AQA A-Level Chemistry Paper 3, 2024: Question 23

Skills & knowledge

• Naming → structure (IUPAC interpretation).
• Correct numbering from the carboxylic acid carbon (C1).
• Placing an alkene using -en- position and substituents using locants.
• Recognising condensed structural formula options and spotting “position swaps”.

Why students lose marks

• Numbering from the wrong end (forgetting COOH is highest priority).
• Putting the C=C in the wrong place (mixing up “pent-2-enoic acid”).
• Attaching substituents (chloro/methyl) to the wrong carbon after mis-numbering.

✅ Correct answer (from mark scheme)

Option C: CH₃CHClCH=C(CH₃)COOH

💡 Key knowledge: decode the name

• Parent: pent- = 5 carbons in the main chain including the carboxyl carbon.
• Functional group: -oic acid means it ends with COOH and this carbon is always C1.
• Alkene position: pent-2-enoic acid means the C=C is between C2 and C3 (counting from COOH).
• Substituents:
  • 2-methyl = CH₃ attached to C2.
  • 4-chloro = Cl attached to C4.

📐 Build it step-by-step (numbering & placement)

1. Start with the acid end: …COOH is C1.
2. Make a 5-carbon chain: C1–C2–C3–C4–C5.
3. Put the double bond at 2: C2=C3.
4. Add methyl at C2: C2 has a CH₃ branch.
5. Add chloro at C4: C4 becomes CH(Cl) .
6. Write condensed formula from the far end: CH₃–CH(Cl)–CH= C(CH₃)–COOH → CH₃CHClCH=C(CH₃)COOH

Note: The double bond carbon at C2 already has COOH attached (via C1) and a CH₃ substituent, so it appears as C(CH₃) next to COOH .

🧠 Exam technique: quick elimination in MCQ

• Anchor point: find the option that clearly ends with …COOH and treat that carbon as C1.
• Then check: is the C=C next to COOH (between C2 and C3)? For pent-2-enoic it must be.
• Finally: check substituents by counting from COOH: methyl must be at C2 and chloro at C4.

❌ Common errors (and why they’re wrong)

• Numbering from the alkyl end: gives the wrong positions for methyl/chloro and often “moves” the double bond.
• Putting Cl on the alkene carbon: “4-chloro” means Cl is on C4, not on C2/C3.
• Misreading pent-2-enoic acid: the “2” refers to the carbon number from COOH, so the C=C must involve C2.
• Trap: options can look similar but swap which carbon bears the substituent (especially around the C=C).

Checklist

• Contains COOH at one end (carboxylic acid).
• Main chain length = 5 carbons including the acid carbon.
• Double bond is between C2 and C3 (next to the acid carbon).
• Methyl substituent at C2 (same carbon as part of the double bond).
• Chloro substituent at C4 (two carbons away from the double bond).

✅ Final (condensed) structural formula

CH₃CHClCH=C(CH₃)COOH

Matches AQA mark scheme: Q23 = C.