Two teaching hours at both levels (outcomes 3.3.1 to 3.3.3). There is no additional higher-level content in this sub-topic.
Guiding question. What happens when a species possesses an unpaired electron?
Reactivity 3.3 · Electron sharing reactions
1Radicals 3.3.1 SL + HL
The syllabus statement: A radical is a molecular entity that has an unpaired electron. Radicals are highly reactive. The skill: identify and represent radicals, for example the methyl radical, •CH3, and the chlorine atom, Cl•.
A mixture of methane and chlorine can be kept in the dark for months without change. Put the same mixture in sunlight and it reacts, sometimes explosively, producing hydrogen chloride and a range of chlorinated methanes. Light does not heat the gases appreciably; it does something more specific. It breaks chlorine molecules into atoms that each carry an unpaired electron, and those atoms start a chain of events that the dark mixture could never begin. This sub-topic is about such species and the reactions they drive.
What makes a species a radical
A radical is a molecular entity (an atom, a molecule or an ion) that has an unpaired electron. The unpaired electron is shown by a dot, •, written next to the atom that carries it.
Almost every stable molecule has all of its electrons in pairs: bonding pairs shared between atoms and lone pairs on single atoms. A pairing of electrons in a bonding orbital is what makes a covalent bond a low-energy arrangement. A species with an unpaired electron has an orbital that is only half filled, and it can lower its energy by gaining a second electron into that orbital, either by forming a new bond with another radical or by taking an atom from a molecule. That is why radicals are highly reactive: most radicals exist for only a tiny fraction of a second before they react.
The dot belongs on the atom that has the unpaired electron, not just anywhere in the formula. In the methyl radical the unpaired electron is on carbon, so it is written •CH3 (or CH3•); in the chlorine atom it is Cl•. Carbon in •CH3 has three bonding pairs and one single electron: seven electrons around it, not an octet.
| Type of entity | Example | Where the unpaired electron is |
|---|---|---|
| Atom | Cl•, Br•, H• | The halogen atom has seven valence electrons: three pairs and one single electron. |
| Molecular fragment | •CH3, •C2H5 | On the carbon atom that has lost a hydrogen atom. |
| Stable molecule | NO, NO2 | An odd total number of valence electrons (11 in NO, 17 in NO2) makes pairing impossible. |
| Ion | O2•− (superoxide); the molecular ion M•+ in a mass spectrometer | Gaining or losing one electron from a species with all electrons paired leaves one unpaired. |
Count the valence electrons. An odd number means at least one electron is unpaired, so the species is a radical. For •CH3: 4 (C) + 3 × 1 (H) = 7, odd. For CH3+: 7 − 1 = 6, even: a carbocation, not a radical. For CH3−: 8, even: a carbanion with a lone pair. The charge is not what makes a radical; the unpaired electron is.
Exam focus · what the published papers show
Reports return to the same weakness session after session: Inconsistency of radical symbols was common. In one paper a mark was also not scored due to inconsistency in use of free radical symbols. Decide on one convention (Cl• and •CH3) and use the dot on every radical, in every step, and never on a molecule such as HCl or CH3Cl.
2Homolytic fission and initiation 3.3.2 SL + HL
The syllabus statement: Radicals are produced by homolytic fission, e.g. of halogens, in the presence of ultraviolet (UV) light or heat. The skill: explain, including with equations, the homolytic fission of halogens, known as the initiation step in a chain reaction, using a single-barbed arrow (fish hook) to show the movement of a single electron.
A covalent bond is a shared pair of electrons. It can break in two ways, and which way it breaks decides what kind of chemistry follows. When the bond breaks evenly, each atom keeps one electron of the pair: this is homolytic fission (homo, the same), and it produces two radicals. When it breaks unevenly, one atom keeps both electrons: this is heterolytic fission, and it produces a cation and an anion (Reactivity 3.4).
Homolytic fission: the breaking of a covalent bond in which each fragment takes one of the two bonding electrons, forming two radicals.
A single-barbed (fish-hook) arrow shows the movement of one electron; a double-barbed curly arrow shows the movement of an electron pair.
Why UV light or heat is needed
Breaking a bond always requires energy. The Cl–Cl bond enthalpy is 242 kJ mol−1, so breaking one Cl–Cl bond requires 242 × 103 ÷ 6.02 × 1023 ≈ 4.0 × 10−19 J. A photon of ultraviolet light carries energy of this size (visible light towards the violet end is just about sufficient for bromine, whose bond is weaker), so absorption of one photon by one molecule can break the bond. Heating to a high temperature achieves the same result by collision. Because both atoms are identical, neither attracts the bonding pair more than the other, and the bond breaks homolytically:
This first step creates radicals from a molecule that has none. It is called initiation, and it is the step that UV light is needed for. The rest of the reaction runs on the radicals that initiation provides.
Exam focus · what the published papers show
For bromine and an alkane the awarded answer was: Br2 → 2Br• ✔ and «sun»light/UV/hv or high temperature ✔, with homolytic fission of bromine accepted for the first mark. Reports note omission of UV in the first step as one of the most common errors.
Connecting to bond strength: CFCs and ozone
The ease of homolytic fission follows bond enthalpy. In a chlorofluorocarbon such as CCl2F2, the C–Cl bond (about 324 kJ mol−1) is much weaker than the C–F bond (about 492 kJ mol−1). High-energy UV light in the stratosphere therefore breaks the C–Cl bond, releasing chlorine radicals, while the C–F bond survives: CFCs release Cl• but typically not F•.
A chlorine radical then removes an oxygen atom from ozone and is regenerated in a second step, so one radical can destroy many ozone molecules. That Cl• can break down O3 but not O2 shows that the bonds in ozone are weaker than the double bond in O2 (Structure 2.2).
The reverse of homolytic fission is two radicals combining, each supplying one electron to a new shared pair. This releases energy equal to the bond enthalpy and is what happens in the termination steps below.
3Radical substitution of alkanes 3.3.3 SL + HL
The syllabus statement: Radicals take part in substitution reactions with alkanes, producing a mixture of products. The skill: explain, using equations, the propagation and termination steps in the reactions between alkanes and halogens, with reference to the stability of alkanes due to the strengths of the C–C and C–H bonds and their essentially non-polar nature.
Why alkanes are so unreactive
Alkanes burn readily, so they are thermodynamically unstable with respect to oxidation, yet they do not react with acids, alkalis, oxidizing agents such as acidified dichromate(VI), or halogens in the dark. Two features explain this kinetic stability:
- Strong bonds. C–H (414 kJ mol−1) and C–C (346 kJ mol−1) bonds need a large input of energy to break, so the activation energy of most reactions is high.
- Non-polar bonds. Carbon (2.6) and hydrogen (2.2) have similar electronegativities, so C–H bonds are essentially non-polar. There is no region of partial positive charge to attract a nucleophile and no region of high electron density to attract an electrophile. Reagents that react by donating or accepting electron pairs find nothing to attack.
Only a species that does not need a polar site can attack an alkane: a radical, which reacts by pairing its single electron with one from a C–H bond.
The chain mechanism
The reaction of methane with chlorine in UV light is a free-radical substitution: a hydrogen atom of the alkane is replaced by a halogen atom, and the reaction is carried by radicals. It happens in three kinds of step.
| Initiation | Cl2 → 2Cl• (UV light). Radicals are formed from a molecule. Net radicals: 0 → 2. |
| Propagation | Cl• + CH4 → HCl + •CH3, then •CH3 + Cl2 → CH3Cl + Cl•. One radical in, one radical out, in each step. The Cl• produced in the second step starts the first step again. |
| Termination | Cl• + Cl• → Cl2; •CH3 + Cl• → CH3Cl; •CH3 + •CH3 → C2H6. Two radicals combine to form a molecule. Net radicals: 2 → 0. |
Adding the two propagation steps gives the overall equation, because Cl• and •CH3 cancel:
This is why the reaction is called a chain reaction: a single initiation can be followed by thousands of propagation cycles before two radicals happen to meet and terminate the chain. Very little light is needed to convert a large amount of reactant. The small amount of ethane found among the products is the evidence for the termination step •CH3 + •CH3 → C2H6: ethane could not otherwise be formed from methane.
Electron sharing and electron-pair sharing — step through a mechanism
Radicals share one electron each and go in threes: initiation, propagation, termination. Nucleophiles supply both, and whether that happens in one step or two depends on the carbon they attack.
A half-headed arrow moves one electron, a full-headed arrow moves a pair. Which arrow the mechanism uses is the classification.
Why the first propagation step forms HCl, not •H
In the first propagation step the chlorine radical removes a hydrogen atom, forming H–Cl (bond enthalpy 431 kJ mol−1) while one C–H bond (414 kJ mol−1) breaks, so the step is slightly exothermic. The alternative, Cl• + CH4 → CH3Cl + H•, would break the same C–H bond but form a C–Cl bond of only about 324 kJ mol−1: strongly endothermic, and it does not occur. Hydrogen radicals never appear in this mechanism.
Exam focus · what the published papers show
This is the single most frequent error in the mechanism. Reports record that Many opined the production of •H in the first propagation step, that The students who gained some marks often lost marks for creating hydrogen radicals, and that It is a common misconception that a bromine radical can displace a hydrogen radical. Other recurring errors: only one propagation step was given, the initiation reaction included as a propagation step, and a mechanism involving ions despite free radical begin stated in the stem.
A mixture of products
The chain does not stop neatly at chloromethane. Chloromethane still contains C–H bonds, so a chlorine radical can remove a hydrogen from it, and the new radical can react with Cl2:
Further substitution gives CHCl3 and CCl4. The product is always a mixture of mono- and multiply-substituted compounds, together with small amounts of longer alkanes from termination. For alkanes with three or more carbon atoms the hydrogens are not all equivalent, so even monosubstitution gives isomers: propane gives both 1-chloropropane and 2-chloropropane. An excess of the alkane favours monosubstitution (a chlorine radical is then most likely to meet an alkane molecule rather than a chloroalkane); an excess of halogen favours multiple substitution. Radical substitution is therefore a poor way to make one pure product, which is why the separation of the mixture is an important practical consideration.
Question. Ethane reacts with bromine in UV light. Write equations for the initiation step, the two propagation steps and two different termination steps, and give the overall equation for the formation of bromoethane.
Reasoning. Initiation breaks the weakest bond, Br–Br, homolytically. In propagation, Br• must remove a hydrogen atom (forming HBr, not H•); the ethyl radical then removes a bromine atom from Br2, regenerating Br•. Termination combines any two radicals present.
Answer. Initiation: Br2 → 2Br• (UV). Propagation: Br• + C2H6 → HBr + •C2H5; •C2H5 + Br2 → C2H5Br + Br•. Termination (any two): Br• + Br• → Br2; •C2H5 + Br• → C2H5Br; •C2H5 + •C2H5 → C4H10. Overall: C2H6 + Br2 → C2H5Br + HBr.
Check. Each propagation step has one radical on each side; each termination step has two radicals on the left and none on the right; the propagation steps add to the overall equation.
Question. Deduce the number of different monochlorinated products formed when 2-methylpropane, (CH3)3CH, reacts with chlorine in UV light.
Reasoning. Identify the different hydrogen environments. The nine hydrogens on the three CH3 groups are equivalent; the single hydrogen on the central carbon is different.
Answer. Two: 1-chloro-2-methylpropane, (CH3)2CHCH2Cl, and 2-chloro-2-methylpropane, (CH3)3CCl.
For chloroethane from ethane the awarded steps were Cl• + C2H6 → •C2H5 + HCl ✔ and •C2H5 + Cl2 → Cl• + C2H5Cl ✔, with any one of three termination steps ✔. An incorrectly placed radical dot (C2H5•) was not penalized, but a missing or inconsistent dot is a risk. For the type of reaction, “free radical substitution” is required: electrophilic or nucleophilic substitution is not accepted.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
Formulate equations for the two propagation steps and one termination step in the formation of chloroethane from ethane. [3]
Bromine reacts with alkanes. Identify the initiation step of the reaction and its conditions. [2]
(i) State the type of reaction occurring when ethane reacts with chlorine to produce chloroethane. [1] (ii) Predict, giving a reason, whether ethane or chloroethane is more reactive. [1]
Propane reacts with chlorine in UV light. (a) Explain why propane does not react with chlorine in the dark. [2] (b) Give the structural formulas of the two monochlorinated isomers formed. [1] (c) Suggest why a small amount of hexane is found among the products. [1]
Solutions and mark-scheme guidance · Set R3I
R3I.1 A
A hydrogen of ethane is replaced by chlorine (substitution) and the reaction is carried by radicals formed by UV light.
R3I.2 A
The halogen–halogen bond breaks homolytically (each atom takes one electron), and the chain is carried by free radicals.
R3I.3 A
Br• + C2H6 → HBr + CH3CH2•. Butane (B) forms in a termination step; H• (C) is never formed; Br− (D) is an ion, not part of a radical mechanism.
R3I.4 B
One radical in, one radical out. A is initiation, C is termination, D wrongly produces H•.
R3I.5 B
Dichlorobutanes with the chlorines on carbons 1,1; 1,2; 1,3; 1,4; 2,2; 2,3 (2,3 and 3,2 are the same; 1,3 and 2,4 are the same compound). Six in all.
R3I.6 A
CH4 + Cl2 reacts by free-radical substitution, which starts with homolytic fission of Cl–Cl. B and C are nucleophilic substitutions and D is electrophilic substitution: all heterolytic.
R3I.7 [3]
Cl• + C2H6 → •C2H5 + HCl ✔. •C2H5 + Cl2 → Cl• + C2H5Cl ✔. Any one of: •C2H5 + Cl• → C2H5Cl; Cl• + Cl• → Cl2; •C2H5 + •C2H5 → C4H10 ✔.
R3I.8 [2]
Br2 → 2Br• ✔; (sun)light / UV / hν, or high temperature ✔. “Homolytic fission of bromine” is accepted for the first mark.
R3I.9 [2]
(i) (Free-radical) substitution ✔; electrophilic or nucleophilic substitution is not accepted. (ii) Chloroethane, because the C–Cl bond (324 kJ mol−1) is weaker than the C–H bond (414 kJ mol−1), or because chloroethane contains a polar bond ✔.
R3I.10 [4]
(a) Initiation needs the Cl–Cl bond to break homolytically, which requires the energy of UV light (or high temperature) ✔; without radicals, chlorine cannot attack propane because its C–H and C–C bonds are strong and non-polar ✔. (b) CH3CH2CH2Cl and CH3CHClCH3 ✔. (c) Termination: two propyl radicals combine, •C3H7 + •C3H7 → C6H14 ✔.
Review · Reactivity 3.3
4Misconceptions, the examiner’s view, and the question types
- “A radical is a charged species.” Why it is wrong: the defining feature is an unpaired electron; Cl• and •CH3 are neutral. Consequence: drawing Cl− or CH3+ in a radical mechanism scores nothing.
- “The chlorine radical replaces a hydrogen radical.” Cl• abstracts a hydrogen atom to form HCl; H• is never formed, because forming C–Cl in place of H–Cl would make the step strongly endothermic.
- “Initiation needs UV at every step.” Only initiation needs UV or heat; propagation runs on the radicals initiation produced.
- “The reaction gives only the monosubstituted product.” Further substitution and isomeric products always give a mixture.
- “Alkanes are unreactive because they are stable.” They are thermodynamically unstable (they burn) but kinetically stable: strong, non-polar C–H and C–C bonds give high activation energies.
- “A fish-hook arrow and a curly arrow mean the same.” A single barb moves one electron (homolytic); a double barb moves a pair (heterolytic).
Evidence base: the IB Diploma chemistry subject reports quoted in this chapter.
Answered well: naming the reaction type (85 % and 74 % chose free-radical substitution for methane and chlorine in sunlight on two papers); naming the three stages; the initiation step, which was the best known; and, in stronger scripts, the whole mechanism, listed among the areas where candidates appeared well prepared.
Found difficult: (1) the propagation steps, where the hydrogen radical was repeatedly produced and only one step was often given; (2) consistent use of the radical dot; (3) stating the need for UV light in initiation; (4) confusing the reaction type with the mechanism, or introducing ions, SN1 or heterolytic fission into a radical question; (5) choosing the free-radical reaction from a list (48 % on one item).
What successful answers did: wrote each propagation step with one radical on each side and the halogen radical forming the hydrogen halide; showed two propagation steps that add up to the overall equation; used a dot on every radical and on no molecule; and stated UV light for initiation.
Four question types cover the sub-topic.
| If the question asks… | …then |
|---|---|
| Identify or represent a radical | Count valence electrons; odd = radical; put the dot on the atom with the unpaired electron. |
| Initiation | X2 → 2X• with UV light (or high temperature); homolytic fission; fish-hook arrows. |
| Propagation and termination | X• + RH → HX + R•; R• + X2 → RX + X•; termination joins any two radicals. |
| Products and reactivity | Count H environments for isomers; further substitution gives a mixture; alkanes are unreactive because C–H and C–C are strong and non-polar. |
5Quick check
- State what is meant by a radical and write the formula of the ethyl radical.
- Explain why the methyl cation, CH3+, is not a radical.
- Write the equation for the initiation step when bromine reacts with methane, and state the condition.
- Write the two propagation steps for the formation of bromomethane.
- Explain why ethane is found among the products of the chlorination of methane.
- Explain why CFCs release chlorine radicals but not fluorine radicals.
- A species with an unpaired electron; •C2H5 (or CH3CH2•).
- It has six valence electrons, all paired; it is a carbocation.
- Br2 → 2Br•; UV light (or high temperature).
- Br• + CH4 → HBr + •CH3; •CH3 + Br2 → CH3Br + Br•.
- Termination: •CH3 + •CH3 → C2H6.
- The C–Cl bond is much weaker than the C–F bond, so UV light breaks C–Cl homolytically.
6Summary and knowledge organiser
Essential knowledge
- A radical has an unpaired electron (shown •); radicals are highly reactive; they can be atoms, molecules or ions.
- Homolytic fission: each fragment takes one bonding electron; fish-hook arrows; halogens need UV light or heat.
- Initiation X2 → 2X•; propagation X• + RH → HX + R• and R• + X2 → RX + X•; termination: any two radicals combine.
- Alkanes are kinetically stable: strong C–C and C–H bonds, essentially non-polar.
- Radical substitution gives a mixture: further substitution, isomers, and longer alkanes from termination.
Examination checklist
- A dot on every radical, none on molecules.
- Two propagation steps; they add up to the overall equation.
- No H• anywhere in the mechanism.
- State UV light for initiation.
- Name the reaction “free-radical substitution”.
Knowledge organiser
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| Radicals 3.3.1 | Unpaired electron; •CH3, Cl•; odd electron count. | Radicals need not be charged; ions need not be radicals. |
| Homolytic fission 3.3.2 | Cl2 → 2Cl• (UV); fish-hook arrows; weakest bond breaks (C–Cl in CFCs). | Single barb = one electron. |
| Radical substitution 3.3.3 | Initiation, propagation (×2), termination; mixture of products. | HCl forms in propagation, never H•. |
Four teaching hours at both levels (outcomes 3.4.1 to 3.4.5) and seven further hours at higher level (3.4.6 to 3.4.13). At standard level, the details of the mechanisms are not required and the mechanisms of the alkene reactions are not assessed.
Guiding question. What happens when reactants share their electron pairs with others?
Reactivity 3.4 · Electron-pair sharing reactions
7Nucleophiles 3.4.1 SL + HL
The syllabus statement: A nucleophile is a reactant that forms a bond to its reaction partner (the electrophile) by donating both bonding electrons. The skill: recognize nucleophiles in chemical reactions, including both neutral and negatively charged species.
Warm 1-bromobutane with aqueous sodium hydroxide, then acidify and add silver nitrate: a cream precipitate of silver bromide appears. Bromine that was covalently bonded to carbon has become free bromide ions, and an –OH group has taken its place. The radicals of Reactivity 3.3 played no part: every bond made and broken in this reaction involves a pair of electrons. This sub-topic follows those electron pairs, from the species that donate them to the species that accept them.
A nucleophile (“nucleus-loving”) is a species that forms a new covalent bond by donating a pair of electrons to an electron-deficient atom. Every nucleophile has a lone pair of electrons; many, but not all, are negatively charged.
A nucleophile is attracted to a region of positive or partial positive charge, most often a carbon atom bonded to a more electronegative atom. The electron pair it donates becomes the bonding pair of the new bond. Common nucleophiles:
| Negatively charged | Neutral |
|---|---|
| hydroxide, :OH− (lone pair on O) | water, H2O: (lone pair on O) |
| cyanide, :CN− (lone pair on C) | ammonia, :NH3 (lone pair on N) |
| halide ions, :Cl−, :Br−, :I− | amines, RNH2; alcohols, ROH |
A negative charge makes a nucleophile stronger (hydroxide reacts faster than water, because the full negative charge on oxygen is attracted more strongly to the partially positive carbon), but the charge is not the requirement. The lone pair is.
Exam focus · what the published papers show
On one multiple-choice item 72% of the candidates chose B (a nucleophile must have a lone pair of electrons). The most commonly chosen distractor was A (a nucleophile must have a negative charge). Water and ammonia are the counter-examples to keep in mind.
8Nucleophilic substitution 3.4.2 SL + HL
The syllabus statement: In a nucleophilic substitution reaction, a nucleophile donates an electron pair to form a new bond, as another bond breaks producing a leaving group. The skill: deduce equations with descriptions and explanations of the movement of electron pairs in nucleophilic substitution reactions. Further details of the mechanisms are not required at SL.
Why halogenoalkanes are attacked
In a halogenoalkane the halogen is more electronegative than carbon, so the C–X bond is polar: Cδ+–Xδ−. The partially positive carbon is the target for a nucleophile. When the nucleophile’s lone pair forms a bond to that carbon, the C–X bond breaks and both of its electrons go with the halogen, which leaves as a halide ion: the leaving group. One group has been replaced by another, so the reaction is a substitution, and it is started by a nucleophile: a nucleophilic substitution.
| Reactant | a halogenoalkane, R–X (X = Cl, Br, I) |
| Reagent, conditions | aqueous sodium (or potassium) hydroxide, warm |
| Product | an alcohol, R–OH, and a halide ion |
| Reaction type | nucleophilic substitution; the nucleophile is OH−, not NaOH |
In words, the movement of electron pairs is: a lone pair on the oxygen of the hydroxide ion is donated to the δ+ carbon, forming a C–O bond; the bonding pair of the C–Cl bond moves on to the chlorine, which leaves as Cl−. Curly arrows show exactly this (Reactivity 3.4.3). Other nucleophiles give other products: with ammonia a halogenoalkane gives an amine, and with water it gives an alcohol more slowly than with hydroxide.
Exam focus · what the published papers show
For the reaction of chloroethane with sodium hydroxide the nucleophile was credited as hydroxide «ion»/OH−, with the note Do not accept NaOH. For the role of the hydroxide ion, any of: nucleophile / Lewis base, donates a lone pair, or attacks the partially positive carbon. For the type of reaction, “«nucleophilic» substitution” or SN2; the answer was not accepted if “electrophilic” or “free radical” substitution was stated.
9Heterolytic fission and curly arrows 3.4.3 SL + HL
The syllabus statement: Heterolytic fission is the breakage of a covalent bond when both bonding electrons remain with one of the two fragments formed. The skill: explain, with equations, the formation of ions by heterolytic fission, using curly arrows to show the movement of electron pairs.
Heterolytic fission: the breaking of a covalent bond in which one fragment takes both bonding electrons, forming a cation and an anion. The more electronegative atom normally takes the pair.
The contrast with Reactivity 3.3 is the classification of the whole of organic reaction chemistry. Homolytic fission of a non-polar bond (Cl–Cl in UV light) gives radicals; heterolytic fission of a polar bond (C–Br, H–Br) gives ions, and the reactions that follow are driven by electron pairs moving from electron-rich to electron-poor sites.
Rules for curly arrows
A curly (double-barbed) arrow shows the movement of one pair of electrons.
- The tail starts where the electrons are: on a lone pair, a negative charge, or the middle of a bond.
- The head ends where the electrons go: at the atom that will form the new bond, or at the atom that will keep the pair when a bond breaks.
- An arrow from a bond to an atom breaks that bond; an arrow from a lone pair to an atom makes a new bond.
- Charges must balance on every side of every step.
Exam focus · what the published papers show
This is the most persistent weakness in all mechanism questions. Reports: Many candidates did not take due care with the starting and finishing parts of curly arrows. some candidates are not aware that curly arrows must start from non-bonding pairs; correct direction is not sufficient. A common error was a curly arrow originating from the hydrogen atom in the hydroxide ion rather than the oxygen, and Some drew curly arrow showing Br leaving incorrectly starting from C rather than C-Br bond.
10Electrophiles 3.4.4 SL + HL
The syllabus statement: An electrophile is a reactant that forms a bond to its reaction partner (the nucleophile) by accepting both bonding electrons from that reaction partner. The skill: recognize electrophiles in chemical reactions, including both neutral and positively charged species.
An electrophile (“electron-loving”) is a species that forms a new covalent bond by accepting a pair of electrons from an electron-rich species. An electrophile has, or can develop, an electron-deficient atom.
Positive ions are obvious electrophiles: H+, the nitronium ion NO2+, a carbocation such as CH3CH2+. Neutral molecules can also be electrophiles when they contain a δ+ atom: in H–Br the hydrogen is δ+; in Br–Br, a molecule with no permanent dipole, a dipole is induced when the molecule approaches the electron-rich C=C bond of an alkene, and the nearer bromine atom becomes δ+. Electrophiles are the partners of nucleophiles: in every bond formed by electron-pair donation, one species donates (the nucleophile) and one accepts (the electrophile).
An electrophile accepts a pair; it needs an electron-deficient atom, not a lone pair. Br2 has lone pairs but acts as an electrophile through its induced δ+ atom. One animated true-or-false statement tests exactly this: “Electrophiles all contain a lone pair of electrons” is false.
11Electrophilic addition to alkenes 3.4.5 SL + HL
The syllabus statement: Alkenes are susceptible to electrophilic attack because of the high electron density of the carbon–carbon double bond. These reactions lead to electrophilic addition. The skill: deduce equations for the reactions of alkenes with water, halogens, and hydrogen halides. The mechanisms of these reactions will not be assessed at SL.
The electron-rich double bond
A C=C double bond is one σ bond (head-on overlap along the C–C axis) and one π bond (sideways overlap of p orbitals above and below the plane of the molecule). The π electrons are further from the nuclei and less tightly held than the σ electrons, and they sit exposed above and below the molecule. The double bond is therefore a region of high electron density that attracts electrophiles. The π bond is also weaker than the σ bond, so it breaks readily, and each carbon forms a new σ bond instead: the alkene adds the reagent across the double bond, and the product is saturated.
| Reagent | Conditions | Equation | Product |
|---|---|---|---|
| Halogen, Br2 or Cl2 | room temperature; in the dark; bromine water or bromine in an organic solvent | CH2=CH2 + Br2 → CH2BrCH2Br | 1,2-dibromoethane (a dihalogenoalkane) |
| Hydrogen halide, HBr, HCl, HI | room temperature (gas, or concentrated solution) | CH2=CH2 + HBr → CH3CH2Br | bromoethane (a halogenoalkane) |
| Water (steam) | acid catalyst (H+, e.g. phosphoric or sulfuric acid), heat | CH2=CH2 + H2O → CH3CH2OH | ethanol (an alcohol) |
Hydration can also be carried out in two stages: cold concentrated sulfuric acid adds across the double bond to give an alkyl hydrogensulfate, which is then hydrolysed by water to the alcohol, regenerating the acid.
The bromine water test
When orange bromine water is shaken with an alkene at room temperature, in the dark, the colour disappears (the solution becomes colourless) because bromine is consumed by addition. An alkane gives no change in the dark: its C–H and C–C bonds are strong and non-polar and present no electron-rich site, and without UV light no radicals form to start substitution (Reactivity 3.3). Decolourization of bromine water in the dark is therefore a test for C=C unsaturation.
Question. Deduce the structural formula of the organic product when but-2-ene, CH3CH=CHCH3, reacts with (a) bromine, (b) hydrogen bromide, (c) steam with an acid catalyst.
Reasoning. In addition, the two atoms (or groups) of the reagent attach to the two carbons of the double bond, which becomes a single bond. But-2-ene is symmetrical, so the two carbons are equivalent and there is only one possible product in each case.
Answer. (a) CH3CHBrCHBrCH3 (2,3-dibromobutane). (b) CH3CHBrCH2CH3 (2-bromobutane). (c) CH3CH(OH)CH2CH3 (butan-2-ol).
Exam focus · what the published papers show
HBr usually was correctly identified but addition was often stated which was not enough to score the mark for the type of reaction, where electrophilic addition was required. For bromine and but-2-ene, The most common incorrect answers were “substitution” and “nucleophilic addition”. The awarded product was CH3CHBrCHBrCH3, and the molecular formula C4H8Br2 was not accepted.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
1-chloropentane reacts with aqueous sodium hydroxide. (i) Identify the type of reaction. [1] (ii) Outline the role of the hydroxide ion in this reaction. [1]
(iii) Write the equation for the reaction of chloroethane with a dilute aqueous solution of sodium hydroxide. [1] (iv) Deduce the nucleophile for the reaction in d(iii). [1]
But-2-ene reacts with hydrogen bromide. (ii) Write the equation for the reaction between but-2-ene and hydrogen bromide. [1] (iii) State the type of reaction. [1]
Bromine also reacts with but-2-ene. (i) Identify the type of reaction. [1] (ii) Predict the structural formula of the reaction product. [1]
Classify each of the following species as a nucleophile or an electrophile, giving the feature that determines your answer: (a) NH3, (b) NO2+, (c) CN−, (d) the hydrogen atom of HBr. [4]
Solutions and mark-scheme guidance · Set R3J
R3J.1 D
Nucleophiles attack an electron-deficient carbon; in halogenoalkanes the C–X bond is polar and the carbon is δ+. Alkenes and benzene are electron-rich, and alkanes have non-polar bonds.
R3J.2 B
The lone pair forms the new bond. A report notes that The most commonly chosen distractor was A (a nucleophile must have a negative charge).
R3J.3 A
–Br is replaced by –OH: the hydroxide nucleophile substitutes the halogen.
R3J.4 A
Decolourization of bromine water in the dark is the test for C=C: ethene undergoes electrophilic addition. Ethane needs UV light.
R3J.5 C
Hex-1-ene is the only unsaturated compound; cyclohexane is a saturated ring.
R3J.6 C
The π electrons attract Br2, which adds across the double bond: electrophilic addition.
R3J.7 A
Ethene adds reagents across its double bond; benzene keeps its delocalized ring by substituting a hydrogen.
R3J.8 D
Cl–Cl breaks heterolytically when it adds to an alkene (I) and when it forms Cl+ for substitution on benzene (II); the C–Cl bond breaks heterolytically when chloride leaves in nucleophilic substitution (III).
R3J.9 [2]
(i) (Nucleophilic) substitution / SN2 ✔; not accepted if “electrophilic” or “free radical” substitution is stated. (ii) Acts as a nucleophile / Lewis base, or donates a lone pair of electrons, or attacks the partially positive carbon ✔.
R3J.10 [2]
(iii) CH3CH2Cl(l) + OH−(aq) → CH3CH2OH(aq) + Cl−(aq), or with NaOH and NaCl ✔. (iv) Hydroxide ion, OH− ✔ (NaOH is not accepted).
R3J.11 [2]
(ii) CH3CH=CHCH3(g) + HBr(g) → CH3CH2CHBrCH3(l), or C4H8 + HBr → C4H9Br ✔. (iii) (Electrophilic) addition ✔; nucleophilic or free-radical addition is not accepted.
R3J.12 [2]
(i) (Electrophilic) addition ✔; nucleophilic addition is not accepted. (ii) CH3CHBrCHBrCH3 ✔; the molecular formula C4H8Br2 is not accepted.
R3J.13 [4]
(a) Nucleophile: lone pair on N ✔. (b) Electrophile: positive charge on an electron-deficient N; accepts a pair ✔. (c) Nucleophile: lone pair and negative charge on C ✔. (d) Electrophile: H is δ+ because Br is more electronegative; it accepts a pair from the C=C bond ✔.
Additional higher level · outcomes 3.4.6 to 3.4.13
12Lewis acids and bases 3.4.6 HL only
The syllabus statement: A Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor. The skill: apply Lewis acid–base theory to inorganic and organic chemistry to identify the role of the reacting species.
The Brønsted–Lowry theory of Reactivity 3.1 is about the transfer of a proton. Look again at what happens when ammonia accepts a proton: the lone pair on nitrogen forms a bond to H+. The proton is an electron-pair acceptor; the base is an electron-pair donor. G. N. Lewis generalized this in 1923: the essential event in acid–base chemistry is the donation and acceptance of an electron pair, whether or not a proton is involved.
A Lewis acid is an electron-pair acceptor. A Lewis base is an electron-pair donor.
Every Brønsted–Lowry base is a Lewis base, because it needs a lone pair to accept a proton. But many Lewis acids are not Brønsted–Lowry acids at all, because they have no proton to donate. Typical Lewis acids:
- Molecules with an incomplete octet: BF3 and AlCl3 have only six electrons around the central atom and a vacant orbital to accept a pair.
- Metal cations, particularly transition element ions such as Cu2+ and Fe3+, which accept lone pairs from ligands.
- H+, and in organic chemistry every electrophile: carbocations, NO2+.
| Brønsted–Lowry | Lewis | |
|---|---|---|
| Acid | proton donor | electron-pair acceptor |
| Base | proton acceptor | electron-pair donor |
| Scope | reactions in which H+ is transferred | any reaction forming a coordination bond: includes all Brønsted–Lowry reactions, complex formation, and organic reactions of nucleophiles with electrophiles |
| Example not covered by the other theory | — | NH3 + BF3 → H3N→BF3; Cu2+ + 4Cl− → [CuCl4]2− |
Exam focus · what the published papers show
Reports note that Occasionally the word “pair” was missing for the definition of a Lewis base, and that BF3 was not always explained well as students mixed up proton donation and electron pair donation. For Mg2+ reacting with water, the mark scheme required “Lewis acid” and “accepts «a lone» electron pair”, with the note Do not accept electron acceptor without mention of electron pair. Surprisingly, on one paper a quarter of the students incorrectly identified chloride ions acting as Lewis acids.
13Coordination bonds, nucleophiles and electrophiles 3.4.7 HL only
The syllabus statement: When a Lewis base reacts with a Lewis acid, a coordination bond is formed. Nucleophiles are Lewis bases and electrophiles are Lewis acids. The skill: draw and interpret Lewis formulas of reactants and products to show coordination bond formation in Lewis acid–base reactions.
A coordination bond (dative covalent bond) is a covalent bond in which both shared electrons came from the same atom. Once formed, it is identical to any other covalent bond of the same kind: in NH4+ all four N–H bonds are equivalent, and there is no way of telling which one was formed by coordination. In a Lewis formula the coordination bond is often drawn as an arrow pointing from the donor atom to the acceptor, to record where the electrons came from.
Question. Identify the Lewis acid and the Lewis base in each reaction: (a) BF3 + F− → BF4−; (b) C2H5+ + Cl− → C2H5Cl; (c) H2O + H+ → H3O+.
Reasoning. Find the species with the lone pair that forms the new bond (base) and the species with the vacant orbital or electron-deficient atom that receives it (acid).
Answer. (a) F− is the Lewis base; BF3 (incomplete octet) is the Lewis acid. (b) Cl− donates a pair: Lewis base (and nucleophile); the carbocation accepts it: Lewis acid (and electrophile). (c) H2O donates a lone pair: Lewis base (and Brønsted–Lowry base); H+ is the Lewis acid.
The last two examples show the synthesis of the whole sub-topic: a nucleophile is a Lewis base, and an electrophile is a Lewis acid. The different names describe the same electron-pair event from the point of view of organic mechanism or of acid–base theory.
14Complex ions 3.4.8 HL only
The syllabus statement: Coordination bonds are formed when ligands donate an electron pair to transition element cations, forming complex ions. The skill: deduce the charge on a complex ion, given the formula of the ion and ligands present.
Dissolve anhydrous white copper(II) sulfate in water and the solution turns pale blue; add excess ammonia and it turns deep blue. Each colour belongs to a different complex ion: a central metal cation surrounded by ligands, each of which donates a lone pair to form a coordination bond. The ligand is the Lewis base; the metal ion is the Lewis acid. Transition element cations are small, highly charged and have vacant low-energy orbitals, so they form complexes readily.
A ligand is a molecule or ion with a lone pair that it donates to a central metal ion, forming a coordination bond. A complex ion is a central metal ion with ligands bonded to it by coordination bonds. The coordination number is the number of coordination bonds to the central ion.
The charge on a complex ion is the sum of the charge (oxidation state) of the metal ion and the charges of the ligands:
| Ligand | Charge | Donor atom |
|---|---|---|
| H2O | 0 | O |
| NH3 | 0 | N |
| Cl− | −1 | Cl |
| OH− | −1 | O |
| CN− | −1 | C |
Question. (a) Deduce the charge on the complex ion formed by Fe3+ with six cyanide ligands. (b) Deduce the oxidation state of cobalt in [Co(NH3)5Cl]Cl2.
Given → Relationship → Calculation. (a) +3 + 6(−1) = −3: [Fe(CN)6]3−. (b) The compound is neutral and contains two chloride counter-ions outside the square brackets, so the complex ion is [Co(NH3)5Cl]2+. Inside: x + 5(0) + (−1) = +2, so x = +3.
Check. Charges inside the brackets must add to the charge on the complex; ions outside the brackets balance it to zero.
Exam focus · what the published papers show
Only 30% of candidates were able to deduce the charge on the complex ion although the ligands were simple ones (water and hydroxide ions). On another paper, 57% of the candidates were able to use the formula of the compound to deduce the oxidation state of the metal ion and the charge of the complex ion. The most commonly chosen distractor was B where the charge of the complex ion was correct but the charge of the metal ion was not. Candidates also thought incorrectly that iron ions act as ligands; the ligand is the lone-pair donor, never the metal ion.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
Magnesium salts form slightly acidic solutions owing to equilibria such as: Mg2+(aq) + H2O(l) ⇌ Mg(OH)+(aq) + H+(aq). Comment on the role of Mg2+ in forming the Mg(OH)+ ion, in acid–base terms. [2]
(iii) When excess ammonia is added to copper(II) chloride solution, the dark blue complex ion, [Cu(NH3)4(H2O)2]2+, forms. State the molecular geometry of this complex ion, and the bond angles within it. [1] (iv) Examine the relationship between the Brønsted–Lowry and Lewis definitions of a base, referring to the ligands in the complex ion [CuCl4]2−. [2]
Aluminium chloride reacts with chloride ions: AlCl3 + Cl− → AlCl4−. (a) Draw Lewis formulas of AlCl3 and AlCl4−, showing the coordination bond. [2] (b) Identify the Lewis acid and explain your answer. [1] (c) Explain why AlCl3 is not a Brønsted–Lowry acid. [1]
Solutions and mark-scheme guidance · Set R3K
R3K.1 A
AlCl3 has an incomplete octet and accepts an electron pair, but has no proton to donate. CH3CO2H and HF are Brønsted–Lowry acids; CCl4 is neither.
R3K.2 A
Cl− donates a lone pair to the electron-deficient carbon of the carbocation: a Lewis base (and nucleophile).
R3K.3 A
Water donates a lone pair from O (Lewis base, e.g. to H+ or Cu2+); as a Brønsted–Lowry acid its proton accepts an electron pair from a base (Lewis acid). NH4+ has no lone pair; Cu2+ can only accept; CH4 does neither.
R3K.4 C
Electrophiles accept an electron pair: Lewis acids. Nucleophiles donate a pair: Lewis bases.
R3K.5 D
All three are correct: a Lewis base donates a pair, as a nucleophile does; an electrophile accepts a pair; a Lewis acid is an electron-pair acceptor.
R3K.6 A
In the dimer Al2Cl6 a lone pair on each bridging chlorine is donated to an aluminium atom: both Al atoms accept a pair, so both act as Lewis acids.
R3K.7 C
Two Cl− ligands and one Cl− counter-ion: the complex is [Co(NH3)4Cl2]+, so x + 0 − 2 = +1 and x = +3.
R3K.8 D
[PtCl6]2−: +4. [Cu(H2O)4(OH)2]: +2. [Ni(NH3)4(H2O)2]2+: +2. [Co(NH3)4Cl2]+: x − 2 = +1, so +3.
R3K.9 [2]
Lewis acid ✔; accepts a (lone) electron pair (from the hydroxide ion / water) ✔. “Electron acceptor” without mention of a pair is not accepted.
R3K.10 [3]
(iii) Octahedral AND 90° (180° for axial) ✔; square-based bipyramid accepted. (iv) Any two of: the ligand/chloride ion is a Lewis base AND donates an electron pair ✔; it is not a Brønsted–Lowry base AND does not accept a proton ✔; the Lewis definition extends / is broader than the Brønsted–Lowry definition ✔.
R3K.11 [4]
(a) AlCl3: Al with three single bonds to Cl, each Cl with three lone pairs, six electrons around Al ✔; AlCl4−: four Al–Cl bonds with one shown as an arrow from a Cl− lone pair to Al, overall charge 1− ✔. (b) AlCl3: its incomplete octet (vacant orbital) accepts an electron pair ✔. (c) It contains no hydrogen, so it cannot donate a proton ✔.
15SN1 and SN2 mechanisms 3.4.9 HL only
The syllabus statement: Nucleophilic substitution reactions include the reactions between halogenoalkanes and nucleophiles. The skill: describe and explain the mechanisms of the reactions of primary and tertiary halogenoalkanes with nucleophiles, distinguishing between the concerted one-step SN2 reaction of primary halogenoalkanes and the two-step SN1 reaction of tertiary halogenoalkanes, including the stereospecific nature of SN2 reactions. Both mechanisms occur for secondary halogenoalkanes.
Kinetics experiments show that nucleophilic substitution happens in two different ways. For bromoethane with hydroxide ions, doubling either [CH3CH2Br] or [OH−] doubles the rate: rate = k[CH3CH2Br][OH−]. For 2-bromo-2-methylpropane, doubling [OH−] has no effect: rate = k[(CH3)3CBr]. A mechanism must explain its rate equation (Reactivity 2.2), so these two results require two mechanisms. The names record the evidence: SN is substitution, nucleophilic; the number is the molecularity of the rate-determining step.
SN2: one concerted step (primary halogenoalkanes)
- Electron-rich: the lone pair (negative charge) on the oxygen of OH−. Electron-deficient: the δ+ carbon bonded to Br.
- Arrow 1 starts on the O lone pair and ends on the carbon: the C–O bond begins to form.
- Arrow 2 starts in the middle of the C–Br bond and ends on Br: the bond pair moves to bromine.
- Both happen together in one step (concerted) through a transition state [HO···CH2(CH3)···Br]−, drawn in square brackets with partial bonds, a negative charge, and the five groups around carbon.
- Products: CH3CH2OH and Br− (the leaving group).
Because two species (the halogenoalkane and the nucleophile) take part in the only step, the rate depends on both concentrations: rate = k[RX][OH−], second order overall; the step is bimolecular. The nucleophile must approach from the side opposite the leaving group (backside attack): that is where the δ+ carbon is least shielded and where the empty σ* orbital of the C–Br bond points. In a primary halogenoalkane the carbon carries only one alkyl group, so this approach is not blocked.
Because attack is always from the back, the three other groups are pushed through to the opposite side, and the configuration at carbon is inverted. If the carbon is a chiral centre, a single enantiomer of the reactant gives a single enantiomer of the product with the opposite configuration. SN2 is therefore stereospecific: the stereochemistry of the product is determined by that of the reactant.
SN1: two steps through a carbocation (tertiary halogenoalkanes)
- Step 1 (slow): an arrow from the C–Br bond to Br. The bond breaks heterolytically: (CH3)3CBr → (CH3)3C+ + Br−. The carbocation intermediate is trigonal planar (sp2 carbon, three bonding domains).
- Step 2 (fast): an arrow from the lone pair on OH− to the C+: (CH3)3C+ + OH− → (CH3)3COH.
Only the halogenoalkane is involved in the slow step, so rate = k[RX]: first order, unimolecular. Two features of tertiary halogenoalkanes favour this route. The three bulky alkyl groups crowd the back of the carbon (steric hindrance), so SN2 attack is very slow. And the three alkyl groups release electron density towards C+ (the positive inductive effect), spreading the charge and making the tertiary carbocation relatively stable, so it forms readily. Because the carbocation is planar, a nucleophile attacks either face with equal probability: from a chiral reactant, SN1 gives an (approximately) racemic mixture, not a single enantiomer.
| SN2 | SN1 | |
|---|---|---|
| Typical substrate | primary (and methyl) halogenoalkanes | tertiary halogenoalkanes |
| Steps | one, concerted | two: ionization (slow), then attack (fast) |
| Species on the energy profile | one transition state | two transition states, one carbocation intermediate |
| Rate equation | rate = k[RX][Nu−], second order | rate = k[RX], first order |
| Stereochemistry at a chiral carbon | inversion; single enantiomer (stereospecific) | racemic mixture |
| Why favoured | little steric hindrance at the back of carbon | stable tertiary carbocation; steric hindrance blocks SN2 |
Secondary halogenoalkanes lie between the two: they react by both mechanisms, and kinetics data decide which predominates under given conditions. A rate that depends on the concentration of the nucleophile points to SN2.
Exam focus · what the published papers show
curly arrow going from lone pair/negative charge on O in −OH to C ✔; curly arrow showing Cl (or Br, I) leaving ✔; representation of transition state showing negative charge, square brackets and partial bonds ✔. Notes: Do not accept curly arrows originating on H in OH−. Do not award M3 if OH–C bond is represented. An SN1 mechanism drawn where SN2 is required scores at most [2].
Many candidates seemed to have very little idea of how to represent an SN2 mechanism. More recently, performance on one SN2 item averaged 33 %, disappointing considering SN2 mechanism has been examined several times before. On stereochemistry, Few were aware that SN2 would give (almost) 100% inversion, whereas SN1 would give (approximately) 50%. When kinetics data are given, use them: many candidates did not relate the kinetics data to the SN1 mechanism.
16The leaving group and the rate 3.4.10 HL only
The syllabus statement: The rate of the substitution reactions is influenced by the identity of the leaving group. The skill: predict and explain the relative rates of the substitution reactions for different halogenoalkanes RCl, RBr and RI. The roles of the solvent and the reaction mechanism on the rate will not be assessed.
When 1-chlorobutane, 1-bromobutane and 1-iodobutane are warmed with aqueous silver nitrate in ethanol, a precipitate of silver halide appears first with the iodide (yellow), then the bromide (cream), and last and slowest with the chloride (white). The halide ion forms only when the C–X bond breaks, so the order of appearance is the order of reaction rate: RI > RBr > RCl.
| Bond | Bond enthalpy / kJ mol−1 | Relative rate of substitution |
|---|---|---|
| C–Cl | 324 | slowest |
| C–Br | 285 | intermediate |
| C–I | 228 | fastest |
The explanation is bond strength. Down group 17 the halogen atom is larger, so the orbital overlap with carbon is poorer and the C–X bond is longer and weaker. A weaker bond is broken with less energy, so the activation energy is lower and the rate is higher. Equivalently, I− is a better leaving group than Cl−: the large iodide ion spreads its negative charge over a greater volume and is more stable once it has left. Bond polarity works in the opposite direction (C–Cl is the most polar bond, so its carbon is the most δ+), but the rates show that bond strength is the deciding factor.
Exam focus · what the published papers show
For 1-iodopentane reacting faster than 1-chloropentane: bond enthalpy C–I lower than C–Cl OR C–I bond weaker than C–Cl ✔ and «weaker bond» broken more easily/with less energy OR lower Ea «for weaker bonds» ✔. For 2-chlorobutane compared with 2-bromobutane: slower AND C-Cl bond is stronger «than C-Br» OR slower AND Br/Br- is a better leaving group ✔. For the C–X bond polarity down group 17: decreases AND the electronegativity of the halogen decreases ✔.
17Mechanisms of electrophilic addition 3.4.11 HL only
The syllabus statement: Alkenes readily undergo electrophilic addition reactions. The skill: describe and explain the mechanisms of the reactions between symmetrical alkenes and halogens, water and hydrogen halides.
Ethene and hydrogen bromide
H–Br is polar, Hδ+–Brδ−. The mechanism has two steps and passes through a carbocation.
- Step 1: an arrow from the C=C (the π bond) to the δ+ H of HBr, and an arrow from the H–Br bond to Br. The π electrons form a new C–H bond; H–Br breaks heterolytically. Products: the carbocation CH3CH2+ and Br−.
- Step 2: an arrow from a lone pair (negative charge) on Br− to the C+, forming the C–Br bond. Product: CH3CH2Br.
Ethene and bromine
Bromine has no permanent dipole; the electron-rich π bond induces one. In step 1 an arrow runs from the C=C to the nearer (δ+) Br and a second arrow from the Br–Br bond to the further Br, forming a carbocation (with Br on the other carbon) and Br−. In step 2, Br− attacks the C+ to give 1,2-dibromoethane.
Ethene and water (acid-catalysed)
Water itself is too weak an electrophile to attack C=C; an acid catalyst supplies H+. (1) The π bond attacks H+ to give CH3CH2+. (2) A lone pair on the oxygen of a water molecule attacks C+, forming CH3CH2OH2+. (3) This ion loses H+ to give ethanol, and the catalyst is regenerated.
Exam focus · what the published papers show
One of the common mistakes was drawing the curly arrow from H to the double bond in step one, and another was neglecting to initiate the curly arrow from the lone pair or negative charge on the iodide ion in the second step. Mechanisms were answered better in some sessions, though some careless errors were sometimes seen such as curly arrows not originating on the lone pair of the bromide ion, and In the first step of the mechanism, the curly arrow showing Br leaving was often missing.
18Unsymmetrical alkenes and carbocation stability 3.4.12 HL only
The syllabus statement: The relative stability of carbocations in the addition reactions between hydrogen halides and unsymmetrical alkenes can be used to explain the reaction mechanism. The skill: predict and explain the major product of a reaction between an unsymmetrical alkene and a hydrogen halide or water.
Propene reacts with hydrogen bromide to give mainly 2-bromopropane, with only a small amount of 1-bromopropane. The two products come from two different carbocations. If H adds to the end carbon (C1), the positive charge is on C2, which carries two alkyl groups: a secondary carbocation, CH3CH+CH3. If H adds to C2, the charge is on C1, which carries one alkyl group: a primary carbocation, CH3CH2CH2+.
Alkyl groups are electron-releasing: they push electron density along their σ bonds towards the C+. This positive inductive effect reduces the charge density on the positive carbon and so stabilizes the ion. The more stable carbocation has the lower-energy transition state leading to it, forms faster, and gives the major product. The empirical Markovnikov rule (“hydrogen adds to the carbon that already has more hydrogens”) is a prediction; carbocation stability is its explanation, and only the explanation earns marks. The same reasoning applies to acid-catalysed hydration: 2-methylpropene and water give mainly 2-methylpropan-2-ol, (CH3)3COH, through the tertiary carbocation (CH3)3C+.
Question. Predict and explain the major product when 2-methylbut-2-ene, (CH3)2C=CHCH3, reacts with hydrogen chloride.
Reasoning. Adding H+ to the CH carbon puts the positive charge on the carbon bearing two methyl groups and the rest of the chain: a tertiary carbocation, (CH3)2C+CH2CH3. Adding H+ to the other carbon gives a secondary carbocation, (CH3)2CHCH+CH3.
Answer. The tertiary carbocation is more stable (three electron-releasing alkyl groups spread the positive charge), so it forms faster, and the major product is 2-chloro-2-methylbutane, (CH3)2CClCH2CH3.
Exam focus · what the published papers show
The majority of candidates did not offer an explanation – they simply stated Markovnikov’s rule - and hence did not score any marks on this part-question. Other recurring errors: Many candidates recognized that the secondary carbocation was more stable than the primary carbocation but failed to explain this in terms of the electron-releasing alkyl groups; A common error was comparing a secondary with a primary carbocation instead of comparing a tertiary carbocation (more stable) with a secondary carbocation; and Another common error is referring to the stability of the final product rather than the intermediate carbocation.
19Electrophilic substitution of benzene 3.4.13 HL only
The syllabus statement: Electrophilic substitution reactions include the reactions of benzene with electrophiles. The skill: describe and explain the mechanism of the reaction between benzene and a charged electrophile, E+. The formation of the electrophile will not be assessed.
Why benzene substitutes rather than adds
Benzene, C6H6, is highly unsaturated, yet it does not decolourize bromine water. Its six p electrons are delocalized in a continuous π system above and below the planar ring (Structure 2.2): all six C–C bonds are the same length, intermediate between single and double bonds. The evidence is thermochemical as well as structural. Hydrogenating cyclohexene releases 120 kJ mol−1; a Kekulé structure with three C=C bonds would release about 360 kJ mol−1, but benzene releases only about 208 kJ mol−1. Benzene is about 150 kJ mol−1 more stable than a structure with three localized double bonds. Addition would destroy the delocalized system and lose this stability; substitution replaces a hydrogen and keeps the ring intact. Benzene therefore undergoes electrophilic substitution: the delocalized ring is electron-rich (attracting electrophiles and repelling nucleophiles), but a stronger electrophile than for alkenes, usually a positive ion, is needed.
The general mechanism
- Attack: an arrow from the delocalized ring (inside the circle) to E+. Two of the π electrons form a C–E bond. The intermediate is a positive ion (an arenium ion or carbocation) in which the attacked carbon is sp3, bonded to both H and E; the remaining four π electrons are delocalized over the other five carbons, drawn as a horseshoe (an incomplete circle open towards the sp3 carbon) with a + inside.
- Loss of H+: an arrow from the C–H bond into the ring. The pair returns to the π system, restoring full delocalization, and H+ leaves.
Nitration of benzene
Benzene is nitrated by warming (about 50 °C) with a mixture of concentrated nitric and sulfuric acids. In this mixture sulfuric acid, the stronger acid, protonates nitric acid, which acts as a Brønsted–Lowry base; the protonated nitric acid loses water to form the nitronium ion, NO2+ (Reactivity 3.1). The formation of the electrophile is not assessed, but it is a useful connection:
Exam focus · what the published papers show
For the carbocation intermediate in nitration, any correct structure with NO2 and H on the same sp3 carbon and a delocalized positive ring was accepted, with the note Do not accept structures missing the positive charge. For why benzene reacts with electrophiles rather than nucleophiles: delocalized electrons / a region of high electron density ✔; electrophiles are attracted AND nucleophiles repelled ✔.
On one item 21% incorrectly identified benzene being unsaturated as the reason it undergoes substitution rather than addition. Only higher-scoring candidates managed to describe the nitronium ion in the nitration of benzene as an electrophile and a Lewis acid. In a later mechanism question About half the candidates correctly identified the nitronium ion as the electrophile, the nitration mechanism produced an average mark of ~2/4, and many marks were lost by a lack of precision in the start and end points of curly arrows.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
Explain the mechanism of the reaction between chloroethane and aqueous sodium hydroxide, NaOH(aq), using curly arrows to represent the movement of electron pairs. [3]
2-Bromobutane can react with cyanide, CN−, in a nucleophilic substitution reaction. (b) Suggest, with a reason, whether the reaction follows an SN1 or SN2 mechanism if only one stereoisomer was obtained as a product. [1] (e) State, with a reason, how the rate of reaction of cyanide with 2-chlorobutane differs from its rate of reaction with 2-bromobutane under the same conditions. [1]
Experiments showed that the rate of the reaction between 2-chloropentane and aqueous sodium hydroxide depends on the concentrations of both reactants. (d) Deduce, with a reason, the mechanism of the reaction between 2-chloropentane and sodium hydroxide. [1] (e) Discuss the reason benzene is more reactive with an electrophile than a nucleophile. [2]
Hydrogen bromide, HBr, reacts with but-1-ene. Two products are possible. (i) Explain the mechanism for the formation of the major product, using curly arrows to indicate the movement of electron pairs. [4] (ii) Explain why the mechanism results in one product being formed in greater quantities than the other. [2]
Compound B, (CH3)3COH, can be prepared by reacting an alkene with water. Explain why the reaction produces more (CH3)3COH than (CH3)2CHCH2OH. [2]
A mixture of nitric acid and sulfuric acid can be used to convert benzene to nitrobenzene, C6H5NO2. (i) Write an equation for the reaction between the acids to produce the electrophile, NO2+. [1] (ii) Draw the structural formula of the carbocation intermediate produced when this electrophile attacks benzene. [1]
(R)-2-iodobutane is warmed with aqueous sodium hydroxide. (a) State the rate equation if the reaction proceeds by SN2. [1] (b) Explain what would be observed if the product were examined with a polarimeter, for an SN2 reaction and for an SN1 reaction. [2] (c) Predict, with a reason, how the rate would change if (R)-2-chlorobutane were used instead. [1]
Solutions and mark-scheme guidance · Set R3L
R3L.1 C
(CH3)3CBr is tertiary: it forms a stable tertiary carbocation and is too hindered for SN2.
R3L.2 D
2-bromo-2-methylbutane (tertiary, SN1) reacts faster than 1-bromopentane (primary, SN2). A and B reverse the mechanisms; C is wrong because C–Br is weaker than C–Cl, so the bromide is faster.
R3L.3 C
Two peaks with an intermediate between them; the first barrier is higher (rate-determining ionization); products lower than reactants (exothermic). D has the higher barrier second; A and B show one step.
R3L.4 D
Doubling [NaOH] doubles the rate (5.0 × 10−7 → 1.0 × 10−6); doubling [RCl] doubles it again (1.0 × 10−6 → 1.9 × 10−6, within experimental error). First order in each, second order overall: SN2.
R3L.5 B
H adds to the CH2 end, giving the secondary carbocation CH3CH+CH3, so the major product is 2-chloropropane.
R3L.6 C
Addition would destroy the delocalized π system, which confers extra stability (the C–C bonds are stronger than in a localized structure); substitution keeps it. A report notes that 21 % chose “unsaturated” (A), despite this being given in the stem.
R3L.7 A
A hydrogen on the ring is replaced by NO2: substitution; NO2+ accepts an electron pair from the ring: an electrophile.
R3L.8 D
NO2+ accepts an electron pair: it is an electrophile and a Lewis acid.
R3L.9 [3]
Curly arrow from the lone pair / negative charge on O in −OH to C ✔; curly arrow showing Cl leaving (from the C–Cl bond to Cl) ✔; transition state with negative charge, square brackets and partial bonds [HO···CH2(CH3)···Cl]− ✔. Arrows from H in OH− are not accepted; an O–C full bond in the transition state loses M3.
R3L.10 [2]
(b) SN2, because SN2 occurs with inversion of configuration (or: SN1 would create a racemic mixture) ✔. (e) Slower, because the C–Cl bond is stronger than C–Br (or: Br− is a better leaving group) ✔.
R3L.11 [3]
(d) SN2 AND the rate depends on both [OH−] and [2-chloropentane] ✔. (e) Benzene has delocalized electrons / π bonds around the ring, a region of high electron density ✔; electrophiles (positive) are attracted AND nucleophiles (negative / electron-rich) are repelled ✔. “Nucleophiles less attracted” alone is not accepted.
R3L.12 [6]
(i) Curly arrow from C=C to H of HBr AND curly arrow from the H–Br bond to Br ✔; secondary carbocation CH3CH+CH2CH3 ✔; curly arrow from a lone pair / negative charge on Br− to C+ ✔; product CH3CHBrCH2CH3 ✔ ([3 max] for the mechanism to the minor product). (ii) The secondary carbocation is more stable than the primary one ✔; because of the greater electron-releasing (inductive) effect of two alkyl groups compared with one ✔. Quoting Markovnikov’s rule without reference to carbocation stability is not accepted.
R3L.13 [2]
The carbocation formed on the way to (CH3)3COH, (CH3)3C+, is more stable than (CH3)2CHCH2+ ✔; because it has more alkyl groups, which are electron-releasing (greater positive inductive effect) ✔. Simply quoting Markovnikov’s rule scores nothing.
R3L.14 [2]
(i) HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4− ✔ (HNO3 + H2SO4 → NO2+ + H2O + HSO4− also accepted). (ii) A six-membered ring with one sp3 carbon bearing H and NO2, and a horseshoe of delocalized electrons with a positive charge over the other five carbons ✔; the positive charge must be shown.
R3L.15 [4]
(a) rate = k[C4H9I][OH−] ✔. (b) SN2: inversion gives a single enantiomer (the S isomer), so the product rotates plane-polarized light ✔; SN1: attack on either face of the planar carbocation gives a racemic mixture, with no net rotation ✔. (c) Slower, because the C–Cl bond (324 kJ mol−1) is stronger than C–I (228 kJ mol−1), giving a higher activation energy ✔.
Review · Reactivity 3.4
20Misconceptions, the examiner’s view, and the question types
- “A nucleophile must be negatively charged.” Why it is wrong: the requirement is a lone pair; H2O and NH3 are neutral nucleophiles. Consequence: the most-chosen wrong answer on a nucleophile question.
- “NaOH is the nucleophile.” The nucleophile is OH−; Na+ is a spectator. “NaOH” is not accepted.
- “The curly arrow starts at the H of OH−.” Arrows start where the electrons are: the lone pair on O.
- “Addition” is a full answer. The type must be “electrophilic addition”, and “substitution” must be qualified as nucleophilic, electrophilic or free-radical.
- “Markovnikov’s rule explains the major product.” It predicts it; the explanation is the greater stability of the more substituted carbocation, due to the electron-releasing alkyl groups.
- “RCl reacts fastest because C–Cl is most polar.” Rate is controlled by bond strength: C–I is weakest, so RI is fastest.
- “SN1 and SN2 give the same stereochemistry.” SN2 inverts (single enantiomer); SN1 gives a racemic mixture.
- “Benzene reacts by addition because it is unsaturated.” Addition would destroy the stable delocalized system; benzene substitutes.
- “The charge on a complex equals the oxidation state of the metal.” Only when all ligands are neutral; anionic ligands must be counted.
Evidence base: the IB Diploma chemistry subject reports quoted in this chapter.
Answered well: recognizing that a nucleophile needs a lone pair (72 %); identifying a Lewis acid; recognizing that NH4+ cannot act as a Lewis base (78 %); deducing the oxidation state of a metal and the charge of a complex (72 % on one paper); knowing that benzene does not react in the same way as alkenes (73 %); the reagents and conditions for hydrolysis of halogenoalkanes; and, in the best sessions, full mechanisms with accurate arrows.
Found difficult: (1) the start and end points of curly arrows, above all arrows from H of OH−, from C rather than the C–X bond, and not from the lone pair of the halide ion; (2) the SN2 transition state (33 % on one recent item); (3) linking kinetics data and stereochemistry to SN1 and SN2; (4) explaining the major product through carbocation stability and the inductive effect instead of quoting Markovnikov’s rule, and comparing the right pair of carbocations; (5) the charge on a complex ion (30 % on one item); (6) naming reaction types precisely; (7) benzene: the reason for substitution and the role of the nitronium ion.
What successful answers did: started every arrow on a lone pair, a negative charge or a bond and ended it on an atom; drew the SN2 transition state in square brackets with partial bonds and a negative charge; named the nucleophile as the ion; compared the stabilities of the intermediate carbocations and explained them with electron-releasing alkyl groups; and used “electron pair” in every Lewis definition.
Seven question types cover the sub-topic.
| If the question asks… | …then |
|---|---|
| Identify nucleophile or electrophile | Lone pair donor = nucleophile (Lewis base); electron-deficient pair acceptor = electrophile (Lewis acid). |
| Nucleophilic substitution equation | RX + OH− → ROH + X−; warm aqueous NaOH; nucleophile is OH−. |
| Alkene + reagent | Add across C=C; electrophilic addition; bromine water decolourized in the dark. |
| HL · Lewis acid/base; complex charge | Pair acceptor/donor; charge = metal oxidation state + ligand charges; ions outside brackets balance. |
| HL · SN1 or SN2? | Primary SN2 (rate = k[RX][Nu]), inversion; tertiary SN1 (rate = k[RX]), racemic; draw arrows and TS or intermediate. |
| HL · Relative rates | RI > RBr > RCl: weaker C–X bond, lower Ea, better leaving group. |
| HL · Addition or substitution mechanism | Two-step via carbocation; more stable (more substituted) carbocation gives major product; benzene: attack, arenium ion, loss of H+. |
21Quick check
- State what feature every nucleophile must have, and give one neutral nucleophile.
- Write an equation for the reaction of 1-bromopropane with aqueous sodium hydroxide, and identify the nucleophile.
- Write the equation for the reaction of propene with bromine, and name the product.
- Explain why alkenes decolourize bromine water in the dark but alkanes do not.
- HL: Identify the Lewis acid in the reaction of NH3 with BF3, and state the type of bond formed.
- HL: Deduce the charge on the complex formed by Cr3+ with four water molecules and two chloride ions.
- HL: State the rate equation and the stereochemical outcome for the SN1 hydrolysis of a tertiary halogenoalkane.
- HL: Explain which reacts faster with hydroxide ions, 1-chlorobutane or 1-iodobutane.
- HL: Explain why the major product of 2-methylpropene and HBr is 2-bromo-2-methylpropane.
- HL: Describe the two steps of the mechanism of benzene with a charged electrophile, E+.
- A lone pair of electrons; H2O or NH3.
- CH3CH2CH2Br + OH− → CH3CH2CH2OH + Br−; OH−.
- CH2=CHCH3 + Br2 → CH2BrCHBrCH3; 1,2-dibromopropane.
- The electron-rich C=C undergoes electrophilic addition with Br2; alkanes have strong non-polar bonds and need UV to form radicals.
- BF3 (incomplete octet, accepts the pair); a coordination (dative covalent) bond.
- +3 + 4(0) + 2(−1) = +1: [Cr(H2O)4Cl2]+.
- rate = k[RX]; a racemic mixture (if the carbon is chiral), because the planar carbocation is attacked from either face.
- 1-iodobutane: the C–I bond is weaker than C–Cl, so the activation energy is lower.
- H+ adds to the CH2 carbon to give the tertiary carbocation (CH3)3C+, stabilized by three electron-releasing methyl groups; Br− attacks it.
- The delocalized π electrons attack E+, giving a positively charged intermediate with one sp3 carbon; the C–H bond electrons return to the ring as H+ leaves, restoring delocalization.
22Summary and knowledge organiser
Essential knowledge
- Nucleophile: electron-pair donor (lone pair; neutral or negative). Electrophile: electron-pair acceptor (neutral or positive).
- Nucleophilic substitution: Nu donates a pair to the δ+ carbon of R–X; X− leaves. RX + OH− → ROH + X−.
- Heterolytic fission: one fragment keeps both electrons; ions form; curly arrows move pairs from lone pairs or bonds to atoms.
- Alkenes: electron-rich π bond; electrophilic addition of X2, HX and H2O (acid catalyst); bromine water decolourized in the dark.
- HL Lewis acid = electron-pair acceptor; Lewis base = donor; coordination bond; nucleophiles are Lewis bases, electrophiles Lewis acids.
- HL Complex ion: ligands donate pairs to a metal cation; charge = metal oxidation state + Σ ligand charges.
- HL SN2 (primary): one step, TS, rate = k[RX][Nu], inversion. SN1 (tertiary): carbocation, rate = k[RX], racemic. Secondary: both.
- HL Rate RI > RBr > RCl: bond strength decreases down the group.
- HL Addition mechanisms go through carbocations; the more stable (tertiary > secondary > primary) gives the major product.
- HL Benzene: delocalized, substitutes; attack on E+, arenium ion, loss of H+; nitration uses NO2+ from conc. HNO3/H2SO4.
Examination checklist
- Name the nucleophile as the ion (OH−), not the compound.
- Qualify every reaction type: nucleophilic, electrophilic, free-radical.
- Arrows start on a lone pair, a negative charge or a bond; they end on an atom.
- HL SN2 transition state: square brackets, partial bonds, negative charge.
- HL Explain the major product with carbocation stability and electron-releasing alkyl groups, not Markovnikov alone.
- HL Use “electron pair”, not “electron”, in Lewis definitions; show the + in the arenium ion.
Knowledge organiser
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| Nucleophiles 3.4.1 | Lone-pair donors: OH−, CN−, H2O, NH3. | Charge not required; lone pair is. |
| Nucleophilic substitution 3.4.2 | RX + OH− (warm, aq) → ROH + X−. | Nucleophile OH−, not NaOH. |
| Heterolytic fission 3.4.3 | Both electrons to one fragment; ions; curly arrows. | Arrow from bond or lone pair to atom. |
| Electrophiles 3.4.4 | Pair acceptors: H+, NO2+, carbocations, HBr, Br2 (induced dipole). | No lone pair needed. |
| Electrophilic addition 3.4.5 | Alkene + X2, HX, H2O (H+) → saturated product. | “Electrophilic addition”, not “addition”. |
| Lewis theory 3.4.6–3.4.8 HL | Acid accepts pair, base donates; coordination bond; complex charge. | BF3, AlCl3, metal ions are Lewis acids. |
| SN1/SN2 3.4.9–3.4.10 HL | 1° SN2 inversion; 3° SN1 racemic; RI > RBr > RCl. | Rate equations from the slow step. |
| Addition mechanisms 3.4.11–3.4.12 HL | Two steps via carbocation; 3° > 2° > 1°. | Stability of the intermediate, not the product. |
| Benzene 3.4.13 HL | Electrophilic substitution; arenium ion; loss of H+. | Addition would lose delocalization energy. |