AQA A-Level Chemistry: Halogenoalkanes (7405) – Revision Notes

These notes condense 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 June 2016 onwards. They cover 3.3.3.1 Nucleophilic substitution, 3.3.3.2 Elimination and 3.3.3.3 Ozone depletion. Everything here is AS content, so none of it is A-level only. It is examined on Paper 2 at AS and A-level, and A-level Paper 3 can test any content. For full explanations and worked examples, read the halogenoalkanes study guide. To test yourself, use the halogenoalkanes practice questions.

Course hub: AQA A-Level Chemistry. Tick off each point on the printable checklist. Electronegativity and bond polarity are in bonding; silver halide colours are also in Group 7, the halogens.

Definitions and key ideas

Term	Meaning
Halogenoalkane	An alkane with one or more H atoms replaced by halogen atoms
Nucleophile	An electron pair donor that attacks an electron-deficient (δ+) carbon
Nucleophilic substitution	A nucleophile replaces the halogen, which leaves as a halide ion
Elimination	H and X are removed from neighbouring carbons, forming a C=C
Base (in elimination)	OH⁻ accepting a proton (H⁺) from the carbon next to C–X
CFC	Chlorofluorocarbon: a compound of carbon, chlorine and fluorine only
Curly arrow	Movement of an electron pair, from a lone pair or bond to where the pair ends up

Why halogenoalkanes react. The halogen is more electronegative than carbon, so the C–X bond is polar: C is δ+ and X is δ−. Nucleophiles attack the δ+ carbon. Alkanes have no δ+ carbon, so halogenoalkanes are much more reactive. Uses: refrigerants, solvents and pharmaceuticals. Some uses are restricted because of the effect of CFCs on the atmosphere.

Reagents, conditions and products

Nucleophile	Reagent and conditions	Product	Example
OH⁻	NaOH or KOH(aq), warm (reflux)	alcohol	1-chloropropane → propan-1-ol
CN⁻	KCN in aqueous ethanol, reflux	nitrile (chain +1 C)	1-bromopropane → butanenitrile
NH₃	excess NH₃ in ethanol, heat in sealed container	primary amine	1-iodobutane → butylamine
OH⁻ as base	KOH in ethanol, hot, reflux	alkene	2-bromopropane → propene

Equations to know the pattern of:

R–X + OH⁻  → R–OH + X⁻
R–X + CN⁻  → R–CN + X⁻
R–X + 2NH3 → R–NH2 + NH4X

3.3.3.1 Mechanisms in steps

Method: nucleophilic substitution (OH⁻ or CN⁻)

1. Draw the halogenoalkane and label C as δ+ and X as δ−.
2. Arrow 1: from a lone pair on the nucleophile (O of OH⁻, C of CN⁻) to the δ+ carbon.
3. Arrow 2: from the C–X bond to X.
4. Products: the substituted compound and X⁻.

Method: nucleophilic substitution with NH₃

1. Arrow from the N lone pair to the δ+ carbon; arrow from C–X to X.
2. Show the intermediate R–NH₃⁺ (charge on N) and X⁻.
3. A second NH₃: arrow from its lone pair to an H on N⁺; arrow from that N–H bond onto N.
4. Products: R–NH₂ and NH₄⁺ (NH₄X overall).

Why excess ammonia? The amine has a lone pair and can itself act as a nucleophile. Excess NH₃ reduces further substitution. Formation of secondary and tertiary amines and quaternary ammonium salts is in 3.3.11 Amines (A-level only).

Naming nitriles. Count all carbons, including the C of C≡N. CH₃CH₂CH₂CN is butanenitrile.

Worked reminder. 1-chloro-2-methylpropane, (CH₃)₂CHCH₂Cl, with KCN gives (CH₃)₂CHCH₂CN. The longest chain now has four carbons including the nitrile carbon (numbered C1), with a methyl group on C3, so the product is 3-methylbutanenitrile.

Bond enthalpy and rate

C–X bond enthalpy:  C–F > C–Cl > C–Br > C–I
rate of substitution:  iodo > bromo > chloro > fluoro

• The C–X bond must break, so the weaker bond reacts faster.
• Polarity predicts the opposite order (C–F most polar), so polarity is not what controls the rate.

Test-tube hydrolysis (suggested activity)

Halogenoalkane	Precipitate with AgNO₃	Appears
iodoalkane	yellow (AgI)	first
bromoalkane	cream (AgBr)	second
chloroalkane	white (AgCl)	last

• Water is the nucleophile; the released X⁻ forms AgX with Ag⁺.
• Ethanol is used as a common solvent so the halogenoalkane and aqueous silver nitrate mix.
• Control: same temperature (water bath), same volumes, same chain length.

Preparing a chloroalkane (suggested activity)

Method: purify a liquid organic product

1. React (e.g. 2-methylpropan-2-ol with concentrated HCl) in a separating funnel; vent often.
2. Remove the aqueous layer.
3. Wash with sodium hydrogencarbonate solution to remove acid (vent the CO₂), then remove the aqueous layer.
4. Dry with an anhydrous salt such as anhydrous sodium sulfate until clear.
5. Distil and collect the fraction near the product’s boiling point.

3.3.3.2 Elimination

Method: elimination mechanism

1. Arrow 1: from the OH⁻ lone pair to an H on a carbon next to the C–X carbon.
2. Arrow 2: from that C–H bond to between the two carbons (new C=C).
3. Arrow 3: from the C–X bond to X.
4. Products: alkene + H₂O + X⁻.

Substitution and elimination happen concurrently. Conditions shift the balance:

Favours	Solvent	Temperature	OH⁻ acts as
Substitution (alcohol)	water	warm	nucleophile
Elimination (alkene)	ethanol	hot	base

If the neighbouring carbons are different, more than one alkene can form. Check each for E–Z isomerism.

3.3.3.3 Ozone depletion

• Ozone forms naturally in the upper atmosphere and absorbs UV radiation.
• UV breaks the C–Cl bond in CFCs (C–F is stronger), giving Cl• atoms.
• Cl• catalyses ozone decomposition:

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

• Cl• is regenerated, so one atom destroys many O₃ molecules. This contributes to the ozone hole.
• Research by different groups in the scientific community gave evidence for legislation banning CFCs as solvents and refrigerants. Chemists developed chlorine-free alternatives such as HFCs.

Must-know distinctions

• Nucleophile vs base. A nucleophile donates a pair to carbon; a base accepts H⁺. OH⁻ can do both.
• Polarity vs bond enthalpy. Polarity explains why nucleophiles attack; bond enthalpy explains the rate order.
• Substitution vs elimination products. Same reagent (KOH), different products: alcohol vs alkene.
• Aqueous vs ethanolic KOH. Water favours substitution; ethanol and heat favour elimination.
• C–Cl vs C–F in CFCs. Only the weaker C–Cl bond is broken by UV to release radicals.

Quick self-test

1. Name the product when 1-bromopropane is refluxed with KCN in aqueous ethanol.
2. Calculate Mr of that product (C = 12.0, H = 1.0, N = 14.0).
3. Give the reagent and conditions to convert 1-chloropropane into propan-1-ol.
4. Write the equation for bromoethane with ammonia.
5. Which of 1-chloropentane and 1-iodopentane hydrolyses faster? Explain in one sentence.
6. State the role of OH⁻ when 2-bromopropane forms propene.
7. Where does the first curly arrow start in the reaction of CN⁻ with a halogenoalkane?
8. Write the equation for the formation of a chlorine atom from CCl₃F.
9. Write the two equations showing Cl• catalysing ozone decomposition.
10. In the overall reaction 2O₃ → 3O₂, how many moles of O₂ form from 0.030 mol of O₃?
11. What colour precipitate forms when 1-bromobutane is warmed with silver nitrate in ethanol?

Answers

1. Butanenitrile, CH₃CH₂CH₂CN.
2. C₄H₇N: 4(12.0) + 7(1.0) + 14.0 = 69.0.
3. Aqueous NaOH (or KOH), warm, under reflux.
4. CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br.
5. 1-iodopentane: the C–I bond has a lower bond enthalpy than C–Cl, so it breaks more easily.
6. Base: it accepts a proton from a carbon next to the C–Br carbon.
7. On the lone pair on the carbon of CN⁻, pointing to the δ+ carbon.
8. CCl₃F → •CCl₂F + Cl•.
9. Cl• + O₃ → ClO• + O₂; ClO• + O₃ → 2O₂ + Cl•.
10. 0.030 × 3/2 = 0.045 mol.
11. Cream (AgBr).

Where marks are usually lost

• Arrow from the H of OH⁻, or from the negative sign with no lone pair shown, instead of from the oxygen lone pair.
• Arrow drawn from X to C on the C–X bond (wrong direction).
• Missing δ+ and δ− labels when the question asks for the dipole.
• Ammonia mechanism stopped at R–NH₃⁺, or the second NH₃ step left out.
• Using 1 NH₃ instead of 2 NH₃ in the balanced equation.
• Giving “ethanolic” conditions for substitution, or “aqueous” for elimination.
• Explaining the rate order by electronegativity or polarity instead of C–X bond enthalpy.
• Removing an H from the C–X carbon itself in elimination, rather than from the neighbouring carbon.
• Writing ozone equations where Cl• is not regenerated, or with ClO without the radical dot.
• Saying UV breaks the C–F bond in CFCs.

Next steps

Work through the practice questions, go back to the study guide for anything you missed, and try the free 10-minute diagnostics.

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).