AQA A-Level Chemistry Paper 2, 2019: Question 6

✅ Mark Scheme Answers

Reagent: Acidified potassium dichromate(VI) / K₂Cr₂O₇ / H₂SO₄ (or K₂Cr₂O₇ / H⁺ )

Observation with 2-methylbutan-1-ol: Orange solution turns green.

Observation with 2-methylbutan-2-ol: No visible change / remains orange.

💡 Chemical Classification

Classify the alcohol types first before picking a test:

• 2-methylbutan-1-ol: Primary (1°) alcohol ( -CH₂OH ). Can be oxidized to an aldehyde and then to a carboxylic acid. Chromium is reduced: Cr(VI) (orange) → Cr(III) (green).
• 2-methylbutan-2-ol: Tertiary (3°) alcohol ( -C(CH₃)(OH)- ). Has no hydrogen atom on the carbinol carbon, so it resists oxidation under mild conditions.

❌ Common Errors

• Omitting the acid: Writing just "potassium dichromate" or "K₂Cr₂O₇" without specifying an acid scores 0 for the reagent (and prevents observation marks).
• Suggesting Tollens' or Fehling's directly: These test for aldehydes, not directly for alcohols in a single-step test-tube reaction.
• Writing "no reaction occurs": For tertiary alcohols, observations must be visible, e.g. "remains orange" or "no visible change".

🧠 Exam Technique

Always state the initial and final appearance or clearly specify the colour change:

"Solution turns from orange to green"

Always check whether the question specifies a single test-tube reaction. Reagents must be fully formulated (name or formula) including oxidation state or acid partner.

✅ Correct Structures

Compound A: 2,2-dibromobutane

Compound B: 1,2-dibromo-2-methylpropane

📐 Structural Logic Breakdown

Compound A (singlet, triplet, quartet):

• Triplet + Quartet: Classic signature of an isolated ethyl group ( -CH₂-CH₃ ) adjacent to a carbon with 0 protons.
• Singlet: A methyl group ( -CH₃ ) attached to a carbon with 0 protons.
• Combining these gives a quaternary central carbon bearing both Br atoms: CH₃-C(Br)₂-CH₂CH₃ .

Compound B (only two singlets):

• Only two proton environments, and no splitting between them (no adjacent non-equivalent H atoms).
• Two identical methyl groups attached to a quaternary carbon: (CH₃)₂C(Br)- (6H singlet).
• An isolated bromomethyl group: -CH₂Br (2H singlet).

🧠 Exam Tip: The (n+1) Rule

Remember that a singlet requires 0 adjacent protons. Whenever you have an ethyl group alongside a methyl singlet, suspect a quaternary carbon (like C2 in 2,2-dibromobutane) acting as a barrier to spin-spin coupling.

❌ Common Trap for Compound B

Students often try symmetrical structures like 1,4-dibromobutane or 2,3-dibromobutane. However:

• 1,4-dibromobutane gives two triplets (not singlets).
• 2,3-dibromobutane gives a doublet and a multiplet.
• Only 1,2-dibromo-2-methylpropane gives exactly two uncoupled singlets for a C₄ chain.

✅ Correct Structures

Compound C (2 peaks): 1,3,5-tribromobenzene

Compound D (4 peaks): 1,2,3-tribromobenzene

💡 Carbon Environment Analysis

The number of peaks in ¹³C NMR equals the number of chemically unique carbon environments:

• 1,3,5-tribromobenzene (3-fold symmetry):
  • 3 equivalent C–Br carbons (Environment 1)
  • 3 equivalent C–H carbons (Environment 2)
  → Exactly 2 peaks.
• 1,2,3-tribromobenzene (Plane of symmetry through C2 & C5):
  • C2 ( C-Br )
  • C1 and C3 ( C-Br , equivalent pair)
  • C4 and C6 ( C-H , equivalent pair)
  • C5 ( C-H )
  → Exactly 4 peaks.

❌ Common Examiner Penalties

• Missing aromatic ring circle: Drawing a plain hexagon without the alternating double bonds (Kekulé) or central delocalized circle is heavily penalized (represents cyclohexane, not benzene).
• Selecting 1,2,4-tribromobenzene: This isomer lacks symmetry and has 6 unique carbon environments, thus producing 6 peaks in ¹³C NMR.

✅ Correct Sequence (Top to Bottom)

💡 Spectral Feature Key

• F (Ester): Has a sharp C=O peak at ~1740 cm⁻¹, but no O–H band above 3000 cm⁻¹. The baseline stays high above 3000 cm⁻¹ except for small sharp C–H stretches.
• G (Alcohol + Ketone): Has a distinct, smooth, rounded alcohol O–H trough centered around 3350 cm⁻¹ clearly separate from the C–H absorptions, plus a ketone C=O at ~1715 cm⁻¹.
• E (Carboxylic Acid): Shows the extremely broad, jagged acid O–H absorption stretching across 2500–3000 cm⁻¹, which swallows up and distorts the C–H stretches, alongside a C=O peak at ~1715 cm⁻¹.