OxfordAQA A-Level Chemistry: Halogenoalkanes (9620) – Revision Notes

Use these notes in the final weeks before your exam. They boil down section 3.3.3 Halogenoalkanes (points 3.3.3.1 Nucleophilic substitution and 3.3.3.2 Elimination) from the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, which applies to exams from May/June 2020. No part of the section is International A-level only. You meet it on the Unit 2 paper, Organic 1 and Physical 1, and the Unit 5 practical and synoptic paper may also draw on it. Full explanations and worked mechanisms are in the study guide for this topic.

Then answer the halogenoalkane questions. Find every 9620 topic on the OxfordAQA chemistry hub, print the 9620 checklist to track progress, and take a short free diagnostic to see which topics need work. Curly-arrow rules and E–Z isomers are summarised in the organic basics summary.

Scope at a glance

Point	What you must do
3.3.3.1	Know that halogenoalkanes contain polar bonds; outline substitution mechanisms with OH⁻, CN⁻ and NH₃; explain how C–X bond enthalpy affects rate
3.3.3.2	Show that substitution and elimination run side by side (spec example: 2-bromopropane and KOH); say when OH⁻ is a nucleophile and when a base; outline both mechanisms

Not in 9620 3.3.3: a learning point on ozone (CFCs appear only in the overview, which is not directly assessed) and any required practical. Multiple substitution of amines is 3.3.11.3 (International A2).

Key definitions

• Halogenoalkane: an alkane in which one or more hydrogen atoms have been replaced by halogen atoms.
• Polar C–X bond: the halogen is more electronegative than carbon, so carbon is δ+ and the halogen δ−.
• Nucleophile: an electron pair donor; it attacks an electron-deficient (δ+) atom.
• Nucleophilic substitution: the halogen is exchanged for the nucleophile; X⁻ is released.
• Elimination: a small molecule (overall H–X) is removed from neighbouring carbons, forming C=C.
• Base (here): a proton acceptor; OH⁻ removes H⁺ from the carbon next to the C–X carbon.
• Concurrent: substitution and elimination happen in the same mixture at the same time.

The three nucleophiles

Nucleophile	Atom with the lone pair	Typical conditions	Organic product	By-product
OH⁻ (NaOH or KOH)	O	aqueous, warm	alcohol, R–OH	X⁻
CN⁻ (KCN)	C	ethanol and water, heat under reflux	nitrile, R–C≡N (one carbon longer)	X⁻
NH₃	N	excess, in ethanol, heated in a sealed vessel	primary amine, R–NH₂	NH₄X

Equations in general form:

RX + OH⁻   → ROH + X⁻          (alcohol)
RX + CN⁻   → RCN + X⁻          (nitrile)
RX + 2NH₃  → RNH₂ + NH₄X       (primary amine)

Method in steps: substitution mechanism

1. Draw the carbon bonded to X with δ+, and X with δ−.
2. Draw the nucleophile with its lone pair (and charge, if any).
3. Arrow from that lone pair to the δ+ carbon.
4. Arrow starting on the C–X bond and finishing on X.
5. Show products: the substituted compound and X⁻.
6. Ammonia only: the first product is R–NH₃⁺. Bring in another NH₃ as a base. One arrow runs from that molecule’s lone pair to a hydrogen on N⁺; a second takes the N–H electrons back to nitrogen. Products: R–NH₂ and NH₄⁺.

Arrow count: OH⁻ and CN⁻ need 2; NH₃ needs 4 (2 per stage).

Method in steps: elimination mechanism

1. Find the carbon bonded to X. Mark the neighbouring carbon whose H will be removed.
2. First arrow: oxygen lone pair (in OH⁻) to that hydrogen.
3. Arrow from the C–H bond to the space between the two carbons (the new π bond).
4. Arrow from the C–X bond to X.
5. Products: alkene, H₂O and X⁻.

Arrow count: 3.

Finding every elimination product

1. List each carbon next to the C–X carbon that carries at least one H.
2. Draw the alkene formed by removing H from each one in turn. Identical structures count once.
3. Check each alkene for E–Z isomerism: both carbons of C=C must carry two different groups.

Bond enthalpy and rate

Halogenoalkane type	Bond broken	Bond enthalpy	Bond polarity	Rate of substitution
iodoalkane	C–I	lowest	least polar	fastest
bromoalkane	C–Br	low		fast
chloroalkane	C–Cl	higher		slow
fluoroalkane	C–F	highest	most polar	slowest

• Rate rises as bond enthalpy falls: less energy is needed to break the bond.
• Polarity falls in the same direction as rate rises, so polarity cannot explain the order.
• One-line exam answer: “C–I is the weakest carbon–halogen bond, so it needs the least energy to break, and iodoalkanes are substituted fastest.”

Must-know distinctions

• OH⁻ as nucleophile vs OH⁻ as base. As a nucleophile it bonds to carbon (substitution, alcohol). As a base it bonds to hydrogen (elimination, alkene).
• Aqueous vs ethanolic hydroxide. Warm, aqueous hydroxide favours the alcohol. Hot hydroxide in ethanol gives a larger share of alkene. Both reactions still happen side by side.
• CN⁻ vs NH₃. Cyanide is charged and attacks through C; ammonia is neutral and attacks through N, so its first product is a positive ion that must lose H⁺.
• Bond enthalpy vs polarity. Only bond enthalpy explains the reactivity order.
• Nitrile names count the CN carbon. It becomes carbon 1, so branch locants can change.

Small worked reminders

Nitrile name. 2-Iodopentane, CH₃CHICH₂CH₂CH₃, with CN⁻ gives CH₃CH(CN)CH₂CH₂CH₃. Counting from the nitrile carbon: five carbons in the chain, methyl on carbon 2. Name: 2-methylpentanenitrile.

Atom economy with ammonia. 2-Iodobutane + 2NH₃ → butan-2-amine + NH₄I. With C = 12.0, H = 1.0, N = 14.0, I = 126.9:

Mr C₄H₉I = 183.9   2 × Mr NH₃ = 34.0   total = 217.9
Mr C₄H₉NH₂ = 73.0
atom economy = 73.0 ÷ 217.9 × 100 = 33.5%

Elimination products. 2-Bromo-3-methylbutane, CH₃CHBrCH(CH₃)₂. Neighbouring carbons: carbon 1 (CH₃) and carbon 3 (CH). Removing H from carbon 1 gives CH₂=CHCH(CH₃)₂, 3-methylbut-1-ene. Removing H from carbon 3 gives CH₃CH=C(CH₃)₂, 2-methylbut-2-ene. Neither has E–Z isomers, because one carbon of each C=C carries two identical groups.

Quick self-test

1. Define the term nucleophile.
2. Which atom of the cyanide ion forms the new bond to carbon?
3. Why does a halogenoalkane have a δ+ carbon atom?
4. Put R–F, R–Br, R–Cl and R–I in order of increasing rate of substitution.
5. Explain in one sentence why polarity cannot account for that order.
6. Why is excess ammonia used to make a primary amine?
7. How many curly arrows does the full ammonia mechanism need?
8. In elimination, what does the hydroxide ion accept, and from which carbon?
9. Give the substitution product of 2-bromo-3-methylbutane with aqueous hydroxide ions.
10. Which conditions give a larger proportion of alkene from a halogenoalkane and hydroxide?
11. Name the product when 1-bromo-4-methylpentane is heated with potassium cyanide.
12. Why does pent-2-ene show E–Z isomerism but but-1-ene does not?

Answers

1. A species that donates an electron pair to form a new covalent bond with an electron-poor atom.
2. The carbon atom.
3. Halogens attract bonding electrons more strongly than carbon, so the shared pair lies nearer X.
4. R–F < R–Cl < R–Br < R–I.
5. C–F is the most polar bond, yet fluoroalkanes react slowest, so polarity predicts the wrong order.
6. The amine product is also a nucleophile; excess NH₃ makes the halogenoalkane more likely to react with ammonia than with the amine.
7. Four (two for substitution, two for removing H⁺).
8. A proton (H⁺), from a carbon adjacent to the carbon bonded to the halogen.
9. 3-Methylbutan-2-ol, CH₃CH(OH)CH(CH₃)₂.
10. Hydroxide dissolved in ethanol, heated.
11. 5-Methylhexanenitrile, (CH₃)₂CHCH₂CH₂CH₂CN.
12. Both double-bonded carbons in pent-2-ene hold two unlike groups; in but-1-ene one C=C carbon carries two hydrogen atoms.

Where marks are usually lost

• Beginning the hydroxide arrow at its hydrogen, or the cyanide arrow at its nitrogen; the arrows must leave a lone pair on oxygen and on carbon respectively.
• Arrows from the halogen to the carbon, or arrows that float between atoms with no clear start.
• Missing δ+ and δ− on the C–X bond when the question says “outline the mechanism”.
• Stopping the ammonia mechanism at R–NH₃⁺, or writing only one NH₃ in the overall equation.
• Explaining reactivity with electronegativity or polarity instead of C–X bond enthalpy.
• In elimination, sending the hydroxide arrow to a carbon atom rather than to a hydrogen.
• Taking H from the C–X carbon, or listing duplicate alkenes as different products.
• Forgetting E and Z isomers when the alkene product has them.
• Naming a nitrile without counting the CN carbon.

Official syllabus

These notes follow section 3.3.3 Halogenoalkanes (International AS) of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for exams from May/June 2020. Points covered: 3.3.3.1 (nucleophilic substitution) and 3.3.3.2 (elimination).