Why alkanes barely react
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understand the general unreactivity of alkanes, including towards polar reagents in terms of the strength of the C–H bonds and their relative lack of polarity.
Fuels that ignore almost everything
Alkanes are fuels: petrol, diesel and natural gas are all alkane mixtures. Yet hexane can be mixed with concentrated sulfuric acid, sodium hydroxide, potassium manganate(VII) or bromine water at room temperature and nothing happens. Most other organic compounds you will meet have at least one reagent that attacks them. Alkanes have almost none.
The reason is examinable, and it also explains the rest of the alkane half of this note: alkanes only react when forced, either by burning (a flame supplies a lot of energy) or by radicals made with UV light.
Two properties, one conclusion
The explanation rests on two properties of the C–H (and C–C) bonds:
- They are strong. Breaking a C–H bond needs about 413 kJ/mol and a C–C bond about 347 kJ/mol (bond energies from the Chemical Energetics note). Ordinary reagents at room temperature cannot supply that much energy.
- They are (almost) non-polar. The electronegativity of carbon is 2.5 and of hydrogen 2.2. The difference, about 0.3, is too small to make a real dipole, so an alkane has no δ+ site and no δ− site.
Polar reagents (acids, alkalis, oxidising agents, nucleophiles, electrophiles) are attracted to charge. A nucleophile looks for a δ+ atom; an electrophile looks for an electron-rich (δ−) region. An alkane offers neither, and it has no weak bond to break either.
Strong bonds and non-polar bonds together leave nothing for a reagent to attack; only burning and UV-light radical reactions work.
"Explain the general unreactivity of alkanes" needs BOTH points: strong C–H bonds AND non-polar C–H bonds. Each is a separate mark. Strong but polar bonds could still be attacked at the δ+ end; non-polar but weak bonds could still be broken by heat.
Predicting which reagents leave an alkane alone
State, with a reason in each case, what happens when hexane at room temperature is mixed separately with: (i) aqueous ; (ii) bromine water in the dark; (iii) concentrated ; (iv) a lighted splint.
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(i) Aqueous NaOH: no reaction. is a nucleophile, so it needs a δ+ carbon. Every carbon in hexane is bonded only to C and H, so no carbon is δ+.
Name what the reagent looks for (a δ+ carbon), then check whether the alkane has it.
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(ii) Bromine water in the dark: no reaction. can only react with an alkane by the radical route, and that needs UV light to make the first radicals (see “Free-radical substitution”). In the dark no radicals form, so the bromine water stays orange.
Students often write 'decolourises' out of habit. That is the alkene result, not the alkane one.
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(iii) Concentrated H₂SO₄: no reaction. It is a strong acid and an oxidising agent. Both need a polar site or a weak bond, and hexane has neither.
The same two-property test works for any polar reagent.
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(iv) A lighted splint: hexane burns. A flame supplies enough energy to break the strong C–H and C–C bonds. Once burning, the heat released keeps the reaction going (see “Combustion and its pollutants”).
Alkanes react only when forced: a flame (lots of energy) or UV light (radicals). No ordinary polar reagent works.
(i) no reaction (ii) no reaction; stays orange (iii) no reaction (iv) burns (combustion).
For any 'does it react?' question about an alkane, ask two things: is there a polar site to attack (no), and is there a flame or UV light? Only a flame or UV makes it react.
The two-mark unreactivity explanation
Give two reasons to explain the general unreactivity of alkanes.
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Reason 1: bond strength. The C–H bonds are strong (high bond energy), so they are not easily broken.
Mark 1 is 'strong C–H (bonds)'. Name the bond: 'alkanes are strong' is not the point.
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Reason 2: polarity. The C–H bonds are non-polar, because C and H have almost the same electronegativity. There are no δ+ or δ− sites for polar reagents to attack.
Mark 2 is 'non-polar (C–H / molecule / bonds)'. Either reason alone scores only 1 of the 2 marks.
1: the C–H bonds are strong (hard to break). 2: the C–H bonds are non-polar (no sites for polar reagents to attack).
Your turn
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A student adds a few drops of bromine water to cyclohexane in a test-tube, shakes it and leaves it in a dark cupboard. State and explain what is observed.
Stuck? Show hint
Cyclohexane is an alkane. Is there a polar site, and is there UV light?
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No decolourisation. Cyclohexane is an alkane: strong, non-polar C–H and C–C bonds and no C=C. Bromine can only attack it by the radical route, which needs UV light. There is no UV light here, so the bromine water stays orange.
A ring does not change anything: cyclohexane is still an alkane (only single bonds).
AnswerNo visible change: the bromine water stays orange. Alkane + Br₂ needs UV light (radical route).
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Explain why alkanes are described as unreactive "towards polar reagents", quoting the two bond properties the explanation rests on and linking each to a mode of attack it defeats.
Stuck? Show hint
One property defeats attack that needs to BREAK bonds; the other defeats attack that needs to FIND a charge.
Show solution
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Property 1: the C–H bonds are strong. Any reaction must start by breaking a bond in the alkane. C–H needs about 413 kJ/mol and C–C about 347 kJ/mol, which ordinary reagents at room temperature cannot supply.
This property defeats any reagent that must break a bond to react.
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Property 2: the C–H bonds are almost non-polar (electronegativity C 2.5, H 2.2). Electrophiles look for δ− regions and nucleophiles look for δ+ atoms. With no real dipole anywhere, neither has a target.
This property defeats any reagent that must find a charge. Quote the two electronegativities if the question asks you to explain the lack of polarity.
AnswerStrong C–H bonds defeat energy-limited attack; non-polar C–H bonds defeat charge-seeking (polar) attack. Together: general unreactivity.
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The rest of this note
Can you do all of these?
Explain alkane unreactivity: strong C–H bonds AND non-polar C–H bonds.
Recall hydrogenation (H₂, Pt/Ni, heat) and cracking (heat, Al₂O₃); balance a cracking equation and say why cracking is useful.
Balance complete and incomplete combustion equations; name the three engine pollutants, their harms, and which the catalytic converter oxidises (CO, hydrocarbons) or reduces (NOₓ).
Write initiation / propagation / termination equations that obey the radical-counting rule (0→2, 1→1, 2→0), quoting UV as the essential condition; count possible monochloro products.
Recall the three routes to alkenes WITH conditions: ethanolic NaOH + heat; hot Al₂O₃ or concentrated acid; cracking.
Recall the four electrophilic additions with conditions (H₂/Ni-Pt heat; steam/H₃PO₄; HX room temp; X₂) and state the bromine-water observation.
Draw the electrophilic-addition mechanism for HX/propene and Br₂/ethene: dipole, π→electrophile arrow, bond→anion arrow, carbocation, lone-pair attack.
Apply the KMnO₄ fates table in both directions — products from an alkene, and alkene structure from products.
Draw repeat units with brackets and dangling bonds, recover monomers from units, and explain polymer unreactivity via the loss of the π bond.
Predict major HX-addition products by counting alkyl groups on the competing carbocations and citing the +I effect.