OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2019: Question 3

1 mark · Medium difficulty · Multiple Choice

Determine the IUPAC name of the branched alkene shown in the structural formula.

Practise this question

Question

Multiple choice question 3 asking for the name of the chemical structure shown. The structure features a central carbon-carbon double bond. The top carbon is bonded to an H and a CH3 group. The bottom carbon is bonded to an ethyl group (H3C-CH2-) on the left and a propyl group (-CH2-CH2-CH3) on the right. Four options are given: A, 3-Propylpent-2-ene; B, 3-Propylpent-3-ene; C, 3-Ethylhex-2-ene; D, 4-Ethylhex-4-ene. A box is provided for the answer.
Question text

3 What is the name of the compound below?

H CH3

C

C

H3C CH2 CH2 CH2 CH3

A 3-Propylpent-2-ene

B 3-Propylpent-3-ene

C 3-Ethylhex-2-ene

D 4-Ethylhex-4-ene

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 3 is C, worth 1 mark.

3 C 1 AO1.2

How to answer it

Naming Alkenes using IUPAC Rules

What this question tests: This question assesses your ability to apply IUPAC organic nomenclature rules to branched alkenes. You must identify the longest continuous carbon chain containing the functional group (the double bond), number the chain from the end that gives the lowest locant (number) to the alkene functional group, correctly name and position alkyl side-chains, and combine them alphabetically.
Question 3

Determining the IUPAC Name of a Branched Alkene

✅ Correct Answer

C: 3-Ethylhex-2-ene

The correct IUPAC name identifies a 6-carbon principal chain (hex), a double bond starting at carbon 2 (-2-ene), and an ethyl substituent at carbon 3 (3-ethyl).

💡 Key Knowledge

  • Principal Chain: Find the longest carbon chain that includes the carbon-carbon double bond (C=C). Here, tracing through gives 6 carbons (hexane derivative).
  • Lowest Locant Rule: Number the chain from the end closest to the double bond. Left-to-right gives the double bond at carbon 2, whereas right-to-left gives it at carbon 4. Therefore, number from left to right.
  • Substituents: Identify groups attached to the main chain. Carbon 3 holds a two-carbon chain, making it an ethyl group.

🧠 Exam Technique

Don't be fooled by horizontal straight lines! Carefully trace all possible continuous pathways through the molecule to find the absolute longest carbon chain containing the double bond. Always prioritize giving the functional group (-ene) the lowest possible number.

❌ Common Errors

  • Selecting Option A/B: Choosing a 5-carbon chain ( pent ) by stopping at the rightmost methyl group instead of continuing into the propyl chain.
  • Incorrect Numbering: Numbering from the wrong end, leading to incorrect locants like hex-4-ene (Option D).
Mark Scheme Allocation: 1 mark available for choosing C (AO1.2 - Demonstrating knowledge and application of nomenclature rules).

Topics

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

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.