AQA A-Level Chemistry Paper 2, 2018: Question 11

✅ Correct Answers

• Number of ¹³C peaks: 4 [1 Mark]
• Splitting pattern: Triplet [1 Mark]
• Explanation: Two H on adjacent C (or equivalent reference to coupling with the adjacent CH₂ protons) [1 Mark]

💡 Key Knowledge

• Carbon Environments: The molecule is (CH₃-CH₂)₂N-CH₂-CH₂-NH₂. Because the two ethyl groups attached to the tertiary nitrogen are equivalent:
  1. Both methyl carbons: ( C H₃-CH₂)₂N- (1 peak)
  2. Both ethyl -CH₂- carbons: (CH₃- C H₂)₂N- (1 peak)
  3. Bridge carbon a: - C H₂-CH₂-NH₂ (1 peak)
  4. Bridge carbon next to primary amine: -CH₂- C H₂-NH₂ (1 peak)
  Total = 4 unique carbon environments.
• (n + 1) Rule: Hydrogens labelled a are bonded to a carbon adjacent to a -CH₂- group (2 protons). Since n = 2, splitting = 2 + 1 = 3 (triplet).

🧠 Exam Technique

M3 depends entirely on getting M2 correct. Always give a concise reason: state the exact number of protons on the adjacent carbon. Protons attached to heteroatoms (like -NH₂) typically do not couple under routine conditions, so focus strictly on adjacent C-H protons.

❌ Common Errors

• Counting 6 separate ¹³C peaks by failing to spot the symmetry of the diethylamino group.
• Stating "quartet" by mistakenly adding coupling to the nitrogen or the amine protons.
• Vague explanations like "because there are 2 hydrogens" without specifying they are on the adjacent carbon.

✅ Correct Answer

Any unambiguous representation of hexane-1,6-diamine (1,6-diaminohexane):

💡 Key Knowledge

Nylon 6,6 is a polyamide formed by a condensation reaction between a 6-carbon dicarboxylic acid (hexanedioic acid) and a 6-carbon diamine (hexane-1,6-diamine).

Both monomers contain 6 carbon atoms, which gives the polymer its name "6,6".

❌ Common Errors

• Drawing an ambiguous condensed group such as -C₆H₁₂- . The mark scheme explicitly states: Not -C₆H₁₂-.
• Miscounting the carbon chain (drawing 5 or 7 carbons instead of 6).

🧠 Exam Technique

When drawing condensed structural formulae, write the repeating methylene units clearly as -(CH₂)₆- . If drawing skeletal formula, count the vertices carefully (6 carbons in the chain between the two NH₂ groups).

✅ Correct Answer

2,3-dimethylbutane-2,3-diamine

or skeletal representation with two central branching carbons each bearing an -NH₂ and two methyl groups.

📐 Deduction Strategy

1. Condition 1: Contains two primary amine groups (-NH₂). This accounts for 2 × N and 4 × H.
2. Condition 2: Molecular formula is C₆H₁₆N₂. Remaining fragment = C₆H₁₂.
3. Condition 3: Only 2 peaks in the ¹³C NMR spectrum. The molecule must possess very high symmetry.
4. A symmetrical split gives two identical halves of 3 carbons each: each half has a central carbon attached to -NH₂ and two methyl groups.
5. The 4 methyl carbons are in 1 environment; the 2 central carbons are in the 2nd environment. Total = 2 peaks.

❌ Common Errors

• Drawing 1,6-diaminohexane (which has 3 ¹³C environments, not 2).
• Drawing secondary or tertiary amines instead of primary amines (e.g., bonding alkyl groups to nitrogen).

🧠 Exam Technique

Always double-check the classification: a primary amine must have its nitrogen bonded to only ONE carbon atom (i.e. strictly -NH₂ groups).

✅ Correct Answer

N,N,N',N'-tetramethylethylenediamine (or 1,2-bis(dimethylamino)ethane):

or represented as (CH₃)₂N-CH₂-CH₂-N(CH₃)₂

📐 Deduction Strategy

1. Condition 1: Two tertiary amine groups. Each nitrogen atom must be bonded to 3 carbons (no N-H bonds).
2. Condition 2: Total formula = C₆H₁₆N₂; only 2 peaks in ¹³C NMR.
3. To have 2 tertiary nitrogens with 6 carbons total and high symmetry:
   • Four methyl groups distributed equally: two on each nitrogen: 4 × CH₃ = 4 carbons. (Peak 1)
   • A central symmetrical ethylene linker: -CH₂-CH₂- connecting the two nitrogens = 2 carbons. (Peak 2)
4. Total carbons = 4 + 2 = 6 carbons; total hydrogens = 12 + 4 = 16. Matches C₆H₁₆N₂ exactly!

❌ Common Errors

• Writing condensed ambiguous bridges like -C₂H₄- (specifically rejected in the mark scheme).
• Drawing secondary amines like (CH₃-CH₂)NH-(CH₂-CH₂)NH₂ which do not have two tertiary amine centres.

🧠 Exam Technique

Contrast 11.3 and 11.4 directly: 11.3 demanded two primary amines (-NH₂), forcing methyl branches onto the carbon skeleton. 11.4 demanded two tertiary amines, meaning the methyl groups must be attached directly to the nitrogen atoms.