AQA A-Level Chemistry Paper 3, 2024: Question 25

Position isomer check for an ester

What this question tests 1 mark

Recognising a position isomer vs functional-group/chain isomer in esters
Using the structural formula CH₃COOCH₂CH(CH₃)₂ to identify the same functional group and carbon skeleton
Comparing ester “split” into RCOO–R′ fragments to spot a change in position of the ester linkage

Examiner focus (AO2): apply structural understanding of isomerism, not recall of definitions alone.

Part (a) Multiple choice: Which is a position isomer? [1 mark]

Step 1: Decode compound Y

💡 Key knowledge: what “position isomer” means

Same molecular formula
Same functional group (here: ester, –COO–)
Same carbon skeleton overall, but the position of the functional group differs (i.e. where the –COO– linkage sits along the chain/branches)

🧠 Exam technique: “split the ester”

Write ester as RCOO–R′ . This makes it much easier to compare options quickly.

Y: CH₃COOCH₂CH(CH₃)₂

Acid part: CH₃CO– (ethanoate)
Alcohol part: –OCH₂CH(CH₃)₂ (2-methylpropyl / isobutyl group)

Step 2: Test each option against “position isomer”

❌ Option A: 5-hydroxyhexan-3-one

Not an ester at all (it’s a hydroxyketone: –OH and C=O in ketone)
So it cannot be a position isomer of an ester

Typical mark loss: students only count carbons and ignore functional group type.

❌ Option B: butyl ethanoate

This is an ester, but it is a different structure from Y’s alcohol side.
“Butyl” implies a straight-chain C₄H₉– group, whereas Y has a branched C₄H₉– group (2-methylpropyl).
That change is best described as a chain isomer difference, not position.

Examiner insight: “butyl” vs branched C₄ group is a classic trap—students treat all C₄H₉ groups as the same.

❌ Option C: hexanoic acid

Different functional group: carboxylic acid (–COOH), not an ester
So it cannot be a position isomer (this would be functional-group isomerism if formulas matched)

✅ Option D: propyl propanoate

This is the correct answer in the mark scheme.

It is an ester (same functional group as Y).
It keeps the overall ester composition but changes how the carbon chain is arranged around the –COO– linkage.
In exam terms, this is treated as the ester group being in a different “position” within the isomeric possibilities.

Mark scheme: D (1 mark, AO2)

How to write the full-mark response

✅ Correct answer (what to select)

D — propyl propanoate

Mark breakdown: 1 mark for choosing D. No working required, but correct structural reasoning helps avoid traps.

💡 Key knowledge: ester naming + fragments

Esters are named alkyl alkanoate
Alkyl comes from the alcohol side (attached to O)
Alkanoate comes from the acid side (contains the C=O)

🧠 Exam technique: quick elimination checklist

First eliminate anything that is not an ester (A and C)
Then check whether the remaining ester option keeps the right “type” of carbon skeleton (straight vs branched) and the placement around –COO–
If you’re unsure, sketch both esters as RCOO–R′ and compare

❌ Common errors (examiner-style)

Confusing position isomerism with functional group isomerism (picking a carboxylic acid or ketone-containing compound).
Not noticing that butyl usually means straight-chain, while CH₂CH(CH₃)₂ is branched.
Trying to decide purely by the number of carbons in the name, instead of checking the functional group and connectivity.

Examiner insight: what distinguishes top responses

Strong students quickly identify Y as an ester and use RCOO–R′ to compare structures, avoiding “name-only” guessing. Weaker answers often spot “ester” but miss the straight vs branched detail or confuse isomer types.