OxfordAQA A-Level Chemistry: Halogenoalkanes (9620)

Here you learn section 3.3.3 Halogenoalkanes from scratch, following the OxfordAQA specification for International AS and A-level Chemistry (9620), Version 5.3, which applies to International AS and A-level exams from May/June 2020. Both of its points are included: 3.3.3.1 Nucleophilic substitution and 3.3.3.2 Elimination. Every part is International AS content; none of it is labelled International A-level only. Its written-paper home is Unit 2, Organic 1 and Physical 1, though the Unit 5 practical and synoptic paper is allowed to test any content.

Afterwards, shorten it into the condensed halogenoalkane notes and check yourself on the halogenoalkane question set. Other 9620 topics are on the OxfordAQA chemistry hub; mark your progress on the 9620 checklist; and try a short, free diagnostic quiz if you are not sure where your gaps are.

Background you will use: curly arrows and E–Z isomers (organic basics), bond polarity (bond polarity page), plus atom economy (moles and yields).

What section 3.3.3 asks of you

Spec point	Stage	What you need to do
3.3.3.1	International AS	Recognise that halogenoalkanes contain polar bonds
3.3.3.1	International AS	Outline the nucleophilic substitution mechanisms with OH⁻, CN⁻ and NH₃
3.3.3.1	International AS	Explain why the carbon–halogen bond enthalpy influences the rate of reaction
3.3.3.2	International AS	Describe the concurrent substitution and elimination reactions of a halogenoalkane (the specification’s example is 2-bromopropane with potassium hydroxide)
3.3.3.2	International AS	Explain the role of the reagent as both nucleophile and base
3.3.3.2	International AS	Outline the mechanisms of these reactions

What 9620 leaves out. The section’s overview mentions uses (refrigerants, solvents, pharmaceuticals) and the restriction of CFCs because of their effect on the atmosphere. The specification says overviews are not directly assessed, and 9620 has no separate learning point on ozone. No required practical is listed under 3.3.3. Further substitution of amines by halogenoalkanes, giving secondary and tertiary amines and quaternary ammonium salts, sits in 3.3.11.3, which is International A2 content.

3.3.3.1 Polar bonds and nucleophiles

A halogenoalkane has a halogen atom (F, Cl, Br or I) joined to a saturated carbon atom in an alkyl chain. Each halogen attracts shared electrons more strongly than carbon does, so the C–X pair sits nearer the halogen:

       δ+   δ−
R — C ——— X        (X = Cl, Br or I)

That electron-poor carbon is the reactive site. Alkanes have no comparable δ+ carbon.

A nucleophile donates a lone pair of electrons to an electron-deficient atom, and that pair becomes a new covalent bond. The three nucleophiles named in 3.3.3.1 are:

• the hydroxide ion, :OH⁻ (lone pair on oxygen)
• the cyanide ion, :C≡N⁻ (lone pair on carbon)
• ammonia, :NH₃ (lone pair on nitrogen, no charge)

In nucleophilic substitution the halogen is swapped for the nucleophile and departs as X⁻.

Curly arrow rules (from 3.3.1.2)

The specification states that forming a covalent bond is shown by a curly arrow starting from a lone pair or from another covalent bond, and breaking a bond is shown by an arrow starting from the bond. So for substitution you always need two arrows:

1. From the nucleophile’s lone pair to the δ+ carbon.
2. From the middle of the C–X bond to X.

Substitution by hydroxide ions: making an alcohol

Warming a halogenoalkane with aqueous sodium or potassium hydroxide (water as the solvent) gives an alcohol.

Worked example 1: 1-chlorohexane and hydroxide ions

Outline the mechanism for the reaction of 1-chlorohexane with hydroxide ions, and name the organic product.

                  H
                  |  δ+    δ−
 HO:⁻  ⟶   C₅H₁₁— C ——— Cl        arrow a: O lone pair → C
                  |                arrow b: C–Cl bond → Cl
                  H
        ⟹   C₅H₁₁CH₂—OH  +  :Cl⁻

• Arrow a starts on a lone pair of the oxygen in OH⁻ and ends at that δ+ carbon.
• Arrow b starts in the middle of the C–Cl bond and ends on the chlorine.
• Products: hexan-1-ol, CH₃(CH₂)₅OH, and a chloride ion.

Equation: CH₃(CH₂)₅Cl + OH⁻ → CH₃(CH₂)₅OH + Cl⁻

Worked example 2: atom economy of the hydrolysis

Calculate the percentage atom economy for making hexan-1-ol from 1-chlorohexane and sodium hydroxide. (Ar values: Cl 35.5, Na 23.0, O 16.0, C 12.0, H 1.0.)

C₆H₁₃Cl + NaOH → C₆H₁₃OH + NaCl
Mr: C₆H₁₃Cl = 6(12.0) + 13(1.0) + 35.5 = 120.5
    NaOH    = 40.0          total reactants = 160.5
    C₆H₁₃OH = 6(12.0) + 14(1.0) + 16.0 = 102.0
atom economy = 102.0 ÷ 160.5 × 100 = 63.6%

The rest of the reactant mass ends up in the by-product, sodium chloride.

Substitution by cyanide ions: making a nitrile

Heating a halogenoalkane under reflux with potassium cyanide dissolved in a mixture of ethanol and water gives a nitrile, R–C≡N. The nucleophile attacks through the carbon lone pair, so a new C–C bond forms and the carbon chain becomes one atom longer.

Worked example 3: naming the nitrile

1-Bromo-2-methylbutane is heated with potassium cyanide. Give the structure and name of the organic product, and describe the two curly arrows.

• Structure of the reactant: BrCH₂–CH(CH₃)–CH₂–CH₃.
• Arrow 1: from the lone pair on the C of :CN⁻ to the δ+ CH₂ carbon. Arrow 2: from the C–Br bond to Br.
• Product: NC–CH₂–CH(CH₃)–CH₂–CH₃.
• Naming: the nitrile carbon is carbon 1 of the main chain. Counting from it gives five carbons, with the methyl branch on carbon 3. Name: 3-methylpentanenitrile.

Equation: CH₃CH₂CH(CH₃)CH₂Br + CN⁻ → CH₃CH₂CH(CH₃)CH₂CN + Br⁻

Watch the numbering. In the reactant the branch was on carbon 2; in the product it is on carbon 3, because the new carbon becomes carbon 1.

Substitution by ammonia: making a primary amine

Ammonia is a neutral nucleophile. The halogenoalkane is heated with excess ammonia in ethanol in a sealed container. Two ammonia molecules are used for each halogenoalkane, so the mechanism has two stages.

Worked example 4: 1-iodo-2-methylpropane and ammonia

Outline the mechanism for the reaction of 1-iodo-2-methylpropane, (CH₃)₂CHCH₂I, with ammonia.

Stage 1 (substitution).

• Draw an arrow starting at nitrogen’s lone pair in :NH₃ and ending at the δ+ carbon of CH₂–I.
• Arrow from the C–I bond to I.
• This gives the ion (CH₃)₂CHCH₂–NH₃⁺ (the positive charge sits on nitrogen) and I⁻.

Stage 2 (deprotonation).

• Another ammonia molecule acts as a base. Its lone pair heads for one of the hydrogens bonded to N⁺.
• The electrons of that N–H bond move back onto the nitrogen, which is then neutral.
• Products: the primary amine (CH₃)₂CHCH₂NH₂, named 2-methylpropylamine, and NH₄⁺.

Overall: (CH₃)₂CHCH₂I + 2NH₃ → (CH₃)₂CHCH₂NH₂ + NH₄I

Why excess ammonia? The amine product still has a lone pair on nitrogen, so it can itself attack another halogenoalkane molecule. Using a large excess of ammonia makes the halogenoalkane far more likely to meet NH₃ than the amine. (The follow-on reactions are International A2, in 3.3.11.3.)

Bond enthalpy and the rate of substitution

Substitution cannot happen unless the C–X bond breaks. Going down Group 7 the halogen atom gets larger, the C–X bond gets longer, the bonding pair is held less tightly and the bond enthalpy falls.

Polarity follows the opposite trend. C–F is the most polar bond, and the electronegativities of carbon (2.55) and iodine (2.66) on the Pauling scale are so close that C–I is barely polar at all. If polarity decided the rate, fluoroalkanes would react fastest. They are in fact the least reactive, and iodoalkanes the most reactive. So the bond enthalpy, not the polarity, controls the rate.

Worked example 5: chloro against bromo

Under the same conditions, 1-bromohexane reacts with hydroxide ions faster than 1-chlorohexane does. Explain this, using these mean bond enthalpies: C–Cl 339 kJ mol⁻¹, C–Br 276 kJ mol⁻¹.

• The C–Br bond enthalpy is lower than the C–Cl bond enthalpy by 63 kJ mol⁻¹.
• Less energy is needed to break the C–Br bond, so the activation energy is lower.
• A larger fraction of collisions have enough energy to react, so 1-bromohexane reacts faster.
• The C–Cl bond is more polar, so this result shows that polarity is not the deciding factor.

Mean bond enthalpies differ slightly between data sources; use whatever values a question gives you.

3.3.3.2 Elimination

The hydroxide ion can act in two ways at once. In the specification’s example, 2-bromopropane heated with potassium hydroxide, two reactions run concurrently in the same flask:

• Some OH⁻ ions behave as nucleophiles: they bond to the δ+ carbon, the bromine leaves, and the alcohol propan-2-ol forms.
• Other OH⁻ ions behave as bases: they pull H⁺ off a carbon adjacent to the one carrying bromine. The overall loss of HBr leaves propene, plus water and a bromide ion.

Conditions shift the balance. Aqueous hydroxide, warm, gives mainly the alcohol. Hydroxide dissolved in ethanol and heated gives a larger share of the alkene.

Worked example 6: 3-bromopentane

3-Bromopentane, CH₃CH₂CHBrCH₂CH₃, reacts with potassium hydroxide. Name the substitution product, name every elimination product, and outline the elimination mechanism.

Substitution product. OH⁻ replaces Br: pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃.

Elimination products. The C–Br carbon is carbon 3. Its neighbours, carbons 2 and 4, are equivalent, so removing H from either gives the same structure, CH₃CH=CHCH₂CH₃, pent-2-ene. Both alkene carbons have two unlike groups attached (H with CH₃, and H with C₂H₅), so pent-2-ene exists as E and Z isomers. Both form.

Mechanism (three arrows).

         H    Br
         |    |
 CH₃ — C —— C — CH₂CH₃          arrow 1: O lone pair of :OH⁻ → H on carbon 2
         |    |                  arrow 2: C–H bond (carbon 2) → between C2 and C3
         H    H                  arrow 3: C–Br bond → Br
 ⟹  CH₃CH=CHCH₂CH₃  +  H₂O  +  :Br⁻

• Arrow 1 shows OH⁻ acting as a base: it goes to a hydrogen.
• Arrow 2 moves the C–H bonding pair into the C2–C3 region, where it becomes the extra bond of C=C.
• Arrow 3 takes the C–Br bonding pair onto bromine.

When the two neighbouring carbons are different, you can get structurally different alkenes. Check every carbon next to the C–X carbon that carries a hydrogen, then check each alkene for E–Z isomerism.

Common errors

• Starting the cyanide arrow on the nitrogen. The lone pair used is on the carbon of CN⁻.
• Drawing arrow b from the halogen towards carbon. The pair moves from the bond to the halogen.
• Forgetting stage 2 of the ammonia mechanism, or leaving the charge off the N⁺ intermediate.
• Numbering a nitrile from the wrong end, so a branch ends up with the reactant’s locant.
• Using polarity to explain the reactivity order. Credit goes to C–X bond enthalpy.
• In elimination, drawing the OH⁻ arrow to the carbon; as a base it attacks a hydrogen.
• Removing a hydrogen from the C–X carbon itself instead of an adjacent carbon.
• Missing E–Z isomers among elimination products.

Next steps

For fast recall, move on to the halogenoalkane notes; for exam-style work, use the question set with answers. How chlorine and methane react is covered in the 9620 alkanes topic. Track progress on the checklist and use the diagnostics to find what to revise next.

Official syllabus

Section 3.3.3 Halogenoalkanes (International AS), points 3.3.3.1 Nucleophilic substitution and 3.3.3.2 Elimination, in the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3 (published by OxfordAQA for exams from May/June 2020).