DP Chemistry HL R3.3 Electron Sharing Reactions

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DP Chemistry HL | R3.3 Electron Sharing Reactions

DP CHEMISTRY · REACTIVITY

R3.3

Electron Sharing Reactions

Full notes for HL · IB-style

Contents

1	Radicals and homolytic fission
2	Alkanes: structure and low reactivity
3	Free-radical substitution of alkanes
4	Products and limitations of radical substitution
5	Radicals in the atmosphere: ozone depletion
6	Radicals in everyday life
7	Radicals: definition and representation
8	Homolytic and heterolytic fission
9	Energy for homolytic fission
10	Radical substitution of alkanes
11	Multiple substitution
12	Isomeric products
13	Ozone and the ozone layer
14	CFCs and ozone depletion
15	Radicals in combustion and the atmosphere
16	Radicals in biology and materials
17	Electron sharing in reactions: summary
18	Mixed review: worked problems
19	Additional exam-style practice
20	Lewis structures of radicals
21	Practical: observing radical reactions
22	Radicals and energy: bond enthalpy data
23	Common misconceptions and links
24	Exam-style question with markscheme
25	Check your understanding
26	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
define radical and explain homolytic fission
explain why alkanes are relatively unreactive
describe the mechanism of free-radical substitution: initiation, propagation and termination
explain why radical substitution gives a mixture of products
explain how CFCs and radicals deplete stratospheric ozone
outline the roles of radicals in combustion, polymerization and biology

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How do radicals form and react? Why do radical reactions often produce mixtures of products?

How this topic is examined

R3.3 is examined through the mechanism of free-radical substitution (writing each step with equations), identification of steps, and ozone depletion.

Show radicals with a dot (Cl•), balance each step, and state the conditions (UV light) for initiation.

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 Radicals and homolytic fission

Breaking a bond so that each atom keeps one electron.

Think about it

Methane and chlorine can be stored together in the dark for years, but in sunlight they react explosively. What does light do?

A covalent bond can break in two ways. In heterolytic fission, one atom takes both electrons, forming ions (R3.4). In homolytic fission, the bond breaks evenly: each atom takes one electron, forming two radicals.

A radical is a species with an unpaired electron, shown by a dot: Cl•, CH3•. Radicals are very reactive, because they tend to pair their unpaired electron by forming a new bond.

Homolytic fission requires energy, often supplied by ultraviolet (UV) light or high temperature. The Cl–Cl bond (242 kJ mol−1) is relatively weak and is broken by UV light: Cl2 → 2Cl•.

Quick check

1. Define radical.

2. What is homolytic fission?

Answers: 1. a species with an unpaired electron 2. bond breaks so each atom keeps one electron

2 Alkanes: structure and low reactivity

Strong, non-polar bonds.

Alkanes are saturated hydrocarbons with the general formula CnH2n+2. They are relatively unreactive because their C–C and C–H bonds are strong (346 and 414 kJ mol−1) and non-polar (C and H have similar electronegativities), so they are not attacked by nucleophiles or electrophiles.

Alkanes undergo combustion (R1.3) and free-radical substitution with halogens in UV light.

3 Free-radical substitution of alkanes

A chain reaction in three stages.

In free-radical substitution, a hydrogen atom in an alkane is replaced by a halogen atom. Example: methane and chlorine in UV light: CH4 + Cl2 → CH3Cl + HCl. The mechanism has three stages.

Initiation: UV light causes homolytic fission of chlorine, producing radicals: Cl2 → 2Cl•.

Propagation: a radical reacts with a molecule to form a product and another radical, so the reaction continues as a chain reaction: CH4 + Cl• → CH3• + HCl, then CH3• + Cl2 → CH3Cl + Cl•. The Cl• regenerated in the second step can react with another methane molecule; one initiation can lead to thousands of propagation cycles.

Termination: two radicals combine, removing radicals and ending the chain: Cl• + Cl• → Cl2; CH3• + Cl• → CH3Cl; CH3• + CH3• → C2H6.

Figure 1. The free-radical substitution of methane.

Stage	Feature	Example
Initiation	radicals formed from a molecule	Cl2 → 2Cl•
Propagation	radical + molecule → product + radical	CH3• + Cl2 → CH3Cl + Cl•
Termination	two radicals → molecule	CH3• + CH3• → C2H6

Common mistake

Writing H• in propagation (CH4 + Cl• → CH3Cl + H•). The chlorine radical removes a hydrogen atom to form HCl and a methyl radical.

Quick check

1. What condition is needed for initiation?

2. Why is it called a chain reaction?

Answers: 1. UV light 2. propagation steps regenerate radicals

4 Products and limitations of radical substitution

Hard to control, so mixtures form.

Radical substitution produces a mixture of products. Further substitution: chloromethane can react further: CH3Cl + Cl• → CH2Cl• + HCl, eventually giving CH2Cl2, CHCl3 and CCl4. Termination produces other products, such as ethane. With longer alkanes, substitution can occur at different positions, giving isomers: propane gives 1-chloropropane and 2-chloropropane.

Using an excess of methane favours monosubstitution (CH3Cl); an excess of chlorine favours further substitution (CCl4). Because of the mixtures, radical substitution is of limited use for making specific products in high yield; products must be separated by fractional distillation.

WORKED EXAMPLE 1 Writing propagation steps

Write the propagation steps for the formation of bromoethane from ethane and bromine.

Solution

C2H6 + Br• → C2H5• + HBrA1

C2H5• + Br2 → C2H5Br + Br•A1

Quick check

1. Why does radical substitution give a mixture?

2. How can monosubstitution be favoured?

Answers: 1. further substitution, termination products, different positions 2. use excess alkane

5 Radicals in the atmosphere: ozone depletion

Chlorine radicals destroy ozone in a chain reaction.

Ozone (O3) in the stratosphere absorbs harmful UV radiation, protecting living things from skin cancer, cataracts and crop damage. Ozone is formed and broken down naturally: O2 → 2O• (UV); O• + O2 → O3; O3 → O2 + O• (UV).

Chlorofluorocarbons (CFCs), once used in refrigerants and aerosols, are very stable and reach the stratosphere. There, UV light breaks C–Cl bonds homolytically, releasing chlorine radicals: CCl2F2 → CClF2• + Cl•. Chlorine radicals then catalyse ozone breakdown in a chain reaction: Cl• + O3 → ClO• + O2; ClO• + O• → Cl• + O2. Overall: O3 + O• → 2O2. The Cl• is regenerated, so one chlorine radical can destroy many thousands of ozone molecules.

The Montreal Protocol (1987) phased out CFCs; replacements such as hydrofluorocarbons (HFCs) contain no chlorine. The ozone layer is slowly recovering. Nitrogen oxides from aircraft also form radicals (NO•) that destroy ozone.

Figure 2. Antarctic ozone fell sharply and is now slowly recovering after CFCs were banned.

Quick check

1. How do CFCs produce chlorine radicals?

2. Why does one Cl• destroy many ozone molecules?

Answers: 1. UV causes homolytic fission of C–Cl 2. it is regenerated in the chain reaction

6 Radicals in everyday life

Radicals are everywhere.

Radicals are involved in combustion (flames contain many radical species), in the polymerization of alkenes to make poly(ethene), and in biology: reactive oxygen species formed in respiration can damage DNA, lipids and proteins, contributing to ageing and disease. Antioxidants such as vitamins C and E react with radicals to stop chain reactions. Food rancidity also involves radical oxidation of fats.

7 Radicals: definition and representation

Unpaired electrons.

A radical is a species with an unpaired electron, shown by a dot: Cl•, •CH3, HO•. Radicals are very reactive because they tend to pair their electron by forming a bond.

WORKED EXAMPLE 2 Identifying (basic)

Which are radicals: Cl−, Cl•, •NO, CH4?

Solution

Cl• and •NO (unpaired electrons)A1

Quick check

1. Why is NO (15 electrons) a radical?

2. Draw the methyl radical.

Answers: 1. odd number of electrons: one must be unpaired 2. •CH3 (carbon with three H and an unpaired electron)

8 Homolytic and heterolytic fission

Splitting bonds.

In homolytic fission, a covalent bond breaks so that each atom keeps one electron, forming two radicals (shown with single-headed 'fish-hook' arrows). In heterolytic fission, one atom takes both electrons, forming ions (double-headed arrows).

WORKED EXAMPLE 3 Fission (standard)

Write equations for homolytic and heterolytic fission of Br–Br.

Solution

Homolytic: Br2 → 2Br•A1

Heterolytic: Br2 → Br+ + Br−A1

Quick check

1. What conditions favour homolytic fission?

Answers: 1. UV light or high temperature; non-polar bonds; gas phase

9 Energy for homolytic fission

UV light and bond enthalpy.

UV light can break bonds if the photon energy exceeds the bond energy per molecule. E per bond = bond enthalpy/L. The wavelength needed is λ = hc/E.

WORKED EXAMPLE 4 Wavelength (exam level)

The Cl–Cl bond enthalpy is 242 kJ mol−1. Calculate the maximum wavelength that can break it.

Solution

E = 242 000/6.02 × 1023 = 4.02 × 10−19 JM1 A1

λ = hc/E = 495 nmA1

Quick check

1. Find the maximum wavelength to break the O=O bond (498 kJ mol−1).

Answers: 1. 240 nm

10 Radical substitution of alkanes

Initiation, propagation, termination.

Methane reacts with chlorine in UV light. Initiation: Cl2 → 2Cl•. Propagation: Cl• + CH4 → HCl + •CH3; •CH3 + Cl2 → CH3Cl + Cl•. Termination: two radicals combine, such as Cl• + •CH3 → CH3Cl or 2•CH3 → C2H6.

WORKED EXAMPLE 5 Mechanism (exam level)

Write the mechanism for the reaction of ethane with bromine in UV light.

Solution

Initiation: Br2 → 2Br•A1

Propagation: Br• + C2H6 → HBr + •C2H5; •C2H5 + Br2 → C2H5Br + Br•A2

Termination: •C2H5 + Br• → C2H5BrA1

Common mistake

Writing H• as a propagation product (Cl• + CH4 → CH3Cl + H•). The H atom is removed by Cl• to form HCl.

Quick check

1. Explain why butane is formed in small amounts in the chlorination of ethane.

2. Why is it called a chain reaction?

Answers: 1. termination: 2•C2H5 → C4H10 2. propagation steps regenerate the radical

11 Multiple substitution

A mixture of products.

With excess chlorine, further substitution gives CH2Cl2, CHCl3 and CCl4. Using excess alkane favours monosubstitution. The product mixture must be separated, so radical substitution has a low atom economy for any single product.

WORKED EXAMPLE 6 Further substitution (standard)

Write the propagation steps for forming CH2Cl2 from CH3Cl.

Solution

Cl• + CH3Cl → HCl + •CH2ClA1

•CH2Cl + Cl2 → CH2Cl2 + Cl•A1

Quick check

1. How can the yield of chloromethane be maximized?

Answers: 1. use excess methane

12 Isomeric products

Different hydrogens.

In alkanes with different types of hydrogen, substitution gives isomers. Propane gives 1-chloropropane and 2-chloropropane.

WORKED EXAMPLE 7 Isomers (standard)

Name the monochloro products of 2-methylpropane.

Solution

1-chloro-2-methylpropaneA1

2-chloro-2-methylpropaneA1

13 Ozone and the ozone layer

Protecting life from UV.

Ozone (O3) in the stratosphere absorbs harmful UV radiation. It forms and decomposes naturally in a dynamic balance: O2 → 2O• (UV); O• + O2 → O3; O3 → O2 + O• (UV).

WORKED EXAMPLE 8 Bond strength (standard)

Explain why ozone absorbs UV of longer wavelength than O2.

Solution

O3 bonds (bond order 1.5) are weaker than O=OR1

Less energy needed: longer wavelengthR1

Quick check

1. Why is stratospheric ozone beneficial but ground-level ozone harmful?

Answers: 1. stratosphere: absorbs UV; ground level: toxic and forms smog

14 CFCs and ozone depletion

Catalytic destruction.

CFCs are broken down by UV in the stratosphere: CCl2F2 → •CClF2 + Cl•. Chlorine radicals catalyse ozone breakdown: Cl• + O3 → ClO• + O2; ClO• + O• → Cl• + O2. One Cl• can destroy thousands of ozone molecules.

WORKED EXAMPLE 9 Catalytic cycle (exam level)

Explain why Cl• is described as a catalyst in ozone depletion.

Solution

It is consumed in the first stepR1

and regenerated in the second, so not used up overallR1

In the real world

The Montreal Protocol (1987) phased out CFCs; the ozone layer is now recovering and is expected to return to 1980 levels by around 2066 over Antarctica.

Quick check

1. Why are C–F bonds not broken in the stratosphere?

Answers: 1. C–F bonds are stronger than C–Cl bonds

15 Radicals in combustion and the atmosphere

Fast chain reactions.

Combustion proceeds through radical chain reactions, which is why it is so fast once initiated. Nitrogen oxides from engines produce radicals that form photochemical smog. Hydroxyl radicals (HO•) clean the atmosphere by oxidizing pollutants.

WORKED EXAMPLE 10 Smog (standard)

Explain why photochemical smog forms on sunny days.

Solution

UV light breaks NO2 into NO and O•R1

O• reacts with O2 to form ground-level ozone; radicals react with hydrocarbonsR1

16 Radicals in biology and materials

Antioxidants and polymers.

Radicals in the body can damage DNA and cell membranes; antioxidants (vitamins C and E) react with radicals to stop chain reactions. Radical initiators are used in the addition polymerization of ethene to poly(ethene).

WORKED EXAMPLE 11 Antioxidants (standard)

Explain how an antioxidant protects cells.

Solution

Reacts with radicals to form stable speciesR1

Terminates radical chain reactionsR1

17 Electron sharing in reactions: summary

Radical versus ionic mechanisms.

Feature	Radical reactions	Ionic (polar) reactions
bond breaking	homolytic	heterolytic
conditions	UV light, high T, gas phase	polar solvents
intermediates	radicals	carbocations, carbanions
example	alkane halogenation	nucleophilic substitution

WORKED EXAMPLE 12 Classifying (standard)

Classify the reaction of bromoethane with hydroxide ions.

Solution

Ionic: heterolytic; nucleophilic substitutionA1

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 13 Mixed problem (exam level)

Methane reacts with excess bromine in sunlight. (a) State the type of reaction. (b) Write the initiation step and two propagation steps leading to CH2Br2. (c) Write a termination step forming ethane.

Solution

(a) radical substitutionA1

(b) Br2 → 2Br•; Br• + CH3Br → HBr + •CH2Br; •CH2Br + Br2 → CH2Br2 + Br•A2

Quick check

1. Explain why the reaction does not occur in the dark at room temperature.

Answers: 1. energy (UV) needed for homolytic fission of Br2

19 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 14 Paper 1 question (standard)

Which step is a propagation step? A. Cl2 → 2Cl• B. Cl• + CH4 → HCl + •CH3 C. 2Cl• → Cl2 D. •CH3 + Cl• → CH3Cl

Solution

BA1

WORKED EXAMPLE 15 Paper 2 question (exam level)

(a) Explain why CFCs were widely used. (b) Write the equations showing how Cl• destroys ozone. (c) Suggest why HFCs are used as replacements.

Solution

(a) unreactive, non-toxic, volatileR1

(b) Cl• + O3 → ClO• + O2; ClO• + O• → Cl• + O2A2

20 Lewis structures of radicals

Odd-electron species.

Radicals with an odd number of electrons cannot give every atom an octet. In NO2, nitrogen has seven electrons around it. Radicals often dimerize to pair electrons: 2NO2 ⇌ N2O4.

WORKED EXAMPLE 16 Dimerization (standard)

Explain why NO2 dimerizes on cooling.

Solution

NO2 has an unpaired electronR1

Forming an N–N bond pairs the electrons; exothermic, favoured at low TR1

21 Practical: observing radical reactions

Bromine decolourization in light.

Hexane mixed with bromine stays orange in the dark but decolourizes slowly in sunlight or UV light, with HBr fumes formed (turning damp blue litmus red). This contrasts with alkenes, which decolourize bromine rapidly in the dark (electrophilic addition).

WORKED EXAMPLE 17 Interpreting (standard)

Explain the different behaviour of hexane and hex-1-ene with bromine in the dark.

Solution

Hexane: radical substitution needs UVR1

Hex-1-ene: addition across C=C occurs without lightR1

Try this

Compare two test tubes of hexane with a few drops of bromine solution, one wrapped in foil and one in bright light (in a fume cupboard, with your teacher). Which decolourizes?

22 Radicals and energy: bond enthalpy data

Which bonds break?

Initiation breaks the weakest bond. In chlorination of methane, Cl–Cl (242 kJ mol−1) breaks rather than C–H (414 kJ mol−1). Propagation steps are often exothermic or nearly thermoneutral, which keeps the chain going.

WORKED EXAMPLE 18 Enthalpy of propagation (exam level)

Calculate ΔH for Cl• + CH4 → HCl + •CH3 (C–H 414, H–Cl 432 kJ mol−1).

Solution

414 − 432M1

= −18 kJ mol−1A1

23 Common misconceptions and links

Students often think…	Actually…
propagation produces H•	Cl• removes H to form HCl
termination makes radicals	termination removes radicals
heterolytic and homolytic fission are the same	homolytic gives radicals; heterolytic gives ions

How R3.3 connects to the rest of the course

Topic	Connection
S3.2 Functional groups	alkanes and halogenoalkanes
R3.4 Electron-pair sharing	heterolytic fission
R1.2 Energy cycles	bond enthalpies
R1.3 Energy from fuels	combustion

24 Exam-style question with markscheme

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

Text A. Ethane reacts with chlorine in sunlight to form chloroethane and other products.

(a) State the type of reaction and the condition required. [2]

(b) Write equations for the initiation and two propagation steps. [3]

(d) Explain why the reaction produces a mixture of products. [2]

(e) Explain how chlorine radicals deplete ozone. [3]

Markscheme

Part	Answer	Marks
(a)	free-radical substitution	A1
	UV light	A1
(b)	Cl2 → 2Cl•	A1
	C2H6 + Cl• → C2H5• + HCl	A1
	C2H5• + Cl2 → C2H5Cl + Cl•	A1
(c)	2C2H5• → C4H10	A1
(d)	further substitution of H atoms	R1
	termination products	R1
(e)	Cl• + O3 → ClO• + O2	A1
	ClO• + O• → Cl• + O2	A1
	Cl• regenerated: catalytic chain	R1

Examiner's comments. In (b), radicals must be shown with dots and equations balanced.

25 Check your understanding

Section A: multiple choice

1. A radical has

A. an unpaired electron B. a positive charge C. a lone pair only D. no electrons

2. Homolytic fission produces

A. two ions B. a cation only C. two radicals D. a molecule

3. Initiation in chlorination requires

A. a catalyst B. UV light C. darkness D. high pressure only

4. Which is a termination step?

A. Cl2 → 2Cl• B. CH4 + Cl• → CH3• + HCl C. CH3• + Cl2 → CH3Cl + Cl• D. CH3• + Cl• → CH3Cl

5. Alkanes are unreactive because bonds are

A. weak and polar B. strong and non-polar C. ionic D. double

6. CFCs were replaced because they

A. release chlorine radicals in the stratosphere B. are toxic to breathe C. are flammable D. cause acid rain

Section B: structured

7. Describe the mechanism of chlorination of methane. [5]

8. Explain why alkanes are relatively unreactive. [2]

9. Explain the role of the Montreal Protocol. [2]

Answers

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

7. Initiation Cl2 → 2Cl• (UV); propagation two steps; termination radical combinations.

8. Strong C–C and C–H bonds; non-polar, no attraction for nucleophiles or electrophiles.

9. Banned CFCs; reduced Cl• in stratosphere; ozone recovering.

26 Summary and topic map

Idea	Remember
Radical	unpaired electron
Homolytic fission	UV light; Cl2 → 2Cl•
Alkanes	strong non-polar bonds; unreactive
Mechanism	initiation, propagation, termination
Products	mixture; further substitution
Ozone	Cl• catalyses O3 destruction; Montreal Protocol

Before the exam, make sure you…

•show radicals with dots
•write balanced propagation steps
•explain chain reactions
•link CFCs to ozone depletion

Electron 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 3. Topic map for Electron Sharing Reactions.

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