Edexcel A-Level Chemistry Paper 3, June 2019: Question 3

8 marks · Hard difficulty · Open Response

Devise a synthetic pathway to prepare propyl propanoate from 1-bromopropane, and deduce the structures of alcohol B and ester A based on their reactions.

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Question

Question 3 presents two parts about esters with formula C6H12O2. Part (a) shows the displayed formula of propyl propanoate and asks to devise a multi-step synthetic pathway starting from 1-bromopropane as the only organic compound, including reagents and intermediate structures (5 marks). Part (b) states that an ester A is hydrolysed to give ethanoic acid and an alcohol B (C4H10O0), which undergoes elimination to form but-1-ene and but-2-ene, and asks to deduce and justify the structures of A and B (3 marks).
Question text

3 This question is about esters with the molecular formula C6H12O2.

(a) Propyl propanoate has the structure shown.

H H O H H H

H C C C O C C C H

H H H H H

Devise a synthetic pathway to prepare propyl propanoate starting with

1-bromopropane as the only organic compound.

Include the reagents for each step in the synthesis, and the names or structures of

the intermediate compounds.

(5)

(b) Another ester, A, with molecular formula C6H12O2, was hydrolysed.

It produced ethanoic acid, and an alcohol, B, with molecular formula C4H10O.

Alcohol B undergoes an elimination reaction to produce a mixture of

but-1-ene and but-2-ene.

Deduce the structures of B and A. Justify your structure of B.

(3)

*P58308A0632*

(Total for Question 3 = 8 marks)

Mark scheme

Show the mark scheme The mark scheme provides acceptable answers and guidance for both parts. Part (a) outlines two routes (via propanoic acid or propanoyl chloride) involving conversion of 1-bromopropane to propan-1-ol, oxidation to propanoic acid/acyl chloride, and reaction to form the ester. Part (b) identifies alcohol B as butan-2-ol, ester A, and provides the justification based on elimination products.

How to answer it

Edexpert Study Guide: Esters & Synthetic Routes (C6H12O2)

What this question tests

This question assesses your core knowledge of organic synthesis pathways, functional group interconversions (halogenoalkanes to alcohols to carboxylic acids/acyl chlorides), esterification reactions, alcohol structural isomerism, and elimination mechanisms to form alkenes.

Question Part (a) - Synthetic Pathway (5 Marks)

Devising a Synthesis for Propyl Propanoate from 1-Bromopropane

💡 Key Knowledge

  • Starting Material: CH₃CH₂CH₂Br (1-bromopropane).
  • Target Product: Propyl propanoate ( CH₃CH₂COOCH₂CH₂CH₃ ), a 6-carbon ester. Since the starting material only has 3 carbons, you must split your supply: convert half into an alcohol ( propan-1-ol ) and half into a carboxylic acid ( propanoic acid ) or acyl chloride.

✅ Correct Answer (Route 1)

  • Step 1 (Alcohol): Aqueous sodium/potassium hydroxide ( NaOH(aq) or KOH(aq) ) under reflux. Produces propan-1-ol ( CH₃CH₂CH₂OH ).
  • Step 2 (Carboxylic Acid): Acidified potassium dichromate( VI ) ( K₂Cr₂O₇ / H⁺ ) under reflux. Produces propanoic acid ( CH₃CH₂COOH ).
  • Step 3 (Esterification): React propan-1-ol and propanoic acid with concentrated sulfuric acid ( conc. H₂SO₄ ) catalyst under reflux/warm.

🧠 Exam Technique

  • Mark schemes are standalone per step: Reagent + intermediate pair = 2 marks, next step reagent + intermediate = 2 marks, esterification = 1 mark.
  • Always specify (aq) for the substitution reaction to form the alcohol, otherwise elimination ( alc. KOH ) will occur instead!

❌ Common Errors

  • Using hydrochloric acid ( HCl ) for oxidation (only H₂SO₄ or H₃PO₄ work safely with dichromate).
  • Forgetting that since 1-bromopropane is the only organic starting material, you must explicitly show how both the alcohol and acid portions of the ester are synthesized from it.
Mark Breakdown: 1 mark for reagent to form alcohol + 1 mark for propan-1-ol structure/name | 1 mark for oxidizing reagent + 1 mark for propanoic acid structure/name | 1 mark for esterification conditions/reactants. (Total: 5 marks)
Question Part (b) - Structure Deduction & Justification (3 Marks)

Deducing Structures of Alcohol B and Ester A

✅ Correct Answers

  • Alcohol B: Butan-2-ol ( CH₃CH(OH)CH₂CH₃ ).
  • Ester A: Ethanoate ester formed from ethanoic acid and butan-2-ol, i.e., sec-butyl ethanoate ( CH₃COOCH(CH₃)CH₂CH₃ ).
  • Justification: Butan-2-ol undergoes acid-catalyzed elimination (dehydration) to yield a mixture of but-1-ene and but-2-ene because the adjacent carbons to the -OH group can lose an H from either side.

💡 Key Knowledge

  • Ester hydrolysis splits C₆H₁₂O₂ into a carboxylic acid ( ethanoic acid , CH₃COOH , 2 carbons) and an alcohol B ( C₄H₁₀O , 4 carbons).
  • Elimination of water from an alcohol yields alkenes. Only butan-2-ol yields a mixture of both but-1-ene and but-2-ene . Butan-1-ol only yields but-1-ene .

🧠 Exam Technique

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  • Ensure structural or displayed formulae clearly show connectivity (e.g., attach the -OH to carbon-2 for alcohol B).
  • In your justification, explicitly state why butan-2-ol gives two different alkenes (hydrogen can be eliminated from carbon-1 or carbon-3 relative to the -OH -bearing carbon).

❌ Common Errors

  • Selecting butan-1-ol : students forget that primary alcohols cannot yield positional alkene isomer mixtures upon simple dehydration without skeletal rearrangement.
  • Failing to link the ester components back to the stated products of hydrolysis.
Mark Breakdown: 1 mark for structure of alcohol B | 1 mark for structure of ester A | 1 mark for clear justification naming butan-2-ol and explaining the formation of both but-1-ene and but-2-ene. (Total: 3 marks)

Topics

Organic Chemistry · Topic 17: Organic Chemistry II · Topic 18: Organic Chemistry III · Topic 6: Organic Chemistry I

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.