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

✅ Correct Answer & Diagram Requirements

• Mark 1: Thermometer and bung placed in the neck of the round-bottom flask, with the thermometer bulb positioned directly opposite/level with the T-junction to the condenser side-arm.
• Mark 2: Condenser jacket enclosing the side-arm with water in at the bottom (lowest point) and water out at the top (highest point).

💡 Key Knowledge

• Bulb position: The thermometer bulb must measure the temperature of the vapor entering the condenser to ensure only the fraction boiling at 83 °C is collected.
• Water flow: Entering at the bottom ensures the condenser jacket fills completely without air pockets, providing efficient cooling.

🧠 Exam Technique

• Draw apparatus as a continuous cross-section. Ensure no closed glass lines block vapor flow from flask into condenser.
• Do not seal the collection end at the conical flask; the system must not be airtight, or pressure build-up could cause an explosion.

❌ Common Errors

• Drawing the thermometer bulb dipped in the boiling liquid or placed too high up the neck.
• Water entering at the top and leaving at the bottom (inefficient cooling).
• Gaps between bung and flask neck, or completely blocking the condenser path with a line.

✅ Correct Answer

The liquids are immiscible (or: they do not mix / form two distinct layers / cyclohexene is insoluble in water).

💡 Key Knowledge

Cyclohexene is a non-polar hydrocarbon capable only of London dispersion forces, whereas water forms strong hydrogen bonds. Because cyclohexene cannot form hydrogen bonds with water, it does not dissolve and forms a separate upper layer.

❌ Common Errors

Referring solely to differences in density or differences in boiling point. Density explains which layer floats on top, but immiscibility is what permits separation in a separating funnel.

✅ Correct Answer

The liquid goes clear / is no longer cloudy.

💡 Key Knowledge

Trace amounts of suspended water droplets disperse light, making wet organic liquids appear turbid or cloudy. Once the drying agent (anhydrous CaCl₂) absorbs the water, the liquid turns completely translucent/clear.

❌ Common Errors

Writing that the liquid turns "colourless". Pure cyclohexene is both clear and colourless, but "clear" specifically refers to lack of cloudiness (transparency), whereas "colourless" refers to lack of hue.

🧠 Exam Technique: Significant Figures

Look at the given data in the question prompt:

• Mass of cyclohexanol: 14.4 g (3 sf)
• Volume of cyclohexene: 4.15 cm³ (3 sf)
• Density of cyclohexene: 0.810 g cm⁻³ (3 sf)

The least number of significant figures in the given measurements is 3, so your final answer must be quoted to 3 sf to earn M5.

❌ Common Errors

• Inverting density: dividing volume by density instead of multiplying ( m = d × V ).
• Premature rounding during intermediate steps leading to rounding errors (e.g. 28.4% or 28.6%).
• Quoting the final answer to 2 sf (28%) or 4 sf (28.47%), forfeiting mark M5.

✅ Mechanism Requirements

• Mark 1: Curly arrow from the C=C double bond to the nearer Br atom of the induced dipole Br^δ+–Br^δ- molecule, AND a curly arrow from the Br–Br bond to the outer Br atom.
• Mark 2: Correct structure of the carbocation intermediate: a cyclohexane ring with a single C–C bond where the double bond was, one Br attached to one of the carbons, and a full positive charge ( + ) on the adjacent ring carbon.
• Mark 3: Curly arrow from a lone pair on the bromide ion (:Br⁻) to the positively charged carbon of the carbocation intermediate.

💡 What to Draw

• Step 1: Show partial charges on bromine: Br^δ+–Br^δ-. Draw arrow 1 starting inside the C=C bond pointing directly to Br^δ+. Draw arrow 2 from the center of the Br–Br bond to Br^δ-.
• Intermediate: Six-membered carbon ring. One former alkene carbon has –Br bonded to it; the neighboring ring carbon carries a + charge.
• Step 2: Draw :Br⁻ with two dots representing the lone pair and a negative sign. Arrow originates strictly from the lone pair and points to the C⁺.

❌ Common Errors & Mark Penalties

• Full charges on Br₂: Showing Br⁺–Br⁻ instead of partial dipoles loses M1 immediately.
• Arrow origins: Starting the curly arrow from a carbon atom rather than directly from the C=C bond line.
• Missing lone pair: Drawing the arrow from the negative charge on Br⁻ rather than from an explicit lone pair of electrons (:Br⁻).
• Incorrect Ring Hydrogens: If you draw hydrogen atoms on the ring carbons, ensure the number is chemically correct (one H on each former double-bond C).