AQA A-Level Chemistry Paper 2, 2019: Question 11

✅ Correct Answer

Equation:

(Or showing displayed/structural formulas: propanoic anhydride + ethanol → ethyl propanoate + propanoic acid)

Name of ester: ethyl propanoate

💡 Key Knowledge

• Acid anhydride + alcohol → ester + carboxylic acid.
• Unlike acyl chlorides, acid anhydrides do not produce toxic HCl fumes; they form a safe carboxylic acid by-product.
• Naming Rule: Alkyl group attached to single-bonded oxygen comes first ( ethyl ), followed by acyl chain name ending in '-oate' ( propanoate ).

❌ Common Errors

• Naming the ester propyl ethanoate (swapping the acyl and alkoxy groups).
• Writing H₂O as a by-product instead of propanoic acid (confusing anhydrides with carboxylic acid condensation).
• Incorrect skeletal/structural connectivity, e.g. drawing 5 carbons in the ester chain.

🧠 Exam Technique

Mark Breakdown:
• M1: Structure of ester ( CH₃CH₂COOCH₂CH₃ or displayed equivalent).
• M2: Formula of propanoic acid ( CH₃CH₂COOH ) and correctly balanced equation.
• M3: "ethyl propanoate" only. Spelling must be exact!

✅ Correct Structure

Structure showing glycerol backbone and ester linkages:

(The two C₁₇H₃₁ chains and one C₁₇H₃₃ chain may appear in any vertical order top-to-bottom).

❌ Common Errors & Connectivity Traps

• Reversed ester linkages: Writing -CO₂- or -COO- directly connected to glycerol (implying glycerol is bonded to the carbonyl carbon instead of the oxygen). Acceptable shorthand forms are -O₂C- , -OOC- , or -OCO- .
• Miscounting hydrogen atoms: forgetting that removing CH₃ (and adding back H for free acid) leaves C₁₇H₃₁ and C₁₇H₃₃.

✅ What the Complete Structure Must Look Like

Continue from C8 (already provided pointing downwards-right):

• Bond from C8 to C9.
• Double bond C9=C10 with Z geometry (both C8 and C11 are on the same side of the double bond, giving a "boat/cis-cup" shape).
• Single bond C10 to C11, and single bond C11 to C12 (a -CH₂- spacer between double bonds).
• Double bond C12=C13 also in Z geometry (both C11 and C14 on the same side).
• Single bond extending to C14.

💡 Key Knowledge: Z (Zusammen) Alkene Rules

• Z isomer: The high-priority groups (here, the continuation carbon chains) are on the same side of the C=C double bond plane.
• In skeletal form, a Z double bond looks like a "U-bend" or cusp rather than a zig-zag (which represents an E / trans double bond).
• Watch the numbering: the double bonds begin at C9 and C12, meaning double bonds exist between C9=C10 and C12=C13.

❌ Common Errors

• Drawing conjugated double bonds (e.g. C9=C10-C11=C12). Remember natural fatty acids have a methylene ( -CH₂- ) bridge between double bonds!
• Drawing one or both double bonds in the E (trans / zig-zag) arrangement.
• Drawing explicit hydrogen atoms on a skeletal structure (if drawn as a displayed formula or with H's on carbons, M1 is forfeited).

🧠 Exam Technique

Mark Breakdown:
• M1: Correct skeletal chain from C9 to C14 with double bonds specifically at C9=C10 and C12=C13.
• M2: Both double bonds clearly displaying Z geometry (independent mark).

✅ Balanced Chemical Equation

Also accepted (doubled):

📐 Balancing Step-by-Step

1. Carbons: 19 C on left → 19 CO₂ (needs 38 O atoms).
2. Hydrogens: 34 H on left → 17 H₂O (needs 17 O atoms).
3. Total O required on right: 38 + 17 = 55 oxygen atoms.
4. Account for O in fuel: The ester already contains 2 oxygen atoms.
5. Oxygen needed from O₂: 55 − 2 = 53 oxygen atoms = 26.5 O₂ (or 53/2 O₂ ).

✅ Correct Explanation

The absorption peak in the spectrum is at ~2350 cm⁻¹, which does not correspond to the Data Booklet Table A value for C=O bonds in organic compounds (which is 1680 – 1750 cm⁻¹).

💡 Why does this difference exist?

• In the AQA Data Booklet, Table A gives infrared absorption data primarily for organic compounds (carbonyl groups in aldehydes, ketones, esters, and carboxylic acids: 1680–1750 cm⁻¹).
• Carbon dioxide is a linear inorganic molecule (O=C=O) with two adjacent double bonds (cumulative double bonds). The asymmetric stretching vibration occurs at a much higher frequency (2350 cm⁻¹).

✅ Model Answer

Point 1: The polar C=O bonds in CO₂ absorb infrared radiation (at ~2350 cm⁻¹).

Point 2: This IR radiation emitted by the Earth is prevented from escaping into space OR this energy is transferred to other atmospheric gas molecules via collisions, warming the atmosphere.

💡 Mechanism of the Greenhouse Effect

1. Sunlight (mostly UV and visible) passes through the atmosphere and warms the Earth's surface.
2. The Earth re-emits this energy as longer-wavelength infrared (IR) radiation.
3. Greenhouse gases (CO₂, H₂O, CH₄) have polar bonds that change dipole moment during vibration, allowing them to absorb IR radiation.
4. The molecules re-emit this IR in all directions (including back to Earth) or transfer kinetic energy by collision, trapping heat.

❌ Common Errors & Disallowed Phrasing

• "IR radiation is reflected": The mark scheme explicitly states Ignore IR reflected. Radiation is absorbed and re-emitted, NOT reflected like a mirror!
• Confusing global warming with ozone depletion: mentions of "UV radiation being trapped" or "holes in the ozone layer" score 0.
• Failing to mention bonds: You must state that C=O bonds absorb the radiation.

🧠 Exam Technique

Mark Breakdown:
• M1: C=O bonds in CO₂ absorb IR radiation.
• M2: Trapping mechanism clearly stated: Earth-emitted IR does not escape / re-radiated to Earth / energy transferred to other gas molecules by molecular collisions.