Edexcel A-Level Chemistry AS Paper 2, June 2016: Question 2

✅ Correct Answer

D: Z-2-bromo-1-chloroprop-1-ene

💡 Key Knowledge

• Determine priority using the Cahn-Ingold-Prelog (CIP) priority rules based on atomic number.
• On C1: Br (atomic number 35) outranks Cl (atomic number 17).
• On C2: CH₃ carbon outranks the hydrogen atom.
• Since both high-priority groups are on the same side of the double bond, it is the Z -isomer.

🧠 Exam Technique

Don't default automatically to cis / trans when multiple different substituents are attached to the C=C carbons. Z (zusammen / same side) is the IUPAC system required for ambiguous multi-substituent alkenes.

❌ Common Errors

Confusing cis with Z , or assigning priority incorrectly by looking at the whole chain mass instead of the direct atoms attached to the C=C bond (e.g., comparing Cl vs Br atoms directly).

✅ Correct Answer

A: electrophilic addition

💡 Key Knowledge

• Alkenes are electron-rich due to the pi bond, making them vulnerable to attack by electrophiles (electron-pair acceptors).
• The π bond breaks, adding atoms across the double bond to form a single saturated product molecule.

❌ Common Errors

Mixing up addition with substitution. Substitution involves replacing an atom/group, which happens with alkanes/arenes, not alkenes.

✅ Correct Answer

C

💡 Key Knowledge

• Curly arrows must start from the electron source: start the first arrow from the C=C double bond pointing to the δ+ bromine atom.
• The second arrow shows the Br-Br bond heterolytically breaking, with the pair of electrons moving entirely onto the δ- bromine atom to form a bromide ion ( Br⁻ ).
• The carbocation intermediate formed carries a + charge on the carbon atom.

🧠 Exam Technique

Examine arrow heads and tails carefully. Option C correctly shows the curly arrow originating right at the double bond and the Br-Br bond pair shifting to form Br⁻ alongside a positively charged carbon skeleton.

❌ Common Errors

Drawing curly arrows starting in 'empty space' rather than directly from the bond or lone pair, or drawing the bond-cleavage arrow pointing the wrong way.

✅ Correct Answer

(yield) decreases / lower yield

💡 Key Knowledge

• Look at the enthalpy change: ΔH = -45 kJ mol⁻¹ , meaning the forward reaction is exothermic.
• According to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the endothermic (reverse) direction to absorb the extra heat.

🧠 Exam Technique

Keep your answer direct. State clearly that the yield decreases. Avoid long philosophical explanations about Le Chatelier unless asked, as examiners penalise contradictory statements.

❌ Common Errors

Saying "equilibrium shifts to the left" without answering the actual question asked, which specifically demanded the effect on the yield of ethanol.

✅ Correct Answer

(yield) decreases / lower yield

💡 Key Knowledge

• Count moles of gas on each side:
  Left side: 1 mol C₂H₄ + 1 mol H₂O = 2 moles of gas.
  Right side: 1 mol C₂H₅OH = 1 mole of gas.
• Decreasing pressure shifts the equilibrium towards the side with more moles of gas (the left-hand side), reducing product formation.

🧠 Exam Technique

Always count balancing numbers (stoichiometric coefficients) explicitly for gaseous species before evaluating pressure changes.

❌ Common Errors

Forgetting to count water vapour as a gas because it is written as H₂O(g) , or confusing reaction rate effects with equilibrium position shifts.

✅ Correct Answer

D: [C₂H₅OH(g)] / ([C₂H₄(g)][H₂O(g)])

💡 Key Knowledge

• The general expression for K_c is products divided by reactants, with each concentration raised to the power of its stoichiometric coefficient.
• All species in this reaction are gases (g) , so all terms must be included in the expression.

🧠 Exam Technique

Always use square brackets [ ] to represent equilibrium concentrations, never round brackets ( ) which indicate partial pressures or general states.

❌ Common Errors

Inverting the fraction (putting reactants on top) or adding concentrations together instead of multiplying them in the denominator.