What this chapter covers14
Topic 14 is two homologous series that could hardly be less alike. Alkanes are the least reactive organic compounds there are: strong bonds, no polarity, nothing for a reagent to grip, and essentially one reaction that is not burning. Alkenes have a single extra bond, and that one bond makes them the starting point for most of the organic industry — plastics, antifreeze, ethanol, margarine. The chapter is really the story of what a π bond does, told twice: once by a series that has none, and once by a series that has one.
What topic 14 asks you to do
14.1 Alkanes
14.1.1 recall the reactions by which alkanes are produced:
(a) addition of hydrogen to an alkene in a hydrogenation reaction, H2(g) and Pt/Ni
catalyst and heat (b) cracking of a longer chain alkane, heat with Al2O3
14.1.2 describe: (a) the complete and incomplete combustion of alkanes (b) the
free-radical substitution of alkanes by Cl2 or Br2 in the presence of ultraviolet
light, as exemplified by the reactions of ethane
14.1.3 describe the mechanism of free-radical substitution with reference to the initiation,
propagation and termination steps
14.1.4 suggest how cracking can be used to obtain more useful alkanes and alkenes of lower
Mr from heavier crude oil fractions
14.1.5 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
14.1.6 recognise the environmental consequences of carbon monoxide, oxides of nitrogen and
unburnt hydrocarbons arising from the combustion of alkanes in the internal combustion engine, and of
their catalytic removal
14.2 Alkenes
14.2.1 recall the reactions by which alkenes are produced:
(a) elimination of HX from a halogenoalkane by ethanolic NaOH and heat (b) dehydration of
an alcohol, using a heated catalyst (e.g. Al2O3) or a concentrated acid (e.g.
concentrated H2SO4) (c) cracking of a longer chain alkane
14.2.2 describe the reactions of alkenes: (a) electrophilic addition with
(i) hydrogen, H2(g) and Pt/Ni catalyst and heat (ii) steam,
H2O(g) and H3PO4 catalyst (iii) a hydrogen halide, HX(g), at
room temperature (iv) a halogen, X2; (b) oxidation by cold dilute acidified
KMnO4 to form the diol; (c) oxidation by hot concentrated acidified KMnO4
leading to rupture of the C=C, and the identities of the products used to determine the position of
alkene linkages in larger molecules; (d) addition polymerisation exemplified by ethene and propene
14.2.3 describe the use of aqueous bromine to show the presence of a C=C bond
14.2.4 describe the mechanism of electrophilic addition in alkenes, using bromine/ethene and
hydrogen bromide/propene as examples
14.2.5 describe and explain the inductive effects of alkyl groups on the stability of primary,
secondary and tertiary cations formed during electrophilic addition (used to explain Markovnikov addition)
Read the verbs. 14.1.1 and 14.2.1 say recall, so those are lists to know — but they are lists of reagents and conditions, and the conditions are where the marks are. 14.1.3 and 14.2.4 say describe the mechanism, which means equations and curly arrows on the page, not a sentence about what happens. 14.1.4 says suggest: you will be given a hydrocarbon and asked to propose products, so the balancing has to be something you do rather than something you remember. And 14.2.5 says describe and explain the inductive effect — an answer that says "Markovnikov's rule" without mentioning carbocation stability has answered a different question.
What this page covers, and what it leaves out
This page is topic 14 and nothing else. Two things that often travel with alkanes are deliberately absent, because they are not in these statements: the fractional distillation of crude oil and the uses of its fractions, which this syllabus does not examine; and sulfur dioxide and acid rain, which come from sulfur impurities and belong to topic 12, not to the combustion of alkanes. Cracking stays, because 14.1.4 names it; the catalytic converter stays, because 14.1.6 names it.
Three statements borrow from topics not yet met. 14.2.1(a) and (b) make alkenes from halogenoalkanes and alcohols, which are topics 15 and 16; they are given here as reactions to recall, with their conditions, and their mechanisms are left to those topics. And the hot-KMnO4 products in 14.2.2(c) are ketones and carboxylic acids, which are topics 17 and 18 — you need to recognise them here, not to know their chemistry.
What an alkane is, and what follows from it14.1, 13.1.2
An alkane is a saturated hydrocarbon: carbon and hydrogen only, every bond single, general formula CnH2n+2. Every carbon is sp3, tetrahedral, at 109.5°, and every C–C single bond rotates freely. There is no functional group at all.
| alkane | formula | b.p. / °C | state at room temperature |
|---|---|---|---|
| methane | CH4 | −162 | gas |
| ethane | C2H6 | −89 | gas |
| propane | C3H8 | −42 | gas |
| butane | C4H10 | −0.5 | gas (just) |
| pentane | C5H12 | 36 | liquid |
| octane | C8H18 | 126 | liquid |
| hexadecane | C16H34 | 287 | liquid |
The only forces between alkane molecules are instantaneous dipole–induced dipole forces, because the molecules have no dipole worth speaking of. Those forces get stronger as the electron cloud gets bigger and as the molecules touch over a greater area — which is why boiling point rises with chain length, and why branching lowers it at constant formula.
Why alkanes are unreactive14.1.5
Statement 14.1.5 asks you to understand the general unreactivity of alkanes "in terms of the strength of the C–H bonds and their relative lack of polarity". Those are two separate arguments and a full answer gives both.
The two halves of the answer
Strength. The weakest bond anywhere in an alkane is C–C at about 350 kJ mol−1, and C–H is stronger still at about 410. Breaking one requires a great deal of energy, and an ordinary reagent at room temperature cannot supply it.
Polarity. Carbon and hydrogen have electronegativities of 2.55 and 2.20, a difference of 0.35 — below the 0.4 usually taken as the threshold for calling a bond polar at all. Carbon and carbon differ by zero. So there is no δ+ carbon anywhere for a nucleophile to attack, and no region of high electron density for an electrophile to attack. There is no site.
Put them together: even if a reagent had the energy, it would not know where to start; and even if it knew where to start, it would not have the energy.
"Alkanes are unreactive because they are saturated"
This sounds right and explains nothing. "Saturated" means there is no C=C — it tells you an addition reaction cannot happen, which is true but is not the question. The question is why substitution by a polar reagent does not happen either, and the answer to that is the bond strength and the lack of polarity, not the absence of a double bond.
Note also that "unreactive" is relative. Alkanes burn extremely well — combustion is strongly exothermic and is the basis of the world's fuel supply. What they resist is reacting with ordinary chemical reagents at ordinary temperatures.
Making an alkane14.1.1
Two routes, and both are reactions you will meet again from the other side.
| reaction | reagent and conditions | example | |
|---|---|---|---|
| (a) | hydrogenation of an alkene | H2(g), Pt or Ni catalyst, heat | CH2=CH2 + H2 → CH3CH3 |
| (b) | cracking a longer alkane | heat with Al2O3 | C10H22 → C7H16 + C3H6 |
"Pt/Ni" means either, not both
The syllabus writes the catalyst as "Pt/Ni", which reads as though a mixture were needed. It is not: platinum and nickel are alternatives. Nickel is the cheaper one and is what is used industrially, for example in hardening vegetable oils into margarine; platinum works at a lower temperature and is used where cost matters less. Either earns the mark. The temperature is around 150 °C.
Complete and incomplete combustion14.1.2(a)
Do not memorise combustion equations. There are infinitely many alkanes and the examiner can choose any of them; what you need is a method that produces the equation in fifteen seconds.
The method, on C4H10
1 · Every carbon becomes one CO2: 4CO2.
2 · Every two hydrogens become one H2O: 5H2O.
3 · Count the oxygen atoms you have just used: 4 × 2 + 5 × 1 = 13. That is 6½ O2.
4 · Fractions are allowed but ugly, so double everything.
2C4H10 + 13O2 → 8CO2 + 10H2O
The same three steps work for any CnHm, and for compounds containing oxygen too — you just subtract the oxygen the molecule brought with it.
Incomplete combustion happens when the oxygen supply is restricted. There is not enough to give every carbon two oxygen atoms, so some carbon leaves as carbon monoxide, and in the worst case as carbon — soot. The equations are balanced the same way; only the carbon-containing product changes.
2C4H10 + 9O2 → 8CO + 10H2O
Why the oxygen number is the whole story
Burning one mole of butane completely needs 6.5 mol of O2; producing carbon monoxide instead needs only 4.5; producing soot needs 2.5. So as the air supply falls, the products shift down that list in order. That is not a rule to learn — it falls out of the arithmetic, and it is why a yellow, sooty flame means a badly adjusted burner and a blue flame means a well adjusted one.
Free-radical substitution14.1.2(b)
Apart from burning, this is essentially the only reaction an alkane has — and it needs ultraviolet light to happen at all, which is itself the evidence for everything said in section 3. No polar reagent can start it; only a photon energetic enough to break a bond outright.
CH3CH3 + Cl2 → UV CH3CH2Cl + HCl
A hydrogen has been removed and a chlorine has taken its place, and a molecule of hydrogen chloride has been released. That is a substitution. The syllabus names Cl2 or Br2, and it names ethane as the example.
The conditions, in the words that earn the mark
"In the presence of ultraviolet light" — or sunlight, which contains it. Not "heat", not "a catalyst", and certainly not "in solution". Writing "UV light" is enough.
The mechanism: initiation, propagation, termination14.1.3
The mechanism is not a ritual. It is a description of what particles actually collide with what, and every step is a real event happening billions of times a second in the flask.
The three stages, and the test that tells them apart
Initiation — radicals are made where there were none. Ultraviolet light breaks the halogen bond homolytically. 0 radicals in, 2 out.
Propagation — a radical reacts with a molecule and a new radical comes out. 1 radical in, 1 out, so the chain keeps going.
Termination — two radicals meet and combine. 2 radicals in, 0 out, so the chain stops.
Count the dots on each side of any step you have written. If the pattern is not 0→2, 1→1 or 2→0, the step is wrong.
Breaking the wrong bond in initiation
A very common answer starts with CH4 → CH3• + H•. It is wrong, and the reason is a number: the C–H bond is worth about 410 kJ mol−1 while Cl–Cl is worth 244. The weaker bond breaks. Initiation is always the halogen.
The second common error is writing the two propagation steps in the wrong order, or giving only one of them. Both are needed, and the second one has to regenerate the halogen radical — that regeneration is what makes it a chain.
Why this reaction is nearly useless for making anything
Every product of the substitution still has hydrogen atoms on it, so it is just as vulnerable to attack as the starting material was. Chlorinating methane does not give chloromethane; it gives a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane, in proportions you can shift but not control. And with ethane, once the first chlorine is in place the two carbons are no longer equivalent, so the second substitution gives two different isomers.
This is worth knowing for its own sake and because it makes a contrast: the reactions of alkenes in the second half of this chapter give one product, cleanly, which is exactly why industry builds on alkenes rather than on alkanes.
Cracking14.1.4
Long-chain alkanes from the heavier fractions of crude oil are in surplus; short-chain alkanes for petrol and alkenes for the chemical industry are in demand. Cracking breaks the long molecules into shorter ones, and 14.1.4 asks you to suggest how — which means producing balanced equations for hydrocarbons you have not seen before.
The arithmetic, and why an alkene always appears
Take C10H22 and split it into a C7 and a C3 fragment. Two alkanes of those sizes would be C7H16 and C3H8, which need 24 hydrogens between them. There are only 22.
The two missing hydrogens have to come from somewhere, and the only way to balance is for one fragment to contain a double bond:
C10H22 → C7H16 + C3H6
That deficit is structural, not incidental. Cracking always produces at least one unsaturated fragment, which is precisely why it is the industry's source of alkenes.
| thermal cracking | catalytic cracking | |
|---|---|---|
| conditions | high temperature (up to about 1000 °C) and high pressure | about 500 °C, slight pressure, zeolite catalyst |
| how the bond breaks | homolytically, giving radicals | heterolytically, by way of carbocations on the catalyst surface |
| main products | a high proportion of alkenes, mostly short ones | branched alkanes, cycloalkanes and aromatics, plus alkenes |
| used for | feedstock for the plastics industry | motor fuel — branched alkanes burn more smoothly |
| cost | higher, because of the temperature and pressure | lower, because the catalyst does the work |
The pollutants from an internal combustion engine14.1.6
Statement 14.1.6 names exactly three pollutants and asks for their environmental consequences and their catalytic removal. It is a short statement with a precise list, and the temptation is to answer it with everything you have ever heard about car exhausts.
| pollutant | where it comes from | why it matters |
|---|---|---|
| carbon monoxide, CO | incomplete combustion, when the cylinder does not get enough oxygen | toxic — it binds to haemoglobin far more strongly than oxygen does, so the blood cannot carry oxygen |
| oxides of nitrogen, NO and NO2 | nitrogen and oxygen from the air combine in the heat of the spark, which supplies enough energy to break the N≡N bond | toxic; NO2 dissolves in rain to give acid rain; with hydrocarbons in sunlight they form photochemical smog |
| unburnt hydrocarbons | part of the petrol–air mixture escapes the cylinder without reacting | contribute to photochemical smog, and absorb infrared radiation |
Two things this statement does not include
Sulfur dioxide comes from sulfur impurities in the fuel, not from the combustion of the alkane, and it belongs to topic 12. Bringing it in here answers a question that was not asked.
Carbon dioxide is the product of complete combustion and is a greenhouse gas, but it is not among the three pollutants 14.1.6 names, and a catalytic converter does not remove it — it produces more of it. Mention it if the question is about the greenhouse effect; leave it out if the question is about the three pollutants and their catalytic removal.
Where the nitrogen comes from
This one is worth getting right because it is counter-intuitive. The nitrogen in NO is not an impurity in the petrol. It is the nitrogen of the air drawn into the cylinder, which is normally the most unreactive gas in the atmosphere: the N≡N triple bond is 994 kJ mol−1, the strongest bond in any common molecule. The spark and the flame front are hot enough to break it, and once broken the atoms combine with oxygen. Run the engine cooler and you make less NO — which is one of the things engine designers trade against efficiency.
What an alkene is14.2, 13.3
An alkene is an unsaturated hydrocarbon containing a carbon–carbon double bond. One C=C gives the general formula CnH2n. Everything in the rest of this chapter follows from three facts about that double bond.
The three facts
1 · It is one σ bond and one π bond. Both carbons are sp2, with three σ bonds each, arranged trigonal planar at 120°. The leftover p orbital on each carbon overlaps sideways with its partner to make the π bond, above and below the plane.
2 · The π bond is the weaker half. C=C is about 610 kJ mol−1 against 350 for C–C, so the π is worth roughly 260 — less than the σ it sits on. It is therefore the one that breaks, and an addition reaction trades one π bond for two new σ bonds, which is why addition to an alkene is exothermic.
3 · The π electrons stick out. They sit above and below the molecular plane rather than between the nuclei, which makes them exposed and loosely held. An alkene is a region of high electron density with its electrons on the outside — exactly what an electrophile is looking for.
One structural consequence: no rotation
Twisting one end of a C=C by 90° pulls the two p orbitals out of alignment and destroys the π bond entirely, which costs about 260 kJ mol−1. Room temperature cannot supply that, so the ends are locked — and that is the origin of cis–trans isomerism, examined in 13.4.3. It is also why poly(ethene) is flexible while its monomer is rigid: the double bond becomes a single bond when the polymer forms, and single bonds rotate.
One animation for this section could not be recovered
Its data is missing from the file rather than damaged, so no emulator can repair it. The model below does the same job and rather more: instead of showing a fixed set of isomers, it generates every alkene skeleton for a chosen formula, puts the double bond in each distinct position, removes the duplicates, and then tests each survivor for cis–trans isomerism.
Making an alkene14.2.1
Three routes to recall, and the conditions are what distinguishes them from reactions you will meet alongside them.
| from | reagent and conditions | example | |
|---|---|---|---|
| (a) | a halogenoalkane | ethanolic NaOH, heat under reflux | CH3CH2Br + NaOH → CH2=CH2 + NaBr + H2O |
| (b) | an alcohol | heated Al2O3, or concentrated H2SO4 at about 170 °C | CH3CH2OH → CH2=CH2 + H2O |
| (c) | a longer alkane | cracking — see section 8 | C10H22 → C7H16 + C3H6 |
Ethanolic or aqueous — one word changes the product entirely
Sodium hydroxide reacting with a halogenoalkane does two completely different things depending on the solvent.
Ethanolic NaOH, heated: the hydroxide acts as a base, removing
a hydrogen from the carbon next door. Elimination → an alkene.
Aqueous NaOH, warmed: the hydroxide acts as a nucleophile, replacing the halogen.
Substitution → an alcohol.
Same reagent, same halogenoalkane, different solvent, different product. Writing "NaOH" without the solvent leaves the examiner unable to give the mark.
The two ways of making an alkene from an alcohol are both dehydration — an elimination in which the small molecule removed is water. Both are the reverse of the hydration in 14.2.2(a)(ii), and the conditions decide which direction you get.
The reactions of alkenes14.2.2
Almost everything in 14.2.2 is one idea applied seven ways: the π bond opens, and two new groups join the two carbons. The model below builds the product from whichever alkene and reagent you pick, so you can see the pattern rather than memorising fourteen equations.
| reagent | conditions | product from ethene | type |
|---|---|---|---|
| H2(g) | Pt or Ni catalyst, heat (about 150 °C) | CH3CH3 | electrophilic addition (hydrogenation) |
| H2O(g), steam | H3PO4 catalyst, high temperature and pressure | CH3CH2OH | electrophilic addition (hydration) |
| HX(g) | room temperature | CH3CH2X | electrophilic addition |
| X2 | room temperature, in the dark | CH2XCH2X | electrophilic addition |
| KMnO4/H+ | cold, dilute | CH2(OH)CH2OH | oxidation to the diol |
| KMnO4/H+ | hot, concentrated | 2CO2 + 2H2O | oxidative rupture of the C=C |
| itself | high pressure and a trace of O2, or a Ziegler–Natta catalyst | poly(ethene) | addition polymerisation |
Hydration: the syllabus route, and the one you may be shown instead
Statement 14.2.2(a)(ii) names steam with an H3PO4 catalyst, at high temperature and pressure. That is the industrial route to ethanol and it is the answer to this statement.
You will very often be taught a second route as well: cold concentrated sulfuric acid adds across the double bond to give ethyl hydrogensulfate, which is then warmed with water to give the alcohol and regenerate the acid.
CH2=CH2 + H2SO4 → CH3CH2OSO3H
CH3CH2OSO3H + H2O → CH3CH2OH + H2SO4
It is good chemistry, it is a nice illustration of electrophilic addition with a different electrophile, and it is not what 14.2.2(a)(ii) asks for. Give the phosphoric acid route unless the question specifies sulfuric acid.
The bromine water test14.2.3
The standard test for unsaturation, and the one most often described badly in an exam.
Say "colourless", not "clear"
Bromine water is already clear — clear means transparent, and you can see straight through it. What changes is that it becomes colourless: the orange goes. "Turns clear" describes no change at all, and it is marked as wrong.
For the equation, the syllabus is content with the simple one:
CH2=CH2 + Br2 → CH2BrCH2Br
In bromine water some 2-bromoethanol is also formed, because water competes with the bromide ion for the carbocation. You do not need it, and 14.2.3 does not ask for it.
Oxidation by manganate(VII)14.2.2(b)–(c)
The same reagent does two different things depending on how hard you push it, and the syllabus asks for both.
Cold and dilute: the diol
CH2=CH2 + [O] + H2O → CH2(OH)CH2OH
An OH group is added to each carbon of the old double bond. The purple manganate(VII) is reduced to colourless Mn2+, so the observation is purple to colourless. Ethane-1,2-diol is the product from ethene, and it is what antifreeze is made of.
Use [O], not the full ionic equation
Chemguide's guidance on this statement is worth repeating: for CIE purposes all you need are the simple equations involving an oxygen in square brackets. The full equation with MnO4− and eight hydrogen ions is correct but is not what is being tested here, and it takes five times as long to write.
Hot and concentrated: the molecule is cut in half
Under forcing conditions the C=C is ruptured completely, and each carbon of the old double bond ends up in a separate molecule. What each fragment turns into depends on what that carbon was carrying, and that is the whole point of the statement.
The three outcomes, and how to remember them
Look at the carbon of the C=C and count its hydrogens.
Two alkyl groups, no hydrogen → a ketone. There is no hydrogen
to remove, so oxidation stops there.
One alkyl group and one hydrogen → a carboxylic acid. The hydrogen is replaced by an OH.
Two hydrogens (a terminal =CH2) → carbon dioxide and water. There is nothing
holding that carbon in an organic molecule, so it goes all the way.
Worked example — using the products to locate the double bond
An alkene C5H10 is heated with concentrated acidified KMnO4 and gives propanone and ethanoic acid.
Propanone is (CH3)2C=O, so that carbon of the original double bond carried two methyl groups. Ethanoic acid is CH3COOH, so the other carbon carried one methyl and one hydrogen.
Rejoin them, replacing the two C=O bonds with the original C=C:
(CH3)2C=CHCH3 — 2-methylbut-2-ene
Check the formula: C5H10. It matches, and the position of the double bond has been deduced rather than guessed.
Forgetting that CO2 is evidence
Students often treat carbon dioxide among the products as an uninteresting by-product. It is the single most informative thing in the list: it says that one carbon of the double bond carried two hydrogens, which places the C=C at the end of the chain. An alkene that gives only a ketone and an acid, with no CO2, has its double bond somewhere in the middle.
The electrophilic addition mechanism14.2.4
Statement 14.2.4 names two examples and expects the mechanism drawn out for each: bromine with ethene, and hydrogen bromide with propene. They are the same mechanism; the difference is that the second one has a choice to make.
The shape of every electrophilic addition
Step 1 · The alkene attacks the electrophile. A curly arrow starts at the π bond and ends at the δ+ atom of the reagent. At the same time a second arrow starts at the reagent's own bond and ends on the atom that leaves, which departs as a negative ion.
Step 2 · A carbocation intermediate. One carbon now has only three bonds, six outer electrons and a positive charge.
Step 3 · The anion attacks. An arrow starts at a lone pair on the negative ion and ends at the positive carbon, forming the second new bond.
Bromine and ethene — the alkene makes its own electrophile
Bromine is a symmetrical, non-polar molecule. It has no dipole and no reason to attack anything. What makes it an electrophile is the alkene itself: as the two approach, the π electrons repel the electrons in the nearer bromine atom, which is pushed away from that atom and towards the far one. An induced dipole appears, the near bromine becomes δ+, and the alkene attacks it.
Why this is worth dwelling on
It is the cleanest demonstration in the syllabus that "electrophile" is a description of behaviour, not of charge. Nothing about a bromine molecule sitting in a bottle is electrophilic. It becomes an electrophile in the presence of something electron-rich, and it stops being one when that something is removed. Define an electrophile as "a positive ion" and this reaction becomes impossible to explain.
Hydrogen bromide and propene — the reagent has a dipole already
HBr is permanently polar, so no induction is needed. But propene is unsymmetrical: its two doubly-bonded carbons are not the same, so the hydrogen can join either one, and the two routes give different products. Which one dominates is decided by the intermediate, not by the product.
Carbocations, the inductive effect, and Markovnikov14.2.5
A carbocation is classified by how many carbon atoms are attached to the carbon carrying the charge.
| type | alkyl groups on the positive carbon | example | stability |
|---|---|---|---|
| primary (1°) | 1 | CH3CH2+ | least stable |
| secondary (2°) | 2 | (CH3)2CH+ | more stable |
| tertiary (3°) | 3 | (CH3)3C+ | most stable |
The inductive effect — the explanation 14.2.5 asks for
An alkyl group is electron-releasing: it pushes electron density along the σ bond towards whatever it is attached to. A carbocation is short of electrons, so this push partly relieves the positive charge and spreads it over more atoms.
Spreading a charge always lowers the energy. So the more alkyl groups attached to the positive carbon, the more stable the ion — and a more stable intermediate forms faster, because less energy is needed to reach it.
Worked example — propene + HBr
CH2=CHCH3. The hydrogen can join either carbon of the double bond.
Route 1: H joins the CH2 end. The charge is left on the middle carbon, which is attached to two other carbons → a secondary carbocation. Br− then attacks it, giving CH3CHBrCH3, 2-bromopropane.
Route 2: H joins the middle carbon. The charge is left on the end carbon, attached to one other carbon → a primary carbocation. The product would be CH2BrCH2CH3, 1-bromopropane.
The secondary carbocation is the more stable, so route 1 is faster and 2-bromopropane is the major product. 1-bromopropane is formed too, as the minor product — this is a preference, not an absolute rule.
Markovnikov's rule is the shortcut, not the answer
"The hydrogen adds to the carbon that already has more hydrogens" gets you to the right product in a few seconds and is worth knowing for that. But 14.2.5 says describe and explain the inductive effects of alkyl groups on the stability of primary, secondary and tertiary cations. An answer that names the rule and stops has described the observation and explained nothing.
The full answer names the two possible carbocations, classifies each as primary, secondary or tertiary, says that alkyl groups release electron density and spread the positive charge, concludes that the more substituted ion is more stable and therefore forms faster, and names the major product. Five steps, and each one is a mark.
Addition polymerisation14.2.2(d)
Given the right conditions an alkene will add to itself, over and over, and the molecules join into a chain thousands of units long. The π bond of each monomer opens and its two electrons are used to make the links.
n CH2=CH2 → –[–CH2–CH2–]n–
Drawing a repeat unit — the four things that are marked
1 · The double bond becomes a single bond. Leaving a C=C inside the brackets is the commonest error in this statement.
2 · Both carbons of the old double bond stay in the repeat unit, with all their groups still attached.
3 · A bond comes out of each end of the bracket, showing where the next unit joins.
4 · An n goes outside the bracket, bottom right.
| monomer | polymer | repeat unit |
|---|---|---|
| ethene, CH2=CH2 | poly(ethene) | –[–CH2–CH2–]n– |
| propene, CH2=CHCH3 | poly(propene) | –[–CH2–CH(CH3)–]n– |
| chloroethene, CH2=CHCl | poly(chloroethene), PVC | –[–CH2–CHCl–]n– |
Two things that follow from the mechanism
The atom economy is 100%. Nothing is released, so the repeat unit has exactly the same molecular formula as the monomer and every atom that went in comes out in the product. Contrast condensation polymerisation in topic 20, where a small molecule is lost at every link.
The bond angle changes. In the monomer, the two carbons of the C=C are sp2 and trigonal planar at 120°. In the polymer they have four single bonds, so they are sp3 and tetrahedral at 109.5°. A rigid planar monomer becomes a flexible zig-zag chain, and that is why poly(ethene) can be moulded.
Conditions for poly(ethene) — two processes, two products
Low density poly(ethene), LDPE: very high pressure (about 2000 atm), about 200 °C, a trace of oxygen as initiator. The chains branch, so they pack badly — low density, flexible, low melting point. Carrier bags and squeezy bottles.
High density poly(ethene), HDPE: a Ziegler–Natta catalyst, low pressure, about 60 °C. The chains are unbranched, so they pack closely — higher density, more rigid, higher melting point. Crates, pipes and bottle caps.
Same monomer, same repeat unit, different conditions, different material. Chemguide's note on 14.2.2(d) is that you should be able to quote conditions as well as draw the repeat unit, so it is worth having one set of numbers for each.
Self-test14
Thirty questions across both halves of the topic. Each explains itself after you answer.
Every reaction in topic 14, on one page14
If you learn one table from this chapter, learn this one. The reagent and the conditions are what distinguish the reactions from each other, and they are where the marks sit.
Alkanes
| reaction | reagent and conditions | product | outcome |
|---|---|---|---|
| made by hydrogenation | alkene + H2(g), Pt or Ni, heat | alkane | 14.1.1(a) |
| made by cracking | longer alkane, heat with Al2O3 | shorter alkane + alkene | 14.1.1(b), 14.1.4 |
| complete combustion | excess O2 | CO2 + H2O | 14.1.2(a) |
| incomplete combustion | limited O2 | CO (or C) + H2O | 14.1.2(a) |
| free-radical substitution | Cl2 or Br2, ultraviolet light | halogenoalkane + HX | 14.1.2(b), 14.1.3 |
Alkenes
| reaction | reagent and conditions | product from propene | outcome |
|---|---|---|---|
| made by elimination | halogenoalkane + ethanolic NaOH, heat | alkene | 14.2.1(a) |
| made by dehydration | alcohol + heated Al2O3, or conc. H2SO4 at 170 °C | alkene + H2O | 14.2.1(b) |
| hydrogenation | H2(g), Pt or Ni, heat | propane | 14.2.2(a)(i) |
| hydration | steam, H3PO4, high T and p | propan-2-ol | 14.2.2(a)(ii) |
| addition of HX | HBr(g), room temperature | 2-bromopropane (major) | 14.2.2(a)(iii) |
| addition of X2 | Br2, room temperature, dark | 1,2-dibromopropane | 14.2.2(a)(iv) |
| oxidation, mild | cold dilute acidified KMnO4 | propane-1,2-diol | 14.2.2(b) |
| oxidation, forcing | hot concentrated acidified KMnO4 | ethanoic acid + CO2 + H2O | 14.2.2(c) |
| test for C=C | aqueous bromine, dark | orange → colourless | 14.2.3 |
| addition polymerisation | high p + trace O2, or Ziegler–Natta | poly(propene) | 14.2.2(d) |
Definitions to learn14
| term | definition |
|---|---|
| alkane | a saturated hydrocarbon, general formula CnH2n+2, containing only C–C and C–H single bonds |
| alkene | an unsaturated hydrocarbon containing a carbon–carbon double bond, general formula CnH2n for one C=C |
| free radical | a species with an unpaired electron |
| initiation | the step in which free radicals are produced from a molecule by homolytic fission, here caused by ultraviolet light |
| propagation | a step in which a radical reacts with a molecule and a new radical is produced, so the chain continues |
| termination | a step in which two radicals combine, removing radicals from the mixture and stopping the chain |
| electrophile | a species that accepts a lone pair of electrons to form a new covalent bond |
| carbocation | an ion in which a carbon atom carries a positive charge, having only three bonds and six outer electrons |
| inductive effect | the release or withdrawal of electron density along a σ bond by a group attached to it; alkyl groups release electron density |
| cracking | breaking a long-chain alkane into shorter alkanes and alkenes, by heat alone or over a catalyst |
| addition polymerisation | the joining of many unsaturated monomer molecules into a long chain, with no other product formed |
| repeat unit | the smallest group of atoms that, repeated, makes up the polymer chain |
| Markovnikov addition | the observation that when HX adds to an unsymmetrical alkene, the hydrogen joins the carbon already carrying more hydrogens — because that leaves the more stable carbocation |
Data used on this page14
These are reference values, not the official data booklet
Bond energies and boiling points differ by a few units between sources, and the differences are large enough to change a calculated ΔH by ten or twenty kJ mol−1. The values below are the ones every model on this page computes from, printed so you can see what each answer rests on. In an examination, use the data booklet you are given.
Bond energies
| bond | kJ mol−1 | bond | kJ mol−1 | bond | kJ mol−1 |
|---|---|---|---|---|---|
| C–C | 350 | C–H | 410 | C=C | 610 |
| C–F | 467 | C–Cl | 340 | C–Br | 280 |
| C–I | 240 | C–O | 360 | C=O | 740 |
| H–H | 436 | O=O | 496 | O–H | 460 |
| F–F | 158 | Cl–Cl | 244 | Br–Br | 193 |
| I–I | 151 | H–F | 562 | H–Cl | 431 |
| H–Br | 366 | H–I | 298 | N≡N | 994 |
Two of these carry arguments on this page. Cl–Cl at 244 against C–H at 410 is why initiation breaks the halogen and not the alkane. H–I at 298 against C–H at 410 is why iodine will not sustain a radical chain at all: the first propagation step is endothermic by 112 kJ mol−1.
Boiling points of the alkanes, °C
| carbons | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| straight chain | −162 | −89 | −42 | −0.5 | 36 | 69 | 98 | 126 | 151 | 174 | 196 | 216 |
The three isomers of C5H12: pentane 36.0 °C, 2-methylbutane 27.8 °C, 2,2-dimethylpropane 9.5 °C.
Relative atomic masses
H 1.0 · C 12.0 · N 14.0 · O 16.0 · Cl 35.5 · Br 79.9 · I 126.9