OCR A Level Chemistry A: Haloalkanes (H432) – Revision Notes

Use these notes in the last weeks before your exams for section 4.2.2, Haloalkanes, in OCR’s H432 Chemistry A specification (Version 3.1, May 2026; first assessed in 2017). Each lettered statement, (a) through (e), appears below in short form. Two written components examine this Module 4 content: component 02 (Synthesis and analytical techniques) and component 03 (Unified chemistry), and PAG7 is linked to statement (a). H432 is a full A level, so there is no AS label.

For full explanations and worked examples, read the haloalkanes study guide. Then try the haloalkanes practice questions. Other units sit on the OCR A Level chemistry hub; keep your place with the H432 chemistry checklist. A 10-minute diagnostic gives a mixed check. Haloalkanes are made from alcohols in 4.2.1(e) (see the alcohols study guide), and the bond-fission ideas come from 4.1.1; for electronegativity and polar bonds, see these bonding notes.

Key definitions

Term	Meaning
Haloalkane	A saturated compound in which a halogen atom takes the place of hydrogen on an alkane chain
Primary haloalkane	Halogen on a carbon bonded to one other carbon
Nucleophile	An electron pair donor
Hydrolysis	Breaking a bond by reaction with water (or OH⁻)
Substitution	One atom or group is replaced by another
Heterolytic fission	One bonded atom takes both electrons of the bond
Homolytic fission	Each bonded atom takes one electron, making two radicals
Radical	A species with an unpaired electron, shown with a dot
CFC	Chlorofluorocarbon: contains only C, Cl and F

The polar C–X bond

• Halogens are more electronegative than carbon, so the bond is Cδ+–Xδ−.
• The δ+ carbon attracts nucleophiles.
• Polarity falls from C–F to C–I, but this is not what sets the rate.

4.2.2(b): Nucleophiles

• Must have a lone pair to donate.
• In this section: :OH⁻ (from NaOH(aq) or KOH(aq)) and H₂O (lone pairs on O).
• Not nucleophiles: H⁺ and CH₄ (no lone pair to donate).

4.2.2(a)(i): Aqueous alkali

Reagent and conditions: NaOH(aq), heat under reflux.

R–X + OH⁻ → R–OH + X⁻

Examples to practise writing:

• CH₃CH₂CH(CH₃)CH₂Br + NaOH → CH₃CH₂CH(CH₃)CH₂OH + NaBr (2-methylbutan-1-ol)
• (CH₃)₃CCl + OH⁻ → (CH₃)₃COH + Cl⁻ (2-methylpropan-2-ol)

The product is always an alcohol, and the OH sits where the halogen was.

4.2.2(c): Mechanism in steps

Method for any primary haloalkane with OH⁻:

1. Draw the haloalkane and label Cδ+ and Xδ−.
2. Draw :OH⁻ with a lone pair on O and the negative charge.
3. Arrow 1: lone pair on O to the δ+ carbon.
4. Arrow 2: from the C–X bond to X.
5. Products: the alcohol and :X⁻ (lone pair plus charge).
6. Name it: nucleophilic substitution.

Curly arrows always start at a lone pair, a bond or a negative charge, and always show an electron pair moving.

4.2.2(a)(ii): Water, AgNO₃ and ethanol (PAG7)

R–X + H₂O → R–OH + H⁺ + X⁻, then Ag⁺(aq) + X⁻(aq) → AgX(s)

Method in steps:

1. Equal volumes of ethanol in each tube (common solvent for the haloalkane and aqueous silver nitrate).
2. Equal amounts of each haloalkane, same chain length, different halogen.
3. Warm all tubes and the AgNO₃(aq) in the same water bath.
4. Add equal volumes of AgNO₃(aq), start timing, stop when a precipitate appears.
5. Rate ∝ 1/time.

Halide	Precipitate	Colour	Speed of appearance
Cl⁻	AgCl	white	slowest
Br⁻	AgBr	cream	middle
I⁻	AgI	yellow	fastest

Controlled variables: temperature, volumes, chain length and type of haloalkane (all primary), concentration of AgNO₃(aq).

4.2.2(d): Bond enthalpy and rate

Strongest to weakest: C–F, then C–Cl, then C–Br, then C–I. Rounded averages are roughly 470, 350, 290 and 230 kJ mol⁻¹ in that order. Published figures vary a little, so in an exam use whatever values the question supplies.

Three-point answer:

1. The C–X bond must break for hydrolysis.
2. Bond enthalpy decreases from C–F to C–I (the halogen is larger, so the bond is longer and weaker).
3. Less energy is needed to break C–I, so iodoalkanes hydrolyse fastest. Rate order: C–I > C–Br > C–Cl > C–F.

If you add polarity: C–F is most polar but slowest, so polarity is not the controlling factor.

4.2.2(e): CFCs and ozone

Radical production (i): UV breaks the weakest bond, C–Cl, homolytically.

• CF₂Cl₂ → CF₂Cl• + •Cl
• CFCl₃ → •CFCl₂ + •Cl

Catalysed breakdown (ii):

Radical	Step 1	Step 2
•Cl	•Cl + O₃ → •ClO + O₂	•ClO + O → •Cl + O₂
•NO	•NO + O₃ → •NO₂ + O₂	•NO₂ + O → •NO + O₂
•Br	•Br + O₃ → •BrO + O₂	•BrO + O → •Br + O₂

Overall for every cycle: O₃ + O → 2O₂

• The radical is regenerated, so it is a catalyst.
• One radical can remove a very large number of O₃ molecules.
• •NO comes from high-temperature reactions of N₂ and O₂, such as in lightning and aircraft engines.
• Dots go on every single-unpaired-electron species. O atoms do not need a dot.

How science works: CFCs are non-toxic, non-flammable and unreactive (refrigerants, aerosols, solvents). Evidence linking them to ozone loss was accepted, and governments acted through the Montreal Protocol (1987) to phase them out. Replacements such as hydrofluorocarbons contain no chlorine.

Small worked reminders

Yield from hydrolysis. 6.036 g of 1-bromo-2-methylbutane (Mr 150.9) gives 2.82 g of 2-methylbutan-1-ol (Mr 88.0).

• n(haloalkane) = 6.036 ÷ 150.9 = 0.0400 mol, and the ratio is 1 : 1.
• Theoretical mass = 0.0400 × 88.0 = 3.52 g
• Yield = 2.82 ÷ 3.52 × 100 = 80.1%

Comparing rates from times. If one tube clouds in 16 s and another in 200 s, the first reaction is 200 ÷ 16 = 12.5 times faster. Divide the longer time by the shorter one; do not subtract them.

Building a new cycle. Given any radical R•, write R• + O₃ → RO• + O₂, then RO• + O → R• + O₂. Check that the radical appears on the left of step 1 and the right of step 2.

Must-know distinctions

• OH⁻ vs H₂O as nucleophile: OH⁻ is faster and is used for the mechanism; H₂O is slower and is used in the AgNO₃ rate comparison.
• Heterolytic vs homolytic: hydrolysis breaks C–X heterolytically (ions form); UV breaks C–Cl in CFCs homolytically (radicals form).
• Bond enthalpy vs polarity: bond enthalpy explains the rate trend; polarity explains why the carbon is attacked at all.
• C–Cl vs C–F in CFCs: UV breaks C–Cl, not the stronger C–F.

Quick self-test

1. Define the term nucleophile.
2. Name the organic product when 3-bromohexane is heated under reflux with NaOH(aq).
3. In the mechanism with OH⁻, where does the first curly arrow start and finish?
4. Why is ethanol added in the AgNO₃ rate experiment?
5. Give the colour of the precipitate formed in the test with an iodoalkane.
6. Write the equation for water hydrolysing iodoethane, C₂H₅I.
7. 2.34 g of iodoethane (Mr 155.9) is fully hydrolysed and all the iodide is precipitated as AgI (Mr 234.8). Calculate the mass of AgI.
8. Which bond in CF₃Cl is broken by UV in the stratosphere? Write the equation.
9. Write the two propagation-style steps for •Br catalysing ozone breakdown.
10. Explain why 1-chlorooctane hydrolyses more slowly than 1-iodooctane.
11. Add the two •NO steps and cancel. What is the overall equation, and what is •NO acting as?

Answers

1. An electron pair donor.
2. Hexan-3-ol.
3. From the lone pair on the oxygen of OH⁻ to the δ+ carbon of the C–X bond.
4. It is a common solvent: the haloalkane and aqueous silver nitrate both dissolve in it, so they can react.
5. Yellow (AgI).
6. C₂H₅I + H₂O → C₂H₅OH + H⁺ + I⁻
7. n = 2.34 ÷ 155.9 = 0.0150 mol; mass of AgI = 0.0150 × 234.8 = 3.52 g
8. C–Cl: CF₃Cl → •CF₃ + •Cl
9. •Br + O₃ → •BrO + O₂, then •BrO + O → •Br + O₂
10. The C–Cl bond enthalpy is greater than the C–I bond enthalpy, so more energy is needed to break C–Cl and the reaction is slower.
11. O₃ + O → 2O₂; •NO is a catalyst because it is regenerated in step 2.

Where marks are usually lost

• Defining a nucleophile as “a negative ion”. Water is a neutral nucleophile; the definition is an electron pair donor.
• Drawing the first arrow from the O atom rather than from a lone pair or the negative charge on it.
• Forgetting the dipole on C–X, or giving the halide ion without its charge.
• Leaving “aqueous” or “heat under reflux” out of the reagents and conditions for hydrolysis.
• Explaining the rate order with polarity or electronegativity instead of bond enthalpy.
• Not controlling the chain length, so more than one variable changes in the rate comparison.
• Mixing up cream (AgBr) and yellow (AgI).
• Breaking C–F in a CFC radical equation.
• Leaving out dots on •Cl, •ClO, •NO or •NO₂.
• Giving an overall equation that is not O₃ + O → 2O₂ (for example leaving the catalyst in).

Official syllabus

Matched to: OCR, H432 A Level Chemistry A; Version 3.1 of the specification (May 2026), first assessed in 2017. Section 4.2.2 Haloalkanes, all statements (a) to (e).