Edexcel A-Level Chemistry Paper 2, November 2020: Question 8

✅ Correct Answer

D ( C₅H₈O₂ )

📐 Step-by-Step Calculation

1. Calculate exact mass for option A ( C₇H₁₆ ): (7 × 12.0000) + (16 × 1.0078) = 100.1248
2. Calculate exact mass for option B ( C₆H₁₂O ): (6 × 12.0000) + (12 × 1.0078) + (15.9949) = 100.0885
3. Calculate exact mass for option C ( C₆H₁₄N ): (6 × 12.0000) + (14 × 1.0078) + (14.0031) = 100.1123
4. Calculate exact mass for option D ( C₅H₈O₂ ): (5 × 12.0000) + (8 × 1.0078) + (2 × 15.9949) = 100.0522 (Matches ion peak exactly!)

❌ Common Errors

Using standard rounded relative atomic masses (e.g., C=12, H=1, O=16) instead of precise high-resolution values provided in the table, which makes distinguishing between close molecular formulas impossible.

✅ Correct Answer

Any order of:

• Alkene (or C=C )
• Carboxylic acid (or -COOH )

💡 Key Knowledge

• 3300–2500 cm⁻¹ corresponds to the very broad O-H stretch of a carboxylic acid.
• 1725–1700 cm⁻¹ corresponds to the C=O stretch in a carboxylic acid.
• 1669–1645 cm⁻¹ corresponds to the C=C stretch of an alkene.

❌ Common Errors

Writing vague terms like "double bond" or just "OH" and "C=O" without specifying the functional group class (carboxylic acid). The mark scheme strictly requires identifying the alkene and carboxylic acid groups.

✅ Correct Answer

A skeletal formula showing an unbranched 5-carbon chain with a carboxylic acid group at one end ( -COOH ) and an alkene double bond ( C=C ) that does not allow E/Z (geometric) isomerism (e.g., a terminal alkene like pent-4-enoic acid, CH₂=CH-CH₂-CH₂-COOH ).

🧠 Exam Technique

Read constraints carefully: unbranched carbon chain AND no geometric isomerism. A terminal double bond ( CH₂=C... ) or a 1,1-disubstituted double bond prevents E/Z isomerism because one of the carbon atoms in the double bond is bonded to two identical groups (two hydrogens).

✅ Correct Answers (Indicative Points)

• IP1 (Similarity): Both ¹H NMR spectra have a peak (singlet) for the -OH proton with a relative peak area of 1.
• IP2 (¹³C NMR): Propan-1-ol has 3 peaks; propan-2-ol has 2 peaks (due to molecular symmetry).
• IP3 (¹H Peaks): Propan-1-ol has 4 peaks; propan-2-ol has 3 peaks.
• IP4 (¹H Areas): Propan-1-ol relative peak areas are 3 : 2 : 2 : 1 (or 6 : 1 : 1 for similarity if grouped); propan-2-ol relative peak areas are 6 : 1 : 1 .
• IP5 (Propan-1-ol Splitting): 2 triplets, 1 sextet (or multiplet/pentet depending on exact assignment) and 1 singlet.
• IP6 (Propan-2-ol Splitting): 1 doublet, 1 septet (or multiplet) and 1 singlet.

🧠 Structuring Top-Level Responses

To secure both marks for "structure and lines of reasoning", avoid random bullet points. Group your answer logically: start with the clear similarities, followed by a direct comparative breakdown of their ¹³C NMR, and finish with a structured contrast of their ¹H NMR (number of peaks, integration/relative areas, and splitting patterns).

❌ Common Traps & Deductions

• Penalty: Mixing up splitting patterns or applying splitting rules to ¹³C spectra (remember: standard ¹³C proton-decoupled spectra show singlets, not splitting multiplets!).
• Failing to explicitly state the similarity (IP1) results in a deduction of 1 reasoning mark.