IB DP Chemistry Reactivity 3.3: Electron sharing reactions

Here you will learn Reactivity 3.3, Electron sharing reactions, checked line by line against the Chemistry guide for the IB Diploma Programme (February 2023 edition, first assessment 2025). Nothing in the sub-topic is skipped: all of Reactivity 3.3.1 to 3.3.3 is here. All three are for SL and HL, and there is no AHL extension to this sub-topic.

Afterwards, the shorter revision notes help you recall it fast, and the practice questions show whether it has stuck. Other units sit on the IB DP chemistry hub; every understanding appears on the printable checklist; our free diagnostics suggest a starting point. To compare radical reactions with proton transfer, electron transfer and electron-pair sharing, open the Reactivity 3 mechanisms overview.

What this unit covers

Every row applies to both SL and HL.

Understanding	Skills you need
Reactivity 3.3.1	Identify radicals and represent them with a dot for the unpaired electron; explain why they are highly reactive
Reactivity 3.3.2	Explain homolytic fission of a halogen by UV light or heat, with an equation and single-barbed (fish-hook) arrows; call it the initiation step of a chain reaction
Reactivity 3.3.3	Write the propagation and termination steps for an alkane reacting with a halogen; explain why a mixture of products forms; explain why alkanes are otherwise unreactive

The guiding question for the sub-topic asks what happens when a species has an unpaired electron. Keep that question in mind: every step below is either making radicals, passing the unpaired electron on, or pairing two unpaired electrons up again.

Radicals (Reactivity 3.3.1)

A radical is any molecular entity carrying an unpaired electron. You show the unpaired electron with a dot next to the atom that carries it: Cl•, CH₃•, HO•. Put the dot on the right atom. The hydroxyl radical is HO• (or •OH), because the unpaired electron sits on oxygen, not hydrogen.

Radicals are highly reactive. An unpaired electron is a high-energy arrangement, so a radical reacts readily with almost any molecule it meets: it pulls an atom away to pair its electron in a new bond, or it meets another radical and the two electrons pair to form a covalent bond. Most radicals have very short lifetimes for this reason.

Spotting a radical by counting electrons

Add up the valence electrons of every atom, subtract one for each positive charge and add one for each negative charge. If the total is odd, at least one electron must be unpaired, so the species is a radical. An even total usually means all electrons are paired. Oxygen, O₂, is the classic exception: it has an even count but two unpaired electrons.

Worked example 1. Decide which of these are radicals: OH, OH⁻, NO, CH₃⁺, C₂H₅, Br⁻.

• OH: 6 + 1 = 7, odd, so radical (HO•).
• OH⁻: 6 + 1 + 1 = 8, even, so not a radical (it is the hydroxide ion).
• NO: 5 + 6 = 11, odd, so radical. Nitrogen monoxide is a stable gas, yet it is still a radical.
• CH₃⁺: 4 + 3 − 1 = 6, even, so not a radical (a carbocation).
• C₂H₅: 2 × 4 + 5 = 13, odd, so radical (the ethyl radical, CH₃CH₂•).
• Br⁻: 7 + 1 = 8, even, so not a radical.

Atom, molecule, cation or anion

The guide asks how radicals can include atoms, molecules, cations and anions. The definition only needs one unpaired electron, so the charge does not matter:

Type	Example	Valence electrons
Atom	Cl•	7
Neutral molecule	NO₂•	17
Cation	CH₄•⁺, formed when methane loses one electron in a mass spectrometer	7
Anion	O₂•⁻, the superoxide ion	13

Notice that a radical and an ion are different ideas. Cl• is a radical but not an ion; Cl⁻ is an ion but not a radical.

Homolytic fission and initiation (Reactivity 3.3.2)

In homolytic fission a covalent bond splits evenly: the two bonding electrons are shared out, one to each atom. Two radicals form. For a halogen, X₂:

X–X → X• + X•

The energy comes from ultraviolet (UV) light or from heat. A photon of UV light carries enough energy to break one halogen–halogen bond, which is why the reaction of chlorine with an alkane starts in sunlight or under a UV lamp and does not happen in the dark at room temperature.

Single-barbed (fish-hook) arrows

A full curly arrow shows a pair of electrons moving. A single-barbed arrow, or fish hook, has half an arrowhead and shows one electron moving. To show homolytic fission of Br₂:

1. Draw the Br–Br bond.
2. Draw one fish-hook arrow starting in the middle of the bond and ending on the left bromine atom.
3. Draw a second fish-hook arrow from the middle of the bond to the right bromine atom.
4. Write the products as Br• + Br•, each with its dot.

Two arrows, one electron each, two radicals. If you draw a full double-headed arrow here, you are describing heterolytic fission (Reactivity 3.4), which makes ions, not radicals.

The initiation step

Homolytic fission of the halogen is the initiation step of a chain reaction. Initiation makes radicals from a molecule with all its electrons paired. Overall:

• initiation: radicals are created;
• propagation: a radical reacts and makes a new radical, so the chain carries on;
• termination: two radicals combine and the chain stops.

Worked example 2. Use these values for X–X bond enthalpies in kJ mol⁻¹: F–F 158, Cl–Cl 242, Br–Br 193, I–I 151. Put the halogens in order of the energy needed for homolytic fission and comment on the order.

1. Least to most energy: I–I (151) < F–F (158) < Br–Br (193) < Cl–Cl (242).
2. From chlorine to iodine the bond weakens as the atoms get larger and the bonding pair is further from each nucleus.
3. Fluorine breaks the pattern. Its very small atoms bring the lone pairs on each atom close together, and their repulsion weakens the F–F bond.

Linking questions from the guide

• Reverse of homolytic fission. Two radicals combining, each giving one electron to the new bond, is the reverse process. It is bond making, so energy is released. This is exactly what a termination step is.
• CFCs and chlorine radicals. In the stratosphere, UV breaks the C–Cl bonds in chlorofluorocarbons and releases Cl•. Using C–Cl 324 and C–F 492 kJ mol⁻¹, the C–F bond needs 168 kJ mol⁻¹ more energy, so fluorine radicals are not typically released.
• Ozone and oxygen. Chlorine radicals can break down ozone, O₃, but not O₂. The oxygen–oxygen bonds in ozone have a bond order of about 1.5 (resonance), so they are weaker than the O=O double bond in O₂.

Substitution of alkanes by halogens (Reactivity 3.3.3)

Why alkanes are hard to react

Alkanes are stable, and the guide wants two reasons:

• Strong bonds. C–C (346 kJ mol⁻¹) and C–H (414 kJ mol⁻¹) bonds need a lot of energy to break, so most reactions with alkanes have a high activation energy.
• Essentially non-polar. Carbon and hydrogen have similar electronegativities (about 2.6 and 2.2), so the C–H bond has almost no partial charges. There is no electron-rich or electron-poor site to attract a nucleophile or electrophile.

This is why the linking question calls alkanes kinetically stable but thermodynamically unstable. Burning an alkane is very exothermic, so the products are far more stable (thermodynamically favoured), yet without a spark or flame the high activation energy means nothing happens (kinetically slow). A radical is reactive enough to get round this: it does not need a polar bond to attack.

Propagation and termination

Worked example 3. Cycloheptane, C₇H₁₄, reacts with chlorine in UV light to give chlorocycloheptane, C₇H₁₃Cl. Write the full mechanism.

Initiation:

• Cl₂ → 2Cl• (UV light, homolytic fission)

Propagation (each step uses one radical and makes one radical):

• Cl• + C₇H₁₄ → C₇H₁₃• + HCl
• C₇H₁₃• + Cl₂ → C₇H₁₃Cl + Cl•

Termination (two radicals pair up):

• Cl• + Cl• → Cl₂
• C₇H₁₃• + Cl• → C₇H₁₃Cl
• C₇H₁₃• + C₇H₁₃• → C₁₄H₂₆ (two rings joined by a C–C bond)

Overall: C₇H₁₄ + Cl₂ → C₇H₁₃Cl + HCl

Check each step: in propagation, the dot count is one on each side. In termination, there are two dots on the left and none on the right. The chlorine radical made in the second propagation step goes back into the first, which is why one initiation can lead to many product molecules before the chain ends. All fourteen hydrogens in cycloheptane are equivalent, so only one monochloro product is possible here.

Energy changes in the steps

Worked example 4. Use these average bond enthalpies (kJ mol⁻¹): C–H 414, Cl–Cl 242, C–Cl 324, H–Cl 431. Calculate ΔH for each propagation step in worked example 3 and for the overall reaction.

1. Step 1 breaks C–H and makes H–Cl: ΔH = +414 − 431 = −17 kJ mol⁻¹.
2. Step 2 breaks Cl–Cl and makes C–Cl: ΔH = +242 − 324 = −82 kJ mol⁻¹.
3. Overall: −17 + (−82) = −99 kJ mol⁻¹.
4. Check with the overall equation: bonds broken 414 + 242 = 656; bonds made 324 + 431 = 755; ΔH = 656 − 755 = −99 kJ mol⁻¹. The answers agree.

Every termination step only makes a bond, so each is exothermic, for example −242 kJ mol⁻¹ for Cl• + Cl• → Cl₂.

Why a mixture of products forms

The guide says substitution produces a mixture of products. There are three sources:

1. Further substitution. The product still has C–H bonds. A Cl• can remove a hydrogen from C₇H₁₃Cl as well, giving C₇H₁₂Cl₂, then more highly substituted products.
2. Different hydrogen positions. If an alkane has hydrogens in different environments, the halogen can replace any of them, so structural isomers form.
3. Termination products. Two alkyl radicals can join, as in C₇H₁₃• + C₇H₁₃• → C₁₄H₂₆, giving a larger hydrocarbon that was never in the flask.

Worked example 5. How many monochloro structural isomers can form when 3-methylpentane, CH₃CH₂CH(CH₃)CH₂CH₃, reacts with chlorine in UV light? Name them.

1. Look for different hydrogen environments. The molecule is symmetrical about carbon 3, so carbons 1 and 5 are equivalent, and so are carbons 2 and 4.
2. The environments are: the end CH₃ groups, the CH₂ groups, the single C–H on carbon 3, and the CH₃ branch.
3. The four products are 1-chloro-3-methylpentane, 2-chloro-3-methylpentane, 3-chloro-3-methylpentane and 3-(chloromethyl)pentane.
4. So there are four monochloro structural isomers, before any further substitution.

Common errors

• Writing Cl⁻ or Br⁻ when the radical is meant. A radical has a dot and, for a halogen atom, no charge.
• Drawing a full curly arrow for homolytic fission. Each electron needs its own fish hook.
• Starting the fish-hook arrows on an atom instead of the middle of the bond.
• Writing Cl• + C₇H₁₄ → C₇H₁₃Cl + H•. The chlorine radical removes a hydrogen atom to make HCl; H• is not formed in this mechanism.
• Forgetting the termination step between two alkyl radicals, which explains a longer alkane in the product mixture.
• Saying alkanes are unreactive because they are “saturated” without mentioning strong C–C and C–H bonds and low polarity.
• Giving “light” as the condition without saying ultraviolet, or leaving out heat as the alternative.

Next steps

• Condensed version: Reactivity 3.3 revision notes
• Test yourself: Reactivity 3.3 practice questions
• See all four mechanism types side by side in the Reactivity 3 overview notes; browse other units on the IB DP chemistry hub
• Mark off Reactivity 3.3 on the printable checklist, then look for gaps with the free diagnostics

Official syllabus

This guide follows 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).