AQA A-Level Chemistry: Halogenoalkanes (7405)

This guide teaches section 3.3.3 Halogenoalkanes of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. It covers every point in 3.3.3.1 Nucleophilic substitution, 3.3.3.2 Elimination and 3.3.3.3 Ozone depletion. All of it is AS content, so nothing in 3.3.3 is A-level only. The specification places organic chemistry 3.3.1 to 3.3.6 on Paper 2 at AS, and all of section 3.3 on Paper 2 at A-level; A-level Paper 3 can test any content and any practical skill.

For quick recall, use the revision notes. To test yourself, use the practice questions. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. Bond polarity and electronegativity are taught in bonding, and the silver nitrate test for halide ions is in Group 7, the halogens.

What this unit covers

Spec point	What you must be able to do	Status
3.3.3 introduction	Know that halogenoalkanes are much more reactive than alkanes; uses as refrigerants, solvents and in pharmaceuticals; why some uses are restricted	AS and A-level
3.3.3.1 Nucleophilic substitution	Explain why the C–X bond is polar; outline the substitution mechanisms with OH⁻, CN⁻ and NH₃; explain why C–X bond enthalpy affects the rate	AS and A-level
3.3.3.2 Elimination	Describe concurrent substitution and elimination (e.g. 2-bromopropane with KOH); explain the reagent’s role as nucleophile and base; outline both mechanisms	AS and A-level
3.3.3.3 Ozone depletion	Explain why ozone is beneficial; how UV breaks C–Cl bonds in CFCs; use equations to show chlorine atoms catalysing ozone decomposition; the role of research and of chlorine-free alternatives	AS and A-level

No required practical is listed under 3.3.3. The two suggested activities (test-tube hydrolysis to compare rates, and preparing a chloroalkane) are taught below.

Why halogenoalkanes react: the polar C–X bond

A halogenoalkane has a halogen atom, X, bonded to a carbon in an alkane chain. The halogen is more electronegative than carbon, so it pulls the bonding pair towards itself. The carbon becomes δ+ and the halogen δ−:

   δ+   δ−
H3C—CH2—Br

The δ+ carbon is attacked by nucleophiles. A nucleophile is an electron pair donor: a species with a lone pair that it uses to form a new bond to an electron-deficient carbon. The three nucleophiles in the specification are OH⁻, CN⁻ and NH₃.

Alkanes have no such δ+ site, which is why halogenoalkanes are much more reactive than alkanes.

3.3.3.1 Nucleophilic substitution

In each reaction the nucleophile replaces the halogen, which leaves as a halide ion. In a mechanism you show movement of an electron pair with a curly arrow. Each arrow starts at a lone pair or a bond and ends where the pair goes.

Hydroxide ions: halogenoalkane → alcohol

Reagent and conditions: aqueous sodium or potassium hydroxide, warm (heat under reflux).

CH3CH2CH2Br + OH⁻ → CH3CH2CH2OH + Br⁻

Worked example 1 (outline the mechanism for 1-bromopropane with OH⁻).

            δ+   δ−
CH3CH2–CH2–Br       :OH⁻
  arrow 1: lone pair on O of OH⁻ → the δ+ carbon
  arrow 2: C–Br bonding pair → Br
products: CH3CH2CH2–OH  +  :Br⁻

Label the dipole, draw arrow 1 from the oxygen lone pair (not the hydrogen), draw arrow 2 from the middle of the C–Br bond to Br, then show propan-1-ol and Br⁻.

Cyanide ions: halogenoalkane → nitrile

Reagent and conditions: potassium cyanide in aqueous ethanol, heat under reflux. The product is a nitrile, R–C≡N. The chain gets one carbon longer, which is useful in synthesis. Name the nitrile by counting every carbon, including the one in C≡N: bromoethane gives propanenitrile.

CH3CH2Br + CN⁻ → CH3CH2CN + Br⁻

The mechanism has the same two arrows: lone pair on the carbon of CN⁻ to the δ+ carbon, and C–Br bond to Br.

Worked example 2 (yield). A student heats 13.69 g of 1-bromobutane with excess KCN and obtains 5.81 g of pentanenitrile. Use Ar values C = 12.0, H = 1.0, N = 14.0, Br = 79.9. Calculate the percentage yield.

C4H9Br + CN⁻ → C4H9CN + Br⁻        (1 : 1)
Mr(C4H9Br) = 4(12.0) + 9(1.0) + 79.9 = 136.9
n(C4H9Br)  = 13.69 ÷ 136.9 = 0.100 mol
Mr(C4H9CN) = 5(12.0) + 9(1.0) + 14.0 = 83.0
theoretical mass = 0.100 × 83.0 = 8.30 g
% yield = 5.81 ÷ 8.30 × 100 = 70.0%

Ammonia: halogenoalkane → primary amine

Reagent and conditions: excess ammonia dissolved in ethanol, heated in a sealed container (under pressure). Two molecules of ammonia are used per halogenoalkane.

CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br

Worked example 3 (outline the mechanism for bromoethane with NH₃).

Step 1: arrow from the N lone pair of NH₃ to the δ+ carbon; arrow from the C–Br bond to Br. Step 2: this gives CH3CH2–NH3⁺ (positive charge on N) and Br⁻. Step 3: a second NH₃ acts as a base: arrow from its lone pair to an H on N⁺, and arrow from that N–H bond onto the nitrogen. Step 4: products are ethylamine, CH3CH2NH2, and NH₄⁺ (with Br⁻, NH₄Br).

Why use excess ammonia? The amine product also has a lone pair on nitrogen, so it can attack another halogenoalkane molecule. Excess ammonia makes it more likely that the halogenoalkane meets NH₃ rather than amine, so the primary amine is the main product. Further substitution, giving secondary and tertiary amines and quaternary ammonium salts, is in 3.3.11 Amines, which is A-level only.

Why bond enthalpy controls the rate

To be substituted, the C–X bond must break. Down Group 7 the halogen atom gets bigger, the C–X bond gets longer and the bond enthalpy decreases: C–F is the strongest and C–I the weakest.

bond enthalpy:  C–F  >  C–Cl  >  C–Br  >  C–I
rate:           R–F  <  R–Cl  <  R–Br  <  R–I

Polarity runs the other way. The C–F bond is the most polar, so on polarity alone you would predict that fluoroalkanes react fastest. In fact, they react slowest. So the bond enthalpy, not the polarity, decides the rate: less energy is needed to break a C–I bond, so iodoalkanes react fastest.

Test-tube hydrolysis: comparing rates

The specification suggests a test-tube comparison. A typical method:

1. Put equal volumes of ethanol in three test tubes. Add the same number of drops of 1-chlorobutane, 1-bromobutane and 1-iodobutane, one per tube. Ethanol is a solvent in which both the halogenoalkane and the aqueous reagent dissolve.
2. Warm the tubes, and a separate tube of aqueous silver nitrate, in a water bath at the same temperature.
3. Add the same volume of silver nitrate solution to each tube, start a timer and watch for a precipitate.

Water is the nucleophile. The halide ion released reacts with Ag⁺:

C4H9Br + H2O → C4H9OH + H⁺ + Br⁻
Ag⁺(aq) + Br⁻(aq) → AgBr(s)

Worked example 4 (interpreting results). A yellow precipitate appears first, then a cream one, and a white precipitate appears last.

Step 1: yellow is AgI, cream is AgBr, white is AgCl. Step 2: so the order of rate is 1-iodobutane > 1-bromobutane > 1-chlorobutane. Step 3: explanation: the C–I bond has the lowest bond enthalpy, so it breaks most easily.

Keep temperature, volumes and chain length the same.

Preparing and purifying a chloroalkane

A tertiary alcohol such as 2-methylpropan-2-ol reacts with concentrated hydrochloric acid at room temperature:

(CH3)3COH + HCl → (CH3)3CCl + H2O

1. Shake the alcohol with concentrated hydrochloric acid in a separating funnel, releasing the pressure often by opening the tap with the funnel inverted.
2. Let the layers separate. Run off and discard the aqueous layer.
3. Shake the organic layer with sodium hydrogencarbonate solution to remove acid. CO₂ forms, so release the pressure often. Run off the aqueous layer again.
4. Run the organic layer into a dry flask and add an anhydrous drying agent such as anhydrous sodium sulfate. Swirl until the liquid is clear.
5. Decant or filter off the liquid and distil it, collecting the fraction that boils near the product’s boiling point.

If you are unsure which layer is organic, add a little water and see which layer gets bigger.

3.3.3.2 Elimination

Hydroxide ions can react in two ways at the same time. With 2-bromopropane and potassium hydroxide, both substitution and elimination happen concurrently.

• As a nucleophile, OH⁻ donates its lone pair to the δ+ carbon. Product: propan-2-ol.
• As a base, OH⁻ accepts a proton (H⁺) from a carbon next to the C–Br carbon. Product: propene.

substitution: CH3CHBrCH3 + OH⁻ → CH3CH(OH)CH3 + Br⁻
elimination:  CH3CHBrCH3 + OH⁻ → CH2=CHCH3 + H2O + Br⁻

Changing the conditions changes the main product.

Main product wanted	Solvent	Temperature
Alcohol (substitution)	water (aqueous KOH)	warm
Alkene (elimination)	ethanol (ethanolic KOH)	hot, under reflux

Worked example 5 (outline the elimination mechanism for 2-bromopropane).

  H   H
  |   |
H–C — C — CH3        :OH⁻
  |   |
  H   Br
  arrow 1: lone pair on O of OH⁻ → an H on the left-hand carbon
  arrow 2: that C–H bonding pair → into the C–C bond, forming C=C
  arrow 3: C–Br bonding pair → Br
products: CH2=CH–CH3  +  H2O  +  :Br⁻

The H removed must be on a carbon adjacent to the carbon bearing Br. Arrow 2 starts on that C–H bond and ends between the two carbons, forming the C=C.

If the two carbons next to the C–X carbon are different, elimination can give more than one alkene; check each product for E–Z isomerism (3.3.1.3).

3.3.3.3 Ozone depletion

Ozone, O₃, forms naturally in the upper atmosphere. It is beneficial because it absorbs ultraviolet radiation, reducing the UV reaching the Earth’s surface.

Chlorofluorocarbons (CFCs) contain only carbon, chlorine and fluorine, for example CCl₂F₂. In the upper atmosphere, UV radiation breaks the C–Cl bond, forming chlorine atoms (free radicals). The C–F bond is stronger, so it is the C–Cl bond that breaks.

CCl2F2 → •CClF2 + Cl•

Chlorine atoms then catalyse the decomposition of ozone:

Cl•  + O3 → ClO• + O2
ClO• + O3 → 2O2  + Cl•

Worked example 6 (showing chlorine is a catalyst). Add the two equations and cancel species on both sides.

Cl• + ClO• + 2O3 → ClO• + 3O2 + Cl•
overall: 2O3 → 3O2

Cl• is used in the first step and regenerated in the second, so it does not appear in the overall equation: it is a catalyst. One chlorine atom can go round the cycle many times, destroying many ozone molecules. This contributes to the hole in the ozone layer.

Science and society. You should appreciate that results of research by different groups in the scientific community provided evidence for legislation to ban the use of CFCs as solvents and refrigerants. Chemists have since developed alternative chlorine-free compounds, such as hydrofluorocarbons (HFCs), which contain no C–Cl bond to release chlorine atoms.

Common errors

• Starting the nucleophile’s arrow on the H of OH⁻ or on the C≡N triple bond. It must start on a lone pair (on O for OH⁻, on C for CN⁻, on N for NH₃).
• Drawing the C–X arrow from the halogen to the carbon. The bond pair moves to the halogen.
• Missing the second NH₃ in the ammonia mechanism.
• Naming CH₃CH₂CN as ethanenitrile. It is propanenitrile.
• Explaining the reactivity order by polarity instead of C–X bond enthalpy.
• Calling OH⁻ a nucleophile in elimination. There it acts as a base.
• Writing the ozone steps without regenerating Cl•.

Next steps

Condense this into the revision notes, then try the practice questions. The AQA A-Level Chemistry exam preparation guide covers paper structure, and the free 10-minute diagnostics show where to focus.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.3 Halogenoalkanes (3.3.3.1 to 3.3.3.3).