OCR A-Level Chemistry Unified chemistry (03), June 2024: Question 5

13 marks · Hard difficulty · Structured Questions

Analyze the alkaline hydrolysis of halogenoalkanes using TLC and deduce the structures of isomers F to I using test-tube reactions.

Practise this question

Question

Chemistry exam question with two parts. Part (a) describes an investigation into the alkaline hydrolysis of 1-bromopropane using TLC, showing four chromatograms at different time intervals (Before reflux, After 10 mins, After 20 mins, After 30 mins) with lanes for propan-1-ol and the reaction mixture. It asks to determine an Rf value, write an equation for the reaction, and predict how using 1-chloropropane would affect the results. Part (b) presents a table of test-tube reactions (2,4-DNP, acidified potassium dichromate, bromine water, and Tollens' reagent) for four structural isomers F, G, H, and I with molecular formula C5H10O, asking to identify species causing color changes and deduce possible structures for each isomer.
Question text

5 This question is about the analysis of organic compounds.

(a) A student investigates the alkaline hydrolysis of 1-bromopropane as outlined below.

Step 1 The student adds 1-bromopropane to an excess of aqueous potassium hydroxide,

KOH(aq), in a pear-shaped flask.

Step 2 A TLC chromatogram is run using propan-1-ol and the reaction mixture.

Step 3 The reaction mixture is refluxed.

A TLC chromatogram of the reaction mixture is run every 10 minutes.

The TLC chromatograms are shown below.

Before reflux After 10 minutes After 20 minutes After 30 minutes

solvent

front

origin

(i) Determine the Rf value of propan-1-ol.

Show your working.

Rf = … [1]

(ii) Write an equation for the alkaline hydrolysis of 1-bromopropane.

Show structures of organic compounds.

[1]

(iii) A student investigates the alkaline hydrolysis of 1-chloropropane using the same method as for

1-bromopropane.

Predict, with reasons, how the appearance of the reaction mixture in the chromatogram produced

after 20 minutes would be different when 1-chloropropane is used instead of 1-bromopropane.

Suggest why propan-1-ol is run alongside the reaction mixture.

… [3]

(b) Compounds F, G, H and I are structural isomers.

A student carries out test-tube tests on the compounds.

The student records the observations after carrying out each test.

These are shown in Table 5.1.

In Table 5.1, 2,4-dintrophenylhydrazine has been abbreviated to 2,4-DNP.

Table 5.1

Test

Compound 2,4-DNP Acidified Bromine water Tollens’ reagent

dichromate(VI)

reflux

Colourless Colourless

F Orange solution Green solution

solution solution

Colourless

G Orange solution Green solution Orange solution

solution

Orange Colourless

H Orange solution Orange solution

precipitate solution

Orange

I Green solution Orange solution Silver mirror

precipitate

(i) Write the formula of the species causing the colours after refluxing with acidified dichromate(VI).

Green solution …

Orange solution …

[2]

(ii)* The student is provided with further information about compounds F–I.

• They all have the molecular formula C5H10O.

• One of the compounds is alicyclic.

• The other compounds are unbranched.

Use this further information and the student’s observations in Table 5.1 to answer the following.

• How do the observations provide evidence for the possible functional groups in

compounds F–I?

• Suggest a possible structure for each of the compounds F–I.

Show your reasoning. [6]

Extra answer space if required.

Mark scheme

Show the mark scheme Mark scheme providing the acceptable answers, calculations for Rf values, balanced equations showing organic structures, detailed points for explaining the differences between 1-bromopropane and 1-chloropropane hydrolysis, formulas of chromium species, and level-of-response marking criteria with indicative structures for isomers F, G, H, and I.

Question Answer Marks Guidance

5 (a) (i) 1.4 14 1 ALLOW 0.12 – 0.18 (i.e. ±0.03)

Rf ~ in cm OR in mm = 0.15 ✓

9.1 91

Working required 1.4

DO NOT ALLOW = 0.14

Check for ~ 9.1 as denominator 10.1

10.1 measured from bottom of plate to solvent front

9.1 cm

10.1 cm

1.4 cm 1.4 cm

(a) (ii) 1 ALLOW any combination of skeletal OR structural OR

H H H H H H displayed formula as long as unambiguous

H C C C Br + OH– H C C C OH + Br–

DO NOT ALLOW Missing H atoms

H H H H H H

✓ DO NOT ALLOW H2O and HBr

Correct balanced equation Question asks for alkaline hydrolysis

ALLOW OH– above the arrow DO NOT ALLOW C H , i.e. C H Br OR C H OH

37 3 7 3 7

Structure asked for in Question

DO NOT ALLOW if a CON reagent is present,

e.g. an acid IGNORE connectivity, e.g.

ALLOW

For OH– and Br– OH

ALLOW KOH and KBr OR NaOH and NaBr BUT DO NOT ALLOW —HO

BUT DO NOT ALLOW K–OH

implies covalent bond

(a) (iii) 3 FULL ANNOTATIONS MUST BE USED

ALLOW ECF and ORA throughout

Difference ----------------------------------------------------

propan-1-ol/product/bottom spot is smaller IGNORE references to halogens as elements:

OR 1-chloropropane/reactant/top spot bigger ✓ i.e. chlorine is less reactive than bromine etc.

Reasons DO NOT ALLOW chloride, bromide

C–Cl bond is stronger than C–Br

AND DO NOT ALLOW 1-chloropropane has larger bond enthalpy

1-chloropropane reacts slower/is less reactive ✓ C–Cl bond required

IGNORE 1-chloroproane has different Rf value

Use of propan-1-ol

shows formation of propan-1-ol IGNORE ‘as a control’ OR ‘as a comparison’

OR shows when reaction has finished … with no further explanation

OR monitors course/progress of reaction ✓

(b) (i) Green solution Cr3+ OR [Cr(H O) ]3+ ✓ 2 Green solution

IGNORE H+

ALLOW Cr (SO ) OR CrCl OR Cr+3

24 3 3

Orange solution Cr O 2– ✓ Orange solution

IGNORE H+

Formulae AND charges must be correct ALLOW K2Cr2O7 OR Na2Cr2O7

DO NOT ALLOW Cr6+

ALLOW 1 mark for correct formulae but wrong way round

Question Answer 24 Marks Guidance

(b)* (ii) Please refer to the marking instructions on page 6 of 6 Indicative scientific points may include:

this mark scheme for guidance on how to mark this Identity of F, G, H and I showing CORRECT structures

question.

Level 3 (5–6 marks)

Reaches a comprehensive conclusion to determine

possible correct structures for ALL of F, G, H and I

AND ALL functional groups of F, G, H and I

There is a well-developed line of reasoning which is clear and

logically structured.

The information presented is relevant and substantiated. ALLOW enols for F, e.g.

Level 2 (3–4 marks)

Reaches a conclusion to determine possible correct

structures for two of F, G, H and I

AND most functional groups of F, G, H and I

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by some

evidence.

Level 1 (1–2 marks)

For G, DO NOT ALLOW tertiary –OH. e.g.

Reaches a simple conclusion to determine a possible

correct structure for one of F, G, H and I

OR some functional groups of F, G, H and I

There is an attempt at a logical structure with a line of

reasoning. The information is in the most part relevant.

0 marks No response or no response worthy of credit.

IGNORE names, even if incorrect

For communication, a typical ‘logical structure’ would link

functional groups to SOME of the test results,

e.g.

2,4-DNP

H and I have carbonyl group/aldehyde or ketone

H+/Cr O 2–

F, G and I are primary or secondary alcohols or aldehydes

Bromine

F is unsaturated/has C=C

Tollens

I is aldehyde

*Correct functional groups may be shown in correct

structures*

How to answer it

Analysis of Organic Compounds & Reaction Kinetics

What this question tests

This comprehensive multi-part question evaluates core organic chemistry practical and analytical skills: calculating Rf values from TLC plates, writing balanced organic reaction mechanisms/equations, applying knowledge of halogenoalkane bond enthalpies to reaction rates, identifying inorganic ion species from colour changes, and performing rigorous logical deduction of organic isomer structures (C₅H₁₀O) using chemical test results.

Part (a)(i): Calculating Rf Values

Determining the Rf value of propan-1-ol

✅ Correct Answer

Rf = 0.12 to 0.18 (or exact fractions like 1.4 / 9.1 giving ~0.15)

📐 Calculation Steps

  1. Measure distance moved by substance: From the origin line to the centre of the propan-1-ol spot = 1.4 cm (or 14 mm).
  2. Measure distance moved by solvent: From the origin line to the solvent front = 9.1 cm (or 91 mm).
  3. Apply formula: Rf = (Distance moved by spot) / (Distance moved by solvent front) = 1.4 / 9.1 = 0.15 .

❌ Common Errors

  • Measuring the solvent front from the very bottom of the TLC plate instead of the origin line ( 10.1 cm measurement trap).
  • Inverting the ratio (solvent over spot distance), yielding an impossible Rf value greater than 1.
Marks: [1]

Part (a)(ii): Alkaline Hydrolysis Equation

Writing the equation for alkaline hydrolysis of 1-bromopropane

✅ Correct Answer

CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂OH + Br⁻

(Skeletal, structural, or displayed formulas are all fully accepted as long as they are unambiguous).

💡 Key Knowledge

  • Alkaline hydrolysis uses hydroxide ions ( OH⁻ ), typically sourced from aqueous KOH or NaOH.
  • The hydroxide can be written above the reaction arrow, or as part of the reactant formula.

❌ Common Errors

  • Including water ( H₂O ) or hydrogen bromide ( HBr ) in the equation instead of ionic products. Remember this is alkaline hydrolysis.
  • Omitting hydrogen atoms or drawing covalent bonds incorrectly into the negative charge of OH⁻ or Br⁻ .
Marks: [1]

Part (a)(iii): Comparing Halogenoalkanes via TLC

1-chloropropane vs 1-bromopropane kinetics and TLC monitoring

✅ Correct Answers

  • Difference: The product/bottom spot is smaller for 1-chloropropane (or unreacted 1-chloropropane spot is bigger).
  • Reason: The C-Cl bond is stronger than the C-Br bond, meaning 1-chloropropane reacts slower / is less reactive.
  • Use of propan-1-ol: Shows formation of propan-1-ol, shows when the reaction has finished, or monitors the course/progress of the reaction.

🧠 Exam Technique

When linking reaction rate to bond strength, always explicitly state that C-Cl bond enthalpy is higher / C-Cl bond is stronger than C-Br, requiring more energy to break, hence slower nucleophilic substitution. Vague references to "chlorine is less reactive than bromine" lose marks.

Marks: [3]

Part (b)(i): Identifying Inorganic Ions from Colour Tests

Formulas of species causing specific test tube observations

✅ Correct Answers

  • Green solution: Cr³⁺ OR [Cr(H₂O)₆]³⁺
  • Orange solution: Cr₂O₇²⁻

💡 Key Knowledge

Acidified potassium dichromate(VI) acts as an oxidising agent (containing orange Cr₂O₇²⁻ ). Upon successful oxidation of primary/secondary alcohols or aldehydes, the dichromate is reduced to green chromium(III) ions ( Cr³⁺ ).

Marks: [2]

Part (b)(ii)*: Structural Isomer Deduction (C₅H₁₀O)

Deducing structures for isomers F, G, H, and I using systematic testing

💡 Systematic Functional Group Analysis

  • 2,4-DNP: Forms an orange precipitate with H and I, proving they are carbonyl compounds (aldehydes or ketones). Compounds F and G only show an orange solution, meaning they lack a carbonyl and are likely alcohols or alkenes.
  • Acidified Dichromate(VI): Turns green with F, G, H, and I. This confirms F and G are oxidisable (primary/secondary alcohols) and H is an aldehyde (which can be further oxidised).
  • Bromine Water: Decolourises with F, G, and H, proving the presence of a carbon-carbon double bond (alkene). I remains orange (no C=C bond).
  • Tollens' Reagent: Produces a silver mirror only with I, confirming I is a specific aldehyde, while H is a ketone.
  • Alicyclic vs Unbranched: The prompt notes one compound is alicyclic (cyclic ring structure without aromatic rings), which matches compound G.

✅ Valid Possible Structures (C₅H₁₀O)

  • Compound F (Alcohol + Alkene): Pent-3-en-1-ol, pent-2-en-1-ol, or similar unsaturated straight-chain alcohols with an OH group.
  • Compound G (Alicyclic Alcohol): Cyclopentanol or alkyl-substituted cyclobutanol derivatives (non-tertiary).
  • Compound H (Ketone + Alkene): Unsaturated ketones such as pent-3-en-2-one.
  • Compound I (Saturated Aldehyde): Pentanal ( CH₃CH₂CH₂CH₂CHO ).
Marks: [6] — Level 3 requires comprehensive correct structures for all 4 isomers with a clear, logical, structured line of reasoning linking test observations to functional groups.

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 4.2 Alcohols, haloalkanes and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.3 Analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.