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How do you plan a multi-step synthesis between functional groups?

Synthetic routes for preparing one organic compound from another in several steps. Reagents and conditions for the interconversion of functional groups in aliphatic and aromatic chemistry. Practical techniques for organic preparation, including purification and the determination of percentage yield.

A focused answer to the AQA A-Level Chemistry 3.3.15 specification points on organic synthesis. Covers planning multi-step routes, the key reagents and conditions for functional-group interconversions, and practical preparation, purification and percentage-yield techniques.

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  1. What this dot point is asking
  2. Planning a route
  3. Practical techniques
  4. Percentage yield
  5. Try this

What this dot point is asking

AQA wants you to plan multi-step synthetic routes between functional groups, recall the reagents and conditions for each interconversion, and describe the practical techniques used to prepare, purify and assess the yield of an organic product.

Planning a route

Work backwards from the target functional group, identifying which functional group can be converted into it, then repeat until you reach the starting material. Each arrow needs the reagent and conditions. Adding a carbon to the chain usually goes through a nitrile (from a halogenoalkane and KCNKCN), which is then hydrolysed to an acid or reduced to an amine.

For aromatic synthesis: benzene to nitrobenzene (nitration, HNO3/H2SO4HNO_3 / H_2SO_4), nitrobenzene to phenylamine (Sn / conc. HCl, then NaOHNaOH), and benzene to a phenyl ketone (Friedel-Crafts acylation, RCOCl/AlCl3RCOCl / AlCl_3).

Practical techniques

  • Reflux: heating without loss of volatile reagents, for slow reactions.
  • Distillation: separating a product by boiling point as it forms or afterwards.
  • Separating funnel: removing an immiscible organic layer from aqueous impurities.
  • Drying agent: e.g. anhydrous MgSO4MgSO_4 to remove water.
  • Recrystallisation: purifying a solid by dissolving in hot solvent, then cooling so the product crystallises and soluble impurities stay in solution.
  • Melting point: a pure solid melts sharply at the correct temperature; impurities lower and broaden it.

Percentage yield

Yields fall below 100% for several reasons that AQA expects you to recognise: reactions may be reversible and reach equilibrium (as in esterification), competing side reactions may form other products, the reaction may not go to completion, and product is always lost during transfers, filtration and purification. A long synthetic route suffers because the yields multiply: three steps each at 80% give an overall yield of only about 51%, which is why a shorter route with fewer steps is usually preferred even if each individual step is less efficient. Choosing reagents that give a high atom economy and minimise hazardous by-products is also part of designing a good synthesis.

Try this

Q1. Give the reagents to convert 1-bromopropane into propylamine. [1 mark]

  • Cue. Excess ammonia (in ethanol, heated under pressure).

Q2. Why is recrystallisation used in organic preparation? [1 mark]

  • Cue. To purify a solid product by removing soluble impurities.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksDescribe how propan-1-ol can be converted into propylamine in two steps, giving the reagents and conditions for each step and naming the intermediate.
Show worked answer →

Step 1: convert the alcohol to a halogenoalkane. React propan-1-ol with hydrogen bromide (made in situ from sodium bromide and concentrated sulfuric acid), reflux, to give 1-bromopropane.

Step 2: react 1-bromopropane with excess ammonia in ethanol, heated in a sealed tube, to give propylamine by nucleophilic substitution. (Reduction of a nitrile is not used here because that would add a carbon.)

The intermediate is 1-bromopropane.

Markers reward the alcohol-to-halogenoalkane step with reagents, the excess ammonia step, naming the intermediate, and keeping the carbon count constant.

AQA 20204 marksIn a preparation, 0.200 mol0.200\ \text{mol} of a carboxylic acid is reacted with excess alcohol to make an ester. The maximum (theoretical) yield of ester is 0.200 mol0.200\ \text{mol}, but only 14.6 g14.6\ \text{g} of ester (Mr=102M_r = 102) is obtained after purification. Calculate the percentage yield.
Show worked answer →

Actual moles of ester =14.6102=0.143 mol= \frac{14.6}{102} = 0.143\ \text{mol}.

Theoretical moles of ester =0.200 mol= 0.200\ \text{mol} (1:1 with the limiting acid).

Percentage yield =0.1430.200×100=71.6%= \frac{0.143}{0.200} \times 100 = 71.6\%.

Markers reward converting the mass to moles using MrM_r, identifying the theoretical moles, and the percentage-yield calculation (loss expected from the reversible reaction and purification).

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