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

Bond Fission, Radical Stability & Photon Energy Calculations

Edexcel A-Level Chemistry • Paper 1 / Paper 2 Core Organic & Physical

What This Question Tests

Planck-Einstein Relation: Converting wavelength (nm to m), calculating energy per photon, and scaling by Avogadro's constant ( L ) to J mol⁻¹ or kJ mol⁻¹.
Photochemical Initiation: Comparing radiation energy to covalent bond dissociation enthalpies (Cl—Cl vs C—H).
Free Radical Mechanisms: Drawing structural formulae of intermediate alkyl radicals with correct dot placement.
Intermediate Stability: Explaining isomeric product yields via radical stability analogies (primary vs secondary radicals and inductive effects).
Dot-and-Cross Diagrams: Differentiating between ions (heterolytic product: OH⁻) and neutral radicals (homolytic product: •OH).
Classification of Reactions: Distinguishing covalent heterolytic bond fission from ionic dissociation.

Part (a)(i): Energy of UV Radiation & Bond Fission

Calculation & Comparison of Bond Energies (4 Marks)

📐 Step-by-Step Calculation

Convert wavelength to metres (M1):
λ = 315 nm = 315 × 10⁻⁹ m = 3.15 × 10⁻⁷ m
Calculate the energy of one photon using E = (h × c) / λ (M2):
E = (6.626 × 10⁻³⁴ J s × 3.00 × 10⁸ m s⁻¹) / (3.15 × 10⁻⁷ m)
E = 6.3105 × 10⁻¹⁹ J
Calculate the energy of one mole of photons (M3):
Multiply by Avogadro's constant ( L = 6.02 × 10²³ mol⁻¹ ):
E_mole = 6.3105 × 10⁻¹⁹ J × 6.02 × 10²³ mol⁻¹ = 379 890 J mol⁻¹
Convert to kJ mol⁻¹: 379.9 kJ mol⁻¹ (or 380 kJ mol⁻¹ )
Compare with bond energies and draw conclusions (M4):
• For Cl—Cl: 379.9 kJ mol⁻¹ > 243 kJ mol⁻¹ , therefore UV light has sufficient energy to break the Cl—Cl bond.
• For C—H: 379.9 kJ mol⁻¹ < 413 kJ mol⁻¹ , therefore UV light does not have enough energy to break the C—H bond.

🧠 Alternative Method

You can convert the bond enthalpies per mole into energy per single bond instead:

Cl—Cl bond: (243 × 1000) / (6.02 × 10²³) = 4.04 × 10⁻¹⁹ J
C—H bond: (413 × 1000) / (6.02 × 10²³) = 6.86 × 10⁻¹⁹ J
Since 4.04 × 10⁻¹⁹ < 6.31 × 10⁻¹⁹ < 6.86 × 10⁻¹⁹ , only Cl—Cl breaks.

❌ Common Traps

Unit mismatches: Forgetting that h × c / λ produces Joules per single photon, while bond enthalpies are given in kilojoules per mole.
Missing the conversion of nm: 1 nm = 10⁻⁹ m. Omitting this leads to an answer off by a factor of 10⁹.
Incomplete comparison: You must state both comparisons explicitly (comparing to 243 AND 413) to earn the final evaluation mark.

Part (a)(ii): Structural Formulae of Butyl Radicals

Propagation Intermediates from Butane (2 Marks)

✅ Correct Structural Formulae

Primary butyl radical:
CH₃CH₂CH₂ĊH₂ or CH₃CH₂CH₂CH₂•
Secondary butyl radical:
CH₃ĊHCH₂CH₃ or CH₃CH•CH₂CH₃

🧠 Exam Technique: Radical Dot Precision

The radical dot represents an unpaired electron. Examiners are strict about where this dot sits:

Place the dot directly on or adjacent to the specific carbon atom lacking the hydrogen.
Do not place the dot vaguely above an entire functional group or on a hydrogen atom.
Displayed formulae showing all bonds with the dot on the correct C atom are fully acceptable.

❌ Common Errors from Examiner Reports

• Adding extra dots or omitting the radical dot entirely.
• Drawing chlorinated radicals (e.g. including Cl) rather than the butyl radicals formed in propagation step 1.
• Drawing charged carbocations instead of neutral radicals.

Part (a)(iii): Explaining Relative Product Yields

Radical Stability vs. Carbocation Stability Analogy (2 Marks)

✅ Model Answer

Mark 1 (Linking products to intermediates): 1-chlorobutane is formed from the primary radical, whereas 2-chlorobutane is formed from the secondary radical.
Mark 2 (Explaining relative stability): The secondary radical is more stable than the primary radical (due to the electron-releasing / positive inductive effect of two alkyl groups compared to one).

❌ Key Trap: The Carbocation Confusion

The question prompts: "by application of your knowledge of the stability of carbocations..."

This asks you to apply the same reasoning (tertiary > secondary > primary) to radicals! Do not state that the mechanism forms a carbocation. Stating that 1-chlorobutane is formed from a "primary carbocation" loses Mark 1 immediately.

💡 Scientific Reason

Like carbocations, carbon radicals are electron-deficient (7 electrons in their outer shell). Alkyl groups push electron density towards the radical centre via the positive inductive effect, dispersing electron deficiency and stabilizing secondary intermediates over primary ones.

Part (b): Dot-and-Cross Diagrams

Hydroxide Ion (Heterolytic) vs. Hydroxyl Radical (Homolytic) (2 Marks)

✅ Specification for Hydroxide Ion, OH⁻ (1 Mark)

Outer Shell Configuration:

1 shared pair between O and H (one dot • from H, one cross × from O).
3 lone pairs on Oxygen (6 non-bonding electrons, either 5 crosses and 1 dot from incoming electron, or all crosses). Total 8 electrons around O.
Square brackets with a negative charge: [ :Ö:—H ]⁻

✅ Specification for Hydroxyl Radical, •OH (1 Mark)