IB DP Chemistry Reactivity 3.3: Electron sharing reactions – Revision Notes

These notes compress Reactivity 3.3, Electron sharing reactions into quick-recall form, checked against the Chemistry guide for the IB Diploma Programme (February 2023 edition, first assessment 2025). Nothing in the sub-topic is left out: they span Reactivity 3.3.1 to 3.3.3, all of which are for SL and HL; Reactivity 3.3 has no additional higher level understandings. The study guide has the full explanations and worked examples.

Next, test your recall on the practice questions. Plan the rest of your revision from the printable checklist or the course hub, and take our free diagnostics to find weak spots. The Reactivity 3 mechanisms overview shows how this sub-topic compares with the other three ways reactions happen.

Key definitions

Term	Meaning
Radical	A molecular entity with an unpaired electron, shown by a dot: Cl•, CH₃•
Homolytic fission	A covalent bond splits evenly, one bonding electron going to each atom, giving two radicals
Single-barbed arrow (fish hook)	Half-headed curly arrow showing the movement of one electron
Initiation	The step that creates radicals, here homolytic fission of the halogen
Propagation	A radical reacts with a molecule and produces another radical
Termination	Two radicals combine; radicals are removed and the chain stops
Chain reaction	Radicals made in one propagation step feed the next, so the cycle repeats
Substitution	One atom (here H) in a molecule is replaced by another atom (here a halogen)

Radicals (3.3.1)

• The defining feature is one unpaired electron. Charge is irrelevant.
• Radicals are highly reactive: they pair the electron by taking an atom from a molecule or by joining another radical.
• Put the dot on the atom that carries the electron: HO•, not OH• written as if hydrogen had it.

Method in steps: is it a radical?

1. Add the valence electrons of every atom.
2. Subtract one for each positive charge; add one for each negative charge.
3. Odd total: radical. Even total: usually not a radical (O₂ is an exception, with two unpaired electrons).

Small reminder: NH₂ has 5 + 2 = 7 valence electrons, so it is the amino radical, NH₂•. Add a negative charge and NH₂⁻ has 8, so it is not a radical.

Four kinds of radical

Kind	Example
Atom	Br•
Molecule	NO•, nitrogen monoxide
Cation	a molecular ion such as CH₄•⁺ in a mass spectrometer
Anion	O₂•⁻, superoxide

Homolytic fission and initiation (3.3.2)

• General equation: X₂ → 2X•, with UV light or heat as the energy source.
• Two fish-hook arrows, each from the centre of the X–X bond to one X atom.
• This is the initiation step of the chain reaction.
• The reverse process is two radicals forming a bond by each giving one electron. It releases energy.

Must-know distinctions

Homolytic fission (3.3)	Heterolytic fission (3.4)
Each atom gets one electron	One atom gets both electrons
Makes two radicals	Makes a cation and an anion
Two single-barbed arrows	One double-barbed arrow
Typical trigger: UV light or heat	Typical case: a polar bond, such as C–Br, attacked by a nucleophile

Linking points worth one sentence each

• CFCs: UV breaks C–Cl to release Cl•; the C–F bond is much stronger, so F• is not typically released.
• Ozone: the O–O bonds in O₃ (bond order about 1.5) are weaker than O=O in O₂, so Cl• can break down O₃ but not O₂.

Alkane substitution (3.3.3)

Why alkanes need radicals

• C–C and C–H bonds are strong, so activation energies are high.
• C–H bonds are essentially non-polar (similar electronegativities of C and H), so there is no δ+ or δ− site for nucleophiles or electrophiles.
• Alkanes are kinetically stable but thermodynamically unstable: combustion is very exothermic, but it needs an input of energy to start.

Method in steps: writing the mechanism for alkane RH with halogen X₂

1. Initiation: X₂ → 2X• (UV light).
2. Propagation 1: X• + RH → R• + HX. The halogen radical removes a hydrogen atom.
3. Propagation 2: R• + X₂ → RX + X•. The new X• returns to step 2.
4. Termination: X• + X• → X₂; R• + X• → RX; R• + R• → R–R.
5. Check: one dot on each side of every propagation step; two dots on the left and none on the right of every termination step.

Small reminder for 2,2,3,3-tetramethylbutane and bromine (every hydrogen is equivalent): Br• + (CH₃)₃CC(CH₃)₃ → (CH₃)₃CC(CH₃)₂CH₂• + HBr, then (CH₃)₃CC(CH₃)₂CH₂• + Br₂ → (CH₃)₃CC(CH₃)₂CH₂Br + Br•.

Why a mixture of products forms

• Further substitution: RX still has C–H bonds, so RX₂, RX₃ and so on form.
• Position isomers: hydrogens in different environments are all replaceable.
• Termination: R• + R• gives an alkane with twice the carbon count.

Method in steps: counting monohalogenated isomers

1. Draw the full structural formula of the alkane.
2. Look for symmetry: carbons that are mirror images of each other along the chain carry equivalent hydrogens.
3. Count the different hydrogen environments. Each one gives one monohalogenated structural isomer.
4. Name each product, checking that the longest chain and the lowest locants are used.

Small reminder: heptane, CH₃(CH₂)₅CH₃, has four hydrogen environments (C1 and C7, C2 and C6, C3 and C5, C4), so it gives four monochloro isomers: 1-, 2-, 3- and 4-chloroheptane.

Alkanes and bromine in the dark

A linking question in Reactivity 3.4 asks why alkenes decolourize bromine water in the dark but alkanes do not. The answer for the alkane half belongs here: without UV light (or heat) there is no homolytic fission of Br₂, so no radicals form, and the non-polar alkane offers nothing for bromine to attack. The orange colour stays. An alkane needs UV light (or heat) before it reacts with bromine.

Energy changes

• ΔH of a step = bonds broken − bonds made.
• The two propagation steps add up to the overall equation, so their ΔH values add up to the overall ΔH.
• Every termination step is exothermic, because a bond forms and none breaks.

Small reminder: in the step Cl• + RH → R• + HCl, the only bond broken is C–H and the only bond made is H–Cl, so ΔH = E(C–H) − E(H–Cl). Do not include Cl–Cl here; that bond breaks in the other propagation step.

Quick self-test

1. Define a radical.
2. Which of CH₃O, H₃O⁺ and CCl₃ are radicals?
3. Name the process that turns Cl₂ into two Cl• and give two ways to supply the energy.
4. What does a single-barbed arrow represent?
5. Classify each step: (i) C₇H₁₅• + Cl₂ → C₇H₁₅Cl + Cl•; (ii) Cl₂ → 2Cl•; (iii) C₇H₁₅• + Cl• → C₇H₁₅Cl.
6. Name the alkane formed when two CH₃(CH₂)₅CH₂• radicals combine.
7. Give two reasons why alkanes are unreactive towards most reagents.
8. Use these bond enthalpies (kJ mol⁻¹): C–H 414, F–F 158, C–F 492, H–F 568. Find ΔH for each propagation step of the fluorination of an alkane, and the overall ΔH.
9. How many monobromo structural isomers can 2-methylpentane, (CH₃)₂CHCH₂CH₂CH₃, give? Name them.
10. Why does UV light in the stratosphere release Cl• but not F• from a CFC?
11. Why is every termination step exothermic?

Answers

1. Any molecular entity carrying an unpaired electron.
2. CH₃O: 4 + 3 + 6 = 13, radical. H₃O⁺: 3 + 6 − 1 = 8, not a radical. CCl₃: 4 + 21 = 25, radical.
3. Homolytic fission; UV light or heat.
4. The movement of a single electron.
5. (i) Propagation (one radical in, one out); (ii) initiation (radicals made); (iii) termination (two radicals in, none out).
6. Tetradecane, C₁₄H₃₀.
7. Strong C–C and C–H bonds (high activation energy); the bonds are essentially non-polar, so nothing attracts a nucleophile or electrophile.
8. Step 1: +414 − 568 = −154 kJ mol⁻¹. Step 2: +158 − 492 = −334 kJ mol⁻¹. Overall: −488 kJ mol⁻¹. Both steps are strongly exothermic, which is why fluorine reacts with alkanes so violently.
9. Five: 1-bromo-2-methylpentane, 2-bromo-2-methylpentane, 3-bromo-2-methylpentane, 2-bromo-4-methylpentane and 1-bromo-4-methylpentane. The two CH₃ groups on carbon 2 are equivalent.
10. C–F is much stronger than C–Cl, so the UV photons break C–Cl but typically not C–F.
11. Two radicals pair their electrons to make a bond and no bond is broken; bond making releases energy.

Where marks are usually lost

• Calling Cl• a chloride ion, or adding a charge to a halogen radical.
• Drawing one full curly arrow for homolytic fission instead of two fish hooks.
• Starting fish-hook arrows on the atoms instead of the centre of the bond.
• Writing a propagation step in which the halogen radical removes a hydrogen ion or makes H•.
• Leaving dots off radicals in propagation equations, so the step looks unbalanced.
• Listing initiation as a termination step, or the reverse.
• Giving only one reason for alkane stability when the question asks about bond strength and polarity.
• Saying the reaction needs “light” without stating ultraviolet.
• Forgetting R• + R• when asked to explain an unexpected longer alkane in the products.

Official syllabus

These notes summarise Reactivity 3.3 (Electron sharing reactions), understandings 3.3.1 to 3.3.3, SL and HL. Source: Chemistry guide, IB Diploma Programme, International Baccalaureate, February 2023 (first assessment 2025).