OCR A-Level Chemistry AS Breadth in chemistry (01), June 2023: Question 16

Structural Isomers of Halogenoalkanes

OCR AS Level Chemistry • Multiple Choice Question

What this question tests

This question assesses your ability to define, identify, and systematically construct structural isomers (specifically positional and chain isomers) for a given molecular formula ( C₄H₉Cl ), testing your visualization and methodical approach to organic molecules.

Question 16

Determining the number of structural isomers for C₄H₉Cl

✅ Correct Answer

C (4)

There are 4 distinct structural isomers for C₄H₉Cl .

💡 Key Knowledge

Structural isomers share the same molecular formula but have different structural arrangements of atoms.
Carbon chain skeletons to consider: a straight chain of 4 carbons (butane backbone) and a branched chain of 3 carbons with a methyl branch (isobutane/2-methylpropane backbone).
Halogen atoms (like Cl ) can occupy terminal or internal positions along the carbon skeleton.

🧠 Exam Technique

Methodical approach: Always start by drawing out all possible carbon skeletons first, then systematically move the functional group ( -Cl ) to every unique carbon position without double-counting.
Name each isomer mentally or on rough paper as you go to ensure none are duplicated.

❌ Common Errors

Stereoisomer confusion: Students sometimes confuse structural isomers with stereoisomers (optical/E-Z). Remember, the question specifically asks for structural isomers.
Double counting: Misnaming 1-chlorobutane as a different isomer simply because it is drawn pointing right-to-left.
Valency mistakes: Drawing carbon atoms with 5 bonds or forgetting hydrogen atoms when sketching.

📐 Step-by-Step Isomer Breakdown

Systematically mapping out the 4 structural isomers of C₄H₉Cl :

Step 1: Consider the straight 4-carbon chain ( CH₃-CH₂-CH₂-CH₃ ):
- Place -Cl on carbon 1 → 1-chlorobutane ( CH₃CH₂CH₂CH₂Cl )
- Place -Cl on carbon 2 → 2-chlorobutane ( CH₃CH₂CH(Cl)CH₃ )
Step 2: Consider the branched 3-carbon chain with a methyl group ( (CH₃)₂CH-CH₃ ):
- Place -Cl on one of the terminal methyl carbons → 1-chloro-2-methylpropane ( (CH₃)₂CHCH₂Cl )
- Place -Cl on the central carbon → 2-chloro-2-methylpropane ( (CH₃)₃CCl )

Mark Scheme Note: Answer is C (1 mark). Examiner reports note that top-level students systematically exhaust the straight-chain possibilities before moving to branched chains, avoiding careless omissions.