooth periodic curve showing potential energy against dihedral angle for ethane. Maxima occur at zero, one hundred and twenty, two hundred and forty and three hundred and sixty degrees where the conformation is eclipsed. Minima occur at sixty, one hundred and eighty and three hundred degrees where it is staggered. The barrier between them is about 12.5 kilojoules per mole. eclipsed staggered staggered staggered eclipsed 0° 60° 120° 180° 240° 300° 360° dihedral angle of rotation about the C—C bond 12.5 kJ mol⁻¹ torsional barrier potential energy Torsional energy of ethane through one full rotation
Fig 5Ethane's rotation barrier is only about , small enough that rotation is essentially free at room temperature.
Torsional strain is the extra energy of an eclipsed conformation compared with a staggered one. In ethane the barrier is about , so at room temperature molecules cross it millions of times a second. The conformers cannot be separated and bottled.
Sawhorse and Newman projections of staggered and eclipsed ethane side by side Each conformation of ethane drawn twice. The sawhorse projection looks at the carbon carbon bond from an angle, so both carbons are visible along a diagonal line. The Newman projection looks straight down the carbon carbon bond, so the front carbon is a dot with three bonds and the back carbon is a circle with three bonds. In the staggered form the back bonds sit exactly between the front bonds, a dihedral angle of 60 degrees, and the energy is at a minimum. In the eclipsed form the back bonds line up behind the front bonds, a dihedral angle of 0 degrees, and the energy is at a maximum. Same conformation, two drawings: sawhorse and Newman STAGGERED · DIHEDRAL 60° · ENERGY MINIMUM HHH HHH SAWHORSE look down the C-C bond back bonds land in the gaps between front bonds H H H H H H NEWMAN ECLIPSED · DIHEDRAL 0° · ENERGY MAXIMUM HHH HHH SAWHORSE look down the C-C bond back bonds are drawn a few degrees turned so you can see them; the true angle is 0° H H H H H H NEWMAN
Fig 6The two standard drawings of the same conformation. A sawhorse looks at the bond obliquely; a Newman projection looks straight along it, so the front carbon becomes a dot and the back carbon a circle. Exam questions switch between the two freely, so practise converting in both directions.

Butane adds steric strain

Replace one hydrogen on each carbon of ethane with a methyl group and you get butane. Now the two methyl groups can crowd each other, so the energy profile is no longer symmetric.

Conformational energy profile of n-butane about the central carbon carbon bond Potential energy against dihedral angle for n-butane rotated about the bond between carbon two and carbon three. The anti conformation at one hundred and eighty degrees is the global minimum. Gauche minima appear at sixty and three hundred degrees. Eclipsed maxima appear at one hundred and twenty and two hundred and forty degrees, and the fully eclipsed form with the two methyl groups aligned is the highest maximum at zero and three hundred and sixty degrees. fully eclipsed gauche eclipsed anti eclipsed gauche fully eclipsed 0° 60° 120° 180° 240° 300° 360° Stability: anti > gauche > eclipsed > fully eclipsed the anti form keeps the two methyl groups 180° apart, which minimises steric strain potential energy Conformations of n-butane about the C2—C3 bond
Fig 7Butane adds steric strain to ethane's torsional strain, so the four conformations rank anti gauche eclipsed fully eclipsed.
ConformationDihedral angleStrain presentRelative energy
AntiNoneLowest, the global minimum
Gauche and StericA little above anti
Eclipsed and TorsionalA maximum
Fully eclipsed and Torsional and stericHighest, the global maximum

3Chemical Properties: Why Alkanes Are So Unreactive

Alkanes are stable towards most common reagents at room temperature. The and bonds are strong and nearly non-polar, so an attacking reagent finds no site of high or low electron density to latch onto. Ordinary collisions do not carry enough energy to break anything.

At high temperature, or under ultraviolet light, that changes. Bonds break homolytically and free radicals appear, and every characteristic reaction of alkanes runs through those radicals.

Carbon hydrogen bond dissociation energies for primary, secondary and tertiary hydrogens Bar chart of bond dissociation energies. Breaking a carbon hydrogen bond in methane needs 105 kilocalories per mole, a primary hydrogen 100, a secondary hydrogen 96 and a tertiary hydrogen 93. The weaker the bond the more easily the hydrogen is abstracted, so the order of ease is tertiary greater than secondary greater than primary greater than methane. CH₃—H methane 105 kcal mol⁻¹ CH₃CH₂—H 1° hydrogen 100 kcal mol⁻¹ (CH₃)₂CH—H 2° hydrogen 96 kcal mol⁻¹ (CH₃)₃C—H 3° hydrogen 93 kcal mol⁻¹ Bond dissociation energy of the C—H bond being broken A weaker C—H bond is easier to break, so a 3° hydrogen is abstracted most readily Ease of abstraction: 3° > 2° > 1° > CH₄
Fig 8The bond gets weaker as the carbon gets more substituted, which is exactly why hydrogens react first.
Stability order of alkyl free radicals A scale running from least stable to most stable holding the methyl radical, the primary ethyl radical, the secondary isopropyl radical and the tertiary butyl radical in that order. Stability rises with the number of attached alkyl groups because hyperconjugation and inductive release stabilise the half filled orbital. Allyl and benzyl radicals are more stable still because the odd electron is delocalised. CH₃• methyl CH₃CH₂• 1° radical (CH₃)₂CH• 2° radical (CH₃)₃C• 3° radical least stable most stable More alkyl groups means more hyperconjugation into the half-filled orbital Allyl and benzyl beat all of these, the odd electron is delocalised Stability order of alkyl free radicals
Fig 9Radical stability rises with substitution. This order, together with Figure 8, explains every product ratio in free-radical halogenation.
Four reasons why alkanes are chemically unreactive Four cards giving the reasons alkanes are inert. First, the carbon hydrogen bond is almost non polar because the electronegativity difference is only about 0.4, so there is no partial charge for a reagent to attack. Second, alkanes have no lone pairs, so they cannot donate electrons and are not Lewis bases. Third, they have no pi electrons, only sigma bonds, so there is no exposed electron cloud for an electrophile. Fourth, the bonds are strong, about 348 kilojoules per mole for carbon carbon and 414 for carbon hydrogen. The only way in is homolysis by heat or ultraviolet light to make free radicals. Four reasons alkanes just sit there 1 Bonds are almost non-polar Electronegativity: C = 2.5, H = 2.1 gap of just 0.4, so no δ+ or δ− site nothing attracts a nucleophile or electrophile 2 No lone pairs anywhere Every valence electron is inside a bond nothing to donate, so not a Lewis base unlike ethers, amines and alcohols 3 No π electrons Only σ bonds, short and tightly held nothing for an electrophile to attack this is the whole difference from alkenes 4 The bonds are strong C-C ≈ 348, C-H ≈ 414 kJ mol-1 breaking one costs a lot of energy mild reagents cannot pay that price SO THERE IS ONLY ONE WAY IN Heat or UV light splits a bond homolytically into free radicals. That is why nearly every alkane reaction in your syllabus is a free-radical reaction.
Fig 10Alkanes are inert for four independent reasons, and all four point the same way: there is no polar site, no lone pair and no cloud for an ionic reagent to grab. Only homolysis, driven by heat or UV light, opens a path, which is why free-radical substitution dominates this chapter.

4Halogenation: The Central Reaction

Halogenation replaces a hydrogen of an alkane with a halogen atom. It is a free-radical chain substitution, and it needs light or heat to get started.

Mechanism of free radical chlorination of methane The chain reaction has three phases. Initiation: chlorine molecules undergo homolysis in light or heat to give two chlorine radicals. Propagation: a chlorine radical removes a hydrogen from methane to give a methyl radical and hydrogen chloride, then the methyl radical attacks chlorine to give chloromethane and a new chlorine radical. Termination: any two radicals combine, consuming radicals and stopping the chain. Initiation, one step Cl₂ —— hν or Δ —→ 2Cl• homolysis gives two chlorine radicals Propagation, the chain-carrying steps CH₄ + Cl• → CH₃• + HCl the slow, product-deciding step CH₃• + Cl₂ → CH₃Cl + Cl• regenerates Cl•, so the chain continues Termination, any two radicals combining Cl• + Cl• → Cl₂ CH₃• + CH₃• → CH₃—CH₃ CH₃• + Cl• → CH₃Cl these consume radicals and stop the chain, so they are minor Free-radical chlorination of methane, the three phases
Fig 11Initiation, propagation, termination. Only the propagation steps consume starting material, which is why a few radicals can convert a large amount of alkane.
The first propagation step is the slow step, and it is the step that decides which hydrogen is removed. That is why the product ratio depends on bond strengths and radical stabilities, and not on anything in the second step.

Two competing factors set the product ratio

  1. How many hydrogens of each type are there? This is a statistical factor and it favours the more numerous hydrogens.
  2. How reactive is each type? This is an energetic factor and it favours the weaker and bonds. For chlorination the ratio is .
Calculating the product ratio for chlorination of n-butane n-Butane is written as C H 3, C H 2, C H 2, C H 3. The two end methyl groups carry six primary hydrogens between them and the two middle methylene groups carry four secondary hydrogens. Multiplying six by the primary reactivity of 1.0 gives a relative amount of 6.0 for 1-chlorobutane. Multiplying four by the secondary reactivity of 3.8 gives 15.2 for 2-chlorobutane. The total is 21.2, so the mixture is about 28 per cent 1-chlorobutane and 72 per cent 2-chlorobutane. Predicting the product ratio: chlorination of n-butane six 1° hydrogens = two CH3 groups × 3 H each CH3 CH2 CH2 CH3 four 2° hydrogens = two CH2 groups × 2 H each 1-chlorobutane 6 hydrogens × reactivity 1.0 relative amount = 6.0 6.0 / 21.2 = 28% 2-chlorobutane 4 hydrogens × reactivity 3.8 relative amount = 15.2 15.2 / 21.2 = 72% Total relative amount = 6.0 + 15.2 = 21.2
Fig 12Relative amount number of equivalent hydrogens their reactivity. Divide by the total and you have the percentage.
Energy profiles for hydrogen abstraction by a chlorine radical and a bromine radical Two reaction profiles for the hydrogen abstraction step. With a chlorine radical the barrier is small and the step is only slightly endothermic, about plus 8 kilojoules per mole, so the transition state comes early and resembles the starting alkane. With a bromine radical the barrier is large and the step is strongly endothermic, about plus 73 kilojoules per mole, so the transition state comes late and resembles the product radical. By Hammond's postulate a late transition state feels the difference in radical stability, which is why bromination is highly selective for tertiary hydrogen while chlorination is not. Why chlorine is fast but careless, and bromine slow but fussy Cl· abstracts H small barrier, early transition state Br· abstracts H large barrier, late transition state Ea early TS ΔH ≈ +8 R-H R· the TS resembles the alkane, so all three H look alike Ea late TS ΔH ≈ +73 R-H R· the TS resembles the radical, so the 3° H wins easily CHLORINATION at 298 K 1° : 2° : 3° = 1 : 3.8 : 5 nearly statistical, so mixtures are unavoidable BROMINATION at 400 K 1° : 2° : 3° = 1 : 82 : 1600 essentially one product, the 3° bromide
Fig 13Hammond's postulate in one picture. The step is nearly thermoneutral, so its transition state arrives early and resembles the alkane, and the three kinds of hydrogen look almost alike. The step is strongly endothermic, so its transition state arrives late and resembles the radical, and the stability gap between and radicals is felt in full.

Reactivity and selectivity pull in opposite directions

Chlorine reactivity versus bromine selectivity in alkane halogenation Chlorination of n-butane in light at room temperature gives a mixture of about 28 per cent 1-chlorobutane and 72 per cent 2-chlorobutane. Bromination at 127 degrees gives over 99 per cent 2-bromobutane with only a trace of the primary product. Fluorine is the most reactive halogen and iodine the least, while selectivity runs in the opposite direction. Chlorination Cl₂, light, room temperature 1-chloro 28% 2-chloro 72% Bromination Br₂, light, 127°C 1-bromo trace 2-bromo >99% Reactivity of the halogen: F₂ > Cl₂ > Br₂ > I₂ Selectivity runs the other way: the less reactive the halogen, the more selective it is Same substrate, n-butane: chlorine is reactive, bromine is selective
Fig 14Bromine is far less reactive than chlorine, and that is exactly why it is far more selective for the weaker and bonds.
Fluorination is violently exothermic and hard to control. Iodination is reversible and effectively does not proceed, because reduces the alkyl iodide straight back to the alkane. In practice only chlorination and bromination are synthetically useful.

5Nitration and Sulphonation

Nitration and sulphonation of alkanes In vapour phase nitration, n-butane reacts with nitric acid between 150 and 475 degrees to replace a hydrogen with a nitro group, giving a mixture of nitroalkanes. In sulphonation, isobutane reacts with oleum, a solution of sulphur trioxide in sulphuric acid, to replace the tertiary hydrogen with a sulphonic acid group. Ease of replacement follows tertiary greater than secondary greater than primary. Nitration: a hydrogen is replaced by —NO₂ CH₃CH₂CH₂CH₃ HNO₃ 150°C to 475°C CH₃CH₂CH₂CH₂NO₂ + isomer This is vapour-phase nitration, and it always gives a mixture of nitroalkanes Sulphonation: a hydrogen is replaced by —SO₃H (CH₃)₃CH H₂SO₄ / SO₃ oleum (CH₃)₃C—SO₃H + H₂SO₄ Ease of replacement: 3° > 2° > 1°, the same order as halogenation Two more substitution reactions of alkanes
Fig 15Nitration and sulphonation both replace , and both follow the same preference as halogenation.

6Oxidation and Combustion

Complete combustion is the reaction that matters commercially, and it is strongly exothermic. This is why alkanes are used as fuels.

Combustion and controlled oxidation of alkanes Methane burns in excess oxygen to carbon dioxide and water with a large release of heat. In a copper tube at 100 atmospheres and 200 degrees it gives methanol. Over red hot molybdenum oxide it gives methanal. With manganese acetate on heating, alkanes give carboxylic acids. Alkanes resist potassium permanganate and potassium dichromate except when a tertiary hydrogen is present, which is oxidised to a tertiary alcohol. CH₄ or any alkane excess O₂, complete combustion CO₂ + H₂O, strongly exothermic Cu tube, 100 atm, 200°C CH₃OH, methanol Mo₂O₃, red hot HCHO, methanal (CH₃COO)₂Mn, heat RCOOH, carboxylic acid oxidised products Alkanes resist KMnO₄ and K₂Cr₂O₇, with one exception: a 3° hydrogen is oxidised to the 3° alcohol, (CH₃)₃CH → (CH₃)₃COH Oxidation of alkanes: the product depends entirely on the conditions
Fig 16One substrate, four products. Combustion is complete oxidation; everything else is controlled oxidation and needs a specific catalyst.
  • Incomplete combustion with limited oxygen gives carbon monoxide and water, or carbon black used in inks, paints and tyres
  • LPG is a mixture of propane and butane, chosen because of the large heat released per unit mass
  • The oxyacetylene flame reaches about and is used for welding, though it is an alkyne, not an alkane

7High-Temperature and Industrial Reactions

Isomerisation, aromatisation, steam reforming and cracking of alkanes n-Hexane heated with anhydrous aluminium chloride and hydrogen chloride rearranges into branched isomers of the same formula. Heated over chromium oxide or vanadium oxide at 773 kelvin and 10 to 20 atmospheres it cyclises and loses four molecules of hydrogen to give benzene. Methane with steam over nickel at 1273 kelvin gives carbon monoxide and hydrogen. Heating a long chain alkane without air cracks it into smaller alkanes and alkenes. Isomerisation: the skeleton rearranges, the formula does not change n-hexane anhyd. AlCl₃ / HCl 2-methylpentane, still C₆H₁₄ Aromatisation: a six-carbon chain closes into a ring and loses hydrogen n-hexane Cr₂O₃ or V₂O₅ 773 K, 10 to 20 atm benzene + 4H₂ Two more high-temperature reactions Steam: CH₄ + H₂O → CO + 3H₂, over Ni at 1273 K, the industrial route to H₂ Cracking: heating a long chain without air breaks it into smaller alkanes and alkenes What high temperature and a catalyst do to an alkane
Fig 17These four reactions are all industrial. Aromatisation of -hexane to benzene is the one most often asked in NEET and JEE Main.
Master map of every reaction of alkanes with its reagent and product A nine card summary of alkane reactions grouped into three types. The substitution reactions are halogenation with a halogen under ultraviolet light giving an alkyl halide and hydrogen halide, nitration with concentrated nitric acid at 400 to 475 degrees giving a nitroalkane, and sulphonation with fuming sulphuric acid giving an alkane sulphonic acid. The oxidation reactions are complete combustion in excess oxygen giving carbon dioxide, water and heat, and controlled oxidation over copper at 523 kelvin and 100 atmospheres giving an alcohol. The reactions that change the carbon skeleton are isomerisation with anhydrous aluminium chloride and hydrogen chloride giving a branched alkane, aromatisation over chromium oxide at 773 kelvin turning a six carbon or longer chain into benzene, cracking at high temperature giving a smaller alkane plus an alkene, and steam reforming over nickel at 1273 kelvin giving carbon monoxide and hydrogen, known as syn gas. R–H · the nine reactions of an alkane Halogenation X2, UV light or Δ R–X + HX free-radical substitution Nitration conc. HNO3, 400-475°C R–NO2 + H2O vapour phase only Sulphonation fuming H2SO4, Δ R–SO3H needs C6 or longer Combustion excess O2, ignite CO2 + H2O + heat limited O2 gives CO or soot Controlled oxidation Cu, 523 K, 100 atm R–OH Mo2O3 gives the aldehyde Isomerisation anhyd. AlCl3 / HCl branched alkane raises the octane number Aromatisation Cr2O3, 773 K, 10-20 atm benzene ring needs C6 or longer Cracking 773-873 K, no air smaller alkane + alkene pyrolysis of the chain Steam reforming Ni, 1273 K, steam CO + 3H2 industrial source of hydrogen Orange = substitution · red = oxidation · green = the carbon skeleton changes
Fig 18Every reaction of an alkane on one card, grouped by type. Substitution replaces a bond and leaves the chain intact, oxidation adds oxygen, and the green routes rearrange or break the carbon skeleton itself.

Solved Examples

Solved Example 1
Monochlorination of 2-methylpropane gives two products. Predict the percentage of each, given the relative reactivities .
Solution:

2-Methylpropane is . It has nine equivalent hydrogens and one hydrogen.

Primary product, 1-chloro-2-methylpropane:

Tertiary product, 2-chloro-2-methylpropane:

Total .

Note that the product wins even though the hydrogen is five times more reactive. There are simply nine times as many of them.

Solved Example 2
Arrange in increasing order of boiling point: n-pentane, 2-methylbutane, 2,2-dimethylpropane.
Solution:

All three are , so the difference comes entirely from shape. More branching gives a more compact, more nearly spherical molecule, less surface contact, weaker dispersion forces and therefore a lower boiling point.

Branching order: 2,2-dimethylpropane (two branches) 2-methylbutane (one branch) n-pentane (none).

Increasing boiling point: 2,2-dimethylpropane 2-methylbutane n-pentane, that is .

Solved Example 3
What is the activation energy for the reaction ?
Solution:

Zero. Two radicals combining form a bond and break none, so no energy has to be supplied to reach a transition state. The reaction is purely exothermic and occurs on every collision that brings the two radicals together with the right geometry.

This is exactly why termination steps are so efficient, and why a chain reaction needs a continuous supply of light to keep the radical concentration up.

Solved Example 4
Why does bromination of 2-methylpropane give almost entirely the tertiary product, while chlorination gives a mixture?
Solution:

The hydrogen-abstraction step for bromine is endothermic, because the bond formed is weaker than the bond. By the Hammond postulate, an endothermic step has a late, product-like transition state, so the stability difference between the and radicals is felt almost fully in the transition state.

For chlorine the same step is strongly exothermic, so the transition state is early and reactant-like, and the radical stability difference barely shows up.

Result: bromine discriminates sharply and gives over of the tertiary bromide, while chlorine gives a mixture.

Solved Example 5
Draw the Newman projections of n-butane looking along the bond and arrange them in order of stability.
Solution:

Looking along , the front carbon carries a methyl and two hydrogens, and so does the back carbon.

Anti, dihedral : the two methyls are as far apart as possible. No torsional strain, no steric strain. Most stable.

Gauche, dihedral : staggered, so no torsional strain, but the two methyls are close enough to repel. Slightly higher in energy.

Eclipsed, dihedral : each methyl eclipses a hydrogen. Torsional strain present.

Fully eclipsed, dihedral : the two methyls eclipse each other. Torsional plus steric strain. Least stable.

Order: anti gauche eclipsed fully eclipsed.

Solved Example 6
Write the balanced equation for the complete combustion of propane, and calculate the volume of oxygen at STP needed to burn of it.
Solution:

For with , the oxygen required is .

Molar mass of propane , so is .

Oxygen needed .

Common Mistakes to Avoid

Watch out
  • Forgetting the statistical factor in halogenation. A hydrogen is more reactive, but if there are nine hydrogens and only one , the primary product can still dominate.
  • Assuming the most reactive halogen is the most selective. It is the opposite. Bromine is far less reactive than chlorine and far more selective.
  • Applying the branching rule to melting point. Branching lowers the boiling point every time, but a symmetric branched alkane can have a higher melting point because it packs better.
  • Calling conformers isomers. Conformers interconvert by simple rotation and cannot be separated; they are not isomers in the usual sense.
  • Drawing a Newman projection with front and back bonds indistinguishable. Front bonds meet at the centre, back bonds appear to come from the circle's edge.
  • Writing that alkanes are oxidised by . They are not, unless a hydrogen is present, in which case the alcohol is formed.
  • Mixing up aromatisation and isomerisation. Isomerisation with keeps the molecular formula; aromatisation over loses four and changes it.

Frequently Asked Questions

Why are alkanes called paraffins?

The name comes from the Latin parum affinis, meaning little affinity. Alkanes have strong, almost non-polar and bonds and no site of high or low electron density, so ordinary acids, bases, oxidising agents and reducing agents leave them alone at room temperature.

Why does branching lower the boiling point of an alkane?

Alkanes are held together only by London dispersion forces, and the strength of those forces depends on how much surface the molecules can present to each other. A branched alkane is compact and close to spherical, so it has less surface contact than its straight-chain isomer. Weaker forces mean less energy is needed to separate the molecules, so the boiling point falls.

What is the difference between torsional strain and steric strain?

Torsional strain is the repulsion between bonding electron pairs on adjacent carbons when bonds line up in an eclipsed arrangement. Steric strain is the repulsion between bulky groups forced close together in space. Ethane shows only torsional strain; butane shows both, which is why its energy profile has maxima of different heights.

Why is bromination more selective than chlorination?

The hydrogen-abstraction step is endothermic for bromine and exothermic for chlorine. By the Hammond postulate, the endothermic bromine step has a late, product-like transition state, so the stability difference between a and a radical is felt almost in full. The chlorine step has an early transition state where that difference hardly registers.

How do you calculate the percentage of each product in chlorination?

For each kind of hydrogen, multiply the number of equivalent hydrogens by its relative reactivity, using for . Add the results to get the total, then divide each relative amount by that total and multiply by 100. For n-butane this gives about 28 per cent 1-chlorobutane and 72 per cent 2-chlorobutane.

Can alkanes be oxidised by potassium permanganate?

Generally no. Alkanes resist and at ordinary temperatures. The one exception is an alkane containing a tertiary hydrogen, which is oxidised to the corresponding tertiary alcohol, for example isobutane gives tert-butyl alcohol.

What is the activation energy for two chlorine radicals combining?

Zero. The step forms a bond and breaks none, so no energy has to be supplied to reach a transition state. This is true of all radical combination steps, which is why termination is so efficient and why a free-radical chain needs continuous light to sustain itself.

What happens when n-hexane is heated with a chromium oxide catalyst?

It undergoes aromatisation. At about 773 K and 10 to 20 atmospheres over or , n-hexane cyclises and loses four molecules of hydrogen to give benzene. This is distinct from isomerisation with anhydrous , which only rearranges the skeleton and keeps the molecular formula unchanged.

Previous year questions on Properties of Alkanes

9 questions from past papers, each with a step-by-step solution.

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