What this chapter covers15
A halogenoalkane is an alkane with a halogen bolted on, and that one change turns the least reactive family in organic chemistry into one of the most useful. The alkane had nothing a reagent could grip: strong bonds, no polarity, no lone pairs. Put a chlorine, bromine or iodine on one of those carbons and you get a bond that is both polar and weak, and a molecule that will hand that carbon over to almost any species carrying a lone pair. Halogenoalkanes are rarely the thing you want; they are how you get to the thing you want — alcohols, nitriles, amines, alkenes. This chapter is the junction box of AS organic chemistry.
What topic 15 asks you to do
15.1 Halogenoalkanes — the whole of topic 15 at AS Level
15.1.1 recall the reactions (reagents and conditions) by which
halogenoalkanes can be produced: (a) the free-radical substitution of alkanes by
Cl2 or Br2 in the presence of ultraviolet light, as exemplified by the reactions
of ethane (b) electrophilic addition of an alkene with a halogen, X2, or hydrogen
halide, HX(g), at room temperature (c) substitution of an alcohol, e.g. by reaction with
HX(g); or with KCl and concentrated H2SO4 or concentrated
H3PO4; or with PCl3 and heat; or with PCl5; or with
SOCl2
15.1.2 classify halogenoalkanes into primary, secondary and tertiary
15.1.3 describe the following nucleophilic substitution reactions: (a) the reaction with
NaOH(aq) and heat to produce an alcohol (b) the reaction with KCN in ethanol and heat to
produce a nitrile (c) the reaction with NH3 in ethanol heated under pressure to
produce an amine (d) the reaction with aqueous silver nitrate in ethanol as a method of
identifying the halogen present as exemplified by bromoethane
15.1.4 describe the elimination reaction with NaOH in ethanol and heat to produce an alkene as
exemplified by bromoethane
15.1.5 describe the SN1 and SN2 mechanisms of nucleophilic substitution in
halogenoalkanes including the inductive effects of alkyl groups
15.1.6 recall that primary halogenoalkanes tend to react via the SN2 mechanism;
tertiary halogenoalkanes via the SN1 mechanism; and secondary halogenoalkanes by a mixture
of the two, depending on structure
15.1.7 describe and explain the different reactivities of halogenoalkanes (with particular
reference to the relative strengths of the C–X bonds as exemplified by the reactions of
halogenoalkanes with aqueous silver nitrates)
Read the verbs, because they set the depth. 15.1.1 says recall — three routes to learn, and what is being marked is the reagents and the conditions, not the idea. 15.1.3 and 15.1.4 say describe: equations, reagents, conditions, products, and the product named correctly. 15.1.5 says describe the mechanisms, which means curly arrows drawn on the page, an intermediate or a transition state shown, and the charges right. 15.1.6 says recall which class goes by which mechanism — three lines to know. And 15.1.7 is the one statement that says describe and explain, so an answer there that stops at "iodoalkanes are the most reactive" has done half the job; the explanation is the C–X bond enthalpy, and saying so is where the mark is.
What this page covers, and what it leaves out
This page is AS topic 15, which is sub-topic 15.1 and nothing else — there is no 15.2. The A Level extension, topic 31 Halogen compounds, takes the same functional group onto a benzene ring: how halogenoarenes are made, and why chlorobenzene is so much less reactive than chloroethane. That is a separate unit and it is not on this page.
Two statements borrow from topics on either side of this one. 15.1.1(a) and (b) are the free-radical substitution of an alkane and the electrophilic addition of an alkene, which belong to topic 14; they appear here because they are how the compounds in this chapter get made, and both mechanisms are given in full below. 15.1.1(c) makes halogenoalkanes from alcohols, which is topic 16 — it is given here as a list of reagents and conditions to recall, which is all 15.1.1 asks for.
What a halogenoalkane is15.1
Take an alkane and replace one or more hydrogen atoms with halogen atoms. That is the whole definition. Almost everything this chapter examines involves compounds with a single halogen on a saturated carbon chain, so that is what the models on this page work with.
Halogenoalkane
A compound in which one or more of the hydrogen atoms of an alkane has been replaced by a halogen atom. General formula for the singly-substituted ones: CnH2n+1X, usually written R–X, where X is F, Cl, Br or I.
The chemistry all follows from two facts about that C–X bond, and they pull in opposite directions.
The bond is polar. With the exception of iodine, the halogens are more electronegative than carbon, so the shared pair sits closer to the halogen. The carbon is left δ+ and the halogen δ−. A δ+ carbon is an invitation to anything with a lone pair, and that is the whole of 15.1.3: a nucleophile arrives, takes the carbon, and the halogen leaves.
The bond is weak — weak, that is, compared with the C–H and C–C bonds around it. Breaking it is the slow, expensive part of every reaction in this chapter, so how strong it is decides how fast the compound reacts. That is 15.1.7.
Why the two facts point different ways
Polarity is largest for C–F and smallest for C–I. Bond strength runs the same way: C–F is the strongest, C–I the weakest. So if reactivity were decided by how δ+ the carbon is, fluoroalkanes would be the most reactive; if it is decided by how easily the bond breaks, iodoalkanes would be. They cannot both be right, and the experiment settles it — section 24 does the arithmetic. Keep the question in mind as you read; it is the one place in this topic where the obvious argument is the wrong one.
Primary, secondary and tertiary15.1.2
Statement 15.1.2 is one word long — classify — and it is worth more than it looks, because almost every later statement in the topic sorts its answer by this classification. Which mechanism a compound uses (15.1.6), whether it substitutes or eliminates (15.1.3 and 15.1.4), how fast it hydrolyses (15.1.7): all of them start by asking what class it is.
The rule counts one thing only: how many carbon atoms are attached to the carbon that carries the halogen.
| Class | Carbons on the C–X carbon | General form | Example |
|---|---|---|---|
| primary (1°) | 0 or 1 | RCH2X | 1-bromobutane, CH3CH2CH2CH2Br |
| secondary (2°) | 2 | R2CHX | 2-bromobutane, CH3CH2CHBrCH3 |
| tertiary (3°) | 3 | R3CX | 2-bromo-2-methylpropane, (CH3)3CBr |
The trap: counting the wrong thing
Wrong: "2-bromobutane has four carbons and the halogen is on carbon 2, so it is secondary because of the 2." The number in the name is a position, not a class — 1-bromo-2-methylpropane, (CH3)2CHCH2Br, has its halogen on carbon 1 and is primary, and that has nothing to do with the 1 in the name.
Right: find the carbon bonded to the halogen, then count the carbon atoms bonded to that carbon. Two of them makes it secondary, whatever the numbering says. Chloromethane, CH3Cl, has none at all and is counted as primary.
Naming them13.1.4
The naming rules are topic 13, but the marks are lost here, so they are worth restating for this family. A halogen is always a substituent prefix — fluoro, chloro, bromo, iodo — never a suffix, because it is never the principal functional group in a molecule that also contains an alcohol, an aldehyde or an acid.
- Find the longest carbon chain and name it as the alkane.
- Number the chain from the end that gives the substituents the lowest set of locants. With nothing but halogens and alkyl groups present, that is the only rule you need.
- Put the prefixes in alphabetical order — bromo before chloro before methyl — ignoring the di/tri multipliers when you alphabetise.
- Every substituent gets its own number, even when two are on the same carbon: 1,1-dichloroethane, not 1-dichloroethane.
How CIE writes the structure
Write 2-bromopropane as CH3CHBrCH3. Brackets round a single substituent — CH3CH(Br)CH3 — read as a branch that is not there. Brackets are for a group that genuinely hangs off the chain, as in (CH3)3CBr, or for a repeated unit, as in CH3(CH2)3Br.
Route one: substituting an alkane15.1.1(a)
Mix an alkane with chlorine or bromine in the dark and nothing happens. Shine ultraviolet light on the same mixture and it reacts — sometimes explosively. The light is not warming anything; it is doing one specific job, and the mechanism is built around it.
The reagents and conditions 15.1.1(a) wants
Cl2 or Br2, in the presence of ultraviolet light (sunlight will do). Exemplified by ethane:
C2H6 + Cl2 → C2H5Cl + HCl
The mechanism is free-radical substitution, in three named stages: initiation, propagation, termination.
Initiation. A chlorine molecule absorbs a UV photon. The Cl–Cl bond breaks homolytically — one electron to each atom — giving two chlorine atoms each with an unpaired electron. Those are free radicals, and the dot is part of the formula, not decoration.
Cl2 → 2Cl• (UV light)
Propagation. Two steps, and they must come as a pair. A radical goes in, a different radical comes out, so the chain keeps itself running.
Cl• + C2H6 → C2H5• + HCl
C2H5• + Cl2 → C2H5Cl + Cl•
Termination. Any two radicals meeting combine, and the chain stops. All three combinations happen, and all three are acceptable answers.
Cl• + Cl• → Cl2 · C2H5• + Cl• → C2H5Cl · C2H5• + C2H5• → C4H10
Homolytic, not heterolytic
Free-radical steps use half-headed (fish-hook) arrows, because one electron moves, not a pair. Every other mechanism in this chapter uses full curly arrows for a pair. Mixing them up is the quickest way to lose a mechanism mark: a full arrow in an initiation step says the chlorine molecule split into Cl+ and Cl−, which is not what UV light does to a non-polar bond.
So this route works. It is also, as a preparation, close to useless — and knowing why is the part of 15.1.1(a) that gets asked about.
Why you would not actually make a halogenoalkane this way
The product is a radical's target too. Chloroethane still has five hydrogens, and a chlorine radical cannot tell them from the ones in ethane. So the chain carries on: C2H4Cl2, C2H3Cl3, and onwards to C2Cl6.
And with anything bigger than ethane, the position is random too. Propane has hydrogens on two chemically different carbons, so a single substitution already gives two isomers — 1-chloropropane and 2-chloropropane. You get a mixture with similar boiling points, and separating it costs more than the product is worth.
This is exactly why the syllabus exemplifies the reaction with ethane: every hydrogen in ethane is equivalent, so the first substitution has only one possible product.
Route two: adding to an alkene15.1.1(b)
The second route fixes everything wrong with the first. An alkene has one place where it reacts — the C=C — so there is nothing random about where the halogen ends up, and the product has no double bond left, so it cannot react again. It is the route you would actually use.
The reagents and conditions 15.1.1(b) wants
A halogen, X2, or a hydrogen halide, HX(g), at room temperature. No catalyst, no heat, no light.
CH2=CH2 + Br2 → CH2BrCH2Br
CH2=CH2 + HBr → CH3CH2Br
The mechanism is electrophilic addition.
Why an alkene reacts at all is worth one paragraph, because it is the mirror image of this chapter's own chemistry. The C=C is made of a σ bond and a π bond, and the π electrons sit above and below the line of the nuclei rather than between them. They are exposed and loosely held, so the double bond is a region of high electron density — which makes it attractive to anything electron-deficient. A species that seeks electrons like this is an electrophile. In the rest of this chapter the polarity is the other way round and the reagents are nucleophiles; the alkene is the one place where the organic molecule does the attacking.
With Br2 there is no permanent dipole to start with: the alkene makes its own electrophile. As the bromine molecule approaches, the π electrons repel the electrons of the near bromine atom, leaving it δ+ and the far one δ−. The π pair then attacks that δ+ atom, the Br–Br bond breaks heterolytically, and a bromide ion leaves. What is left is a carbocation, which the bromide ion attacks in the second step. With HBr the dipole is already there — bromine is more electronegative than hydrogen — so the hydrogen is permanently δ+ and no induction is needed.
One of those three reactions is not the one this syllabus teaches
The animation includes the hydration of ethene by cold concentrated sulfuric acid, followed by hydrolysis of the intermediate to ethanol. It is genuine chemistry and it is how the reaction was once run industrially, but statement 14.2.2(a)(ii) names steam with an H3PO4 catalyst as the hydration route, and that is the one to give in an answer. It is also not a route to a halogenoalkane at all, so for this chapter only the bromine and hydrogen bromide stages matter.
With an unsymmetrical alkene, HX can add two ways round and the two products are not formed in equal amounts. Propene and HBr give mostly 2-bromopropane, not 1-bromopropane. The reason is the carbocation: adding the hydrogen to the CH2 end leaves the charge on the middle carbon, which has two alkyl groups feeding electron density towards it, while adding it the other way leaves the charge on an end carbon with only one. The more stable carbocation forms faster, so its product dominates. Hold on to that argument — it comes back in section 21 as the reason tertiary halogenoalkanes react by a completely different mechanism from primary ones.
Route three: substituting an alcohol15.1.1(c)
The third route swaps an –OH for a halogen. Statement 15.1.1(c) says recall, and it lists five sets of reagents, so this section is a list to learn rather than an argument to follow. The chemistry of alcohols is topic 16; what is wanted here is the reagents and the conditions.
| Reagent | Conditions | Equation, for ethanol | Makes |
|---|---|---|---|
| HX(g) | — | C2H5OH + HBr → C2H5Br + H2O | any halogenoalkane |
| KCl + concentrated H2SO4 | the HX is generated in the flask | C2H5OH + HCl → C2H5Cl + H2O | chloroalkane |
| a halide salt + concentrated H3PO4 | the HX is generated in the flask | C2H5OH + HBr → C2H5Br + H2O | bromo- and iodoalkanes |
| PCl3 | heat | 3C2H5OH + PCl3 → 3C2H5Cl + H3PO3 | chloroalkane |
| PCl5 | room temperature | C2H5OH + PCl5 → C2H5Cl + POCl3 + HCl | chloroalkane |
| SOCl2 | room temperature | C2H5OH + SOCl2 → C2H5Cl + SO2 + HCl | chloroalkane |
Why the syllabus offers phosphoric acid as an alternative
Concentrated sulfuric acid is an oxidising agent. Used with a chloride it is fine, because HCl resists oxidation. Used with a bromide or an iodide it partly oxidises the hydrogen halide it has just made — to brown bromine, or to purple iodine vapour — and the yield falls. Concentrated phosphoric acid is just as good at liberating HX and is not an oxidising agent, so it is the one used for bromo- and iodoalkanes. That is the same argument as the concentrated-sulfuric-acid test for halide ions in topic 11, read from the other side.
PCl5 does double duty
The steamy, acidic fumes of HCl given off when PCl5 meets an alcohol at room temperature are the standard test for an –OH group. So this reaction appears twice in the course: here as a way of making a chloroalkane, and in topic 16 as a way of proving an alcohol is present.
Choosing a route15.1.1
Three routes, and a question that asks you to suggest a preparation is asking you to pick one and justify it. The model below takes a target halogenoalkane, works out which routes can actually reach it, and builds the equation and the conditions for each — including the cases where a route gives a mixture rather than the compound you asked for.
The three routes in one sentence each
From an alkane — works, but gives a mixture of positions and of degrees of substitution, so it is only sensible when every hydrogen is equivalent. From an alkene — clean, room temperature, one product with HX on a symmetrical alkene, and the major product predictable on an unsymmetrical one; but you need the alkene, and you cannot put the halogen anywhere except next to where the double bond was. From an alcohol — the halogen goes exactly where the –OH was, so you have complete control of the position; this is the route a synthesis would normally take.
Nucleophiles, and why the C–X bond gives way15.1.3, 13.2.1
Every reaction in statement 15.1.3 is the same reaction with a different reagent. A species carrying a lone pair is attracted to the δ+ carbon; it donates that pair to the carbon; the C–X bonding pair is pushed onto the halogen, which leaves as a halide ion. One group has replaced another, so it is a substitution, and the attacking species was a nucleophile, so it is a nucleophilic substitution.
Nucleophile
A species — an ion or a molecule — that is attracted to a region of positive charge and donates a pair of electrons to form a new covalent bond. Every nucleophile therefore has a lone pair; many, but not all, carry a negative charge.
In this topic there are four: OH−, CN−, NH3 and H2O. The first two are negative ions; the last two are neutral molecules with a lone pair on the nitrogen or the oxygen.
The trap: "a nucleophile is a negative ion"
Wrong. Two of the four nucleophiles in this topic are neutral molecules. What defines a nucleophile is the lone pair it donates, not the charge it carries. Ammonia and water are nucleophiles in exactly the same sense that hydroxide is — they just leave their products carrying a positive charge that has to be removed afterwards, which is why the ammonia and water equations have an extra step that the hydroxide one does not.
Three things are happening at once, and a good mechanism answer shows all three: a bond forming to the nucleophile, a bond breaking to the halogen, and a negative charge ending up on the halide ion. The halogen is called the leaving group, and how readily it leaves is what section 23 is about.
Substitution by hydroxide: the alcohol15.1.3(a)
15.1.3(a) — reagent and conditions
NaOH(aq), heat — in practice, warm under reflux with aqueous sodium hydroxide. The product is an alcohol.
CH3CH2Br + NaOH → CH3CH2OH + NaBr
CH3CH2Br + OH− → CH3CH2OH + Br−
Either equation is acceptable; the ionic one is the honest description, because the sodium ion takes no part. The reaction is also called hydrolysis, and that word is worth keeping, because the same substitution happens slowly with water alone — which is what the silver nitrate test in section 14 depends on.
The word "aqueous" is doing real work here
The same halogenoalkane, the same sodium hydroxide, a different solvent, and you get a different product. Aqueous NaOH gives the alcohol by substitution. Ethanolic NaOH gives an alkene by elimination — statement 15.1.4, section 15. Writing "NaOH and heat" without saying which solvent answers neither statement properly. Section 17 sets the two side by side.
Substitution by cyanide: the nitrile15.1.3(b)
15.1.3(b) — reagent and conditions
KCN in ethanol, heat — heated under reflux with potassium cyanide dissolved in ethanol. The product is a nitrile.
CH3CH2Br + KCN → CH3CH2CN + KBr
This is the most useful reaction in the whole of AS organic chemistry, and the reason is easy to miss: the cyanide ion brings a carbon atom with it. The chain gets one carbon longer. Nothing else you meet at AS does that. If a synthesis question asks you to turn a two-carbon compound into a three-carbon compound, this is the step it is fishing for.
The trap: naming the nitrile
Bromoethane has two carbons. Its nitrile is propanenitrile, not "ethanenitrile" — the carbon of the –C≡N group is carbon 1 of the new chain and counts towards the stem. Count the carbons in the product, not in the starting material.
Why the solvent is ethanol and not water
Cyanide is the conjugate base of a weak acid, so in water a fair proportion of it exists as HCN and the solution also contains hydroxide ions. Those hydroxide ions do the reaction of 15.1.3(a) instead, and you get the alcohol contaminating your nitrile. Ethanol keeps the cyanide as cyanide. This is the same kind of solvent argument as the one in 15.1.4, and examiners expect the solvent to be stated.
Substitution by ammonia: the amine15.1.3(c)
15.1.3(c) — reagent and conditions
NH3 in ethanol, heated under pressure — in practice excess ammonia in a sealed tube. The product is an amine.
CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br
Two ammonia molecules appear in the balanced equation and only one of them attacks the carbon. The first is the nucleophile; its lone pair bonds to the δ+ carbon, and what forms immediately is the protonated amine, CH3CH2NH3+, with the positive charge on the nitrogen. A second ammonia molecule then removes that proton and walks away as NH4+. Leave the second ammonia out and the equation does not balance.
Why "under pressure", and why "excess"
Under pressure, in a sealed tube: ammonia is a gas and the halogenoalkane is volatile, so heating them in an open flask simply loses both.
Excess ammonia: the amine that forms is itself a nucleophile — the nitrogen still has a lone pair — so it can attack another halogenoalkane molecule to give a secondary amine, then a tertiary amine, then a quaternary ammonium salt. Using a large excess of ammonia makes it far more likely that the halogenoalkane meets ammonia than the amine, so the primary amine dominates. This syllabus asks only for the first step, but the reason for the excess is fair game.
The four nucleophiles side by side15.1.3
The model below takes any halogenoalkane, applies each of the four nucleophiles in turn, and builds the equation, the product and its name from the structure rather than looking them up. The fourth nucleophile, water, is not listed in its own sub-statement, but it is the one doing the work in 15.1.3(d), and it behaves exactly like ammonia: a neutral molecule attacks, and a proton has to come off at the end.
| Nucleophile | Reagent and conditions | Product class | Product from bromoethane |
|---|---|---|---|
| OH− | NaOH(aq), heat under reflux | alcohol | ethanol, CH3CH2OH |
| CN− | KCN in ethanol, heat under reflux | nitrile | propanenitrile, CH3CH2CN |
| NH3 | excess NH3 in ethanol, heat under pressure | amine | ethylamine, CH3CH2NH2 |
| H2O | warm with water; AgNO3(aq) in ethanol to follow it | alcohol | ethanol, CH3CH2OH |
The silver nitrate reaction15.1.3(d)
This one experiment is named in two separate statements, and it answers two different questions. Statement 15.1.3(d) uses which colour the precipitate is, to say which halogen was in the compound. Statement 15.1.7 uses how fast it appears, to compare how reactive the compounds are. Same test tube, two readings.
15.1.3(d) — the procedure
Warm the halogenoalkane with aqueous silver nitrate in ethanol — typically in a water bath at about 50 °C. A precipitate of the silver halide appears, and its colour identifies the halogen.
RBr + H2O → ROH + H+ + Br−
Ag+(aq) + Br−(aq) → AgBr(s)
Water is the nucleophile. It is a poor one — neutral, and a weaker electron-pair donor than hydroxide — which is exactly why the reaction is slow enough to time. The silver ion has nothing to do with the substitution itself; it sits in solution waiting for a halide ion to be released, and precipitates it the moment one appears. The precipitate is therefore a visible marker for a step that is otherwise invisible.
Why ethanol has to be there
Halogenoalkanes are essentially insoluble in water and silver nitrate is insoluble in ethanol. In either solvent alone, one reagent sits as a separate layer and the reaction is limited by how fast the two phases mix — which would make the timings meaningless. A mixture of ethanol and water dissolves both, so every molecule is in the same solution and the rate you measure is the rate of the chemistry.
| Halogen in the compound | Precipitate | Colour | How quickly it appears |
|---|---|---|---|
| iodine | AgI | yellow | fastest — often almost at once |
| bromine | AgBr | cream | slower |
| chlorine | AgCl | white | slowest of the three |
| fluorine | — | no precipitate | no reaction under these conditions |
Run the three side by side, at the same temperature and concentration, and the order in which the precipitates appear is the order of reactivity. That order is the evidence for statement 15.1.7, and section 23 explains it.
Where you have met these precipitates before
The same three colours identify halide ions in topic 11.3.2(a), where silver nitrate is added to a solution of the halide and the precipitates are then tested with dilute and concentrated ammonia. The colours are the same because the precipitates are the same compounds. The difference is what is being proved: in topic 11 the halide ion was already there, and here it has to be released from a covalent molecule first — which is the whole point of the experiment.
Elimination: making the alkene15.1.4
Change the solvent and the same two reagents do something else entirely. In ethanol, with heat, the hydroxide ion stops behaving as a nucleophile and behaves as a base: instead of attacking the carbon it removes a hydrogen from the carbon next door. Two things leave the molecule — an H from one carbon, an X from the next — and the pair of electrons left behind becomes a π bond. The product is an alkene.
15.1.4 — reagent and conditions
NaOH in ethanol, heat — hot ethanolic sodium hydroxide, no water. The product is an alkene.
CH3CH2Br + NaOH → CH2=CH2 + NaBr + H2O
CH3CH2Br + OH− → CH2=CH2 + Br− + H2O
Three products, not two. The water is where the removed hydrogen went, and leaving it out is the commonest way to get this equation wrong.
Elimination reaction
A reaction in which a small molecule — here HX — is removed from a single molecule, with no other reactant becoming part of the product, so that a double bond is formed. Compare addition, which does the reverse, and substitution, in which one group replaces another and nothing is lost.
The mechanism has three curly arrows and they must be drawn in the right order of cause:
The trap: which hydrogen comes off
Wrong: removing the hydrogen from the same carbon as the halogen. That would leave both electrons on one carbon and there is nothing for them to do; no double bond can form between a carbon and itself.
Right: the hydrogen comes off an adjacent carbon. That is also why a halogenoalkane with no hydrogen on any adjacent carbon cannot eliminate at all — bromomethane, CH3Br, has no neighbouring carbon, so ethanolic NaOH can only substitute.
The syllabus example is the least representative one
Statement 15.1.4 exemplifies elimination with bromoethane, and that is the equation to be able to write. But bromoethane is primary, and primary halogenoalkanes mostly substitute even under ethanolic conditions — the elimination is real but it is the minor pathway. Tertiary halogenoalkanes are the ones that eliminate readily. Learn bromoethane because the syllabus names it, and do not carry away the idea that primary compounds are good at elimination.
More than one alkene15.1.4
Bromoethane has only one kind of adjacent hydrogen, so there is only one possible alkene. Most halogenoalkanes are not so tidy. 2-bromobutane has hydrogens on the carbons on both sides of the C–Br, and the two choices give different alkenes — and one of those choices gives a pair of stereoisomers as well.
CH3CH2CHBrCH3 → but-1-ene, cis-but-2-ene or trans-but-2-ene
Three products from one starting material and one reagent. The model below finds them by construction: it locates every hydrogen on a carbon adjacent to the C–X, removes each one in turn with the halogen, builds the alkene that results, discards duplicates, and then tests each survivor for cis– trans isomerism by checking whether either carbon of the new double bond carries two identical groups.
Which of them is the major product
In practice the alkene with more alkyl groups attached to the double bond dominates — but-2-ene rather than but-1-ene — because a more substituted double bond is more stable. This syllabus does not name that rule or ask you to apply it, so an answer that lists the possible alkenes has done what 15.1.4 requires. It is worth knowing that the products are not formed in equal amounts, and that a question saying "a mixture of isomeric alkenes is formed" is describing exactly this.
Substitution or elimination?15.1.3, 15.1.4
The same halogenoalkane and the same hydroxide ion give an alcohol under one set of conditions and an alkene under another. Nothing about the reagent changes; what changes is whether the hydroxide behaves as a nucleophile or as a base, and that is decided by the solvent, the temperature, and the class of the halogenoalkane.
| Factor | Favours substitution → alcohol | Favours elimination → alkene |
|---|---|---|
| solvent | water — NaOH(aq) | ethanol — NaOH in ethanol, no water |
| temperature | warm | hotter, and under reflux |
| concentration of NaOH | moderate | concentrated |
| class of halogenoalkane | primary | tertiary |
| role of the OH− | nucleophile — attacks the carbon | base — removes a hydrogen from the next carbon |
None of these is a switch; they are pressures, and in most real mixtures both products form. What an exam question gives you is usually a clean case: aqueous means alcohol, ethanolic means alkene, and a tertiary halogenoalkane in ethanol means alkene with confidence.
The trap: quoting the reagent without the solvent
Wrong: "react with sodium hydroxide and heat". That sentence is true of both reactions and therefore identifies neither. It scores nothing on a question that asks for reagents and conditions.
Right: "heat under reflux with aqueous sodium hydroxide" for the alcohol; "heat under reflux with sodium hydroxide dissolved in ethanol" for the alkene. One word separates full marks from none, and it is the same word both times.
What a mechanism has to show13.2.2, 15.1.5
Statement 15.1.5 says describe the mechanisms, and a mechanism is a drawing, not a sentence. Before the two mechanisms themselves, here is what a marker is looking for on the page — these are the conventions of statement 13.2.2, and they are worth the same marks in every organic topic.
The five things that get marked
1 · Every curly arrow starts at a bond or at a lone pair. The syllabus says so in the statement itself. An arrow starting at a negative sign, or at an atom, or in mid-air, is wrong even when it points somewhere sensible.
2 · An arrow ends where the pair ends up — on an atom that is gaining a lone pair, or between two atoms that are gaining a bond.
3 · The lone pair on the nucleophile is drawn. HO− with no lone pair shown gives the arrow nothing to start from.
4 · Charges are on the page and they balance. Losing the negative charge from the halide ion, or forgetting the positive charge on a carbocation, breaks the bookkeeping.
5 · A transition state is in square brackets with a ‡; an intermediate is not. They are different objects, and the difference is the whole distinction between SN2 and SN1.
Intermediate and transition state
An intermediate is a real species. It sits in a dip in the energy profile, has a finite lifetime, and could in principle be detected. The carbocation in SN1 is an intermediate.
A transition state is the arrangement at the top of an energy barrier. It has no lifetime at all — bonds are part-made and part-broken and it exists only in passing. The five-coordinate species in SN2 is a transition state, and that is why it is drawn with dotted partial bonds inside square brackets.
SN2, the one-step mechanism15.1.5
Read the label first, because it is a summary of the mechanism. S for substitution, N for nucleophilic, and 2 because two species are involved in the rate-determining step. In SN2 there is only one step, so that step is necessarily the rate-determining one, and it needs both the halogenoalkane and the nucleophile. The rate depends on both concentrations.
rate = k[RX][Nu−]
The nucleophile approaches the δ+ carbon from the side directly opposite the halogen. It has to: the halogen is bulky and δ−, so that side of the carbon is both blocked and repellent. As the nucleophile's lone pair begins to bond to the carbon, the electron density arriving pushes the C–X pair out onto the halogen. Bond making and bond breaking happen together, in one continuous motion, and halfway through the carbon is momentarily bonded to five things — the three groups that never moved, plus a half-formed bond to the nucleophile and a half-broken bond to the halogen.
The three unmoved groups get pushed through the plane of the carbon as the reaction passes the transition state, like an umbrella turning inside out in the wind. If that carbon was a chiral centre, the product has the opposite configuration from the starting material.
Drawing the transition state
Square brackets, a ‡ outside the top right corner, and the overall charge outside the brackets too. For OH− attacking a halogenoalkane the transition state carries a single negative charge, spread between the incoming oxygen and the departing halogen — both are drawn as δ−, and the dotted lines are the partial bonds. Drawing a full bond to both is the commonest error; that would be a five-bonded carbon, which does not exist.
SN1, the two-step mechanism15.1.5
Same label, same reading: substitution, nucleophilic, and 1 because only one species takes part in the rate-determining step. Here the two jobs are done one after the other rather than together.
Step 1, slow. The C–X bond breaks on its own, heterolytically — both electrons go to the halogen. What is left is a carbocation: a carbon with only three bonds, six electrons in its outer shell and a full positive charge. Nothing else is involved, so the rate of this step depends only on the concentration of the halogenoalkane.
rate = k[RX]
Step 2, fast. The nucleophile's lone pair attacks the carbocation. Because this step is fast, it has no influence on the overall rate — which is why doubling the hydroxide concentration does nothing to an SN1 reaction and doubles the rate of an SN2 one. That difference is how the two mechanisms were distinguished experimentally in the first place.
The carbocation is flat
The positive carbon has three bonding pairs and no lone pair, so it is trigonal planar, bond angles 120° — the same shape argument as topic 3.5. That matters for two reasons. It is why the nucleophile can attack from either face, so a chiral starting material gives a roughly 50:50 mixture of both configurations rather than a clean inversion. And it is why bulky alkyl groups do not hinder step 2: by the time the nucleophile arrives, the halogen has already gone and the crowding has been relieved.
The trap: what the 1 and the 2 count
Wrong: "SN1 has one step and SN2 has two." It is the other way round — SN1 has two steps and SN2 has one. The number is the molecularity of the rate-determining step: how many species have to come together in the slow step. SN1's slow step involves the halogenoalkane alone; SN2's involves the halogenoalkane and the nucleophile.
Carbocations and the inductive effect15.1.5
SN1 only happens if the carbocation can form, and whether it can form depends on how stable it is. Statement 15.1.5 asks for the explanation by name: the inductive effect of alkyl groups.
The inductive effect
An alkyl group is electron releasing: it pushes electron density along the σ bond towards whatever it is attached to. Attached to a positively charged carbon, it feeds electron density into the electron-deficient centre, so the positive charge is spread over a larger volume instead of being concentrated on one atom. A dispersed charge is a lower-energy arrangement, so the ion is more stable.
More alkyl groups, more of this effect. Hence the stability order:
tertiary > secondary > primary > CH3+
Which mechanism, and why15.1.6
Statement 15.1.6 is three lines of recall, but each line has a reason behind it, and the reasons are what a "why" question wants.
| Class | Mechanism | Why the other one is unavailable |
|---|---|---|
| primary | SN2 | a primary carbocation is too unstable to form — only one alkyl group to spread the charge — so there is no SN1 route; and the carbon is open, so backside attack is easy |
| secondary | a mixture of both, depending on structure | the secondary carbocation is stable enough to form sometimes, and the carbon is still reachable sometimes — neither route is ruled out |
| tertiary | SN1 | three alkyl groups block the approach from behind, so SN2 is hindered; and the tertiary carbocation is stable enough to form on its own |
Notice that two arguments run in parallel and they happen to point the same way. Electronic stability makes SN1 easier as you go from primary to tertiary; steric crowding makes SN2 harder over the same range. An answer that gives both is a complete answer; an answer that gives only the carbocation argument is the more important half.
Reactivity and the C–X bond15.1.7
Everything so far has varied the carbon skeleton and kept the halogen fixed. Statement 15.1.7 does the opposite: same skeleton, different halogen, and asks which reacts fastest and why.
The answer follows from one observation about both mechanisms. In SN1 the slow step is the breaking of the C–X bond. In SN2 the single step includes the breaking of the C–X bond. Either way, the rate-determining step requires the C–X bond to break — so the weaker that bond, the lower the activation energy and the faster the reaction.
| Bond | Bond energy / kJ mol−1 | Bond length / nm | Reactivity | With AgNO3 in ethanol |
|---|---|---|---|---|
| C–F | 467 | 0.138 | essentially unreactive | no precipitate |
| C–Cl | 340 | 0.177 | slowest of the three that react | white, slowly |
| C–Br | 280 | 0.193 | faster | cream |
| C–I | 240 | 0.214 | fastest | yellow, quickly |
The order of reactivity is therefore iodo > bromo > chloro >> fluoro, which is exactly the order in which the precipitates appeared in section 14. The bond energies fall down the group because the halogen atom gets larger: the bonding pair sits further from both nuclei and is more shielded, so it is held less tightly.
Bond strength beats bond polarity15.1.7
Here is the argument the topic is really testing, and it is the one place in this chapter where the obvious reasoning gives the wrong answer.
Nucleophilic substitution begins with a nucleophile attacking a δ+ carbon. The more δ+ that carbon is, the stronger the attraction should be — and the carbon is most δ+ when the halogen is most electronegative. Fluorine is the most electronegative element there is. So fluoroalkanes ought to be the most reactive of the four.
They are the least reactive by a wide margin. Fluoroalkanes give no precipitate with silver nitrate at all under conditions where iodoalkanes react almost at once.
Why the polarity argument fails
Polarity affects how strongly the nucleophile is attracted. It does not change the height of the energy barrier, and the barrier is set by the bond that has to break. Attraction gets you to the starting line; bond enthalpy is the race. Since the rate-determining step of both mechanisms requires the C–X bond to break, the bond energy wins — and it happens to run in the opposite direction from the polarity, so the two arguments make opposite predictions and the experiment chooses between them cleanly.
The model below puts both predictors against the observed order at once, so the failure is quantified rather than asserted: it ranks the four halogenoalkanes by bond enthalpy and by electronegativity difference, and reports how each ranking compares with the order the precipitates actually appear in.
How to write the answer
Statement 15.1.7 says describe and explain, so a complete answer has three parts: the order (iodoalkane fastest, then bromo, then chloro; fluoro does not react), the evidence (the order in which the silver halide precipitates appear when the three are warmed with aqueous silver nitrate in ethanol), and the reason (the C–I bond is the weakest, and the rate-determining step of both mechanisms involves breaking the C–X bond). Adding that bond polarity would predict the opposite order, and is therefore not what controls the rate, is what turns a good answer into a complete one.
Ten things that lose marks in this topic15
Every one of these has appeared somewhere on the page above. Collected here, they make a checklist to run through before a topic 15 question.
| # | The error | What to write instead |
|---|---|---|
| 1 | "React with NaOH and heat" | Name the solvent. Aqueous NaOH gives the alcohol; NaOH in ethanol gives the alkene. Without the solvent the answer fits both statements and scores on neither |
| 2 | Classifying by the number in the name | Count the carbon atoms attached to the carbon that carries the halogen. 1-bromo-2-methylpropane is primary |
| 3 | Calling the product of bromoethane and KCN "ethanenitrile" | Propanenitrile. The nitrile carbon joins the chain, so the chain grows by one |
| 4 | Writing one NH3 in the ammonia equation | Two. The second removes the proton from the protonated amine, leaving NH4Br |
| 5 | Leaving the water out of the elimination equation | CH3CH2Br + NaOH → CH2=CH2 + NaBr + H2O — three products |
| 6 | "SN1 has one step" | SN1 has two steps; the 1 counts the species in the rate-determining step |
| 7 | Drawing a full bond to both the nucleophile and the halogen in the SN2 transition state | Dotted partial bonds, square brackets, a ‡, and the overall charge outside the brackets |
| 8 | Starting a curly arrow at a negative sign or at an atom | Every arrow starts at a bond or at a lone pair — the syllabus says so inside statement 13.2.2 |
| 9 | Explaining reactivity by how polar the C–X bond is | Bond enthalpy. Polarity predicts the opposite order and is contradicted by the experiment |
| 10 | Removing the hydrogen from the carbon that carries the halogen | From the carbon next to it — otherwise no double bond can form |
If you remember one thing
Every reaction in this topic begins by breaking the C–X bond, and the two variables you can change are the halogen and the carbon skeleton. Change the halogen and you change how fast the bond breaks — that is 15.1.7, and the answer is bond enthalpy. Change the skeleton and you change how it breaks — on its own, or under attack — and that is 15.1.5 and 15.1.6, where the answer is the stability of the carbocation. Two questions, two answers, one bond.
Self-test15
Thirty questions across the whole of topic 15. Each one explains itself after you answer, including why the wrong options are wrong.
Definitions to learn15
Halogenoalkane
A compound in which one or more hydrogen atoms of an alkane have been replaced by halogen atoms.
Primary, secondary, tertiary
Classified by the number of carbon atoms bonded to the carbon that carries the halogen: one or none is primary, two is secondary, three is tertiary.
Nucleophile
A species that is attracted to a region of positive charge and donates a pair of electrons to form a covalent bond. It must have a lone pair; it need not be negatively charged.
Electrophile
A species that is attracted to a region of high electron density and accepts a pair of electrons to form a covalent bond.
Nucleophilic substitution
A reaction in which a nucleophile replaces an atom or group — here the halogen — on a carbon atom.
Elimination
A reaction in which a small molecule is removed from a single molecule, forming a double bond, with no other reactant becoming part of the product.
Hydrolysis
A reaction in which a compound is split by water, or by hydroxide ions. The reaction of a halogenoalkane with NaOH(aq) is a hydrolysis.
Homolytic and heterolytic fission
Homolytic: a covalent bond breaks so that one electron goes to each atom, giving two radicals. Heterolytic: it breaks so that both electrons go to one atom, giving a positive and a negative ion.
Free radical
A species with a single unpaired electron.
Carbocation
An ion in which a carbon atom carries a positive charge, having only three bonding pairs and six outer electrons. It is trigonal planar.
Inductive effect
The release or withdrawal of electron density along a σ bond. Alkyl groups release electron density, which spreads the charge of an adjacent carbocation and so stabilises it.
Rate-determining step
The slowest step of a multi-step reaction, which therefore controls the overall rate. Only the species taking part in it appear in the rate equation.
Intermediate and transition state
An intermediate is a real species with a finite lifetime, sitting in a dip in the energy profile. A transition state is the arrangement at the top of a barrier, with partly formed and partly broken bonds, and has no lifetime.
Leaving group
The atom or group that departs with the bonding pair when a nucleophile substitutes. In this topic it is the halogen, which leaves as a halide ion.
Data used on this page15
These are reference values, not the official data booklet
The numbers below are consistent with one another and are what every model on this page computes from, which is why they are printed rather than hidden. They are not taken from the Cambridge data booklet, and small differences between sources are normal: bond energies are averages over many compounds, and published values for C–Cl range from about 338 to 346 kJ mol−1 depending on the source. The order is what this topic rests on, and every source agrees on that. In an examination, use the values supplied with the paper.
Carbon–halogen bonds
| Bond | Bond energy / kJ mol−1 | Bond length / nm | Electronegativity of X | χ(X) − χ(C) | Silver halide | Colour |
|---|---|---|---|---|---|---|
| C–F | 467 | 0.138 | 3.98 | 1.43 | — | no precipitate |
| C–Cl | 340 | 0.177 | 3.16 | 0.61 | AgCl | white |
| C–Br | 280 | 0.193 | 2.96 | 0.41 | AgBr | cream |
| C–I | 240 | 0.214 | 2.66 | 0.11 | AgI | yellow |
Electronegativity of carbon, 2.55. The two columns run in opposite directions, and that is the whole of section 24.
Other bond energies used on this page
| Bond | C–H | C–C | C=C | C–O | O–H | C–N | C≡N | H–Cl | H–Br | Cl–Cl | Br–Br |
|---|---|---|---|---|---|---|---|---|---|---|---|
| kJ mol−1 | 410 | 350 | 610 | 360 | 460 | 305 | 890 | 431 | 366 | 244 | 193 |
Conditions, collected
| Reaction | Reagent | Conditions | Product | Statement |
|---|---|---|---|---|
| from an alkane | Cl2 or Br2 | ultraviolet light | halogenoalkane, as a mixture | 15.1.1(a) |
| from an alkene | X2 or HX(g) | room temperature | halogenoalkane | 15.1.1(b) |
| from an alcohol | HX(g); KCl + conc H2SO4; conc H3PO4; PCl3; PCl5; SOCl2 | heat with PCl3; the others at room temperature | halogenoalkane | 15.1.1(c) |
| substitution | NaOH(aq) | heat under reflux | alcohol | 15.1.3(a) |
| substitution | KCN in ethanol | heat under reflux | nitrile | 15.1.3(b) |
| substitution | excess NH3 in ethanol | heat under pressure, sealed | amine | 15.1.3(c) |
| substitution | AgNO3(aq) in ethanol | warm, about 50 °C | alcohol, and a silver halide precipitate | 15.1.3(d) |
| elimination | NaOH in ethanol | heat under reflux | alkene | 15.1.4 |