OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 17
1 mark · Medium difficulty · Multiple Choice
Determine the number of structural isomers of C5H12O that are secondary alcohols.
Practise this questionQuestion
Question text
17 How many structural isomers of C5H12O are secondary alcohols?
A 1
B 2
C 3
D 4
Your answer [1]
Mark scheme
Show the mark scheme
17 C 1 ALLOW 3
How to answer it
Secondary Alcohol Isomers of C₅H₁₂O
This question tests your ability to determine structural isomerism in alcohols, systematically draw branched and unbranched carbon skeletons, and correctly identify secondary (2°) alcohols based on the connectivity of the carbon atom bonded to the –OH group.
Question 17
Identifying Secondary (2°) Alcohol Isomers
✅ Correct Answer
C (3 isomers)
💡 Key Knowledge
- General Formula: C₅H₁₂O fits CnH2n+2O, representing a saturated acyclic alcohol (or ether).
- Secondary (2°) Alcohol: The –OH group is bonded to a carbon atom that is directly attached to exactly two other carbon atoms and one hydrogen atom: R–CH(OH)–R' .
- Structural Isomers: Molecules with the same molecular formula but different structural arrangements of atoms.
📐 Systematic Deduction of the 3 Isomers
To avoid missing isomers or double-counting duplicates, systematically vary the carbon skeleton:
- Straight 5-Carbon Chain (Pentane skeleton: C–C–C–C–C):
- Pentan-1-ol: CH₃CH₂CH₂CH₂CH₂OH → Primary (1°) ❌
- Pentan-2-ol: CH₃–CH(OH)–CH₂CH₂CH₃ → The C–OH carbon is bonded to 2 carbons (methyl and propyl). Secondary (2°) ✅ [Isomer 1]
- Pentan-3-ol: CH₃CH₂–CH(OH)–CH₂CH₃ → The C–OH carbon is bonded to 2 carbons (two ethyl groups). Secondary (2°) ✅ [Isomer 2]
- 4-Carbon Chain with a Methyl Branch (Butane skeleton: C–C(CH₃)–C–C):
- 3-Methylbutan-2-ol: CH₃–CH(CH₃)–CH(OH)–CH₃ → The C2 carbon holding –OH is bonded to C1 and C3 (2 carbon atoms). Secondary (2°) ✅ [Isomer 3]
- 2-Methylbutan-2-ol: (CH₃)₂C(OH)CH₂CH₃ → The C–OH carbon is bonded to 3 carbons. Tertiary (3°) ❌
- 2-Methylbutan-1-ol: HOCH₂CH(CH₃)CH₂CH₃ → Primary (1°) ❌
- 3-Methylbutan-1-ol: HOCH₂CH₂CH(CH₃)₂ → Primary (1°) ❌
- 3-Carbon Chain with Two Methyl Branches (Dimethylpropane skeleton):
- 2,2-Dimethylpropan-1-ol: (CH₃)₃C–CH₂OH → The C–OH carbon is only bonded to one carbon. Primary (1°) ❌ (The central carbon has 4 methyl groups, so it cannot hold an –OH group).
Total secondary alcohols found = 3 (Pentan-2-ol, Pentan-3-ol, 3-methylbutan-2-ol).
🧠 Exam Technique
- Draw skeletons first: Quickly sketch the carbon backbones on rough paper (5-in-a-line, 4+1 branch, 3+2 branches).
- Test each position: Move the –OH group along each skeleton and classify each carbon as 1°, 2°, or 3° based on how many carbons it is bonded to.
- Check IUPAC names: If two structures produce the same IUPAC name, they are identical molecules drawn from different perspectives.
❌ Common Errors
- Missing the branched isomer: Students frequently only identify pentan-2-ol and pentan-3-ol (selecting option B, 2), forgetting that branched skeletons like 3-methylbutan-2-ol can also be secondary.
- Confusing 2° and 3° alcohols: Incorrectly classifying 2-methylbutan-2-ol as secondary, leading to an answer of 4 (option D). In 2-methylbutan-2-ol, the carbon bearing the –OH group is bonded to 3 carbons, making it tertiary.
- Counting enantiomers: The question asks for structural isomers, not stereoisomers, so optical isomers (such as chiral pentan-2-ol) should not be counted separately.
Topics
Module 4: Core organic chemistry · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis
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.