AS · Practice questions · Organic synthesis and reaction pathways

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Six original Cambridge-style questions on organic synthesis and reaction pathways.

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, carbons, reagents, conditions.

Groups first: name every functional group, then predict its reactions.
Count carbons: one more carbon means a CN⁻ step.
Every step: reagent, conditions and type of reaction.
Overall yield: multiply the step yields.

Analysis

[4 marks]

Devise a three-step synthesis of ethyl ethanoate using ethene as the only organic starting material. Give the reagents and conditions for each step.

Step 1: ethene + steam, H₃PO₄ catalyst, 300 °C, 6 MPa → ethanol (electrophilic addition). ✓
Step 2: oxidise part of the ethanol with excess acidified K₂Cr₂O₇, heat under reflux → ethanoic acid. ✓
Step 3: heat ethanoic acid with the rest of the ethanol and a conc. H₂SO₄ catalyst → ethyl ethanoate + water (esterification). ✓
The ethanol from step 1 must be split: some is oxidised, the rest is the alcohol for step 3. ✓

Analysis

[4 marks]

Bromoethane is converted into propanoic acid in two steps. (a) Give the reagents, conditions and intermediate. (b) 21.8 g of bromoethane (M_r 108.9) is used and the step yields are 75.0% and 80.0%. Calculate the mass of propanoic acid (M_r 74.0) formed.

(a) KCN in ethanol, heat under reflux, gives propanenitrile, CH₃CH₂CN (nucleophilic substitution). ✓
Then dilute HCl(aq), heat under reflux, hydrolyses the nitrile to CH₃CH₂COOH. ✓
(b) n(bromoethane) = 21.8 ÷ 108.9 = 0.200 mol; overall yield = 0.750 × 0.800 = 0.600. ✓
n(acid) = 0.120 mol, mass = 0.120 × 74.0 = 8.88 g. ✓

Analysis

[3 marks]

Prop-2-en-1-ol is CH₂=CHCH₂OH. Name its two functional groups and give the organic product with (a) bromine water, (b) sodium, (c) acidified K₂Cr₂O₇, distilling the product as it forms.

Alkene (C=C) and primary alcohol (–CH₂OH). ✓
(a) CH₂BrCHBrCH₂OH (bromine water decolourised); (b) CH₂=CHCH₂O⁻Na⁺ and H₂. ✓
(c) the aldehyde CH₂=CHCHO (propenal); the C=C is not attacked by dichromate. ✓

Recall

[3 marks]

Ethanal is converted into 2-hydroxypropanoic acid. Give the two steps with reagents and conditions, and name the type of reaction in each.

Step 1: HCN with a KCN catalyst gives 2-hydroxypropanenitrile, CH₃CH(OH)CN. ✓
This is nucleophilic addition (CN⁻ attacks the δ+ carbonyl carbon) and adds one carbon atom. ✓
Step 2: heat under reflux with dilute HCl(aq): hydrolysis of –CN to –COOH, giving CH₃CH(OH)COOH. ✓

Explain

[3 marks]

A student makes propanal from 1-bromopropane: step 1 NaOH, heat; step 2 acidified K₂Cr₂O₇, heat under reflux. Identify two mistakes in the conditions and the unwanted products they would cause.

Step 1 must use NaOH(aq), heated under reflux; NaOH in ethanol would cause elimination and give propene. ✓
Step 2 under reflux oxidises the propanal further to propanoic acid. ✓
Instead, add the oxidant to warm propan-1-ol and distil the propanal off as it forms (b.p. 49 °C). ✓

Calculation

[2 marks]

A three-step synthesis has step yields of 90.0%, 80.0% and 75.0%. Starting with 0.150 mol of reactant, calculate the mass of product (M_r 88.0) obtained, assuming a 1 : 1 ratio in every step.

Overall yield = 0.900 × 0.800 × 0.750 = 0.540 (54.0%). ✓
n(product) = 0.150 × 0.540 = 0.0810 mol; mass = 0.0810 × 88.0 = 7.13 g. ✓

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.