OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 15

1 mark · Medium difficulty · Multiple Choice

Identify the systematic IUPAC name and stereochemistry (E/Z isomerism) of a given fluoroalkene.

Practise this question

Question

Question 15 asks for the systematic name of an alkene shown as a displayed/structural formula. The carbon-carbon double bond has a fluorine atom and an ethyl group (CH3CH2) bonded to the left-hand carbon, with the fluorine pointing upwards and the ethyl pointing downwards. The right-hand carbon is bonded to a methyl group (CH3) pointing upwards and a hydrogen atom (H) pointing downwards. Four multiple-choice options are given: A: E-3-fluoropent-2-ene, B: E-3-fluoropent-3-ene, C: Z-3-fluoropent-2-ene, and D: Z-2-fluoropent-3-ene.
Question text

15 A compound is shown below.

What is the systematic name of this compound?

F CH3

C C

CH3CH2 H

A E-3-fluoropent-2-ene

B E-3-fluoropent-3-ene

C Z-3-fluoropent-2-ene

D Z-2-fluoropent-3-ene

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme excerpt showing that for question 15, the correct answer is C, with 1 mark allocated.

15 C 1

How to answer it

Naming Haloalkenes & Determining E/Z Isomerism

WHAT THIS QUESTION TESTS

Core AS-Level Organic Chemistry Skills:

  • Identifying the longest unbranched carbon chain containing the C=C double bond.
  • Applying IUPAC numbering rules to give functional groups (the alkene double bond) the lowest possible position locants.
  • Applying the Cahn-Ingold-Prelog (CIP) priority rules based on atomic number (Z).
  • Assigning E or Z stereodescriptors to substituted alkenes.
QUESTION 15

Systematic IUPAC Nomenclature of Substituted Haloalkene

Multiple Choice Question (1 Mark)

✅ Correct Answer

C : Z-3-fluoropent-2-ene

Mark Scheme: Option C earns [1 mark].

💡 Key Knowledge

  • Longest Carbon Chain: Must contain both double-bonded carbons. Ethyl (-CH₂CH₃) contains 2 carbons, giving a total of 5 carbons ( pent- ).
  • Principal Functional Group Priority: The alkene C=C takes precedence over the halo substituent in chain numbering. Number from the end that gives the C=C the lower number.
  • CIP Priority Rules: Priority is determined by the atomic number of the atom directly bonded to each C=C carbon:
    • Left carbon: F (atomic number = 9) > C (atomic number = 6).
    • Right carbon: C (atomic number = 6) > H (atomic number = 1).

📐 Step-by-Step Systematic Naming

1 Find the longest continuous carbon chain containing C=C:

The chain runs from the methyl group on the right through the C=C bond to the ethyl group on the left:
CH₃ - CH = C(F) - CH₂CH₃ .
Total number of carbons = 1 + 1 + 1 + 2 = 5 carbons → stem is pent.

2 Number the chain to give the C=C bond the lowest number:

• Numbering right-to-left: C=C starts at C2 ( pent-2-ene ).
• Numbering left-to-right: C=C starts at C3 ( pent-3-ene ).
2 is lower than 3, so we number from right to left:
C1(H₃) - C2(H) = C3(F) - C4(H₂) - C5(H₃) .
The fluoro group is on carbon 3 → 3-fluoropent-2-ene (eliminating options B and D).

3 Assign CIP priorities to each double-bond carbon:

• Carbon 3 (left C of C=C): Attached to -F (atomic no. 9) and -CH₂CH₃ (carbon, atomic no. 6).
Since 9 > 6, -F is high priority and pointing UP.
• Carbon 2 (right C of C=C): Attached to -CH₃ (carbon, atomic no. 6) and -H (atomic no. 1).
Since 6 > 1, -CH₃ is high priority and pointing UP.

4 Determine stereoisomerism (E or Z):

Both high-priority groups (-F and -CH₃) are on the same side (top) of the C=C double bond.
"Same side" = Zusammen → Z.
Full systematic name: Z-3-fluoropent-2-ene.

🧠 Exam Technique & Mnemonic

  • Remember Z vs E:
    • Z = "on ze Zame Zide" (high-priority groups together on top or bottom).
    • E = "Enemies / Entgegen" (high-priority groups on opposite sides).
  • Process of Elimination:
    Alkenes are always numbered so the double bond has the lowest possible locant. A name containing pent-3-ene is immediately invalid when pent-2-ene is possible. Cross out B and D straight away!

❌ Common Student Traps

  • Numbering from the wrong end: Numbering left-to-right gives pent-3-ene . Students incorrectly think the halogen takes priority over the alkene locant. The alkene locant takes precedence!
  • Prioritising group size over atomic number: Mistakenly thinking ethyl (-CH₂CH₃, mass 29) has higher priority than fluorine (-F, mass 19). CIP rules look only at atomic number of the directly attached atom (F = 9 vs C = 6).
  • Confusing cis/trans with E/Z: cis/trans only applies when identical groups are present on both carbons. For tri- or tetra-substituted alkenes, CIP E/Z notation is mandatory.

Topics

Module 4: Core organic chemistry · 4.1 Basic concepts and hydrocarbons

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.