AQA A-Level Chemistry Paper 2, June 2025: Question 2

Correct Answers

• Name: Z-3-methylpent-2-ene [1 mark]
• Explanation:
  • There is restricted/limited rotation about the C=C double bond. [1 mark]
  • There are two different groups attached to each carbon atom of the C=C double bond. [1 mark]

Step-by-Step Naming

1. Find the longest chain containing C=C: 5 carbons → pentene.
2. Number the chain: Start from the left to give the double bond the lowest number → pent-2-ene.
3. Identify substituents: A methyl group is on carbon-3 → 3-methylpent-2-ene.
4. Determine E/Z priority:
   • Left C: -CH₃ (high priority) vs -H (low). High priority is on the bottom.
   • Right C: -CH₂CH₃ (high priority) vs -CH₃ (low). High priority is on the bottom.
   • Both high priority groups are on the same side (Zusammen) → Z.

Key Knowledge: E/Z Conditions

For geometric (E/Z) isomerism to occur, two conditions must be met:
1. π-bond prevents rotation: The overlap of p-orbitals forming the pi-bond locks the carbon atoms in place.
2. Asymmetry on both carbons: Each carbon in the double bond must be bonded to two non-identical groups.

Common Pitfalls & Examiner Tips

• Vague explanations: Do not just say "there are different groups on the double bond". You must specify that there are two different groups attached to each carbon of the C=C double bond.
• Forgetting the stereochemical prefix: Always check if a drawn alkene structure requires an E- or Z- prefix when asked for its IUPAC name.

Correct Answers

• Structural Formula of M:
  CH₂=CHCH(CH₃)CH₂CH₃ [1 mark]
  Skeletal formula showing a 5-carbon chain with a double bond between C1 and C2, and a methyl branch on C3 is also fully acceptable.
• Explanation: It has a chiral centre / asymmetric carbon atom (or a carbon atom bonded to four different groups). [1 mark]

How to Solve This

1. Keep the same carbon skeleton as L (3-methylpentane).
2. Move the double bond to a new position. If you move it to C1-C2, you get 3-methylpent-1-ene.
3. Inspect the C3 carbon: it is bonded to -H , -CH₃ , -CH₂CH₃ , and -CH=CH₂ .
4. Because it has four different groups attached, it is chiral and exhibits optical isomerism.

Common Errors to Avoid

• Drawing the wrong skeleton: Ensure you do not accidentally change the carbon chain connectivity. It must remain a 3-methylpentane derivative.
• Incomplete chirality definition: Simply stating "it is chiral" is not enough for the explanation mark. You must define why it is chiral (i.e., "it has a carbon bonded to four different groups").

Correct Mechanism Drawing

To complete the mechanism for the major product, you must draw:

• The Tertiary Carbocation Intermediate:
  A central carbon with a positive charge ( C⁺ ) bonded to:
  • One methyl group ( -CH₃ )
  • Two ethyl groups ( -CH₂CH₃ ) [1 mark]
• The Attack Step:
  A bromide ion ( Br⁻ ) with a lone pair and a negative charge.
  A curly arrow pointing directly from the lone pair on the Br⁻ to the positive carbon ( C⁺ ). [1 mark]

Key Knowledge: Carbocation Stability

Markovnikov's Rule: The major product of electrophilic addition is formed via the most stable carbocation intermediate.

Stability Order: Tertiary (3°) > Secondary (2°) > Primary (1°)

Alkyl groups are electron-donating and push electron density towards the positive carbon (the inductive effect), stabilizing the charge.

How to Draw the Intermediate

When H⁺ adds to C2 of compound L, it forms a stable tertiary carbocation at C3:

Make sure the positive charge is placed directly on the central carbon atom, not floating in space.

Curly Arrow Pitfalls

• Arrow starting point: The curly arrow must start precisely at the lone pair or the negative charge of the Br⁻ ion. Starting it from the element symbol "Br" will lose the mark.
• Arrow endpoint: The arrow must point directly to the positive carbon atom ( C⁺ ).
• Missing charges: Do not forget the negative charge on the bromide ion ( Br⁻ ) and the positive charge on the carbocation ( C⁺ ).