OCR A Level Chemistry A: Haloalkanes (H432)

Haloalkanes are section 4.2.2 in OCR’s specification for A Level Chemistry A, code H432 (Version 3.1, issued May 2026; first assessment 2017). You meet them in Module 4, which OCR calls Core organic chemistry. This guide works through each lettered outcome, (a) through (e), and gives a worked example for each group of ideas. Because H432 is a full A level, nothing here is labelled AS or A2. Module 4 is examined in two of the written components: Synthesis and analytical techniques (02) and Unified chemistry (03). The specification ties outcome 4.2.2(a) to PAG7, qualitative analysis of organic functional groups.

Browse the rest of the course from the OCR A Level chemistry hub, and record your progress outcome by outcome on the H432 chemistry checklist. If you want to see where your gaps are before reading, a short diagnostic helps. Afterwards, condense this page with the haloalkanes revision notes and test yourself with the haloalkanes practice questions.

Outcomes in 4.2.2

Statement	In plain words	Guidance and links in the specification
4.2.2(a)(i)	Describe hydrolysis of haloalkanes by aqueous alkali as a substitution reaction	PAG7; see also 6.3.1(c)
4.2.2(a)(ii)	Describe hydrolysis by water with AgNO₃ and ethanol, used to compare rates of hydrolysis of different carbon–halogen bonds	PAG7
4.2.2(b)	Define and use the term nucleophile (an electron pair donor)	–
4.2.2(c)	Draw the nucleophilic substitution mechanism for hydrolysis of primary haloalkanes with aqueous alkali	HSW1, 2; see also 4.1.1(h)–(i)
4.2.2(d)	Explain the trend in rates of hydrolysis of primary haloalkanes using C–F, C–Cl, C–Br and C–I bond enthalpies	–
4.2.2(e)(i)	Write equations for the production of halogen radicals when UV acts on CFCs in the upper atmosphere	Simple breakdown equations required
4.2.2(e)(ii)	Write equations for the catalysed breakdown of ozone by Cl• and other radicals such as •NO	Construct similar equations for other stated radicals; HSW9, 10, 11, 12

The carbon–halogen bond

A haloalkane has a halogen atom, X, bonded to an sp³ carbon in an alkane chain. You name them with a prefix: fluoro, chloro, bromo or iodo, plus a position number. Primary haloalkanes have the halogen on a carbon joined to one other carbon, or to none in a halomethane (for example 1-bromo-3-methylbutane). Secondary and tertiary haloalkanes have it on a carbon joined to two or three other carbons.

Halogens are more electronegative than carbon. The bonding pair is pulled towards X, so the bond is polar: Cδ+–Xδ−. That δ+ carbon is the site of attack in every substitution reaction in this section.

4.2.2(b): Nucleophiles

A nucleophile is an electron pair donor. It has a lone pair of electrons, often with a negative charge, and it is attracted to an electron-deficient (δ+) carbon atom. When it donates its lone pair, a new covalent bond forms.

Nucleophiles you meet in this section:

• the hydroxide ion, :OH⁻, from aqueous NaOH or KOH
• the water molecule, H₂O:, which has two lone pairs on oxygen

Ammonia and the cyanide ion are also nucleophiles, but OCR uses them with haloalkanes in Module 6, not here.

A species with no lone pair, such as CH₄ or H⁺, cannot act as a nucleophile.

4.2.2(a)(i): Hydrolysis by aqueous alkali

Hydrolysis means breaking a bond by reaction with water or its ions. A haloalkane is warmed under reflux with aqueous sodium hydroxide. The OH⁻ ion replaces the halogen, so the reaction is a substitution, and an alcohol forms.

General equation: R–X + OH⁻ → R–OH + X⁻

With sodium hydroxide written in full: R–X + NaOH → R–OH + NaX

Reflux is used because haloalkanes are volatile and the reaction is slow at room temperature. Heating under reflux lets you heat for a long time without losing the organic compound as vapour.

Worked example 1: equation and yield

Question: 9.06 g of 1-bromo-3-methylbutane, (CH₃)₂CHCH₂CH₂Br, is heated under reflux with excess NaOH(aq). After purification, 3.96 g of alcohol is collected. Name the alcohol, write the equation and calculate the percentage yield.

1. The OH group replaces Br on carbon 1. The alcohol is 3-methylbutan-1-ol.
2. Equation: (CH₃)₂CHCH₂CH₂Br + NaOH → (CH₃)₂CHCH₂CH₂OH + NaBr
3. Mr of C₅H₁₁Br = (5 × 12.0) + (11 × 1.0) + 79.9 = 150.9
4. Amount of haloalkane = 9.06 ÷ 150.9 = 0.0600 mol
5. One mole of haloalkane gives one mole of alcohol, so at most 0.0600 mol of alcohol can form.
6. Mr of C₅H₁₂O = 88.0, so the theoretical mass = 0.0600 × 88.0 = 5.28 g
7. Percentage yield = 3.96 ÷ 5.28 × 100 = 75.0%

4.2.2(c): The nucleophilic substitution mechanism

The specification asks for the mechanism with aqueous alkali on a primary haloalkane. You show it with curly arrows, and each arrow shows the movement of an electron pair (4.1.1(h)–(i)). Include the dipole on the C–X bond. A curly arrow must start from a lone pair, a bond or a negative charge.

Worked example 2: OH⁻ with bromoethane

The steps, in order:

1. Label the C–Br bond Cδ+–Brδ−.
2. Draw a lone pair on the oxygen of OH⁻ (with its negative charge).
3. Curly arrow 1 starts at that lone pair and ends at the δ+ carbon. This shows a new C–O bond forming.
4. Curly arrow 2 begins on the C–Br bond and points to bromine. Both bonding electrons go to Br, so this is heterolytic fission.
5. Products: ethanol, CH₃CH₂OH, and the bromide ion, :Br⁻ (show it with a lone pair and negative charge).

             H                        H
             |                        |
  HO:⁻   H — C — Br      ⟶     H — C — OH   +   :Br⁻
             |  δ+  δ−                |
            CH₃                      CH₃

  first arrow:  O lone pair  ->  δ+ carbon      (new C–O bond)
  second arrow: C–Br bond     ->  Br             (bond breaks)

The nucleophile attacks from the side opposite the halogen, and the new bond forms as the old one breaks. Name the overall process nucleophilic substitution: a nucleophile has replaced the halogen.

4.2.2(a)(ii): Hydrolysis by water, with AgNO₃ and ethanol (PAG7)

Water is a weaker nucleophile than OH⁻, but it still hydrolyses haloalkanes slowly when warmed:

R–X + H₂O → R–OH + H⁺ + X⁻

The halide ion released meets silver ions and forms a precipitate:

Ag⁺(aq) + X⁻(aq) → AgX(s)

So the time taken for a precipitate to appear tells you how fast the C–X bond is being broken.

Method for comparing rates

1. Put equal volumes of ethanol into three test tubes. Ethanol is a common solvent: haloalkanes do not mix with water, but both the haloalkane and aqueous silver nitrate dissolve in ethanol.
2. Add the same small volume of each haloalkane, one per tube. Keep the carbon chain the same, so only the halogen changes.
3. Warm the three tubes in one water bath held at a set temperature, for example 50 °C. Warm the silver nitrate solution in its own tube in the same bath.
4. Add an equal volume of the warm silver nitrate to each tube, starting a stopwatch for that tube as you add it. Record how long each mixture takes to turn cloudy.

Haloalkane	Precipitate	Colour
Chloroalkane	AgCl	white
Bromoalkane	AgBr	cream
Iodoalkane	AgI	yellow

The same test identifies a haloalkane in an unknown compound (see also 6.3.1(c)).

Worked example 3: interpreting timings

Question: At 50 °C, 1-iodo-2-methylpropane gives a precipitate after 45 s and 1-bromo-2-methylpropane after 540 s. The chloro compound shows only a faint cloudiness after 30 minutes. Compare the rates.

1. Rate is proportional to 1/time, because each tube makes the same small amount of precipitate before it is seen.
2. 1/45 = 0.0222 s⁻¹; 1/540 = 0.00185 s⁻¹.
3. 540 ÷ 45 = 12.0, so the iodo compound hydrolyses about 12 times faster than the bromo compound.
4. Order of rate: iodo > bromo > chloro.

4.2.2(d): Bond enthalpy controls the rate

To hydrolyse a haloalkane, the C–X bond must break. The weaker the bond, the less energy is needed and the faster the reaction.

Approximate average bond enthalpies, rounded (published values vary a little from source to source):

Bond	Bond enthalpy / kJ mol⁻¹	Rate of hydrolysis
C–F	about 470	effectively none
C–Cl	about 350	slow
C–Br	about 290	faster
C–I	about 230	fastest

Bond enthalpy falls from C–F to C–I because the halogen atom gets larger. The bonding pair is further from the halogen nucleus, so the overlap and the bond are weaker.

Polarity points the other way. C–F is the most polar bond, so its carbon is the most δ+. If polarity controlled the rate, fluoroalkanes would hydrolyse fastest. They do not, so you conclude that bond enthalpy, not bond polarity, decides the rate of hydrolysis. A full answer states both the trend and the reason.

4.2.2(e): CFCs and the ozone layer

Why CFCs were used

Chlorofluorocarbons (CFCs) contain carbon, chlorine and fluorine only. They are non-toxic, non-flammable and very unreactive, which made them useful as refrigerants, aerosol propellants, solvents and blowing agents for foams. Their low reactivity is also the problem: they are not broken down in the lower atmosphere, so they drift up to the stratosphere.

4.2.2(e)(i): Making radicals

Ozone in the stratosphere absorbs much of the Sun’s harmful ultraviolet radiation. When a CFC reaches the stratosphere, UV radiation breaks a C–Cl bond homolytically. A C–F bond is stronger, so it stays intact.

Example from the specification: CF₂Cl₂ → CF₂Cl• + •Cl

Use a dot for every species with one unpaired electron (4.1.1(g)).

Worked example 4: a different chlorine source

Question: Chlorodifluoromethane, CHClF₂, can reach the stratosphere in small amounts. Write the equation for radical formation and the catalytic cycle.

1. The weakest bond breaks: CHClF₂ → •CHF₂ + •Cl
2. Step 1 of the cycle: •Cl + O₃ → •ClO + O₂
3. Step 2: •ClO + O → •Cl + O₂ (the O atom comes from UV splitting O₂ or O₃)
4. Overall, add and cancel: O₃ + O → 2O₂

Step 1 uses up the chlorine radical and step 2 gives it back. Because it cancels from the sum, the radical acts as a catalyst, and one radical can destroy a very large number of ozone molecules before it is removed.

4.2.2(e)(ii): Other radicals

Nitrogen monoxide, •NO, forms from nitrogen and oxygen at high temperatures, for example in lightning and aircraft engines. It catalyses ozone breakdown in the same pattern:

• •NO + O₃ → •NO₂ + O₂
• •NO₂ + O → •NO + O₂
• overall: O₃ + O → 2O₂

You may be given another radical, such as •Br from a bromine-containing compound. Write it in place of Cl in the same two steps.

How science works

Scientists linked chlorine radicals to ozone loss in the 1970s, and measurements over Antarctica confirmed severe thinning. The evidence was accepted, and governments agreed to phase out CFCs under the Montreal Protocol (1987). This is a case of weighing the benefits of a useful chemical against its environmental harm, and of chemists developing alternatives such as hydrofluorocarbons, which contain no chlorine.

Common errors

• Writing the curly arrow from the carbon to the hydroxide ion. The arrow goes from the nucleophile’s lone pair to the δ+ carbon.
• Leaving out the δ+ and δ− on the C–X bond, or the lone pair on the hydroxide ion.
• Saying the iodo compound reacts fastest “because C–I is the most polar”. It is the least polar; it reacts fastest because it is the weakest.
• Calling the precipitate “silver chloride” for every tube without naming the colours: white, cream, yellow.
• Missing the dots on radicals (an O atom needs none).
• Writing an overall ozone equation that still contains Cl• or ClO•.
• Breaking a C–F bond in a CFC. UV breaks the weaker C–Cl bond.

Official syllabus

Source document: the OCR specification for A Level Chemistry A, code H432, in its Version 3.1 of May 2026 (first assessment 2017). This guide follows section 4.2.2 Haloalkanes, statements (a) to (e).