A2 · Practice questions · Multi-step organic synthesis

From start to target.

Six original Cambridge-style questions on multi-step organic synthesis.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Groups, reagents, types, yield.

Label every group before choosing a reagent.
Work backwards from the target, one group at a time.
Each step: reagent, conditions, reaction type.
Overall yield: multiply the step yields.

Analysis

[4 marks]

Tyrosine is HOC₆H₄CH₂CH(NH₂)COOH. Name its three functional groups and predict what happens with (a) NaOH(aq), (b) Br₂(aq), (c) dilute HCl.

Phenol, primary amine, carboxylic acid (it also has a chiral carbon). ✓
(a) Both the COOH and the phenol OH are deprotonated, giving a salt. ✓
(b) The phenol-activated ring is substituted by bromine: decolourised, white precipitate. ✓
(c) The NH₂ group is protonated to –NH₃⁺, giving a salt. ✓

Recall

[4 marks]

Give the reagents and conditions for each step of the conversion benzene → nitrobenzene → phenylamine → benzenediazonium chloride → 4-(phenylazo)phenol.

Conc. HNO₃ and conc. H₂SO₄, 25–60 °C. ✓
Sn and conc. HCl, heat under reflux, then NaOH(aq). ✓
NaNO₂ and dilute HCl (HNO₂), below 10 °C. ✓
Phenol in NaOH(aq), below 10 °C. ✓

Analysis

[3 marks]

Suggest a three-step route from methylbenzene to N-ethylbenzamide, C₆H₅CONHC₂H₅. Give reagents and conditions and the type of each reaction.

Hot alkaline KMnO₄, reflux, then dilute acid: benzoic acid (oxidation). ✓
SOCl₂ (or PCl₅): benzoyl chloride (substitution). ✓
Ethylamine, room temperature: N-ethylbenzamide (addition-elimination, condensation). ✓

Explain

[3 marks]

Propanone and ethanal each react with HCN and a KCN catalyst. Explain why only the product from ethanal is chiral, and why that product is formed as a racemic mixture.

Ethanal gives CH₃CH(OH)CN: the central carbon has four different groups (CH₃, H, OH, CN), so it is chiral. Propanone gives (CH₃)₂C(OH)CN, with two identical CH₃ groups, so it is achiral. ✓
The C=O carbon is trigonal planar. ✓
CN⁻ attacks equally from above and below the plane, so equal amounts of the two enantiomers form. ✓

Calculation

[3 marks]

15.6 g of benzene is converted into 4-(phenylazo)phenol (M_r 198.0) in three stages with yields of 85%, 70% and 60%. Calculate the overall yield and the mass of dye obtained.

Overall yield = 0.85 × 0.70 × 0.60 = 0.357 = 35.7%. ✓
n(benzene) = 15.6 ÷ 78.0 = 0.200 mol, so the maximum is 0.200 × 198.0 = 39.6 g. ✓
Mass obtained = 39.6 × 0.357 = 14.1 g. ✓

Explain

[3 marks]

Explain why pharmaceutical companies aim to make a drug as a single enantiomer, and give two ways of doing so.

Usually only one enantiomer binds to the chiral receptor or enzyme; the other may be inactive or cause harmful side effects. ✓
A single enantiomer means a smaller dose, fewer side effects and no wasted half of the product. ✓
Methods: a chiral catalyst or enzyme, a naturally chiral starting material, or separating (resolving) the racemic mixture. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.