AQA A-Level Chemistry Paper 2, 2019: Question 13

✅ Correct Answer & Mark Scheme Breakdown

• M1 (Structure): Correct structure of 2-methylbutanal: CH₃CH₂CH(CH₃)CHO (drawn displayed, structural, or skeletal).
• M2 (Mechanism Arrows):
  • Hydride ion drawn with a lone pair and negative charge: :H⁻
  • Curly arrow starting strictly from the lone pair on :H⁻ to the carbonyl carbon atom.
  • Curly arrow starting from the C=O double bond moving onto the oxygen atom.
• M3 (Explanation - Aldehyde): The nucleophile / H⁻ ion is attracted to the partially positive carbon ( δ+ C ) of the polar carbonyl group.
• M4 (Explanation - Alkene nature): The alkene double bond ( C=C ) is electron-rich / has an area of high electron density.
• M5 (Explanation - Alkene repulsion): The nucleophile / H⁻ is repelled by the electron-rich C=C bond (or: C=C is only attacked by electrophiles).

💡 How to Draw the Mechanism

Since this only asks for the first step, you do not need to show the alkoxide intermediate reacting with H⁺ / H₂O:

• Substrate: Draw the central chain of 4 carbons. Place a methyl group on carbon-2: CH₃-CH₂-CH(CH₃)-CH=O .
• Partial Charges: If included, ensure δ+ is on C and δ- is on O.
• Arrow 1: From the lone pair of :H⁻ pointing directly at the carbonyl carbon atom.
• Arrow 2: From the centre of the C=O double bond pointing directly onto the oxygen atom.

🧠 Exam Technique: Structuring the Explanation

To secure all 3 explanation marks (M3, M4, M5), compare both species systematically:

1. State that H⁻ acts as a nucleophile (an electron-pair donor).
2. Identify the target in the aldehyde: the δ+ carbon of the polar C=O group attracts H⁻ .
3. Identify the electronic nature of the alkene: the C=C bond has high electron density.
4. State the interaction: like charges repel, so the H⁻ nucleophile is repelled by the C=C bond.

❌ Common Errors to Avoid

• Missing the lone pair: Drawing H⁻ with just a negative charge but no lone pair loses M2.
• Inaccurate curly arrows: Starting the arrow from the negative charge sign rather than the lone pair, or terminating vague arrows in empty space.
• Inverted polarities: Writing δ- on carbon and δ+ on oxygen immediately cancels M2.
• Vague comparisons: Stating merely that "alkenes react with electrophiles" without mentioning that the C=C double bond is electron-rich or repels the nucleophile.

✅ Correct Answer Options

Option 1 (Tollens' Reagent):

• Reagent: Tollens' reagent (or ammoniacal silver nitrate solution) [1 mark]
• Observation: Silver mirror OR black precipitate / deposit [1 mark]

Option 2 (Fehling's / Benedict's Solution):

• Reagent: Fehling's solution OR Benedict's solution [1 mark]
• Observation: Red precipitate / solid (allow brown or orange precipitate) [1 mark]

❌ The Major Trap: Why NOT Acidified Dichromate?

Examiner Warning: The mark scheme explicitly states "NOT dichromate".

• The reaction product of reducing 2-methylbutanal is a primary alcohol (2-methylbutan-1-ol).
• If reduction is incomplete, the mixture contains both the remaining aldehyde and the newly formed primary alcohol.
• Acidified potassium dichromate(VI) oxidises both aldehydes and primary alcohols (turning orange to green in both cases). Therefore, it cannot confirm the presence of unreacted aldehyde!
• Only Tollens' or Fehling's reagents selectively oxidise aldehydes without reacting with primary alcohols.

🧠 Exam Technique: Naming Reagents Accurately

• Always state Tollens' reagent or ammoniacal silver nitrate. Simply writing AgNO₃ or [Ag(NH₃)₂]⁺ on its own is ignored by examiners.
• For Fehling's, write Fehling's solution. Writing just Cu²⁺(aq) or CuSO₄ will not gain credit for the reagent mark.
• Always state the state of the product in the observation: say silver mirror or red precipitate, not just "turns red".