Edexcel A-Level Chemistry AS Paper 2, June 2016: Question 7

📐 Step-by-Step Calculation

1. Find the empirical formula: Assume a 100 g sample. Divide each percentage by relative atomic mass ( C = 12.0 , H = 1.0 , O = 16.0 ).
   C: 68.2 / 12 = 5.68
   H: 13.6 / 1.0 = 13.6
   O: 18.2 / 16 = 1.14
2. Find the simplest ratio: Divide all answers by the smallest value ( 1.14 ).
   C: 5.68 / 1.14 = 5
   H: 13.6 / 1.14 = 12
   O: 1.14 / 1.14 = 1
   Empirical formula = C₅H₁₂O
3. Use the molecular ion peak: The molecular ion peak at m/z = 88 matches the relative molecular mass of C₅H₁₂O ((5 × 12) + (12 × 1) + 16 = 88). Therefore, empirical formula equals molecular formula.

❌ Common Errors & Exam Technique

• Missing the final link: Simply calculating the empirical formula ( C₅H₁₂O ) is only worth 1 mark. You must explicitly state how the m/z = 88 peak proves it is also the molecular formula.
• Alternative method: You can also work backwards by calculating the percentage composition of C₅H₁₂O directly to show it matches the question data.

✅ Correct Answer

Alcohol X is a primary ( 1° ) alcohol.

💡 Key Knowledge

Only primary alcohols are oxidised all the way to carboxylic acids (via aldehyde intermediates) using acidified potassium dichromate( VI ). Secondary alcohols form ketones, and tertiary alcohols do not oxidise under normal conditions.

🧠 Exam Technique & Requirements

• Displayed formula rule: Every single bond ( C-C , C-H , O-H ) must be drawn out explicitly. Do not condense groups like CH₃ or OH .
• You must draw four different structural isomers of primary alcohols with formula C₅H₁₂O :
  1. Pentan-1-ol ( CH₃CH₂CH₂CH₂CH₂OH )
  2. 3-methylbutan-1-ol ( (CH₃)₂CHCH₂CH₂OH )
  3. 2-methylbutan-1-ol ( CH₃CH₂CH(CH₃)CH₂OH )
  4. 2,2-dimethylpropan-1-ol ( (CH₃)₃CCH₂OH )

❌ Common Errors

• Drawing secondary or tertiary isomers (e.g., pentan-2-ol) which contradict part (b)(i).
• Omission of hydrogen bonds or accidental double bonds.
• Drawing carbon-oxygen-hydrogen connections backwards (e.g., C-H-O instead of C-O-H ).

✅ Correct Answer

Structure of the fragment: [CH₂OH]⁺ or with explicit displayed bonds showing carbon bonded to two hydrogens, a positive charge, and an OH group.

💡 Mass Spec Insight

A peak at m/z = 45 in primary alcohols typically corresponds to the cleavage of the alpha-carbon, resulting in the [CH₂OH]⁺ fragment (Relative mass: 12 + 2 + 16 = 30? Wait, CH₂OH⁺ is 12 + 2 + 16 = 30... actually, C₂H₅O⁺ or CH₂CH₂OH⁺ gives mass 45: (2×12) + 5 + 16 = 45, specifically [CH₂CH₂OH]⁺ or related stable oxonium/fragment ions depending on branching).

✅ Correct Answer

Alcohol X is 3-methylbutan-1-ol (displayed formula required or clear name).

🧠 Explaining Your Reasoning

• State that X has a branched chain.
• Explain that out of the primary alcohol isomers, 3-methylbutan-1-ol is the only branched isomer that fragments to form the specific peak/species CH₂OHCH₂⁺ or C₂H₄OH⁺ (or explain why the other branched/straight chains do not fit the fragmentation/structural criteria given).