Alkanes

— Chemistry · AS Level (7405) · Crash Course —

★ Exam Tip: Alkanes are the “first functional group” — they are in every organic compound. Examiners love: free-radical substitution mechanism (3 stages, half-arrow heads), cracking (conditions + equations), combustion volumes/ratios and boiling-point explanations. Reactions here are mostly learn-by-heart, so be neat.

1. What are Alkanes?

Definition: Alkanes are saturated hydrocarbons — they contain the maximum number of hydrogens possible (every C has 4 single bonds, no C=C can form). A hydrocarbon = a compound containing only H and C.

General formula: C_nH_2n+2

Obtained from the fractional distillation of crude oil (not by “cracking” — cracking comes later, and hydrogenation is not used to get alkanes).

Remember: Fractional distillation = a separating technique that separates a mixture using different boiling points.

Definition (CIE 3(a)(ii)): Fractional distillation (exam wording): a technique by which compounds with different boiling points are separated from a mixture. Sir: mixture = different species just mixed, no fixed chemical formula.

2. Physical Properties

Property	Alkanes	Result
Intermolecular forces	temporary (induced) dipole–dipole = London forces	⇒ low m.p. / b.p.
Polarity	non-polar	⇒ insoluble in water; soluble in non-polar solvents (alkenes dissolve in alkanes)

Trend: methane (CH₄) → ethane → propane → butane are gases; from pentane upwards liquids, then solids. b.p. and m.p. increase down the series because the number of electrons increases, so the temporary dipoles get stronger (more energy needed to overcome them).

b.p. (K) rises as chain gets longer — more electrons ⇒ stronger temporary dipoles

branching lowers b.p.

Exam wording: Boiling-point answers: “straight chain ⇒ greater surface area over which temporary dipoles act and molecules can pack more closely ⇒ stronger id–id forces ⇒ more energy needed”. Branched isomers have less surface contact ⇒ lower b.p. (board: butane 273 K vs 2-methylpropane 261 K). Sir: don’t write “bonds break” — it’s the intermolecular forces being overcome.

3. Chemical Properties — Why Alkanes Are Unreactive

Alkanes undergo only 3 types of reaction: combustion, cracking, free-radical substitution. They are unreactive because:

C–C and C–H are strong σ bonds (high bond energies; electron density lies along the internuclear axis) — hard to break;
the bonds are non-polar (very similar electronegativity) ⇒ they do not attract nucleophiles or electrophiles — no δ+/δ– sites, so reactive species just can’t “come close”.

So they only react under unfavourable (harsh) conditions: lots of heat or high-energy UV light.

4. Combustion

Hydrocarbons burn in oxygen to give CO₂ and H₂O (complete combustion).

C_xH_y + (x + y4) O₂ → x CO₂ + y2 H₂O

e.g. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O | C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

Tip: Sir: you don’t need to memorise the general equation — you can derive it. Just balance C → H → O. Fractions like 9/2 O₂ are allowed.

Incomplete combustion (limited O₂): products can be CO or C (soot) + H₂O. e.g. C₇H₁₆ + 4O₂ → 7C + 8H₂O (solid pollutant = carbon/soot).

Environmental consequences of combustion

Pollutant	Effect
CO₂	greenhouse effect → global warming
CO	(incomplete combustion) binds irreversibly to haemoglobin (forms carboxyhaemoglobin) ⇒ blood carries less O₂ ⇒ can be fatal
NO, NO₂ (NOₓ)	formed at the very high temperatures in engines (N₂ + O₂ react); cause acid rain / photochemical smog; board Q5: NO — if not removed by the catalytic converter, forms acid rain
SO₂	from sulfur impurities in fuel ⇒ acid rain
Unburnt hydrocarbons	irritation, greenhouse effect, smog

Board note: Refer to the Nitrogen & Sulfur chapter (later) for acid rain, smog and notes on the catalytic converter — sir deliberately covers it there. Smog contains PAN (peroxyacetyl nitrate) — just recognise the name.

5. Cracking

Definition: the process by which large alkane molecules are broken down into smaller alkanes and alkenes.

to obtain motor fuel (e.g. octane / petrol);
to obtain alkenes such as ethene — a very important feedstock.

Remember: Cracking always produces at least one alkane + one alkene. The equation must balance C and H on both sides — check carbon count and hydrogen count so the right-hand side never has more atoms than the left. (Cracking a tiny alkane like ethane gives ethene + H₂.)

Type	Conditions	Main product
Thermal cracking	high temperature 450–700 °C and high pressure ~70 atm	gives a high proportion of alkenes (extreme conditions)
Catalytic cracking	Al₂O₃ and SiO₂ (zeolite) catalyst, with heating (lower T, no extreme pressure)	gives a high proportion of alkanes (e.g. branched, motor fuels)

Equations (board):thermal: C₂₀H₄₂ → C₈H₁₈ + 6C₂H₄catalytic: C₂₀H₄₂ → C₁₀H₂₂ + C₁₀H₂₀ethane: C₂H₆ → C₂H₄ + H₂C₁₁H₂₄ → C₅H₁₂ + C₆H₁₂ (CIE Q3(b)(ii))

Sir’s exam hack: If you forget the exact T / P ranges in the exam, don’t guess wrong numbers — write “high temperature” (thermal) or “zeolite catalyst + heat” (catalytic) instead. Learn them, but give a safe answer if blank.

Lab apparatus for cracking

paraffin vapour passes over the hot catalyst; the alkene gas is collected over water

Safety — suck-back: When finishing: remove the delivery tube from the water FIRST, then stop heating. If you stop heating first, the gas contracts, cold water rushes back into the hot tube and the glass can crack/explode — this is called “suck-back”.

6. Free-Radical Substitution of Halogens

Alkanes react with chlorine / bromine in the presence of UV light (not in the dark). One H is substituted by a halogen each time:

CH₄ + Cl₂ → CH₃Cl + HClCH₃Cl + Cl₂ → CH₂Cl₂ + HClCH₂Cl₂ + Cl₂ → CHCl₃ + HClCHCl₃ + Cl₂ → CCl₄ + HCl

Why substitution, not addition? Alkane bonds are strong, so nothing adds across; but replacing an H keeps the bond energy about the same ⇒ substitution is favourable once enough energy (UV) is given. (Weak π bonds in alkenes ⇒ addition favoured — next chapter.)

Mechanism: free-radical substitution.

Free radical: a species that contains an unpaired electron (Cl•, Br•, CH₃•) — very reactive. They are not charged but want to pair that electron.

Homolytic fission: the breaking of a covalent bond so that each atom receives one electron — shown with a half-arrow (fish-hook) ⌒. (Heterolytic = both electrons go to one atom — full arrow.)

homolytic fission of Cl–Cl by UV light

The three stages (chlorine with methane)

Stage	Equation(s)	Key idea
① Initiation	Cl–Cl →(UV) 2Cl•	free radicals made from a stable molecule. Never break C–H here — only the weak X–X bond (UV energy).
② Propagation	CH₄ + Cl• → •CH₃ + HCl •CH₃ + Cl₂ → CH₃Cl + Cl•	1 stable + 1 radical → 1 stable + 1 radical; the Cl• is regenerated ⇒ chain reaction (keeps going).
③ Termination	Cl• + •Cl → Cl–Cl •CH₃ + Cl• → CH₃Cl •CH₃ + •CH₃ → CH₃CH₃ (trace)	any 2 radicals → 1 stable molecule; chain stops. CH₃–CH₃ is only a trace by-product.

propagation is a cycle: Cl• is regenerated, so one photon can convert thousands of molecules

Same mechanism with bromine

Stage	Equations
Initiation	Br–Br →(UV) 2Br•
Propagation	CH₄ + Br• → •CH₃ + HBr •CH₃ + Br₂ → CH₃Br + Br•
Termination	Br• + •Br → Br₂ ; •CH₃ + Br• → CH₃Br ; •CH₃ + •CH₃ → C₂H₆

Exam checklist: Every radical equation must show the dot on the radical. In the exam name each stage and use half-arrow heads for movement of single electrons. Termination = two unstable ⇒ one stable (that’s how to recognise it). Further substitution (CH₂Cl₂ …) happens because Cl• keeps being made — limit it by using excess alkane.

Link to Chapter 1: Substituting H on a CH₃ / CH₂ carbon (e.g. ethane → chloroethane) can never create a chiral centre; but a monochloro product of a longer chain (e.g. pent-2-ene, board Q23 → D) can have both a chiral C and cis-trans.

7. Past-Paper Questions (board)

MCQ 3 · ethane + Cl₂ (UV)1st stage of mechanism

First stage = initiation: Cl–Cl is split homolytically by UV.

A ✓ Cl–Cl split homolyticallyB ✗ heterolyticC ✗ C–H splits first — neverD ✗ same

Answer: A

MCQ 5 · pollutantengine / catalytic converter

Formed in the engine, if not removed becomes involved in acid rain ⇒ nitrogen oxide.

A C — sootB C₈H₁₈ — unburnt fuelC CO — poisonous, not acid rainD NO ✓ → NO₂ → HNO₃

Answer: D

MCQ 30 · valid propagation stepethane + chlorine

Propagation = 1 radical + 1 molecule → 1 radical + 1 molecule.

A C₂H₆ + Cl• → C₂H₅Cl + H• ✗ H• radical, wrongB C₂H₅• + Cl₂ → C₂H₅Cl + Cl• ✓C C₂H₆ + H• … ✗ H• isn’t madeD C₂H₅• + Cl• → C₂H₅Cl ✗ termination

Answer: B

MCQ 9 · strong heating (cracking)ethene : propane : but-1-ene = 5 : 1 : 1

Balance atoms (cracking conserves C and H):

C_xH_y → 5C₂H₄ + 1C₃H₈ + 1C₄H₈C: 10 + 3 + 4 = 17H: 20 + 8 + 8 = 36formula = C₁₇H₃₆

Answer: B (C₁₇H₃₆)

MCQ 36 · cracking C_nH_2n+2two products, same no. of C, both have isomers

Product alkane + alkene each have n/2 carbons. Test each option:

C₄H₁₀ → C₂H₆ + C₂H₄ : no isomers ✗C₆H₁₄ → C₃H₈ + C₃H₆ : no isomers ✗C₈H₁₈ → C₄H₁₀ + C₄H₈ : both have isomers ✓C₉H₂₀ ✗ odd n can’t split equally

Answer: C (8)

MCQ 23 · monochloro derivativewhich hydrocarbon gives both chirality and cis–trans?

Need C=C with two different groups on each carbon and a chiral C after Cl substitution.

A CH₃CH=CH₂ ✗B (CH₃)₂C=CH₂ ✗ never chiralC CH₃CH=C(CH₃)₂ ✗D CH₃CH=CHCH₂CH₃ ✓ — Cl on C4 makes chiral C* and C2=C3 keeps cis–trans

Answer: D

Combustion calculations — gas volumes and ratios

Key idea: Gas volumes at the same T, P ∝ moles (Avogadro). Use volume ratio = mole ratio. After burning: H₂O condenses (liquid at room T), NaOH / KOH absorbs CO₂ (and acidic SO₂), the rest is unused O₂. Always check: total = remaining O₂ + gaseous products.

MCQ 14 · methanethiolCH₃SH + 3O₂ → CO₂ + SO₂ + 2H₂O(l)

10 cm³ CH₃SH + 60 cm³ O₂. Final volume of gases at room temperature?

O₂ used = 3 × 10 = 30 cm³ ⇒ 30 cm³ O₂ unusedCO₂ = 10 cm³, SO₂ = 10 cm³ (H₂O is liquid)total = 30 + 10 + 10 = 50 cm³

Answer: C

MCQ 27 · CS₂CS₂ + 3O₂ → CO₂ + 2SO₂

10 cm³ CS₂ + 50 cm³ O₂.

O₂ used = 30 ⇒ unused = 20 cm³CO₂ = 10, SO₂ = 20after burning = 20 + 10 + 20 = 50 cm³after NaOH (absorbs CO₂ + SO₂): only O₂ left = 20 cm³

Answer: C (50, 20)

MCQ 40 · propan-1-olvolume of O₂ for 0.1 mol (room T, P)

CH₃CH₂CH₂OH + 92O₂ → 3CO₂ + 4H₂O

n(O₂) = 0.1 × 4.5 = 0.45 molV = 0.45 × 24V = 10.8 dm³

Answer: A

MCQ 51 · limonene C₁₀H₁₆mole ratio CO₂ : H₂O

C₁₀H₁₆ + 14O₂ → 10CO₂ + 8H₂O

10 : 8simplify ⇒ 5 : 4

Answer: B (5, 4)

MCQ 11 · C₆H₁₂ burned; P absorbs H₂O, Q absorbs CO₂ratio M_P / M_Q

Ratio ⇒ the mass chosen cancels (sir assumed 100 g). C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O.

n(C₆H₁₂) = 100 / 84m(CO₂) = 6 × (100/84) × 44 = 314.3 gm(H₂O) = 6 × (100/84) × 18 = 128.6 gM_P/M_Q = 128.6 / 314.3 = 0.41

Answer: A

8. Long CIE-style Questions (board)

CIE Q3 · crude oilfractional distillation, cracking, isomers, b.p.

(a) hydrocarbon = compound containing only hydrogen and carbon · fractional distillation = technique to separate compounds with different boiling points from a mixture.

(b)(i) process 1: thermal cracking, 450–700 °C, 70 atm; process 2: Al₂O₃ & SiO₂ catalyst + heat. (ii) C₁₁H₂₄ → C₅H₁₂ + C₆H₁₂.

(c) 3 isomers of C₅H₁₂: pentane, 2-methylbutane, 2,2-dimethylpropane — the straight-chain pentane has the highest b.p. (greatest surface area, packs closely).

(e)(i) b.p. CH₄ 112 K → C₄H₁₀ 273 K: chain longer ⇒ more electrons ⇒ temporary dipole forces stronger. (ii) 2-methylpropane 261 K < butane: branched ⇒ less surface area/can’t pack closely.

CIE Q5 · Davy & Faraday’s gas Afinding the molecular formula

(a) molecular formula = number and different atoms of an element bonded together in 1 molecule. (b) C_xH_y + (x + y/4)O₂ → xCO₂ + (y/2)H₂O. 10 cm³ A + 50 cm³ O₂ → 40 cm³ left; KOH leaves 30 cm³.

gas left after KOH = O₂ (30 cm³)CO₂ produced = 40 − 30 = 10 cm³O₂ used = 50 − 30 = 20 cm³10 cm³ A : 10 cm³ CO₂ ⇒ x = 1x + y/4 = 2 ⇒ y = (2 − 1) × 4 = 4formula = CH₄

Answer: (c)(i) O₂ (ii) CO₂ (iii) 10 cm³ (iv) 20 cm³ (d) CH₄

CIE Q11 · heptane C₇H₁₆structural isomers, combustion, mechanism

(a) structural isomers = same molecular formula, different structural formula; chiral = a carbon atom bonded to 4 different atoms/groups. Two heptane isomers with a chiral C: 3-methylhexane and 2,3-dimethylpentane. (board wrote “2,3-diethylpentane” by mistake — the structure drawn has two CH₃ ⇒ dimethyl)

(b)(i) C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O(ii) C₇H₁₆ + 4O₂ → 7C + 8H₂O (solid = soot)(iii) unburnt hydrocarbons → greenhouse effect → global warming

(c) mechanism: free-radical substitution. Initiation Cl–Cl → 2Cl• ; propagation (1) C₇H₁₆ + Cl• → C₇H₁₅• + HCl, (2) C₇H₁₅• + Cl₂ → C₇H₁₅Cl + Cl• ; termination Cl• + •Cl → Cl₂ and C₇H₁₅• + Cl• → C₇H₁₅Cl.

✓ Quick Recap — Alkanes

C_nH_2n+2, saturated, non-polar, only id–id forces ⇒ low m.p./b.p., insoluble in water. b.p. ↑ with chain length (more electrons); branching ↓ b.p.
Unreactive: strong, non-polar σ bonds. React only by combustion, cracking, free-radical substitution.
Combustion: C_xH_y + (x+y/4)O₂ → xCO₂ + (y/2)H₂O; incomplete ⇒ CO / C. CO is toxic (haemoglobin); NOₓ from engine heat; SO₂ from impurities; CO₂ ⇒ global warming.
Cracking: large alkane → smaller alkane + alkene. Thermal 450–700 °C, ~70 atm ⇒ alkenes. Catalytic Al₂O₃/SiO₂ + heat ⇒ alkanes. Suck-back: remove delivery tube first.
Free-radical substitution (UV, Cl₂/Br₂): initiation (homolytic fission, 2 radicals), propagation (radical regenerated), termination (2 radicals → 1 molecule). Half-arrows.
Gas-volume sums: ratio of volumes = ratio of moles; H₂O condenses; NaOH/KOH removes CO₂.