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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 ... 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
3.1 Explain, with reference to intermolecular forces, why distillation allows propanal to be separated from the other organic compounds in this reaction mixture.
Answer
The separation of propanal via distillation is influenced by intermolecular forces. Propan-1-ol has stronger hydrogen bonding due to the hydroxyl (–OH) group, while propanal, being an aldehyde, has weaker dipole-dipole interactions. This means that propanol will have a higher boiling point than propanal. Distillation exploits these differences by heating the mixture below the boiling point of propan-1-ol (97 °C) and collecting propanal (boiling point 49 °C) as it vaporizes first.
Step 2
3.2 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 (keep it below 97 °C) to prevent the co-distillation of unwanted compounds.
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Use a distillation apparatus with adequate cooling to ensure efficient condensation of the vaporized propanal, reducing losses.
Step 3
3.3 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, I would add a few drops of the distilled sample to a test tube and introduce a few drops of sodium bicarbonate solution. If effervescence is observed, it indicates the presence of carboxylic acid (propanoic acid). If no bubbles form, then the sample is confirmed to be free of propanoic acid.
Step 4
3.4 Calculate a value, in kJ mol⁻¹, for the enthalpy of combustion of ethanol in this experiment.
Answer
To calculate the enthalpy of combustion, we use the formula:
o ext{q} = m imes c imes riangle T
Where:
- m = mass of water = 150 g
- c = specific heat capacity = 4.18 J K⁻¹ g⁻¹
- riangle T = 40.2 - 25.1 = 15.1 °C
Calculating q:
o ext{q} = 150 imes 4.18 imes 15.1 = 9427.5 ext{ J}
Converting to kJ = 9.4275 kJ. Now, since this amount of heat was produced by 457 mg of ethanol:
o ext{number of moles of ethanol} = rac{457 ext{ mg}}{46.07 ext{ g mol}^{-1}} = 0.00990 ext{ mol}
Finally, the enthalpy of combustion per mole = rac{9.4275 ext{ kJ}}{0.00990 ext{ mol}} = 952.9 ext{ kJ mol}^{-1}.
Step 5
3.5 Name and outline a mechanism for the reaction producing pent-1-ene.
Answer
Name of mechanism: Elimination. The mechanism involves the loss of water (H₂O) from pentan-2-ol, resulting in the formation of a double bond. The steps include:
- Protonation of the alcohol group by sulfuric acid.
- Loss of a water molecule (dehydration), resulting in a carbocation intermediate.
- Simultaneous loss of a proton, leading to the formation of pent-1-ene.
Step 6
3.6 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. E-isomers have higher molecular weight groups on opposite sides, resulting in a less dipole moment compared to Z-isomers where similar groups are on the same side.