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The oxidation of propan-1-ol can form propanal and propanoic acid - AQA - A-Level Chemistry - Question 3 - 2018 - Paper 2
Question 3
The oxidation of propan-1-ol can form propanal and propanoic acid. The boiling points of these compounds are shown in Table 1. Table 1 | Compound | Boiling poi... show full transcript
Worked Solution & Example Answer:The oxidation of propan-1-ol can form propanal and propanoic acid - AQA - A-Level Chemistry - Question 3 - 2018 - Paper 2
Step 1
Explain, with reference to intermolecular forces, why distillation allows propanal to be separated from the other organic compounds in this reaction mixture.
Answer
Distillation is effective due to the differences in boiling points among the compounds. Propan-1-ol and propanoic acid have stronger hydrogen bonds due to their hydroxyl and carboxylic groups, which lead to higher boiling points compared to propanal. Consequently, when the mixture is heated, propanal, having the lowest boiling point at 49 °C, evaporates first, allowing it to be separated from the other components based on the principle that the vapor of a component can be condensed back into liquid form while other components remain in the liquid phase due to their higher boiling points.
Step 2
Give two ways of maximising the yield of propanal obtained by distillation of the reaction mixture.
Answer
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Maintain the temperature of the reaction mixture below the boiling point of propan-1-ol (below 97 °C) to prevent its evaporation.
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Collect the distillate at low pressure to facilitate the evaporation of propanal while minimizing the loss of other components.
Step 3
Describe how you would carry out a simple test-tube reaction to confirm that the sample of propanal obtained by distillation does not contain any propanoic acid.
Answer
To confirm the absence of propanoic acid, add a small amount of sodium bicarbonate to the propanal sample in a test tube. If effervescence is observed, this indicates the presence of an acid, confirming that propanoic acid is still present. If no bubbles form, it suggests that the sample is free from propanoic acid.
Step 4
Calculate a value, in kJ mol⁻¹, for the enthalpy of combustion of ethanol in this experiment.
Answer
Using the formula:
where:
- = heat absorbed by water
- = mass of water (150 g)
- = specific heat capacity of water (4.18 J/g °C)
- = temperature change (40.2 °C - 25.1 °C = 15.1 °C)
Calculating the heat:
Next, convert the mass of ethanol burned to moles:
ext{Molar mass of ethanol} = 46.07 ext{ g/mol} \ n = rac{457 ext{ mg}}{1000 imes 46.07 ext{ g/mol}} = 0.00990 ext{ mol} \
Now calculate the enthalpy of combustion:
ext{Enthalpy of combustion} = rac{9.4773 ext{ kJ}}{0.00990 ext{ mol}} = 955.0 ext{ kJ/mol}
Thus, the enthalpy of combustion of ethanol is approximately 955 kJ/mol.
Step 5
Name and outline a mechanism for the reaction producing pent-1-ene.
Answer
Name of mechanism: Elimination (E1 or E2)
In the presence of sulfuric acid, the hydroxyl (-OH) group of pentan-2-ol is protonated, making it a better leaving group. This facilitates the subsequent loss of water (H₂O), leading to the formation of a carbocation. In the case of E1 mechanism, the loss of water occurs first, followed by deprotonation to form the alkene. The E2 mechanism involves simultaneous loss of H₂O and a proton to form the double bond directly, hence producing pent-1-ene.
Step 6
Name the less polar stereoisomer formed.
Answer
Name: E-pent-2-ene
Explanation: The less polar stereoisomer arises due to the arrangement of substituents around the double bond. The E-isomer has more bulky groups on opposite sides of the double bond, reducing dipole interactions and hence showing less polarity compared to the Z-isomer where bulkier groups are on the same side.