AQA A-Level Chemistry Paper 3, 2024: Question 26

Title: Identifying E–Z isomerism from an alkene name

What this question tests

Skills & knowledge assessed (AO2)

Recognising the condition for E–Z isomerism in alkenes: each carbon of the C=C must have two different groups.
Interpreting IUPAC names (positions of double bond + methyl substituents) to infer groups attached to each alkene carbon.
Eliminating options where one alkene carbon has two identical substituents (so no E–Z).

Mark scheme outcome: Q26 = D (1 mark, AO2): 3‑methylpent‑2‑ene

Part (a) — Which compound shows E–Z isomerism? (1 mark)

Use the “two different groups on each C of C=C” test

✅ Correct answer (1/1)

D: 3‑methylpent‑2‑ene

This matches the mark scheme exactly.

💡 Key knowledge (what must be true)

E–Z isomerism requires restricted rotation (C=C) AND
Both double-bond carbons must each have two different substituents.
If either alkene carbon has two identical groups (e.g. two CH₃ groups), E–Z is impossible.

🧠 Exam technique (fast elimination)

Find where the double bond is from the name (e.g. pent‑2‑ene means C2=C3).
Write the carbon skeleton quickly (even as a line formula).
On each C of C=C, list the two attached groups. Ask: “Are they different?”
If both sides pass → E–Z possible.

❌ Common errors (why marks are lost)

Thinking “any alkene has E–Z” — false. Many alkenes only show cis–trans/E–Z if substituents allow it.
Forgetting to check both alkene carbons (students sometimes check only one side).
Confusing “branched name” with “E–Z present” — branching alone doesn’t guarantee E–Z.

📐 Worked reasoning (no maths, but step-by-step logic)

Option D: 3‑methylpent‑2‑ene
- Main chain: pent‑2‑ene → C2=C3.
- At C2: attached groups are CH₃ (C1 end) and H → different.
- At C3: attached groups are CH₃ (the 3‑methyl substituent) and CH₂CH₃ (continuation to C4–C5) → different.
- Both alkene carbons have two different groups → E–Z isomerism possible.
Why the others do not show E–Z (quick checks)
- A: 2,3‑dimethylbut‑1‑ene → one end of C=C is CH₂ (two H attached) → identical groups on that carbon → no E–Z.
- B: 2,3‑dimethylbut‑2‑ene → each C of C=C has two CH₃ groups (identical) → no E–Z.
- C: 2‑methylpent‑2‑ene → at C2 there are two CH₃ groups (one from C1 and one methyl substituent) → identical on one alkene carbon → no E–Z.

🧠 Examiner-style commentary (what distinguishes full marks)

This is a 1-mark multiple-choice AO2 question, so examiners reward the ability to apply the E–Z condition correctly from the name alone. Strong responses (and efficient candidates) typically:

spot that options A, B, C each contain an alkene carbon with two identical groups (two H, or two CH₃), and eliminate them rapidly,
select D because it is the only one where both alkene carbons have two different substituents.

✅ Mark breakdown (linked to the mark scheme)

1 mark for choosing D only.

Mark scheme: D = 3‑methylpent‑2‑ene (AO2).