Covers Lessons 17–20: Carboxylic acids, esters (esterification, saponification, soaps), amines and amides, reaction pathways, and multi-step organic synthesis.
1. Why does ethanoic acid (BP 118 °C) have a higher boiling point than ethanol (BP 78 °C) despite ethanol having a lower molecular mass?
2. Solutions A, B, and C are tested: A reacts with NaHCO₃ → CO₂ produced. B — no reaction with NaHCO₃; reacts with NaOH. C — no reaction with either. Which identification is correct?
3. The saponification of glyceryl trioleate (a fat) with NaOH produces which products?
4. Which is the correct balanced equation for esterification of propanoic acid with ethanol?
5. Which of the following is NOT achievable in a single Module 7 reaction step?
MC Answers: 1-B | 2-B | 3-A | 4-B | 5-B
1-B: Both have –OH and form H-bonds, but ethanoic acid forms cyclic dimers with TWO simultaneous H-bonds per pair — breaking both requires more energy than breaking single H-bond interactions in ethanol → higher BP.
2-B: Only carboxylic acids (pKa ~5) are strong enough to react with NaHCO₃ (H₂CO₃ pKa ~6.4) → CO₂. Phenol (pKa ~10) reacts with NaOH but not NaHCO₃. Alcohol (pKa ~16) reacts with neither.
3-A: Saponification: fat + 3NaOH → glycerol + 3 × sodium oleate (soap) + 3H₂O. One mole of water per ester bond hydrolysed. Option C omits water.
4-B: Esterification is a reversible equilibrium (⇌ required). Product = ethyl propanoate (ethanol provides "ethyl", propanoic acid provides "propanoate"). Option A uses single arrow (wrong). Option C has wrong acid part (ethanoate, not propanoate).
5-B: No direct single-step alkane → alcohol exists in Module 7. Required two steps: alkane → haloalkane (UV light), then haloalkane → alcohol (NaOH aq). Options A, C, D are all single-step reactions.
SA1 (6 marks):
Step 1: CH₃CH₂CH₂CH₂I + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaI. Reagent: NaOH(aq); reflux. Product: butan-1-ol. [1]
Step 2: Butan-1-ol + [O] → CH₃CH₂CH₂CHO + H₂O. Reagent: K₂Cr₂O₇/H₂SO₄; mild, DISTILLATION. Product: butanal. [1]
Step 3: CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH. Reagent: K₂Cr₂O₇/H₂SO₄ (excess); REFLUX. Product: butanoic acid. [1]
Step 4: CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂O. Reagent: butan-1-ol (from Step 1); catalyst: conc. H₂SO₄; reflux; ⇌ arrow required. Product: butyl butanoate. [2 marks] Note: butan-1-ol is used in both oxidation steps AND as the ester alcohol — batch must be split from Step 1. [1]
SA2 (4 marks):
Ethanamine: N lone pair in sp³ orbital, freely available. CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻. Kb ~4 × 10⁻⁴ → measurable OH⁻ → pH > 7. [1+1 marks]
Ethanamide: N lone pair participates in resonance with adjacent C=O (CH₃CO–NH₂ ↔ CH₃C(O⁻)=NH₂⁺). Lone pair is delocalised into the C=O π system → not available in a non-bonding orbital → cannot accept a proton from water. Kb ~10⁻¹⁵ → no measurable OH⁻ → pH ≈ 7. [1+1 marks]
Tick when you have finished all questions.