9701 · 21.1

Organic synthesis — practice questions

Practice and worked examples for 9701 Organic synthesis. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Devise a two-step synthesis for 2-hydroxypropanoic acid ( $CH_3CH(OH)COOH$ ) starting from ethanal ( $CH_3CHO$ ). State all reagents and conditions.

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Retrosynthetic Analysis:

Target: 2-hydroxypropanoic acid, $CH_3CH(OH)COOH$ . It contains both an alcohol and a carboxylic acid group.
Step Back 1: The carboxylic acid group ( $-COOH$ ) can be formed by the acid hydrolysis of a nitrile group ( $-CN$ ). This suggests the immediate precursor is 2-hydroxypropanenitrile. $CH_3CH(OH)COOH \Rightarrow CH_3CH(OH)CN$ .
Step Back 2: The precursor, 2-hydroxypropanenitrile, can be formed by the nucleophilic addition of hydrogen cyanide to a carbonyl compound. The structure $CH_3CH(OH)CN$ points to the starting carbonyl being ethanal, $CH_3CHO$ . This matches our given starting material. The plan is complete.

Forward Synthesis:

Step 1: Ethanal to 2-hydroxypropanenitrile
- Reagents: Potassium cyanide ( $KCN(aq)$ ) and dilute sulfuric acid ( $H_2SO_4(aq)$ ). (This generates $HCN$ in situ).
- Conditions: Room temperature.
- Equation: $CH_3CHO + HCN \rightarrow CH_3CH(OH)CN$
Step 2: 2-hydroxypropanenitrile to 2-hydroxypropanoic acid
- Reagents: Dilute hydrochloric acid ( $HCl(aq)$ ).
- Conditions: Heat under reflux.
- Equation: $CH_3CH(OH)CN + 2H_2O + H^+ \rightarrow CH_3CH(OH)COOH + NH_4^+$

Worked example 2

Devise a synthetic route to convert propan-1-ol ( $CH_3CH_2CH_2OH$ ) into ethyl propanoate ( $CH_3CH_2COOCH_2CH_3$ ). You may use any other organic or inorganic reagents. Identify all intermediates.

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Analysis:

The target molecule is an ester, ethyl propanoate. Esters are formed from a carboxylic acid and an alcohol via esterification.
The 'propanoate' part of the name indicates it is derived from propanoic acid ( $CH_3CH_2COOH$ ).
The 'ethyl' part of the name indicates it is derived from ethanol ( $CH_3CH_2OH$ ).
Our starting material is propan-1-ol. We can use this to make the propanoic acid component.

Route Plan:

Part A: Synthesise propanoic acid from propan-1-ol.
Part B: React the synthesised propanoic acid with ethanol (which the question allows us to use) to form the final ester product.

Forward Synthesis:

Step 1: Synthesis of Propanoic Acid
- Reaction: Oxidation of a primary alcohol to a carboxylic acid.
- Reagents: Acidified potassium dichromate(VI), $K_2Cr_2O_7(aq) / H_2SO_4(aq)$ .
- Conditions: Heat under reflux.
- Intermediate: Propanoic acid, $CH_3CH_2COOH$ .
- Equation: $CH_3CH_2CH_2OH + 2[O] \rightarrow CH_3CH_2COOH + H_2O$
Step 2: Synthesis of Ethyl Propanoate
- Reaction: Esterification.
- Reagents: Ethanol ( $CH_3CH_2OH$ ) and a few drops of concentrated sulfuric acid catalyst.
- Conditions: Heat (often under reflux).
- Final Product: Ethyl propanoate, $CH_3CH_2COOCH_2CH_3$ .
- Equation: $CH_3CH_2COOH + CH_3CH_2OH \rightleftharpoons CH_3CH_2COOCH_2CH_3 + H_2O$

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