Edexcel A-Level Chemistry Paper 2, November 2020: Question 4
8 marks · Medium difficulty · Open Response
Identify organic compounds based on reactions involving hydrolysis, sodium hydrogencarbonate, and functional group tests, and deduce the structures of unknown compounds T, U, and V with justifications.
Practise this questionQuestion
Question text
4 This question is about the identification of some organic compounds.
(a) The skeletal formulae of four organic compounds are shown.
Compound P Compound Q
O O
OH O
Compound R Compound S
O OH
O
(i) Which of these compounds can be hydrolysed to form methanol as one of
the products?
(1)
A Compound P
B Compound Q
C Compound R
D Compound S
(ii) Which of these compounds produces carbon dioxide when it reacts with
aqueous sodium hydrogencarbonate?
(1)
A Compound P
B Compound Q
C Compound R
D Compound S
(b) Compound T, C4H10O, is oxidised by acidified potassium dichromate(VI) to form
compound U, C4H8O.
U gives an orange precipitate with 2,4-dinitrophenylhydrazine (Brady’s reagent)
but does not give a red precipitate when heated with Fehling’s solution.
T reacts with ethanoyl chloride to form compound V, C6H12O2.
Deduce the structures of compounds T, U and V. Justify your answers.
10 (6)
*P62669A01032*
(Total for Question 4 = 8 marks)
Mark scheme
Show the mark scheme
How to answer it
Identification of Organic Compounds Study Guide
This question assesses your knowledge of functional group chemistry, including ester hydrolysis, acidity of carboxylic acids vs. phenols, reactions of carbonyl compounds (Brady's reagent and Fehling's solution), alcohol oxidation, and acylation reactions using acyl chlorides. You will need to interpret analytical/chemical test data to deduce unknown structures and justify your reasoning using clear chemical terminology.
Question 4(a)(i): Hydrolysis to Form Methanol
Which of these compounds can be hydrolysed to form methanol as one of the products?
✅ Correct Answer
B (Compound Q)
Compound Q is a methyl ester (methyl benzoate). Hydrolysis of Q cleaves the ester bond to produce benzoic acid and CH₃OH (methanol).
💡 Key Knowledge
- Esters hydrolyse in acidic or alkaline conditions to form a carboxylic acid (or carboxylate salt) and an alcohol.
- The alkyl chain attached to the oxygen atom of the ester group determines which alcohol is formed. Compound Q features a -OCH₃ group, yielding methanol.
❌ Common Errors & Distractor Analysis
- Compound A (P): A carboxylic acid; it does not undergo hydrolysis to form alcohols.
- Compound C (R): An ester, but it has an ethyl/alkyl group attached via the oxygen derived from ethanol/phenol pathways yielding different fragments, not methanol.
- Compound D (S): A phenol derivative; phenols do not hydrolyse in this manner.
Question 4(a)(ii): Reaction with Sodium Hydrogencarbonate
Which of these compounds produces carbon dioxide when it reacts with aqueous sodium hydrogencarbonate?
✅ Correct Answer
A (Compound P)
Compound P is benzoic acid ( C₆H₅COOH ). Carboxylic acids are sufficiently strong acids to react with weak bases like aqueous sodium hydrogencarbonate ( NaHCO₃ ), producing carbon dioxide gas ( CO₂ ).
🧠 Exam Technique
- Use the NaHCO₃ test as your go-to chemical test for carboxylic acids. Observation: effervescence / gas turns limewater cloudy.
- Remember that phenols (like compound S) are weaker acids than carboxylic acids and do not react with hydrogencarbonate to give CO₂ . Esters (Q, R) are neutral and do not react either.
Question 4(b): Structure Deduction and Justification
Deduce the structures of compounds T, U and V. Justify your answers. (6 marks)
✅ Correct Answers (Structures)
- Compound T: butan-2-ol ( CH₃CH₂CH(OH)CH₃ )
- Compound U: butanone ( CH₃CH₂COCH₃ )
- Compound V: 1-methylpropyl ethanoate (or standard structural/skeletal ester variant matching C₆H₁₂O₂ )
💡 Step-by-Step Analytical Breakdown
- Formula Clues: T is C₄H₁₀O (an alcohol or ether). It oxidises to U ( C₄H₈O ), losing 2 hydrogens, which confirms T is a secondary alcohol.
- Functional Group Tests on U: U gives an orange precipitate with 2,4-DNP (confirming a carbonyl: aldehyde or ketone). It fails to give a red precipitate with Fehling's solution, proving U is a ketone (not an aldehyde). Combined with the carbon chain length, U is butanone.
- Deducing T: Since oxidation of secondary alcohol T yields butanone, T must be butan-2-ol.
- Deducing V: T reacts with ethanoyl chloride ( CH₃COCl ) to form ester V ( C₆H₁₂O₂ ). Alcohols react with acyl chlorides to form esters and HCl.
❌ Common Errors & Where Students Lose Marks
- Primary vs. Secondary confusion: Choosing butan-1-ol for T. If T were butan-1-ol , oxidation would yield butanal (an aldehyde), which would react positively with Fehling's solution.
- Incomplete Justifications: Simply drawing structures without explicitly linking the test results (e.g., "U is a ketone because it reacts with Brady's but not Fehling's") loses the justification marks. You must state both observations and what they prove.
Topics
Organic Chemistry · Topic 6: Organic Chemistry I · Topic 17: Organic Chemistry II · Topic 18: Organic Chemistry III
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.