Part 1: Planning routes and predicting reactions

Work backwards from the required functional group, retain the carbon skeleton and select a complete reagent-and-condition set for each step.

Use the reactions already learned

Worked two-step route

To prepare ethanoic acid from bromoethane: first heat CH₃CH₂Br with aqueous NaOH under reflux to give CH₃CH₂OH. Then heat the ethanol under reflux with excess acidified dichromate to give CH₃COOH. Each stage preserves two carbons; the second stage requires excess oxidant rather than immediate aldehyde distillation.

To convert propene to propanone: hydration with steam and an acid catalyst gives predominantly propan-2-ol under ordinary regioselective conditions; oxidation with acidified dichromate gives propanone. Account for selectivity and purification instead of pretending every route is automatically quantitative.

Inspect every reactive site

A molecule can contain more than one functional group. HOCH₂CH=CH₂ contains a primary alcohol and an alkene. Bromine can react at C=C; an oxidant may react with the alcohol and, depending on reagent/conditions, other groups too. Use the information given to judge selectivity instead of assuming one reagent affects only the group you are thinking about.

For unfamiliar routes, mark which bonds change and which atoms remain. Reagents that only add, substitute, eliminate or oxidise the groups here generally preserve the carbon skeleton; a proposed extra carbon needs a justified source. Avoid importing Year 13 chain-extension reagents as assumed AS knowledge.

Overall yield compounds the losses

For successive isolated steps, multiply the fractional yields. A route with 80.0% then 75.0% yield gives 0.800 × 0.750 = 0.600, or 60.0% overall, not 77.5%. Check mole ratios if they are not 1:1.

H032/02 June 2025 Q5(d) assessed structures and reagents in a route. Audit carbon count, hydrogen count and the exact reagent arrangement before writing a mechanism name. A chemically incompatible extra reagent can undermine an otherwise correct stage.

Work backwards from the final functional group

To prepare butan-2-one from 2-bromobutane, first ask which familiar group can form the ketone: a secondary alcohol. The intermediate is therefore butan-2-ol. Hydrolyse the haloalkane with aqueous NaOH or KOH and heat; isolate the alcohol as needed, then oxidise it with acidified dichromate under suitable heating.

This backwards reasoning selects a meaningful intermediate. A route that starts by making but-1-ene and then blindly hydrating it adds avoidable selectivity and separation issues. Check that each proposed arrow is a transformation in the AS toolkit or one supplied by the question.

For an aldehyde target, choose a primary alcohol and controlled oxidation with distillation. For a carboxylic acid target, choose the primary alcohol with excess oxidant and reflux. A reagent name without the product-controlling conditions does not finish the route.

Audit atoms and reagent compatibility at every arrow

Write the full intermediate structure, not just “alcohol”. CH₃CH₂CH₂OH and CH₃CH(OH)CH₃ lead to different oxidation products. Keep branches and carbon count in the same positions unless the stated transformation changes them.

Give each stage its own reagents and conditions. Aqueous hydroxide hydrolysis and acidified oxidation normally require appropriate separation or work-up between stages; writing all reagents together can neutralise the hydroxide or expose several groups to conflicting conditions.

For a molecule with more than one group, inspect every site. For example HOCH₂CH₂CH=CH₂ contains a primary alcohol and an alkene. It can show alcohol-related hydrogen bonding and alkene addition, but a claimed selective preparation must be supported by the specified reagent and conditions. Identifying a possible reaction is not a guarantee of one pure product.

Worked two-stage amount and final mass

An illustrative route converts 0.100 mol of a haloalkane to an alcohol at 80.0% isolated yield, then to a ketone at 75.0% yield. Both steps are 1:1. After step 1 there are 0.0800 mol alcohol; after step 2 there are 0.0600 mol ketone. Overall yield = 0.800 × 0.750 × 100 = 60.0%.

If the ketone has M = 72.0 g mol⁻¹, the isolated mass is 0.0600 × 72.0 = 4.32 g. For a target of 8.64 g under those same assumptions, work backwards: required final amount = 8.64/72.0 = 0.120 mol; starting amount = 0.120/0.600 = 0.200 mol.

Yield losses multiply because the second percentage applies to the material actually entering that step. Averaging 80.0 and 75.0 gives no valid measure of the overall conversion. Atom economy is a different quantity, based on the balanced equation rather than the amount isolated.

Quick checks

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

Sources