Part 1: Hydrolysis and nucleophilic substitution

Explain why a polar C–X bond is attacked by a nucleophile, then compare hydrolysis rates using bond strength.

Make an alcohol using aqueous hydroxide

The more electronegative halogen draws electron density towards itself, giving Cδ+–Xδ−. OH⁻ is a nucleophile: it donates an electron pair to the electron-deficient carbon. Heating a haloalkane with aqueous NaOH or KOH, usually under reflux, replaces X with OH and forms an alcohol.

The aqueous condition matters. Do not silently replace it with an ethanolic alkali condition, which can favour a different reaction. The core AS mechanism here is substitution by OH⁻.

Show both electron-pair movements

For a primary haloalkane such as bromoethane, draw an arrow from the oxygen lone pair of OH⁻ to the δ+ carbon attached to Br, and a second from the C–Br bond to Br. Show OH bonded through oxygen in the product and a separate Br⁻ ion.

This is nucleophilic substitution: the incoming electron-pair donor replaces the leaving group. Total charge −1 is conserved. The hydroxide attacks carbon, not the hydrogen of the alkyl chain or the halogen atom.

Use the same carbon skeleton

In hydrolysis with water, R–X + H₂O → R–OH + H⁺ + X⁻. Add aqueous silver nitrate so released halide ions precipitate as AgX. Ethanol acts as a co-solvent to help mix haloalkane and water. Keep concentrations, volumes, temperature and carbon skeleton the same; compare time to a defined cloudiness.

For otherwise similar chloro-, bromo- and iodoalkanes, C–I is weakest and hydrolysis is fastest, followed by C–Br then C–Cl. C–F is very strong and fluoroalkanes are resistant. Do not predict faster hydrolysis simply from greater bond polarity.

AgI is yellow, AgBr cream and AgCl white. The precipitate reports halide release; silver nitrate is not being added to test an already free halide in the original covalent molecule. A thermostatic water bath and repeats improve the comparison; ethanol makes a naked flame unsuitable.

Bond polarity explains the site; bond strength explains this trend

Carbon is δ+ in C–X because the halogen attracts the bonding electrons more strongly. That polarity explains why an electron-pair donor attacks carbon. It does not by itself give the observed rate order down the halogen series: the highly polar C–F bond is also very strong.

For comparable primary haloalkanes, decreasing C–X bond enthalpy from C–F to C–I makes breaking the bond easier, so iodoalkanes hydrolyse most readily. Keep the carbon skeleton and substitution class the same when isolating the effect of halogen identity. Comparing a primary chloroalkane with an unrelated tertiary iodoalkane changes more than one variable.

Classify a haloalkane at the carbon bonded to X: one carbon neighbour is primary, two secondary and three tertiary. The specified AS drawn hydroxide-substitution mechanism is for a primary haloalkane; do not invent a carbocation intermediate in its single substitution step.

Plan the silver-nitrate hydrolysis comparison

Prepare equal-concentration solutions of the comparable haloalkanes in the same ethanol/water mixture. Bring them and the silver nitrate reagent to the same water-bath temperature. Mix the same measured volumes, start timing consistently and record the time to the same defined turbidity endpoint. Repeat rather than trusting a single visual judgement.

Ethanol helps the organic and aqueous components mix; water supplies the hydrolysis reactant; Ag⁺ captures the released halide as AgX. These are different roles. Do not add aqueous NaOH indiscriminately to this silver-nitrate comparison: it changes the reaction conditions and can produce additional silver-containing precipitate.

A shorter time to a fixed endpoint suggests a faster reaction under the controlled conditions. The reciprocal 1/t can be a comparative rate proxy, but it is not automatically a measured rate in mol dm⁻³ s⁻¹ or a rate constant. Precipitate colour, nucleation and visual judgement limit the comparison.

Worked illustrative timing data

Suppose matched primary chloro-, bromo- and iodoalkanes reach the agreed cloudiness at mean times of 180 s, 60 s and 15 s. Their reciprocal-time values are 0.00556, 0.0167 and 0.0667 s⁻¹. Relative to the chloroalkane these are 1:3:12. This is an endpoint-based comparison, not evidence for a particular rate equation.

The order is consistent with weakest C–I hydrolysing fastest and strongest of these three C–Cl hydrolysing slowest. If a tube never clouds within the observation window, report “no endpoint observed within the time” rather than assigning an infinite time or asserting that no reaction is possible.

For greater consistency use a standard background, lighting and endpoint criterion, or suitable instrumental turbidity monitoring. Control temperature before mixing because temperature differences also alter collision energies and rates.

Quick checks

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

Sources