OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2021: Question 19

14 marks · Hard difficulty · Structured Questions

Name prenal, complete a synthetic flowchart involving its reactions, and plan a two-stage synthesis of 2-phenylethanoic acid from (chloromethyl)benzene including percentage yield calculations.

Practise this question

Question

Question 19 starts with prenal, an alpha,beta-unsaturated aldehyde ((CH3)2C=CHCHO). Part (a)(i) asks for its systematic name. Part (a)(ii) shows a reaction flowchart starting from prenal with missing structures and reagents leading to two different esters via oxidation, reduction with NaBH4, hydrogenation, and esterification. Part (b)* is an extended response question asking students to plan a two-stage synthesis of 5.44 g of Compound A (2-phenylethanoic acid) starting from (chloromethyl)benzene with a 25% overall yield, including calculation of the starting mass, reagents, and balanced equations.
Question text

19 This question is about organic synthesis.

(a) Prenal, shown below, is used in the synthesis of some pharmaceuticals.

O

prenal

(i) What is the systematic name for prenal?

… [1]

(ii) Complete the flowchart below for the synthesis of two compounds starting from prenal.

O

NaBH4

prenal

reagent(s): …

reagent(s): …

catalyst: …

reagent(s): … reagent(s): …

catalyst: H+ catalyst: H+

O O

O O

[7]

(b)* A student intends to synthesise compound A using the two-stage route below.

O

Stage 1 Stage 2

CH2 Cl Intermediate CH2 C

OH

(chloromethyl)benzene Compound A

Plan a two-stage synthesis to prepare 5.44 g of compound A starting from

(chloromethyl)benzene, C6H5CH2Cl. Assume that the overall percentage yield of compound

A from (chloromethyl)benzene is 25%.

In your answer, include the mass of C6H5CH2Cl required, reagents, and equations where

appropriate.

Purification details are not required. [6]

Mark scheme

Show the mark scheme Mark scheme for question 19: (a)(i) identifies prenal as 3-methylbut-2-enal. (a)(ii) credits the structure of 3-methylbut-2-en-1-ol from NaBH4 reduction, oxidation of prenal using Cr2O7 2-/H+ to 3-methylbut-2-enoic acid, followed by reaction with (CH3)2CHOH to make the first ester; on the right route, hydrogenation using H2/Ni gives 3-methylbutan-1-ol, which reacts with CH3COOH to form the second ester. (b)* level of response mark scheme provides criteria for 1-6 marks: calculating moles of A = 0.0400 mol, required mass of C6H5CH2Cl = 20.2 g (accounting for 25% yield), specifying CN- in ethanol for stage 1 (forming phenylacetonitrile), and aqueous acid hydrolysis (H+/H2O) for stage 2, accompanied by balanced equations.

AO

Question Answer Marks Guidance

element

19 (a) (i) 3-methylbut-2-enal 1 AO1.2 IGNORE lack of hyphens, or addition of commas

(ii) 7 AO1.2 ALLOW any combination of skeletal OR structural

×4 OR displayed formula as long as unambiguous

AO2.5 ALLOW names of reagents and catalyst

×3

For oxidation,

ALLOW K Cr O for Cr O 2–

22 7 2 7

ALLOW H SO for H+

For left hand side esterification

IGNORE C3H7OH

IF esterification is given instead of hydrogenation

contact your Team Leader

14 AO

element

(b)* Refer to marking instructions on page 5 of mark scheme for 6 AO2.4 Indicative scientific points may include:

guidance on marking this question. ×2

Calculation of mass of C6H5CH2Cl

Level 3 (5-6 marks) AO2.7

Correct calculation of the mass of C6H5CH2Cl ×2 Using moles

AND 5.44

• n(A) =

Planned synthesis to form the intermediate C6H5CH2CN AO3.3 136

followed by hydrolysis to form A with most of the reagents ×2 = 0.04(00) (mol)

identified and equations are mostly correct. 100

• n(C6H5CH2Cl) = 0.0400 ×

There is a well-developed line of reasoning which is clear = 0.16(0) (mol)

and logically structured. The information presented is • Mass of C6H5CH2Cl = 126.5 × 0.16

relevant and substantiated. = 20.2(4) g

Level 2 (3-4 marks) Using mass

Correct calculation of the mass of C6H5CH2Cl 100

AND • Theoretical mass of ester = 5.44 ×

Planned synthesis to form the intermediate C6H5CH2CN = 21.76 (g)

with most of the reagents identified and equation is mostly 21.76

correct • Theoretical n(C6H5CH2Cl) =

OR = 0.16(0) (mol)

Calculation of the mass of C6H5CH2Cl is partly correct • Mass of C6H5CH2Cl = 126.5 × 0.160

AND = 20.2(4) g

Planned synthesis includes formation of the intermediate

C6H5CH2CN followed by hydrolysis to form A with some of ALLOW small slip/rounding errors such as errors in

the reagents identified Mr e.g. use of 137 instead of 136 for

OR C H CH COOH

65 2

Attempts to calculate mass of C6H5CH2Cl but makes little -----------------------------------------------------------------

progress Examples of partly correct calculations

AND 25

Planned synthesis includes formation of the intermediate Mass = 1.265 g from 0.0400 × ×126.5

C6H5CH2CN followed by hydrolysis to form A with most of

(% yield inverted)

the reagents identified and equations are mostly correct

Mass = 5.06 g from 0.0400 × 126.5

(% yield omitted)

15 AO

element

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by Synthesis: reagents and conditions

some evidence.

Stage 1: Formation of intermediate, C6H5CH2CN

Level 1 (1-2 marks) • Reagents: CN–(/ethanol)

Calculation of the mass of C6H5CH2Cl is partly correct • Equation:

OR C H CH Cl + CN– → C H CH CN + Cl–

65 2 6 5 2

Attempts to calculate mass of C6H5CH2Cl but makes little OR C6H5CH2Cl + NaCN → C6H5CH2CN + NaCl

progress (OR use of KCN)

AND

Planned synthesis includes formation of the intermediate Stage 2: Formation of A, C6H5CH2COOH

C6H5CH2CN with the reagent identified • Reagents: H+/H2O (ALLOW ‘acid hydrolysis’

OR • Equation:

Planned synthesis includes both steps with some of the C H CH CN + 2H O + H+ → C H CH COOH +

65 2 2 6 5 2

reagents identified +

NH4

OR OR C H CH CN + 2H O + HCl → C H CH COOH +

65 2 2 6 5 2

Attempts equations for both steps but these may contain NH4Cl

errors

OR

Describes one step of the synthesis with reagent(s) and

equation mostly correct

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.

Total 18

How to answer it

Module 6: Organic Chemistry & Analysis

Multi-Step Organic Synthesis & Quantitative Planning

What this question tests

A rigorous assessment spanning systematic IUPAC nomenclature, selective functional group interconversions (oxidation, reduction, hydrogenation, esterification), and a full level-of-response multi-stage synthetic plan with stoichiometric percentage yield calculations.

Core Knowledge Tested:
  • IUPAC naming of unsaturated carbonyl compounds (alkenals).
  • Selective reduction: NaBH₄ reduces C=O but does not affect C=C.
  • Catalytic hydrogenation of alkene bonds using H₂/Ni.
  • Esterification of carboxylic acids and alcohols using H⁺.
  • Nitrile formation via nucleophilic substitution (extending carbon chains).
  • Acid hydrolysis of nitriles to carboxylic acids.
Key Skills Assessed:
  • Predicting reaction pathways and intermediate structures.
  • Deducing required reagents and catalysts from target structures.
  • Multi-step backward stoichiometric calculation involving percentage yield.
  • Constructing balanced stoichiometric organic and ionic equations.
Part (a)(i) • 1 Mark

Systematic Nomenclature of Prenal

✅ Correct Answer

3-methylbut-2-enal

Mark scheme: 1 mark for the complete systematic name. (Lack of hyphens or addition of commas is ignored).

💡 Key Knowledge

  • Principal functional group: Aldehyde ( -CHO ), which always takes C1 priority.
  • Longest continuous carbon chain: 4 carbons containing both the C=C and C=O = but.
  • Alkene position: Double bond begins at C2 = but-2-en-.
  • Aldehyde suffix: -al (the '1' is implied, so but-2-enal ).
  • Substituent: Methyl branch at carbon 3 = 3-methyl.

❌ Common Errors

  • Numbering from the wrong end (e.g. 2-methylbut-2-enal). The carbonyl carbon must be C1.
  • Omitting the double bond locator (e.g. 3-methylbutenal).
  • Calling the parent chain pentanal by incorrectly counting the branched methyl into the main stem.

🧠 Exam Technique

Always number the chain so the highest priority functional group receives the lowest number. For aldehydes, carboxylic acids, acyl chlorides, and nitriles, the functional carbon is definitively C1.

Part (a)(ii) • 7 Marks

Synthesis Flowchart from Prenal

Complete the flowchart identifying intermediates, reagents, and catalysts across both synthetic pathways.

✅ Right-Hand Pathway (Reduction & Esterification)

  • Intermediate 1 (top-right box):
    (CH₃)₂C=CHCH₂OH (3-methylbut-2-en-1-ol)
    NaBH₄ selectively reduces the aldehyde (-CHO) to a primary alcohol (-CH₂OH), leaving the C=C bond unchanged.
  • Reagents & Catalyst (going down):
    Reagent: H₂
    Catalyst: Ni (nickel)
  • Intermediate 2 (middle-right box):
    (CH₃)₂CHCH₂CH₂OH (3-methylbutan-1-ol)
    Hydrogenation saturates the C=C double bond.
  • Reagent (bottom-right line):
    CH₃COOH (ethanoic acid)
    Reacts with the alcohol in the presence of H⁺ catalyst to give 3-methylbutyl ethanoate.
4 marks awarded on this branch (1 mark per box/line).

✅ Left-Hand Pathway (Oxidation & Esterification)

  • Reagent(s) (going down from prenal):
    Cr₂O₇²⁻ AND H⁺ (or acidified potassium dichromate, K₂Cr₂O₇ / H₂SO₄ ).
    Must include both the oxidising agent and acid!
  • Intermediate (middle-left box):
    (CH₃)₂C=CHCOOH (3-methylbut-2-enoic acid)
    Aldehyde oxidised to carboxylic acid; alkene is preserved.
  • Reagent (bottom-left line):
    (CH₃)₂CHOH or propan-2-ol.
    Esters require the matching alcohol: propan-2-ol forms an isopropyl ester.
3 marks awarded on this branch (1 mark per box/line).

💡 Key Knowledge: Selectivity of Reagents

  • NaBH₄ in aqueous/alcoholic solution: Nucleophilic addition of hydride (H⁻) only attacks polar carbonyl groups (C=O). It cannot attack electron-dense non-polar alkene C=C double bonds.
  • H₂ with Ni catalyst: Heterogeneous catalytic hydrogenation reduces C=C double bonds under typical conditions.
  • Esterification: R-COOH + R'-OH ↔ R-COOR' + H₂O (catalysed by concentrated acid, H⁺). Look carefully at which oxygen is part of the original acid and which comes from the alcohol!

❌ Common Errors

  • Forgetting the acid catalyst: Writing only K₂Cr₂O₇ without H⁺ or H₂SO₄ loses the oxidation mark immediately.
  • Reducing both groups with NaBH₄: Drawing a saturated alcohol in the top-right box.
  • Formula ambiguity for alcohols: Writing molecular formula C₃H₇OH instead of structural/displayed formula for propan-2-ol. The mark scheme explicitly states: IGNORE C₃H₇OH because it does not distinguish between propan-1-ol and propan-2-ol.
Part (b)* • 6 Marks (Level of Response)

Two-Stage Synthesis of Compound A from (Chloromethyl)benzene

Prepare 5.44 g of Compound A (2-phenylethanoic acid) starting from (chloromethyl)benzene with an overall yield of 25%.

📐 Calculation Breakdown: Mass of (Chloromethyl)benzene Required

  1. Calculate the Molar Mass of Compound A (C₆H₅CH₂COOH, C₈H₈O₂):
    M(C₈H₈O₂) = (8 × 12.0) + (8 × 1.0) + (2 × 16.0) = 136.0 g mol⁻¹
  2. Calculate Moles of Compound A Produced:
    n(Compound A) = mass / Mᵣ = 5.44 g / 136.0 g mol⁻¹ = 0.0400 mol
  3. Account for the 25% Overall Yield:
    Since the stoichiometric ratio from starting material to product is 1:1, theoretical moles needed:
    n(C₆H₅CH₂Cl) = 0.0400 mol × (100 / 25) = 0.160 mol
  4. Calculate Molar Mass of Starting Material (C₆H₅CH₂Cl, C₇H₇Cl):
    M(C₇H₇Cl) = (7 × 12.0) + (7 × 1.0) + 35.5 = 126.5 g mol⁻¹
  5. Calculate Required Mass:
    Mass = n × Mᵣ = 0.160 mol × 126.5 g mol⁻¹ = 20.24 g (or 20.2 g to 3 s.f.)

✅ Synthetic Route & Balanced Equations

Stage 1: Formation of Intermediate (Phenylethanenitrile)

  • Intermediate: C₆H₅CH₂CN
  • Reagents & Conditions: NaCN (or KCN / CN⁻ ) in ethanol / ethanolic, under reflux.
  • Equation:
    C₆H₅CH₂Cl + CN⁻ → C₆H₅CH₂CN + Cl⁻
    OR: C₆H₅CH₂Cl + NaCN → C₆H₅CH₂CN + NaCl

Stage 2: Formation of Compound A (2-Phenylethanoic Acid)

  • Reagents & Conditions: Aqueous acid ( H⁺ / H₂O , e.g. dilute HCl or dilute H₂SO₄ ), heated under reflux.
  • Equation:
    C₆H₅CH₂CN + 2H₂O + H⁺ → C₆H₅CH₂COOH + NH₄⁺
    OR: C₆H₅CH₂CN + 2H₂O + HCl → C₆H₅CH₂COOH + NH₄Cl

🧠 Grading Criteria for Level 3 (5–6 Marks)

To secure top marks, your answer must be a well-developed line of reasoning containing all three strands:

  • Correct numerical answer: 20.2 g or 20.24 g with fully shown logical working.
  • Complete synthetic plan: Correct intermediate ( C₆H₅CH₂CN ) identified with suitable reagents for both steps.
  • Balanced equations: Accurate equations for both stages (ensuring water and acid are balanced for nitrile hydrolysis).

❌ Common Pitfalls in Part (b)

  • Inverting the yield calculation: Multiplying moles by 25/100 instead of 100/25 (giving 1.265 g). Remember: you need more starting material if the reaction isn't 100% efficient!
  • Omitting the yield completely: Calculating 0.0400 × 126.5 = 5.06 g.
  • Unbalanced hydrolysis: Forgetting the stoichiometric 2H₂O or producing ammonia ( NH₃ ) instead of ammonium ion ( NH₄⁺ ) under acidic conditions.
  • Skipping the nitrile: Trying to convert chloroalkane directly to a carboxylic acid with one extra carbon (impossible in one step without a carbon chain extension).

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.