AQA A-Level Chemistry Paper 3, June 2022: Question 8

1 mark · Easy difficulty · Multiple Choice

Identify which substance out of graphite, methylbenzene, poly(propene), and sodium contains no delocalised electrons.

Practise this question

Question

Question 08 asks: 'Which substance has no delocalised electrons?' followed by four multiple-choice options with checkboxes: A graphite, B methylbenzene, C poly(propene), D sodium. It is worth 1 mark.
Question text

08 Which substance has no delocalised electrons?

[1 mark]

A graphite

B methylbenzene

C poly(propene)

D sodium

Mark scheme

Show the mark scheme Mark scheme table row for question 8 shows: Question '8', Answer 'C (AO1)', Mark '1', Additional Comments 'poly(propene)'.

8 C (AO1) 1 poly(propene)

How to answer it

Delocalised Electrons Across Structure & Bonding

📋 What this question tests

This question assesses your ability to recall and distinguish the bonding and electron distribution across four distinct types of chemical structures: giant covalent macromolecules, arenes (aromatic compounds), addition polymers (saturated hydrocarbons), and metallic lattices.

  • Bonding Types: Identifying where delocalised π systems or electron 'seas' exist versus localised σ bonds.
  • Polymer Structure: Understanding that addition polymerisation converts double bonds (with π electrons) into single σ bonds.
Question 08

Identifying Substances Without Delocalised Electrons

Multiple Choice — 1 Mark (AO1)

✅ Correct Answer

C — poly(propene)

Mark Scheme: Award 1 mark for option C [AO1].

Poly(propene) is an addition polymer formed from propene. In poly(propene), the polymer chain consists entirely of single covalent C–C and C–H (σ) bonds. All valence electrons are fixed (localised) in covalent bonds, so there are no delocalised electrons.

💡 Key Knowledge: Why the Others Have Delocalised Electrons

  • A graphite: Giant covalent structure where each carbon atom forms 3 σ bonds. The remaining p electron on each carbon is delocalised across the hexagonal planar sheets, allowing electrical conductivity.
  • B methylbenzene: Contains a benzene ring. The 6 p-orbitals overlap side-on to form a delocalised π electron ring system above and below the plane of carbon atoms.
  • D sodium: Giant metallic lattice consisting of Na⁺ cations surrounded by a sea of delocalised electrons.

🧠 Exam Technique: Elimination Strategy

For questions asking which substance has no delocalised electrons:

  1. Check for metals first: Sodium (D) is a group 1 metal, so it automatically possesses a metallic sea of delocalised electrons. Eliminate immediately.
  2. Check for standard allotropes: Graphite (A) is the classic non-metal conductor with delocalised p-electrons between layers. Eliminate immediately.
  3. Check for aromatic rings: Methylbenzene (B) has an aromatic benzene ring with a delocalised π cloud. Eliminate.
  4. Confirm the remaining choice: Poly(propene) is a long-chain alkane-like molecule. The C=C double bond in the propene monomer opened up during addition polymerisation, leaving only localised σ bonds.

❌ Common Misconceptions & Traps

  • Confusing monomer with polymer: Students see the name propene and think of an alkene with a C=C double bond (which has a π bond). However, addition polymerisation breaks the π bond to form poly(propene), which is saturated.
  • Overlooking the methyl group: Thinking methylbenzene's aliphatic –CH₃ group prevents delocalisation, forgetting that the aromatic ring contains the delocalised π system.
  • Not reading the negative: Rushing through multiple-choice questions often leads students to read "has delocalised electrons" instead of "no delocalised electrons", prompting them to incorrectly choose A, B, or D.

Topics

Physical Chemistry · Organic Chemistry · 3.1.3 Bonding · 3.3.4 Alkenes · 3.3.10 Aromatic Chemistry

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.