OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2023: Question 17

✅ Correct Answer

C₆H₁₁OH + 8½ O₂ → 6 CO₂ + 6 H₂O

Mark breakdown: 1 mark for correct molecular formula (or C₆H₁₂O); 1 mark for balanced equation (multiples accepted).

💡 Key Knowledge

• An unsaturated alcohol with 6 carbons and one double bond has the general formula C₆H₁₁OH (or C₆H₁₂O ).
• Complete combustion always produces carbon dioxide ( CO₂ ) and water ( H₂O ).

❌ Common Errors

• Balancing oxygen incorrectly by forgetting to account for the oxygen atom already present within the alcohol functional group ( -OH ).

✅ Correct Answer

Compound C contains three -OH groups (one secondary, two primary). Under reflux with excess oxidising agent, secondary alcohols oxidize to ketones and primary alcohols oxidize to carboxylic acids.

Balanced Equation:
Compound C + 3 [O] → Tricarboxylic/diketone product + 2 H₂O

Mark breakdown: 2 marks for correct organic product structure; 1 mark for balancing with 3 [O] and 2 H₂O .

🧠 Exam Technique

Pay close attention to functional group transformations under reflux conditions. Primary groups become -COOH and secondary groups become C=O .

❌ Common Errors

• Stopping the oxidation prematurely at the aldehyde stage despite the question stating reflux with excess.
• Miscounting the stoichiometric requirement of oxygen atoms ( [O] ) and water molecules produced.

✅ Correct Answer

Dehydration via H₂SO₄ eliminates H₂O to form alkenes with the molecular formula C₇H₁₀ . The three possible isomer structures are:

• 1. A methyl-substituted cyclohexene ring with a double bond positions relative to the methyl group.
• 2. An alternative positional isomer on the ring ring system.
• 3. An exocyclic double bond or ring-opened diene/alkene structure as defined by the mark scheme.

Mark breakdown: 3 marks total (1 mark for each correct unique isomer structure).

💡 Key Knowledge

Elimination of water from a cyclic diol yields unsaturated ring structures or dienes maintaining the carbon skeleton backbone.

✅ Correct Answer

NaI / KI AND H₂SO₄ (or H₃PO₄ / HI )

Mark breakdown: 1 mark for stating both the iodide source and the acid catalyst.

🧠 Exam Technique

Memorize standard substitution reagents used to swap hydroxyl groups ( -OH ) for halogens. For iodination in situ, a metal iodide combined with a strong acid is required.

✅ Correct Answer

1. The 4 structural alcohol isomers of C₄H₁₀O are:

• Butan-1-ol ( CH₃CH₂CH₂CH₂OH ) - 4 environments
• Butan-2-ol ( CH₃CH₂CHOHCH₃ ) - 4 environments
• 2-methylpropan-1-ol ( (CH₃)₂CHCH₂OH ) - 3 environments
• 2-methylpropan-2-ol ( (CH₃)₃COH ) - 2 environments

2. Statement mark: (CH₃)₂CHCH₂OH or (CH₃)₃COH can be distinguished from other isomers because they possess unique numbers of carbon environments (3 and 2 peaks respectively), whereas butan-1-ol and butan-2-ol cannot be distinguished from each other solely by peak count (both have 4 peaks).

Mark breakdown: 1 mark for all 4 structures; 3 marks for correct peak/environment counts; 1 mark for the comparative statement.

📐 Step-by-Step Breakdown

1. Identify Isomers: Draw out all primary, secondary, and tertiary C₄ alcohols.
2. Count Environments: Look for planes of symmetry to group equivalent carbon atoms.
3. Compare: Note that Carbon-13 NMR alone cannot tell apart isomers if they share the exact same number of non-equivalent carbon environments.

❌ Common Errors

• Failing to explicitly compare at least two isomers when attempting to earn the final statement mark.
• Confusing Carbon-13 NMR peak splitting (which does not occur due to low abundance of C-13 coupling) with proton ( ¹H ) NMR multiplicity.