Edexcel A-Level Chemistry AS Paper 2, June 2025: Question 4

✅ Exact Marking Points

• Mark 1: Curly arrow starting specifically from a lone pair on the oxygen of :OH⁻ pointing directly to the δ+ carbon of the C–Br bond.
• Mark 2: Curly arrow starting from the C–Br bond and pointing directly to (or just beyond) the bromine atom.
• Mark 3: Partial charges correct ( C^δ+—Br^δ- ) on the halogenoalkane AND the bromide ion product correctly shown as Br⁻ (or KBr if K⁺ was included on reactants).

💡 Mechanism Layout Details

🧠 Exam Technique & Examiner Commentary

• Lone pair is essential: The arrow MUST start from an explicit lone pair on the oxygen of OH⁻. Starting from the negative charge or oxygen symbol without a drawn pair loses Mark 1.
• Do not draw K–OH: Aqueous KOH dissociates. The attacking nucleophile is OH⁻. If you draw a covalent K–OH bond attacking, Mark 1 cannot be awarded.
• S_N1 vs S_N2: 1-bromobutane is a primary halogenoalkane and reacts predominantly via an S_N2 pathway (single concerted step). While the mark scheme permits a fully correct S_N1 mechanism, primary carbocations are unstable—always draw the direct one-step S_N2 mechanism for primary halogenoalkanes.

❌ Common Errors to Avoid

• Vague arrow origins: Drawing the arrow floating in empty space rather than precisely from the lone pair of electrons.
• Forgetting product charges: Leaving out the negative charge on the bromide product ( Br⁻ ).
• Arrow landing: Ending the C–Br cleavage arrow on the bond rather than specifically pointing to the Br atom.

✅ Required Diagram Features

• Mark 1: The vertical y-axis labelled specifically as "Enthalpy" or "Energy", with the horizontal line for the reactants drawn lower than the horizontal line for the products.
• Mark 2: A single-headed vertical arrow pointing upwards from the reactant level to the product level labelled "ΔH", with correct chemical formulae written on the levels:
  • Reactants: CH₃CH₂CH₂CH₂Br + H₂O (or C₄H₉Br + H₂O )
  • Products: CH₃CH₂CH₂CH₂OH + HBr (or C₄H₉OH + HBr )

🧠 Visualizing the Profile

❌ Common Errors & Instant Mark Losers

• Labelling the y-axis "Enthalpy Change": The axis represents absolute Enthalpy or Energy. "Enthalpy change" is the difference (ΔH) between levels, not the axis itself!
• Downward arrow: An arrow pointing downwards indicates an exothermic process and scores 0 for Mark 2. Endothermic changes require an upward arrow.
• Missing inorganic product: Forgetting HBr on the product line. Both products must be stated.
• Double-headed arrows: The ΔH arrow must be single-headed showing the direction of enthalpy change (↑).

💡 Examiner Guidance

• State symbols are not required (as stated in the question stem).
• You do not need to draw activation energy (E_a) or an energy hump unless asked for a reaction profile; this is an enthalpy level diagram. However, if drawn, an E_a arrow pointing down will be penalised.

✅ Model Answer

As the carbon chain length increases, the molecules have more electrons, resulting in stronger London forces (instantaneous dipole–induced dipole forces).

Therefore, more energy is required to overcome / separate these intermolecular forces, increasing the melting temperature until they are solid at room temperature.

💡 Why London Forces Increase

• Every additional —CH₂— group adds 8 electrons.
• A greater number of electrons leads to larger, more easily polarised electron clouds.
• This creates stronger temporary (instantaneous) dipoles and induced dipoles between adjacent chains.

❌ Major Pitfalls

• "Breaking covalent bonds": Writing that "more energy is needed to break the bonds in the alcohol" scores 0 for Mark 2. You must state intermolecular forces are overcome.
• Saying hydrogen bonding increases: Each unbranched mono-alcohol has only one —OH group, so the extent of hydrogen bonding remains roughly constant. The change in state is driven entirely by increasing London forces!
• Vague naming: Avoid just "dipole-dipole forces". Name them explicitly as London forces, dispersion forces, or instantaneous dipole–induced dipole forces.

🧠 Two-Step Exam Structure

1. Cause: More electrons → stronger London forces. (1 mark)
2. Consequence: More energy needed to separate molecules / break intermolecular forces. (1 mark)

✅ Model Answer

• Trend: Boiling temperature decreases as the amount of branching increases (pentan-1-ol is unbranched, 2-methylbutan-1-ol has one branch, 2,2-dimethylpropan-1-ol has two branches).
• Reason: Increased branching makes the molecules more spherical/compact, leading to a smaller surface area (less surface contact between molecules), resulting in weaker London forces that require less thermal energy to overcome.

💡 The Mechanism of Branching

All three isomers share the same molecular formula ( C₅H₁₂O ) and identical electron counts (42 electrons), so the difference is purely molecular shape:

• Straight-chain (pentan-1-ol): Long, linear molecules pack closely together with large surface contact → maximal London forces.
• Branched isomers: Alkyl branches prevent molecules from packing closely together → fewer points of contact → weaker London forces.

❌ Common Errors

• Claiming electron numbers change: These compounds are structural isomers. They have identical numbers of electrons and protons. Stating branched isomers have fewer electrons loses the mark.
• Attributing difference to hydrogen bonds: All three are primary alcohols with one —OH group. The differences in boiling temperatures are due to London forces, not hydrogen bonds.
• Confusing boiling point with covalent bond strength: Boiling involves overcoming intermolecular forces, not C–C or C–H bonds.

🧠 Key Examiner Phrases to Use

• "Increased branching decreases surface contact between molecules"
• "London forces / dispersion forces are weaker"
• "Less energy required to separate molecules"