OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2023: Question 19

6 marks · Hard difficulty · Extended Response

Plan a multi-stage organic synthesis to convert 3-bromopropene into compound Z, identifying reagents, structures of intermediates, and equations.

Practise this question

Question

An exam question featuring a reaction flowchart starting with a halogenoalkene, going through Intermediate 1 and Intermediate 2 in three stages to form Compound Z. The student is asked to plan this synthesis showing reagents, structures of intermediates, and equations, worth 6 marks.
Question text

A student intends to synthesise compound Z, as shown in the flowchart below.

H

H C Br

C C

H H H

Stage 1

Intermediate 1

Stage 2

Intermediate 2

Stage 3

Br H H

H

H C C

C C NH2

H

H H H

Compound Z

Plan this synthesis showing reagents, the structures of intermediate 1 and intermediate 2, and

equations. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme A mark scheme detailing a three-level response for a 6-mark synthesis question, including guidance on reagents, equations, and structures for stages 1, 2, and 3 involving nucleophilic substitution with cyanide, addition of HBr, and reduction.

AO

Question Answer Marks Guidance

element

Refer to marking instructions on page 4 of mark scheme for 6 AO3.3 Mark second page as SEEN

guidance on marking this question. 6 Indicative scientific points may include:

Level 3 (5-6 marks) IGNORE conditions

A three stage synthesis in the correct order

AND Stage 1: Reaction with CN–

Equations for each stage are mostly correct • Reagents: CN– (in ethanol)

AND • Equation:

Most reagents correct H C=CHCH Br + CN– → H C=CHCH CN + Br–

22 2 2

Intermediate 1

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

and logically structured. The information presented is

relevant and substantiated.

Level 2 (3-4 marks) Stage 2: Addition of HBr to C=C

Synthesis includes at least two stages in any order OR • Reagents: HBr

uses NH3 and HBr in the correct order (without chain • Equation:

extension) H2C=CHCH2CN + HBr → CH3CHBrCH2CN

AND Intermediate 2

some of the reagents and some equations correct

There is a line of reasoning presented with some

structure. The information presented is relevant and

supported by some evidence. Stage 3: Reduction of CN

• Reagents: H2 (with Ni)

Level 1 (1-2 marks) • Equation:

Planned synthesis includes reagents for any two stages H3CCHBrCH2CN + 2H2 → CH3CHBrCH2CH2NH2

OR

Describes one stage with reagents and equation mostly Needs CN– before HBr

correct – CN– would react with both Br atoms

There is an attempt at a logical structure with a line of Needs HBr before H2

reasoning. The information is in the most part relevant. – H2 would react with C=C

AO

Question Answer 21 Marks Guidance

element

Alternative three stage syntheses:

0 marks

No response or no response worthy of credit. Alternative using LiAlH4

Caution - Can be done as stage 2 or 3

• Reagents: LiAlH4

• Equation:

H2C=CHCH2CN + 4[H] → H2C=CHCH2CH2NH2

OR

H3CCHBrCH2CN + 4[H] → CH3CHBrCH2CH2NH2

Needs CN– before HBr and LiAlH

Can have HBr and LiAlH4 in any order

Alternative using radical substitution:

Stage 1: Reaction with CN–

• Reagents: CN– (in ethanol)

• Equation:

H C=CHCH Br + CN– → H C=CHCH CN + Br–

22 2 2

Stage 2: Reduction of CN and C=C

• Reagents: H2 (with Ni)

• Equation:

H2C=CHCH2CN + 3H2 → CH3CH2CH2CH2NH2

Stage 3: Reaction with Br2

• Reagents: Br2 (with UV)

• Equation:

CH3CH2CH2CH2NH2 + Br2 →

CH3CHBrCH2CH2NH2 + HBr

Needs CN– before H

Needs H2 before Br2

AO

element

Two stage synthesis using NH3 and HBr forming

product with no lengthening of carbon chain

Stage 1: Reaction of NH3

• Reagents: NH3 (in ethanol)

• Equation:

H2C=CHCH2Br + NH3 → H2C=CHCH2NH2 + HBr

OR 2 NH3 → NH4Br

Stage 2: Addition of HBr to C=C

• Reagents: HBr

• Equation:

H2C=CHCH2NH2 + HBr → CH3CHBrCH2NH2

Needs NH3 before HBr

– HBr would react with C=C

How to answer it

Multi-Stage Organic Synthesis Planning

What this question tests

This 6-mark extended response question assesses your ability to plan a multi-step organic synthesis. You must coordinate functional group interconversions (halogenoalkanes, alkenes, nitriles, amines), maintain strict control over reaction sequence ordering (chain extension vs. functionalisation), and write balanced chemical equations with correct structural intermediate formulas.

Question 19 (6 Marks)

Strategic Synthesis of Compound Z

To successfully convert H₂C=CHCH₂Br into Compound Z ( CH₃CH(Br)CH₂CH₂CH₂NH₂ ), you must deduce the missing intermediates, select appropriate reagents for each step, and write balanced equations.

💡 Key Knowledge: Pathway Logic

  • Carbon Chain Extension: Compound Z has 4 carbons, while the starting material has 3 carbons. A nitrile group ( -CN ) must be introduced via nucleophilic substitution using CN⁻ to lengthen the carbon chain.
  • Functional Group Interconversions: Alkenes undergo electrophilic addition (e.g., adding HBr ). Nitriles undergo catalytic hydrogenation ( H₂ / Ni ) to form primary amines.

✅ Correct Answer (Standard Pathway)

  • Stage 1 Reagents: KCN or NaCN in ethanol ( CN⁻ )
  • Intermediate 1 Structure: H₂C=CHCH₂CN (but-3-enenitrile)
  • Stage 2 Reagents: HBr
  • Intermediate 2 Structure: CH₃CH(Br)CH₂CN
  • Stage 3 Reagents: H₂ with a nickel ( Ni ) catalyst

🧠 Exam Technique: Level-Based Marking

  • Level 3 (5–6 marks): Requires a complete 3-stage synthesis in the correct order, with mostly correct equations and reagents.
  • Reagent Ordering Constraint: You must introduce CN⁻ before adding HBr , otherwise CN⁻ would substitute both bromine atoms. Similarly, H₂ must be added after HBr to avoid reducing the C=C double bond prematurely.

❌ Common Errors & Pitfalls

  • Wrong Reagent Order: Introducing HBr before CN⁻ means the initial C=C bond is lost or reacts incorrectly.
  • Unbalanced Equations: Forgetting byproduct molecules (like Br⁻ or H₂O ) or miscounting hydrogen atoms during reduction steps.

📐 Step-by-Step Equations for the Preferred Pathway

Ensure your equations clearly show molecular or displayed structures for intermediates:

  • Stage 1 Equation: H₂C=CHCH₂Br + CN⁻ → H₂C=CHCH₂CN + Br⁻
  • Stage 2 Equation: H₂C=CHCH₂CN + HBr → CH₃CH(Br)CH₂CN
  • Stage 3 Equation: CH₃CH(Br)CH₂CN + 2H₂ → CH₃CH(Br)CH₂CH₂CH₂NH₂
Examiner Tip: Always double-check that your intermediate structures match the chain length and functional groups required for the next step in the sequence.

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 6.2 Nitrogen compounds, polymers and synthesis · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.