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Questions on alcohols: hydration, fermentation and classification
Alcohols and the hydroxyl group, primary, secondary and tertiary alcohols, boiling points and solubility from hydrogen bonding, ethanol by fermentation of glucose and the reasons for its conditions, ethanol by hydration of ethene with steam and a phosphoric acid catalyst and the mechanism, fermentation and the hydration of ethene compared on rate, purity, cost and renewability, the masses of raw material each route needs, biofuels, why bioethanol is described as carbon neutral and why it is not quite, and for Cambridge the reaction of alcohols with sodium and why ethanol is a weaker acid than water.
22 original questions · 65 marks · the alcohols: hydration, fermentation and classification notes · Alkenes and alcohols · Flashcards for this unit
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Evaluate the claim that ethanol produced by fermentation is a carbon neutral fuel. Include equations in your answer.
Mark scheme
Indicative content
- photosynthesis, 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(aq) + 6O₂(g), absorbs six molecules of CO₂ for each glucose the crop makes
- fermentation, C₆H₁₂O₆(aq) → 2CH₃CH₂OH(aq) + 2CO₂(g), releases two molecules of CO₂ for each glucose fermented
- combustion, 2CH₃CH₂OH(l) + 6O₂(g) → 4CO₂(g) + 6H₂O(l), releases four molecules of CO₂ when the two ethanol burn
- the 2 + 4 = 6 molecules of CO₂ released equal the 6 absorbed in photosynthesis, which is the basis of the claim
- but fossil fuels are burned in farming, making fertiliser, transporting the crop and in the fractional distillation, releasing CO₂ the crop did not absorb
- so the claim is not fully valid: bioethanol gives lower net CO₂ emissions than petrol but is not carbon neutral; ignore answers about cost
Levels
- Level 3 (5–6 marks) all three equations are correct and balanced, the six molecules of carbon dioxide absorbed are shown to equal the two plus four released, the fossil-fuel inputs of farming, fertiliser, transport and distillation are identified, and the claim is judged partly valid with the reason
- Level 2 (3–4 marks) the photosynthesis and combustion equations are given and the carbon dioxide balance is argued; the fermentation equation or the molecule count is incomplete, or fossil-fuel inputs are listed without a judgement following from them
- Level 1 (1–2 marks) the claim is accepted because the crop absorbs the carbon dioxide released on burning, with one equation or none and no thought for the fuel used in farming, transport or distillation
Calculate the percentage yield of ethanol from a fermentation, using the results given in the table.
Quantity Value Mass of glucose fermented 2.50 kg Volume of solution after fermentation 10.0 dm³ Ethanol in the solution, by volume 12.0% Density of ethanol 0.789 g cm⁻³ Mr: glucose, ethanol 180.0, 46.0 Mark scheme
- M1 volume of ethanol = 0.120 × 10 000 cm³ = 1200 cm³, so mass = 1200 × 0.789 = 947 g
- M1 amount of glucose = 2500 ÷ 180.0 = 13.89 mol, giving 2 × 13.89 = 27.78 mol of ethanol
- M1 maximum mass of ethanol = 27.78 × 46.0 = 1278 g; error carried forward from the amount of ethanol
- A1 percentage yield = 947 ÷ 1278 × 100 = 74.1%, to three significant figures
Compare the production of ethanol by fermentation with its production by the hydration of ethene.
Mark scheme
- B1 fermentation uses a renewable raw material, sugar from crops, while hydration uses ethene from crude oil, which is finite
- B1 fermentation is a slow batch process, while hydration is fast and continuous
- B1 fermentation gives dilute ethanol, about 15%, that must be concentrated by fractional distillation, while hydration gives nearly pure ethanol
- B1 fermentation needs only a low temperature and cheap equipment but has an atom economy of 51.1%, while hydration needs high temperature and pressure in costly plant but has an atom economy of 100%
Outline the mechanism for the acid-catalysed hydration of ethene to ethanol.
Mark scheme
- B1 a curly arrow from the C=C double bond to an H⁺ ion, giving the carbocation CH₃C⁺H₂
- B1 a curly arrow from a lone pair on the oxygen of a water molecule to the positively charged carbon
- B1 the protonated alcohol, CH₃CH₂OH₂⁺, with the positive charge on the oxygen
- B1 a curly arrow from an O–H bond to the oxygen, releasing H⁺ and regenerating the catalyst
Identify each of these alcohols as primary, secondary or tertiary: 2-methylpropan-1-ol, pentan-3-ol and 2-methylbutan-2-ol.
Mark scheme
- B1 2-methylpropan-1-ol is primary: the carbon bonded to the OH is bonded to one other carbon
- B1 pentan-3-ol is secondary: the carbon bonded to the OH is bonded to two other carbons
- B1 2-methylbutan-2-ol is tertiary: the carbon bonded to the OH is bonded to three other carbons
Give the conditions used to make ethanol by fermentation.
Mark scheme
- B1 yeast, whose enzymes catalyse the reaction, with glucose in aqueous solution
- B1 a temperature of 30 to 40 °C, typically about 35 °C
- B1 anaerobic conditions, with air excluded
Explain why fermentation is carried out at about 35 °C rather than at 20 °C or at 60 °C.
Mark scheme
- B1 at 20 °C the enzymes in the yeast work slowly, because fewer collisions have enough energy, so the rate is too low
- B1 at 60 °C the enzymes denature, their active sites changing shape, so the rate falls and the yeast dies
- B1 35 °C gives a reasonable rate without denaturing the enzymes
Give an equation, with state symbols, for the industrial hydration of ethene, and give the temperature and the catalyst used.
Mark scheme
- B1 CH₂=CH₂(g) + H₂O(g) ⇌ CH₃CH₂OH(g)
- B1 a temperature of about 300 °C
- B1 a phosphoric acid catalyst, H₃PO₄, on a silica support
Calculate the volume of carbon dioxide, at room temperature and pressure, released when 90.0 g of glucose is fully fermented. Mr of glucose is 180.0, and one mole of gas occupies 24.0 dm³ at room temperature and pressure.
Mark scheme
- M1 amount of glucose = 90.0 ÷ 180.0 = 0.500 mol
- M1 two moles of CO₂ form per mole of glucose, so 1.00 mol of CO₂
- A1 volume = 1.00 × 24.0 = 24.0 dm³, to three significant figures
Explain why a temperature of about 300 °C is used for the hydration of ethene, although the forward reaction is exothermic.
Mark scheme
- B1 the forward reaction is exothermic, so a lower temperature would shift the equilibrium to the right and give a higher equilibrium yield
- B1 but at a lower temperature the rate would be too slow, even with the catalyst
- B1 300 °C is a compromise that gives an acceptable yield at an acceptable rate
Explain why ethanol (Mr 46.0) boils at 78 °C while propane (Mr 44.0) boils at −42 °C.
Mark scheme
- B1 the O–H bond in ethanol is polar, because oxygen is much more electronegative than hydrogen
- B1 hydrogen bonds form between the δ+ H of one ethanol molecule and a lone pair on the O of another; reject any answer that breaks covalent bonds
- B1 propane has only van der Waals forces, of similar strength to ethanol's, so more energy is needed to separate ethanol molecules
Suggest the major product formed when propene reacts with steam in the presence of an acid catalyst, and explain your answer.
Mark scheme
- B1 the major product is propan-2-ol, CH₃CH(OH)CH₃
- B1 H⁺ adds to the CH₂ end of the double bond, giving the secondary carbocation CH₃C⁺HCH₃
- B1 the secondary carbocation is more stable than the primary one, having two electron-releasing alkyl groups, so it forms faster and water bonds to the middle carbon
Explain why sodium reacts less vigorously with ethanol than with water.
Mark scheme
- B1 in both reactions the O–H bond breaks and hydrogen is given off, ethanol forming the ethoxide ion, CH₃CH₂O⁻, and water the hydroxide ion
- B1 the ethyl group releases electron density towards the oxygen, so the negative charge on ethoxide is more concentrated than on hydroxide
- B1 ethoxide therefore attracts H⁺ more strongly, so ethanol releases H⁺ less readily: it is a weaker acid and reacts more slowly
Calculate the volume of hydrogen, at room temperature and pressure, given off when 0.230 g of sodium reacts with excess ethanol. Ar of Na is 23.0, and one mole of gas occupies 24.0 dm³ at room temperature and pressure.
Mark scheme
- M1 amount of sodium = 0.230 ÷ 23.0 = 0.0100 mol
- M1 2Na gives 1H₂, so 0.00500 mol of hydrogen forms
- A1 volume = 0.00500 × 24.0 = 0.120 dm³ (120 cm³), to three significant figures
Calculate the mass of glucose that must be fermented to make 2.30 tonnes of ethanol, assuming all the glucose is fermented. Mr: glucose 180.0, ethanol 46.0.
Mark scheme
- M1 amount of ethanol = 2.30 × 10⁶ g ÷ 46.0 = 5.00 × 10⁴ mol
- M1 each glucose gives two ethanol, so 2.50 × 10⁴ mol of glucose is needed
- A1 mass = 2.50 × 10⁴ × 180.0 = 4.50 × 10⁶ g, 4.50 tonnes, to three significant figures
Show that the percentage atom economy for making ethanol by the fermentation of glucose is 51.1%. Mr: glucose 180.0, ethanol 46.0, carbon dioxide 44.0.
Mark scheme
- M1 C₆H₁₂O₆(aq) → 2CH₃CH₂OH(aq) + 2CO₂(g), so one glucose gives two ethanol and two carbon dioxide
- M1 atom economy = mass of desired product ÷ total mass of reactants × 100 = (2 × 46.0) ÷ 180.0 × 100
- A1 atom economy = 92.0 ÷ 180.0 × 100 = 51.1%, the other 48.9% leaving as carbon dioxide, 2 × 44.0 = 88.0
State what is meant by a secondary alcohol, and give the name of a secondary alcohol with three carbon atoms.
Mark scheme
- B1 an alcohol in which the carbon atom bonded to the OH group is bonded to two other carbon atoms
- B1 the three-carbon secondary alcohol is propan-2-ol, CH₃CH(OH)CH₃
Explain why air is kept out of a fermentation vessel.
Mark scheme
- B1 with oxygen present, yeast respires aerobically, converting glucose to carbon dioxide and water instead of ethanol
- B1 oxygen also lets bacteria oxidise the ethanol to ethanoic acid, which lowers the yield
Explain why the solubility of alcohols in water decreases as the carbon chain gets longer.
Mark scheme
- B1 the OH group forms hydrogen bonds with water molecules, but the hydrocarbon chain cannot
- B1 a longer chain disrupts more of the hydrogen bonding between water molecules while forming only weak van der Waals forces with water, so less alcohol dissolves
Suggest why Brazil makes most of its fuel ethanol by fermentation rather than by the hydration of ethene.
Mark scheme
- B1 Brazil has the land and climate to grow sugar cane, a cheap and renewable source of sugar
- B1 fermentation does not use crude oil, so the fuel supply does not depend on oil prices or imports
Suggest why small scraps of waste sodium are destroyed in the laboratory by adding them to ethanol rather than to water.
Mark scheme
- B1 ethanol is a weaker acid than water, so it reacts with sodium more slowly and steadily
- B1 heat is released more slowly, so the sodium does not melt and the hydrogen does not ignite
State what is meant by the term biofuel.
Mark scheme
- B1 a fuel produced from renewable biological material, such as crops
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