Part 1: Nucleophilic substitution and hydrolysis rates

Predict alcohol, nitrile and amine products, track the electrons in each mechanism and explain the evidence from a hydrolysis experiment.

Start with the carbon–halogen bond

A halogenoalkane has a halogen bonded to a saturated carbon. Use fluoro-, chloro-, bromo- or iodo- with the alkane name and the required locant. For classification, count carbon neighbours of the carbon carrying the halogen: one is primary, two secondary and three tertiary. This is not the total number of carbons in the molecule.

The halogen attracts the C–X bonding pair more strongly than carbon, leaving Cδ+–Xδ−. A nucleophile donates an electron pair to the electron-deficient carbon, replacing the halogen. The starting carbon has a partial positive charge, not a full carbocation charge. OH⁻ and CN⁻ are charged nucleophiles; NH₃ is a neutral one.

Three substitutions to recognise

Conditions belong in the answer alongside the reagent. Cyanide reagents are highly toxic and sealed pressure reactions require specialist supervised equipment; these are syllabus descriptions, not instructions for an unsupervised practical.

OH⁻ and CN⁻: the two-arrow pattern

For a primary halogenoalkane, start one full-headed curly arrow at the nucleophile’s lone pair and end at the carbon bonded to the halogen. Start the other at the C–X bond and end at X. The leaving species is X⁻. Do not begin an arrow at the negative-charge symbol or at an atom without identifying its electron pair.

In cyanide substitution the carbon end attacks, making a new C–C bond. Bromoethane has two carbons; CH₃CH₂CN is propanenitrile and has three, including the nitrile carbon. The C≡N triple bond remains intact.

Ammonia needs an extra proton-transfer step

NH₃ attacks using its nitrogen lone pair while the C–Br bond pair goes to Br. Nitrogen now has four bonds, giving CH₃CH₂NH₃⁺. A second NH₃ molecule removes a proton: its lone pair points to an N–H hydrogen of the organic ion, while that N–H bond pair returns to the organic nitrogen. The products are ethylamine and NH₄⁺.

Excess ammonia favours the primary amine because an incoming halogenoalkane is more likely to encounter NH₃ than the amine product. It does not make further alkylation impossible: the amine also has a nitrogen lone pair.

Explain the bond, then connect it to the measurement

Compare compounds with the same carbon skeleton under the same conditions. C–I is weaker than C–Br, which is weaker than C–Cl; breaking the weaker bond is easier, so iodoalkanes hydrolyse faster than corresponding bromoalkanes and chloroalkanes. C–F is especially strong. Polarity alone predicts the wrong order here.

A typical comparison uses aqueous silver nitrate with ethanol as a co-solvent, since halogenoalkanes dissolve poorly in water. Water performs hydrolysis; released halide ions then precipitate with Ag⁺. AgCl is white, AgBr cream and AgI yellow. The covalently bonded halogen does not begin as a free halide ion.

Keep temperature, carbon skeleton, reagent concentrations, volumes and the cloudiness endpoint comparable. A shorter time to the same endpoint suggests a faster reaction; 1/time is a comparative rate measure, not automatically a concentration rate in mol dm⁻³ s⁻¹. On a halide-concentration graph, compare gradients at equivalent points.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources