Covers Lessons 13–16: Alcohols (production, classification, reactions — dehydration, substitution, oxidation), aldehydes, ketones, Tollens'/Fehling's tests, and combustion calorimetry.
1. A student heats butan-1-ol with K₂Cr₂O₇/H₂SO₄ under reflux. What is the organic product and what colour change is observed?
2. A student burns 0.45 g of propan-1-ol (M = 60.09 g/mol) under a calorimeter with 150 g of water. The temperature rises from 20.0 °C to 33.6 °C. What is the experimental molar enthalpy of combustion? (c = 4.18 J g⁻¹ K⁻¹)
3. Which conditions produce butanal (not butanoic acid) from butan-1-ol?
4. Compound A (C₄H₁₀O) gives no colour change with K₂Cr₂O₇/H₂SO₄. Compound B (C₄H₁₀O) gives orange → green with K₂Cr₂O₇/H₂SO₄ and the product gives a positive Tollens' test. What are A and B?
5. The fermentation of glucose is: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Which of the following is NOT a required condition?
MC Answers: 1-B | 2-A | 3-B | 4-A | 5-D
1-B: Butan-1-ol is primary → reflux + excess K₂Cr₂O₇/H₂SO₄ → full oxidation to butanoic acid. Butanal would require distillation, not reflux. Colour: orange → green (Cr₂O₇²⁻ → Cr³⁺).
2-A: q = 150 × 4.18 × 13.6 = 8527 J. n = 0.45/60.09 = 0.00749 mol. ΔHc = −8.527/0.00749 = −1139 kJ/mol. Negative sign because combustion is exothermic. Note: this is significantly below the theoretical value (−2021 kJ/mol) due to heat loss and incomplete combustion.
3-B: Distillation removes butanal (BP 75 °C) from the reaction flask before excess oxidant can oxidise it to butanoic acid. Reflux keeps the aldehyde in contact with oxidant → over-oxidation to acid.
4-A: No reaction with K₂Cr₂O₇ → A is a tertiary alcohol (cannot be oxidised — no H on C–OH). K₂Cr₂O₇ → green + positive Tollens' → B oxidises to an aldehyde → B is a primary alcohol (primary → aldehyde). C₄H₁₀O tertiary = 2-methylpropan-2-ol; primary = butan-1-ol.
5-D: High pressure (~65 atm) is required for ALKENE HYDRATION — not fermentation. Fermentation operates at atmospheric pressure with yeast, ~35 °C, and anaerobic conditions.
SA1 (5 marks):
(a) CH₃CHBrCH₃ + NaOH(aq) → CH₃CHOHCH₃ + NaBr. Reagent: NaOH(aq) (aqueous); reflux. [1 mark]
(b) Step 1 as above (propan-2-ol). Step 2: CH₃CHOHCH₃ + [O] → CH₃COCH₃ + H₂O. Reagent: K₂Cr₂O₇/H₂SO₄; reflux; orange → green. [2 marks]
(c) CH₃CHBrCH₃ + NaOH(alc) → CH₃CH=CH₂ + NaBr + H₂O. Reagent: NaOH in ETHANOL (alcoholic); reflux. Key: AQUEOUS NaOH gives substitution (alcohol); ALCOHOLIC NaOH gives elimination (alkene). [2 marks]
SA2 (4 marks):
(a) 680/1367 × 100% = 49.7% ≈ 50% [1 mark]
(b) Reason 1 — Heat loss to surroundings: the calorimeter walls, air, and bench absorb heat that is not captured in q = mcΔT → q_measured < q_actual → |ΔHc_exp| < |ΔHc_theoretical| [1 mark]. Reason 2 — Incomplete combustion: limited O₂ in inner flame → CO and soot (C) produced instead of CO₂ → less energy released per mole burned → smaller |ΔHc_exp| [1 mark]. Both effects cause ΔHc_exp to be systematically less negative than the theoretical value [1 mark].
Tick when you have finished all questions.