AS Chemistry: Halogenoalkanes — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Halogenoalkanes revision notes

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Questions

1. Explain why the carbon atom in a halogenoalkane is susceptible to nucleophilic attack. [2]

2. Define a nucleophile. [1]

3. 1-bromobutane is heated under reflux with aqueous sodium hydroxide.

(a) Name the product and the mechanism. [2] (b) Describe the mechanism, stating where each curly arrow starts and ends. [3]

4. 1-bromobutane is heated with ethanolic sodium hydroxide instead.

(a) Name the product and the mechanism. [2] (b) Explain why the solvent determines which reaction occurs. [2]

5. The rate of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane is compared using silver nitrate in ethanol.

(a) Predict the order of reaction rate. [1] (b) Explain this order. [3] (c) Explain why the fluoroalkane is the least reactive despite having the most polar bond. [2] (d) State the observation for each and why silver nitrate is used. [3]

6. (Background only — CFCs and stratospheric ozone depletion are not part of the current 9701 2025–2027 syllabus; included for context, no marks awarded.) Explain why CFCs damage the ozone layer, naming the type of mechanism.

7. 2-bromobutane, CH₃CH₂CHBrCH₃, a secondary halogenoalkane, is hydrolysed with water under conditions that favour the SN1 mechanism.

(a) State the mechanism, and explain, in terms of carbocation stability, why this mechanism can compete effectively here. [3] (b) Describe, qualitatively, how the rate of this mechanism would compare if a primary halogenoalkane were used instead (no rate equation is required — this is an AS-level, qualitative comparison). [1] (c) The starting halogenoalkane is optically active (shows optical isomerism — “optically active” is strictly an A Level term, used here for brevity). State what happens to this property in the product, explaining your answer in terms of the reaction’s intermediate. [2]

8. A halogenoalkane is heated under reflux with ethanolic potassium cyanide.

(a) Name the type of reaction and the functional group in the product. [2] (b) Explain, in terms of the atoms involved, why the product has one more carbon atom than the starting halogenoalkane. [2] (c) State the reagent and conditions needed to convert the same halogenoalkane into a primary amine instead. [2]

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Answers

1. The halogen is more electronegative than carbon [1], so the C–X bond is polar and the carbon carries a partial positive charge (δ+) that attracts electron-rich species [1].

2. An electron-pair donor, attracted to an electron-deficient (δ+) centre [1].

3. (a) Butan-1-ol [1]; nucleophilic substitution [1]. (b) An arrow from the lone pair on the OH⁻ ion to the δ+ carbon [1]; an arrow from the C–Br bond to the bromine atom [1]; Br⁻ leaves and butan-1-ol forms [1].

4. (a) But-1-ene [1]; elimination [1]. (b) In aqueous solution OH⁻ acts as a nucleophile, attacking the carbon [1]; in ethanolic solution it acts as a base, removing a hydrogen from the adjacent carbon [1].

5. (a) iodo > bromo > chloro (fastest to slowest) [1]. (b) The rate depends on the strength of the carbon–halogen bond [1]. C–I has the lowest bond enthalpy, so it breaks most readily [1]; C–Cl has the highest of the three, so it breaks least readily [1]. (c) The C–F bond is the strongest of the carbon–halogen bonds [1], and bond enthalpy — not polarity — determines the rate [1]. (d) Iodo gives a yellow precipitate fastest, bromo a cream precipitate, chloro a white precipitate slowest [1] [1]. Silver nitrate reacts with the halide ion released, so the precipitate forms only as hydrolysis occurs, allowing the rate to be compared [1].

6. (Background, no marks.) UV light causes homolytic fission of the C–Cl bond, producing chlorine radicals. A chlorine radical reacts with ozone: Cl• + O₃ → ClO• + O₂. The ClO• then reacts with another ozone or oxygen radical, regenerating Cl•. Because the radical is regenerated, one radical destroys many ozone molecules — a radical chain mechanism.

7. (a) SN1 [1]. The intermediate is the secondary carbocation CH₃CH₂CH⁺CH₃, stabilised by electron donation from two alkyl groups (the inductive effect) [1]; this is a more stable carbocation than a primary one would give, so SN1 can compete effectively with SN2 for a secondary halogenoalkane [1]. (b) SN1 would be slower for a primary halogenoalkane, because the resulting primary carbocation has only one alkyl group donating electron density and so is less stable than the secondary carbocation formed here [1]. (c) The product, butan-2-ol, forms as a racemic mixture — equal amounts of both enantiomers, so no net optical rotation [1] — because the reaction goes via a planar carbocation that can be attacked by the nucleophile (water) from either face with equal probability [1].

8. (a) Nucleophilic substitution [1]; the product contains a nitrile group, –C≡N [1]. (b) The cyanide ion attacks through its carbon atom, forming a new C–C bond between the original halogenoalkane’s carbon skeleton and the carbon of the CN group [1], so the nitrile carbon becomes an additional carbon in the chain, one more than the starting halogenoalkane had [1]. (c) Excess ethanolic ammonia, heated in a sealed tube [1], giving a primary amine as the major product [1].

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Where marks are usually lost

• Explaining the C–X reactivity order by electronegativity instead of bond enthalpy.
• Curly arrows starting at the atom rather than the lone pair or bond.
• Not stating that the CFC chlorine radical is regenerated (background material — not examinable on the current 9701 syllabus, but a useful worked example of a radical chain mechanism).
• Forgetting that the solvent, not the reagent, decides substitution vs elimination.
• Trying to write a rate equation or name a rate-determining step for SN1/SN2 at AS — this is A Level extension content (Topic 26); at AS, compare rates qualitatively via carbocation stability instead.
• Explaining racemisation in SN1 without mentioning the planar carbocation and attack from either face — “it just loses its optical activity” scores nothing on its own.
• Forgetting the cyanide ion attacks through its carbon, not its nitrogen — this is exactly why the chain lengthens by one carbon rather than staying the same length.
• Naming ethanolic ammonia at room temperature instead of a sealed tube — without the sealed tube, ammonia gas simply escapes rather than reacting under pressure.

Questions 7 and 8 draw on the mechanism-selection table and the “three routes” and “reactions to know” sections of the Halogenoalkanes revision notes — material the questions above don’t reach, since they focus on primary-halogenoalkane SN2 substitution, elimination and the aqueous silver nitrate rate test.