DP Chemistry HL R3.4 Electron-Pair Sharing Reactions

DP Chemistry HL R3.4 Electron-Pair Sharing Reactions - Full notes

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DP Chemistry HL | R3.4 Electron-Pair Sharing Reactions

DP CHEMISTRY · REACTIVITY

R3.4

Electron-Pair Sharing Reactions

Full notes for HL · IB-style

Contents

1	Heterolytic fission and curly arrows
2	Nucleophiles and electrophiles
3	Lewis acids and bases
4	Nucleophilic substitution of halogenoalkanes
5	Electrophilic addition to alkenes
6	Testing for unsaturation and uses of addition reactions
7	SN1 and SN2 mechanisms (HL only)
8	Markovnikov's rule and carbocation stability (HL only)
9	Electrophilic substitution of benzene and complex ions (HL only)
10	Nucleophiles
11	Electrophiles
12	Lewis acids and bases
13	Complex ions (introduction)
14	Nucleophilic substitution of halogenoalkanes
15	Rates of hydrolysis of halogenoalkanes
16	Electrophilic addition to alkenes
17	Testing for alkenes
18	Curly arrows
19	SN1 and SN2 mechanisms (HL)
20	Markovnikov's rule and carbocation stability (HL)
21	Electrophilic substitution of benzene (HL)
22	Comparing reaction types
23	Practical: comparing halogenoalkane hydrolysis rates
24	Mixed review: worked problems
25	Additional exam-style practice
26	Organic synthesis routes
27	Electron-pair sharing in biology
28	Common misconceptions and links
29	Exam-style question with markscheme
30	Check your understanding
31	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
explain heterolytic fission and use curly arrows to show electron-pair movement
define nucleophile and electrophile with examples
define Lewis acids and bases and identify them in reactions
describe nucleophilic substitution of halogenoalkanes with hydroxide
describe electrophilic addition of halogens, hydrogen halides and water to alkenes
use bromine water to test for C=C bonds
compare SN1 and SN2 mechanisms and explain the effect of halogenoalkane class
apply Markovnikov's rule using carbocation stability
describe nitration of benzene and the formation of complex ions

Sections marked HL only (Sections 7 to 9) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How do electron-rich and electron-poor species react with each other? How can reaction mechanisms be represented and predicted?

How this topic is examined

R3.4 is examined through mechanisms with curly arrows, identification of nucleophiles and electrophiles, and predicting products of substitution and addition reactions. HL questions include SN1/SN2, Markovnikov's rule and nitration of benzene.

Curly arrows must start from a lone pair or a bond and point to where the electron pair goes.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Heterolytic fission and curly arrows

One atom takes both electrons.

Think about it

Why does a hydroxide ion attack the carbon atom in chloromethane, rather than the hydrogen atoms?

In heterolytic fission, a covalent bond breaks unevenly: one atom takes both electrons of the shared pair, forming a negative ion, while the other forms a positive ion. The more electronegative atom usually takes the electrons: H–Br → H+ + Br−. Compare homolytic fission (R3.3), which forms radicals.

Reaction mechanisms show electron-pair movement with curly arrows. A curly arrow represents the movement of a pair of electrons. It starts from a lone pair or a bond (where the electrons are) and points to an atom (forming a new bond) or to an atom that will take the pair (breaking a bond).

Quick check

1. What does a curly arrow represent?

2. What is formed by heterolytic fission?

Answers: 1. movement of a pair of electrons 2. a cation and an anion

2 Nucleophiles and electrophiles

Electron-rich attacks electron-poor.

A nucleophile is an electron-rich species with a lone pair that it donates to an electron-deficient atom, forming a new covalent bond. Nucleophiles are often negative ions or molecules with lone pairs: OH−, CN−, Cl−, H2O, NH3. ('Nucleus-loving': attracted to positive centres.)

An electrophile is an electron-deficient species that accepts a pair of electrons to form a new covalent bond. Electrophiles are often positive ions or molecules with a δ+ atom: H+, NO2+, Brδ+ in a polarized Br2 molecule, the δ+ hydrogen in HBr. ('Electron-loving'.)

In a halogenoalkane, the C–X bond is polar (Cδ+–Xδ−), so the carbon is attacked by nucleophiles. In an alkene, the C=C double bond is a region of high electron density (the π bond), so it is attacked by electrophiles.

	Nucleophile	Electrophile
Nature	electron-rich; has a lone pair	electron-deficient
Role	donates an electron pair	accepts an electron pair
Examples	OH−, CN−, NH3, H2O	H+, NO2+, Br2 (polarized)
Attacks	Cδ+ in halogenoalkanes	C=C in alkenes

Quick check

1. Define nucleophile.

2. Why do electrophiles attack alkenes?

Answers: 1. species that donates a lone pair to form a bond 2. C=C is a region of high electron density

3 Lewis acids and bases

A broader definition based on electron pairs.

A Lewis acid is an electron-pair acceptor; a Lewis base is an electron-pair donor. A Lewis acid-base reaction forms a coordination (dative) bond. Example: NH3 + BF3 → H3N→BF3: ammonia donates its lone pair (Lewis base) to boron, which has an incomplete octet (Lewis acid).

The Lewis definition is broader than Brønsted-Lowry (R3.1): every Brønsted-Lowry base is a Lewis base (it donates a lone pair to H+), and H+ is a Lewis acid. But Lewis acids include species with no protons, such as BF3, AlCl3 and metal ions. Nucleophiles are Lewis bases; electrophiles are Lewis acids.

Quick check

1. Define Lewis acid.

2. Identify the Lewis base in H2O + H+ → H3O+.

Answers: 1. electron-pair acceptor 2. H2O

4 Nucleophilic substitution of halogenoalkanes

A nucleophile replaces the halogen.

Halogenoalkanes react with nucleophiles in nucleophilic substitution: the nucleophile replaces the halogen atom. With aqueous sodium hydroxide (heated under reflux), an alcohol forms: CH3CH2Br + OH− → CH3CH2OH + Br−.

Mechanism (outline): the lone pair on OH− is attracted to the δ+ carbon; a curly arrow from the lone pair to the carbon shows a new C–O bond forming; a curly arrow from the C–Br bond to the Br shows the bond breaking heterolytically; Br− leaves (the leaving group).

Rate and the halogen: the rate depends on the strength of the C–X bond: iodoalkanes react fastest and fluoroalkanes slowest, because the C–I bond is the weakest (238 kJ mol−1) and C–F the strongest (492 kJ mol−1). Bond strength is more important than bond polarity here.

Figure 1. Nucleophilic substitution of bromoethane by hydroxide.

Figure 2. Weaker C–X bonds react faster in substitution.

Common mistake

Drawing the curly arrow from the carbon to the hydroxide ion. Arrows start where the electrons are: from the lone pair on O.

Quick check

1. What is the product of bromoethane with aqueous NaOH?

2. Why do iodoalkanes react fastest?

Answers: 1. ethanol 2. C–I bond is weakest

5 Electrophilic addition to alkenes

Electrophiles add across the double bond.

Alkenes undergo electrophilic addition: the π bond of C=C breaks and two atoms or groups add across the double bond, forming a saturated product.

With halogens: ethene + bromine → 1,2-dibromoethane. As Br2 approaches the electron-rich C=C, it becomes polarized (Brδ+–Brδ−). The π electrons attack Brδ+; the Br–Br bond breaks heterolytically; a carbocation intermediate forms; Br− then attacks the carbocation.

With hydrogen halides: ethene + HBr → bromoethane. With water (steam): ethene + H2O → ethanol, using a phosphoric acid catalyst at about 300 °C and 60 atm (industrial hydration). With hydrogen: alkene + H2 → alkane, with a nickel catalyst (hydrogenation; used to make margarine).

Figure 3. Products of electrophilic addition to ethene.

Reagent	Product from ethene	Conditions
Br2	1,2-dibromoethane	room temperature
HBr	bromoethane	room temperature
H2O (steam)	ethanol	H3PO4 catalyst, 300 °C, 60 atm
H2	ethane	Ni catalyst, 150 °C

Quick check

1. Why does Br2 become polarized near an alkene?

2. Name the product of propene with HCl (major).

Answers: 1. electron-rich C=C repels electrons in Br2 2. 2-chloropropane

6 Testing for unsaturation and uses of addition reactions

Bromine water decolourizes.

Bromine water is used to test for a C=C double bond: shaken with an alkene, the orange-brown bromine water decolourizes rapidly, because bromine adds across the double bond. With an alkane, there is no change in the dark.

Addition reactions are important industrially: making ethanol from ethene, hydrogenating vegetable oils, and making addition polymers (S2.4).

Quick check

1. Describe the test for an alkene.

2. What is observed with an alkane?

Answers: 1. bromine water decolourizes from orange to colourless 2. no colour change (in the dark)

7 SN1 and SN2 mechanisms (HL only)

Two pathways for nucleophilic substitution.

HL only

This section is Additional Higher Level content. SL students do not need it.

SN2 (substitution, nucleophilic, bimolecular): occurs in one step. The nucleophile attacks the carbon from the side opposite the leaving group, while the C–X bond breaks, through a transition state in which the carbon is partially bonded to both (five groups around C). The configuration is inverted. Rate = k[halogenoalkane][nucleophile] (second order). Favoured by primary halogenoalkanes, where the carbon is not crowded.

SN1 (unimolecular): occurs in two steps. First, the C–X bond breaks heterolytically (slow, rate-determining) to form a carbocation; then the nucleophile attacks the carbocation (fast). Rate = k[halogenoalkane] (first order). Favoured by tertiary halogenoalkanes, because tertiary carbocations are stabilized by the electron-releasing (positive inductive) effect of three alkyl groups, and because steric hindrance blocks SN2 attack.

Secondary halogenoalkanes react by both. Solvent: polar protic solvents (water, ethanol) favour SN1, stabilizing the ions; polar aprotic solvents (propanone) favour SN2.

	SN1	SN2
Steps	two (carbocation intermediate)	one (transition state)
Rate equation	k[RX]	k[RX][Nu−]
Favoured by	tertiary halogenoalkanes; protic solvents	primary halogenoalkanes; aprotic solvents

8 Markovnikov's rule and carbocation stability (HL only)

Predicting the major product.

HL only

This section is Additional Higher Level content. SL students do not need it.

When an unsymmetrical reagent such as HBr adds to an unsymmetrical alkene such as propene, two products are possible: 1-bromopropane and 2-bromopropane. Markovnikov's rule: the hydrogen attaches to the carbon that already has more hydrogens, so the major product is 2-bromopropane.

Explanation: the H+ adds to form the more stable carbocation. Carbocation stability increases from primary < secondary < tertiary, because alkyl groups release electron density (positive inductive effect), spreading the positive charge. Adding H to C1 gives a secondary carbocation on C2, which is more stable than the primary carbocation formed by adding H to C2.

9 Electrophilic substitution of benzene and complex ions (HL only)

Benzene keeps its ring; metal ions accept lone pairs.

HL only

This section is Additional Higher Level content. SL students do not need it.

Benzene has six delocalized π electrons (S2.2) that give it extra stability. It undergoes electrophilic substitution, not addition, because substitution keeps the delocalized ring. Nitration: concentrated nitric and sulfuric acids form the nitronium ion, NO2+ (HNO3 + 2H2SO4 → NO2+ + 2HSO4− + H3O+). NO2+ is attacked by the π electrons; an intermediate forms in which delocalization is partly lost; H+ is lost, restoring the ring, giving nitrobenzene, C6H5NO2.

Complex ions form by Lewis acid-base reactions: ligands (Lewis bases) donate lone pairs to a transition metal ion (Lewis acid), forming coordination bonds, as in [Cu(H2O)6]2+ and [Cu(NH3)4(H2O)2]2+ (S3.1).

Figure 4. Nitration of benzene gives nitrobenzene.

Quick check

1. Why does benzene undergo substitution rather than addition?

2. Which halogenoalkane class reacts mainly by SN1?

Answers: 1. substitution keeps the stable delocalized ring 2. tertiary

10 Nucleophiles

Electron-pair donors seeking positive centres.

A nucleophile has a lone pair (and often a negative charge) that it donates to an electron-deficient carbon to form a new covalent bond. Examples: OH−, CN−, NH3, H2O, Cl−.

WORKED EXAMPLE 1 Identifying (basic)

Which are nucleophiles: NH3, BF3, H2O, CH4?

Solution

NH3 and H2O (lone pairs)A1

Quick check

1. Why is OH− a stronger nucleophile than H2O?

2. Define nucleophile.

Answers: 1. negative charge: electron pair more available 2. an electron-pair donor that attacks an electron-deficient centre

11 Electrophiles

Electron-pair acceptors.

An electrophile is an electron-deficient species that accepts an electron pair: H+, NO2+, Brδ+ in a polarized Br2 molecule, carbocations. Electrophiles attack electron-rich regions such as C=C double bonds and benzene rings.

WORKED EXAMPLE 2 Polarization (standard)

Explain how a non-polar Br2 molecule acts as an electrophile towards ethene.

Solution

The C=C electron density repels electrons in Br2R1

Induced dipole: Brδ+ end is attacked by the π electronsR1

Quick check

1. Give two examples of electrophiles.

Answers: 1. H+, NO2+ (or carbocations, Br2)

12 Lewis acids and bases

Electron pairs, not protons.

A Lewis acid accepts an electron pair; a Lewis base donates one. The reaction forms a coordination (dative) bond. All Brønsted–Lowry bases are Lewis bases, but Lewis acids include species without protons, such as BF3, AlCl3 and metal ions.

WORKED EXAMPLE 3 Lewis reaction (standard)

Identify the Lewis acid and base in NH3 + BF3 → H3NBF3.

Solution

Lewis base NH3 (donates lone pair)A1

Lewis acid BF3 (accepts into empty orbital)A1

Quick check

1. Explain why Fe3+ is a Lewis acid in [Fe(H2O)6]3+.

Answers: 1. it accepts lone pairs from six water ligands

13 Complex ions (introduction)

Ligands donate electron pairs.

Transition metal ions form complex ions by accepting lone pairs from ligands (Lewis bases) through coordination bonds: [Cu(H2O)6]2+, [Cu(NH3)4(H2O)2]2+. The number of coordination bonds is the coordination number.

WORKED EXAMPLE 4 Coordination number (standard)

State the coordination number and charge of the metal in [Fe(CN)6]3−.

Solution

Coordination number 6A1

6(−1) + x = −3: Fe +3A1

In the real world

Haemoglobin contains Fe2+ complexed by a porphyrin ring; oxygen binds as a ligand, while carbon monoxide binds more strongly, which makes CO toxic.

Quick check

1. Find the oxidation state of Co in [Co(NH3)6]3+.

Answers: 1. +3

14 Nucleophilic substitution of halogenoalkanes

Replacing the halogen.

The C–X bond is polar (Cδ+). A nucleophile attacks the δ+ carbon and the halide leaves. With aqueous NaOH, halogenoalkanes form alcohols: CH3CH2Br + OH− → CH3CH2OH + Br−.

WORKED EXAMPLE 5 Equation (standard)

Write the equation for 1-chloropropane with warm aqueous sodium hydroxide, and name the product.

Solution

CH3CH2CH2Cl + OH− → CH3CH2CH2OH + Cl−A1

propan-1-olA1

Quick check

1. Name the product of 2-bromobutane with aqueous NaOH.

2. What type of reaction is this?

Answers: 1. butan-2-ol 2. nucleophilic substitution

15 Rates of hydrolysis of halogenoalkanes

Bond strength matters.

Rates of hydrolysis: iodoalkanes > bromoalkanes > chloroalkanes. Although C–Cl is the most polar, the C–I bond is the weakest, and bond strength dominates. Rates can be compared by adding aqueous silver nitrate in ethanol and timing the appearance of the precipitate.

Bond	Bond enthalpy / kJ mol−1	Precipitate with Ag+
C–Cl	324	white, slowest
C–Br	285	cream
C–I	228	yellow, fastest

WORKED EXAMPLE 6 Explaining (exam level)

Explain why 1-iodobutane is hydrolysed faster than 1-chlorobutane.

Solution

C–I bond is weaker than C–ClR1

Less energy needed to break it: lower activation energyR1

16 Electrophilic addition to alkenes

Opening the double bond.

The π electrons of C=C attack an electrophile. Alkenes add hydrogen halides (forming halogenoalkanes), halogens (dihalogenoalkanes), water (alcohols, with acid catalyst) and hydrogen (alkanes, Ni catalyst).

WORKED EXAMPLE 7 Products (standard)

Give the products of propene with (a) Br2; (b) HBr (major); (c) H2O/H+.

Solution

(a) 1,2-dibromopropaneA1

(b) 2-bromopropaneA1

Quick check

1. Write the equation for ethene with hydrogen chloride.

2. Name the product of but-2-ene with bromine.

Answers: 1. CH2=CH2 + HCl → CH3CH2Cl 2. 2,3-dibromobutane

17 Testing for alkenes

Bromine water.

Orange bromine water is decolourized rapidly by alkenes (addition) at room temperature, even in the dark. Alkanes do not react without UV light. This is the standard test for unsaturation.

WORKED EXAMPLE 8 Product (standard)

Ethene is bubbled through bromine water. Name the main product.

Solution

2-bromoethanol (from water attacking the intermediate), with some 1,2-dibromoethaneA1

18 Curly arrows

Showing electron-pair movement.

In mechanisms, a curly arrow shows the movement of an electron pair: it starts at a lone pair or a bond and ends where the new bond forms (or at the atom that receives the pair). Partial charges and lone pairs should be shown.

WORKED EXAMPLE 9 Arrow rules (standard)

Describe the curly arrows in the reaction of OH− with CH3Br.

Solution

Arrow from a lone pair on O of OH− to the δ+ carbonA1

Arrow from the C–Br bond to Br, forming Br−A1

Common mistake

Drawing arrows from the positive centre to the nucleophile. Arrows always start from the electrons (nucleophile or bond).

Quick check

1. Where does the first curly arrow start in electrophilic addition of HBr to ethene?

Answers: 1. at the C=C double bond, pointing to H of HBr

19 SN1 and SN2 mechanisms (HL)

One step or two.

SN2 (primary halogenoalkanes): one step; the nucleophile attacks as the halide leaves, through a transition state with five groups around carbon; rate = k[RX][Nu]; inversion of configuration. SN1 (tertiary): two steps; the C–X bond breaks to form a carbocation (slow), then the nucleophile attacks; rate = k[RX]; racemic mixture from a chiral centre.

WORKED EXAMPLE 10 Choosing (exam level)

Predict the mechanism for 2-bromo-2-methylpropane with OH− and explain.

Solution

SN1: tertiaryA1

stable tertiary carbocation; steric hindrance prevents backside attackR1 R1

20 Markovnikov's rule and carbocation stability (HL)

Major and minor products.

When HX adds to an unsymmetrical alkene, the H attaches to the carbon with more hydrogens, because this gives the more stable (more substituted) carbocation intermediate. Tertiary > secondary > primary carbocation stability, due to the electron-donating (positive inductive) effect of alkyl groups.

WORKED EXAMPLE 11 Major product (exam level)

Explain why HBr with propene gives mainly 2-bromopropane.

Solution

Secondary carbocation (CH3C+HCH3) more stable than primaryR1

Br− attacks the secondary carbocationR1

21 Electrophilic substitution of benzene (HL)

Keeping the ring.

Benzene reacts with electrophiles by substitution, not addition, to keep the stable delocalized ring. Nitration uses concentrated HNO3 and H2SO4, which generate the nitronium ion NO2+: HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−.

WORKED EXAMPLE 12 Nitration (exam level)

Explain why benzene undergoes substitution rather than addition.

Solution

Addition would destroy the delocalized π systemR1

Substitution restores aromatic stabilityR1

22 Comparing reaction types

Addition, substitution, elimination.

Addition: two reactants combine into one product (alkene + Br2). Substitution: one atom or group replaces another (halogenoalkane + OH−). Elimination: a small molecule is removed to form a double bond (dehydration of alcohols). Identify the reaction type from what happens to the functional group.

Reaction	Type	Attacking species
ethene + HBr	electrophilic addition	electrophile (Hδ+)
bromoethane + OH−	nucleophilic substitution	nucleophile
methane + Cl2 (UV)	radical substitution	radical
ethanol → ethene (H2SO4)	elimination	acid catalyst

WORKED EXAMPLE 13 Classifying (standard)

Classify the reaction CH3CH2Br + CN− → CH3CH2CN + Br−.

Solution

Nucleophilic substitution: CN− replaces BrA1

Quick check

1. Classify the hydrogenation of ethene.

2. Classify the reaction of benzene with bromine (FeBr3 catalyst).

Answers: 1. (electrophilic) addition 2. electrophilic substitution

23 Practical: comparing halogenoalkane hydrolysis rates

A silver nitrate experiment.

Add equal volumes of 1-chlorobutane, 1-bromobutane and 1-iodobutane to separate tubes containing ethanol and aqueous silver nitrate at the same temperature (water bath). Time the appearance of each precipitate. Water acts as the nucleophile, releasing halide ions that precipitate with Ag+.

WORKED EXAMPLE 14 Results (standard)

Predict the order in which precipitates appear and their colours.

Solution

Iodo first (yellow), then bromo (cream), then chloro (white)A1

Rate follows C–X bond strengthR1

Common mistake

Thinking the most polar bond (C–Cl) reacts fastest. Bond enthalpy, not polarity, determines the rate here.

Quick check

1. Why is ethanol added?

2. Name two control variables.

Answers: 1. halogenoalkanes and aqueous silver nitrate mix in ethanol 2. temperature, volumes and concentrations

24 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 15 Mixed problem (exam level)

Starting from ethene, outline a two-step synthesis of ethanol via a halogenoalkane, naming the reagents and reaction types.

Solution

Step 1: ethene + HBr → bromoethane (electrophilic addition)A1 A1

Step 2: bromoethane + aqueous NaOH, warm → ethanol (nucleophilic substitution)A1 A1

Quick check

1. Suggest a one-step route from ethene to ethanol.

Answers: 1. steam with phosphoric acid catalyst (electrophilic addition of water)

25 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 16 Paper 1 question (standard)

Which species is an electrophile? A. Cl− B. NH3 C. NO2+ D. H2O

Solution

CA1

WORKED EXAMPLE 17 Paper 2 question (exam level)

(a) Define a Lewis base. (b) Identify the Lewis acid in AlCl3 + Cl− → AlCl4−. (c) State the shape of AlCl4−.

Solution

(a) electron-pair donorA1

(b) AlCl3A1

26 Organic synthesis routes

Linking reactions.

Functional group interconversions build synthesis routes: alkene → halogenoalkane (HX) → alcohol (OH−) → aldehyde/ketone/carboxylic acid (oxidation) → ester (with alcohol). Recognizing the reaction type (addition, substitution, oxidation) helps plan multi-step syntheses.

WORKED EXAMPLE 18 Planning (standard)

Suggest reagents to convert propene into propan-2-ol.

Solution

HBr, then warm aqueous NaOH (or steam/H3PO4)A1

27 Electron-pair sharing in biology

Enzymes and metal ions.

Many enzyme mechanisms involve nucleophiles (such as the OH of serine) attacking electrophilic carbonyl carbons, and metal ions acting as Lewis acids. Cisplatin, an anticancer drug, works when its platinum centre forms coordination bonds to nitrogen atoms in DNA.

In the real world

Cisplatin, [Pt(NH3)2Cl2], loses its chloride ligands inside cells; the platinum then binds to DNA bases, preventing cancer cells from dividing.

Quick check

1. Identify the ligands in cisplatin.

Answers: 1. two NH3 and two Cl−

28 Common misconceptions and links

Students often think…	Actually…
curly arrows show atoms moving	they show electron pairs moving
nucleophiles attack electron-rich centres	they attack electron-deficient centres
the most polar C–X bond reacts fastest	the weakest (C–I) reacts fastest
SN1 is one step	SN1 has two steps via a carbocation

How R3.4 connects to the rest of the course

Topic	Connection
R3.3 Electron sharing	homolytic fission and radicals
S3.2 Functional groups	halogenoalkanes, alkenes, alcohols
R3.1 Proton transfer	Brønsted-Lowry acids
S2.4 Materials	addition polymers
R2.2 How fast? (HL)	rate equations and mechanisms

29 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Text A. Bromoethane is warmed under reflux with aqueous sodium hydroxide.

(a) Identify the type of reaction when bromoethane reacts with aqueous NaOH. [1]

(b) Describe, with reference to curly arrows, how the reaction occurs. [3]

(d) Describe a test to distinguish hexane from hex-1-ene. [2]

(e) Identify the Lewis acid and base in BF3 + F− → BF4−. [2]

(f) Explain why 2-bromo-2-methylpropane reacts by SN1. [2]

Markscheme

Part	Answer	Marks
(a)	nucleophilic substitution	A1
(b)	arrow from lone pair on OH− to Cδ+	R1
	arrow from C–Br bond to Br	R1
	Br− leaves; ethanol forms	R1
(c)	CH2=CHCH3 + Br2 → CH2BrCHBrCH3	A1
	1,2-dibromopropane	A1
(d)	bromine water	A1
	hex-1-ene decolourizes; hexane no change	A1
(e)	BF3 Lewis acid	A1
	F− Lewis base	A1
(f)	tertiary carbocation stabilized by three alkyl groups	R1
	steric hindrance prevents backside attack	R1

Examiner's comments. In (b), arrows starting from the carbon or the hydroxide's negative sign rather than its lone pair were not credited.

30 Check your understanding

Section A: multiple choice

1. A nucleophile

A. donates an electron pair B. accepts an electron pair C. is always positive D. has an unpaired electron

2. Heterolytic fission produces

A. radicals B. atoms C. ions D. molecules only

3. Which is an electrophile?

A. OH− B. NO2+ C. NH3 D. CN−

4. Bromine water with an alkene

A. turns orange B. forms a precipitate C. turns blue D. decolourizes

5. A Lewis acid is

A. a proton donor B. an electron-pair acceptor C. an electron-pair donor D. a hydroxide

6. Which halogenoalkane reacts fastest with OH−?

A. iodoethane B. fluoroethane C. chloroethane D. bromoethane

7. SN2 is favoured by

A. tertiary halogenoalkanes B. carbocations C. radicals D. primary halogenoalkanes

8. Propene + HBr gives mainly

A. 1-bromopropane B. propane C. 2-bromopropane D. 1,2-dibromopropane

Section B: structured

9. Distinguish nucleophiles and electrophiles with examples. [4]

10. Describe the mechanism of addition of Br2 to ethene. [4]

11. Explain why the Lewis theory is broader than Brønsted-Lowry. [2]

12. Compare SN1 and SN2 mechanisms. [4]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Nucleophile: lone pair donor (OH−); electrophile: pair acceptor (H+, Brδ+).

10. Br2 polarized; π electrons attack Brδ+; carbocation; Br− attacks; 1,2-dibromoethane.

11. Includes acids without protons (BF3, metal ions); all Brønsted bases are Lewis bases.

12. SN1: two steps, carbocation, first order, tertiary; SN2: one step, transition state, second order, primary.

31 Summary and topic map

Idea	Remember
Heterolytic fission	both electrons to one atom; ions
Curly arrows	from lone pair or bond to atom
Nucleophiles / electrophiles	pair donors / pair acceptors
Lewis acids and bases	acceptors / donors; coordination bonds
Nucleophilic substitution	halogenoalkane + OH− → alcohol; C–I fastest
Electrophilic addition	alkenes + Br2, HBr, H2O, H2
Test	bromine water decolourizes
HL	SN1/SN2; Markovnikov; nitration of benzene; complex ions

Before the exam, make sure you…

•draw curly arrows from electron pairs
•classify reagents as nucleophiles or electrophiles
•state reaction conditions
•name products systematically

Electron-Pair Sharing Reactions on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 5. Topic map for Electron-Pair Sharing Reactions.

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