OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 16

✅ Correct Answer

2-methylpent-2-ene

💡 Key Knowledge

• Find the longest continuous chain containing the C=C bond: 5 carbons = pent- .
• Number from the end nearest the double bond: C2 has both the C=C and the methyl branch.
• Double bond is at C2 ( -2-ene ), methyl branch is at C2 ( 2-methyl ).

📐 Step-by-Step σ Bond Counting

Hexa-1,5-diene:

• 5 C–C single connections in the skeleton = 5 σ bonds
• 10 C–H bonds = 10 σ bonds
• Total σ = 5 + 10 = 15

Cyclohexene:

• 6 C–C ring connections = 6 σ bonds
• 10 C–H bonds = 10 σ bonds
• Total σ = 6 + 10 = 16

❌ Common Errors

• Forgetting C–H bonds: Counting only visible skeletal lines (C–C bonds) and missing implicit hydrogens.
• Double bond confusion: Counting a double bond as two σ bonds instead of 1 σ + 1 π bond.
• Cyclic formula error: Miscalculating hydrogens in the ring; cyclohexene has two CH carbons and four CH₂ carbons = 10 H's.

✅ Correct Structures

E-isomer:

High priority groups on opposite sides across the C=C double bond:

• Left carbon: –CH₃ (top), –H (bottom)
• Right carbon: –CH₃ (top), –CH₂CH₃ (bottom)
• Result: Left –CH₃ (priority 1) and Right –CH₂CH₃ (priority 1) are on opposite sides.

Z-isomer:

High priority groups on the same side across the C=C double bond:

• Left carbon: –CH₃ (top), –H (bottom)
• Right carbon: –CH₂CH₃ (top), –CH₃ (bottom)
• Result: Left –CH₃ (priority 1) and Right –CH₂CH₃ (priority 1) are on the same side.

🧠 Exam Technique: CIP Priority Rules

• Left C (C2): Compare –CH₃ (atomic number C = 6) with –H (atomic number H = 1).
  👉 –CH₃ is Priority 1, –H is Priority 2.
• Right C (C3): Compare –CH₂CH₃ with –CH₃. Both attach by C; look at next attached atoms: –CH₂CH₃ has C attached to (C, H, H); –CH₃ has C attached to (H, H, H).
  👉 –CH₂CH₃ is Priority 1, –CH₃ is Priority 2.
• Connectivity trap: Ensure ethyl is drawn bonded through the CH₂ group (e.g. –CH₂CH₃ or –C₂H₅ ), NOT through the CH₃ (e.g. –CH₃CH₂ is rejected!).

✅ Required Marking Points

1. Restricted rotation around the C=C double bond (or C=C double bond cannot rotate / π-bond prevents rotation). [1 mark]
2. Each carbon atom of the C=C double bond is attached to two different groups/atoms. [1 mark]

❌ Disqualifying Language Pitfalls

• Imprecise terms: Do NOT say "different functional groups", "different molecules", or "different species". You must say groups or atoms.
• Vague location: Do NOT just say "there are two different groups on the double bond". You must explicitly state each carbon of the C=C bond has two different groups.
• Wrong terminology: Avoid saying "the bond does not bend" or "the molecule is fixed". The precise term is restricted / no rotation.

✅ Correct Organic Products & Conditions

1. Top Reaction (+ Br₂):

• Structure: 1-bromo-1-(bromomethyl)cyclopentane.
• A cyclopentane ring with a –Br atom and a –CH₂Br group attached to the same ring carbon.

2. Right Reaction (+ HCl, Major Product):

• Structure: 1-chloro-1-methylcyclopentane.
• A cyclopentane ring with a –Cl atom and a –CH₃ group attached to the same ring carbon.

3. Bottom Reaction (Hydrogenation):

• Reagent: H₂ (hydrogen gas)
• Catalyst: Ni (nickel) [Pt or Pd also accepted]

💡 Chemical Principles

• Bromination: Addition of Br₂ across the exocyclic double bond breaks the π bond and adds one Br to each C atom of the double bond.
• Markovnikov's Rule (+ HCl): H⁺ adds to the CH₂ group (fewer alkyl substituents) to form the more stable tertiary carbocation intermediate at the ring carbon. Cl⁻ then attacks this carbocation to give 1-chloro-1-methylcyclopentane as the major product.
• Hydrogenation: Addition of H₂ across C=C requires a transition metal catalyst (typically finely divided nickel, Ni, at ~150 °C).