AQA A-Level Chemistry Paper 2, 2017: Question 1

✅ Correct Answer

Propan-1-ol displayed formula showing every single bond and atom explicitly:

🧠 Exam Technique: "Displayed Formula"

• A displayed formula requires all atoms and all covalent bonds to be drawn out.
• Don't forget the bond between oxygen and hydrogen: write - O - H , never -OH . Writing -OH loses the mark immediately!

✅ Correct Answer

• Mechanism: Nucleophilic substitution
• Condition: Excess NH₃ (excess ammonia)

💡 Key Knowledge

The primary amine product ( CH₃CH₂CH₂NH₂ ) still has a lone pair on nitrogen, making it a nucleophile capable of reacting further with unreacted 1-bromopropane to produce secondary and tertiary amines, and quaternary ammonium salts. Using an excess of ammonia ensures that an incoming haloalkane molecule is far more likely to collide with NH₃ than with the formed amine, maximising the yield of the primary amine.

❌ Common Errors

• Writing conditions like "heat", "reflux", or "ethanolic" alone without specifying excess ammonia.
• Misidentifying the mechanism as "electrophilic substitution" or just "substitution".

📐 Step-by-Step Calculation

1. Calculate Molar Masses (M_r):
   M_r(CH₃CH₂CH₂Br) = (3 × 12.0) + (7 × 1.0) + 79.9 = 122.9 g mol⁻¹
   (123 g mol⁻¹ is also accepted)
   M_r(CH₃CH₂CH₂NH₂) = (3 × 12.0) + (9 × 1.0) + 14.0 = 59.0 g mol⁻¹
2. Find theoretical moles of halogenoalkane:
   Moles of CH₃CH₂CH₂Br = 25.2 / 122.9 = 0.20504 mol [M1]
3. Account for 75.0% yield:
   Moles of amine formed = 0.20504 × 0.750 = 0.15378 mol [M2]
   (Or calculate theoretical maximum mass: 0.20504 × 59.0 = 12.10 g, then apply 75%)
4. Calculate actual mass to 3 sig figs:
   Mass = 0.15378 mol × 59.0 g mol⁻¹ = 9.07 g [M3]

❌ Calculation Traps & Exam Tips

• Significant figures: The prompt explicitly asks for the appropriate number of sig figs. Since 25.2 g and 75.0% are both 3 sig figs, the final answer must be given to 3 sig figs (9.07 g). Writing 9.1 g or 9.073 g loses M3!
• Premature rounding: Keep intermediate values in your calculator memory to avoid rounding errors (e.g. 9.09 g).
• Did you calculate 18.9 g? That corresponds to 75% of the starting mass, which ignores the change in molar mass entirely! (Scores only 1 mark).

✅ Correct Answer

Skeletal formula of tripropylamine:

A central nitrogen atom bonded to three 3-carbon (propyl) chains in a skeletal zigzag arrangement:

Functional group: Tertiary amine (or 3° amine)

💡 How to deduce the structure

Each propyl group is C₃H₇. Three propyl groups provide 3 × 3 = 9 carbons and 3 × 7 = 21 hydrogens, matching C₉H₂₁N exactly. This confirms that all 3 hydrogen atoms of ammonia have been substituted: (C₃H₇)₃N .

🧠 Exam Technique

• Skeletal rules: Do not draw C atoms or H atoms attached to carbon. Each vertex and line ending represents a carbon with its required hydrogens. The heteroatom (N) must be explicitly drawn.
• Classification requirement: Stating just "amine" scores 0. The question asks for the functional group including its classification, requiring "tertiary" or "3°".

✅ Correct Answer

NaOH in ethanol (or KOH in ethanol)

Both reagent and solvent are required for this 1 mark.

💡 Reagent vs Solvent: The Classic Comparison

❌ Common Errors

• Writing "aqueous NaOH" — this gives substitution, not elimination! Mark scheme note: "NOT aqueous".
• Writing just "NaOH" without specifying the ethanolic / alcoholic solvent.

✅ Mechanism Name

Elimination (or Base Elimination)

(1 mark)

📐 Mechanism Diagram (E2 mechanism - 3 Curly Arrows)

Draw 1-bromopropane showing the C(2)-H bond and C(1)-Br bond clearly:

1. Arrow 1 (M1): Starts from the lone pair on :OH⁻ to the H atom on carbon-2 (adjacent to C-Br).
2. Arrow 2 (M2): Starts from the C-H single bond to the middle of the C-C single bond (forming the C=C double bond).
3. Arrow 3 (M3): Starts from the C-Br single bond to the Br atom (breaking the bond to release Br⁻).

❌ Where Students Lose Marks on Mechanisms

• Arrow origin: Arrows must start precisely from a lone pair or the centre of a covalent bond. Starting from a minus charge or somewhere in empty space loses the mark.
• Attacking the wrong hydrogen: The OH⁻ must abstract a hydrogen from the carbon adjacent to the halogen-bearing carbon (C-2), NOT from C-1!
• Arrow destination: Arrow 2 must clearly point between the two carbon atoms to represent the forming double bond.

💡 Alternative Valid Pathway: E1 Mechanism

The mark scheme also accepts the two-step E1 mechanism:

• Step 1: Loss of Br⁻ via arrow from C-Br bond to Br, forming the carbocation intermediate CH₃-CH⁺-CH₂ or CH₃-CH₂-CH₂⁺ .
• Step 2: Arrow from lone pair on :OH⁻ to adjacent H, followed by arrow from C-H bond into the C-C bond to form the alkene.