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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 po... 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
03.1 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 from the reaction mixture during distillation is based on the differences in boiling points of the compounds involved. Propan-1-ol, which has hydrogen bonding due to its hydroxyl (-OH) group, exhibits stronger intermolecular forces compared to propanal, which primarily experiences dipole-dipole interactions. As a result, propanal has a lower boiling point (49 °C) and can be vaporized and condensed back to liquid form more readily than propan-1-ol. Therefore, maintaining the temperature below 97 °C allows propanal to be distilled out while leaving the more strongly bonded propan-1-ol in the flask.
Step 2
03.2 Give two ways of maximising the yield of propanal obtained by distillation of the reaction mixture.
Answer
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Maintain a constant temperature close to the boiling point of propanal (49 °C) to ensure efficient condensation without allowing higher boiling compounds to vaporize.
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Use a fractionating column during distillation to increase the separation efficiency, allowing for repeated condensation and re-vaporization, which can enhance the purity and yield of propanal.
Step 3
03.3 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 in the distilled sample, add a few drops of sodium bicarbonate solution to a test tube containing the propanal sample. If effervescence (bubbling) is observed, this indicates the presence of a carboxylic acid, such as propanoic acid, since it would react with sodium bicarbonate to produce carbon dioxide. The lack of fizzing implies that propanoic acid is not present.
Step 4
03.4 Calculate a value, in kJ mol⁻¹, for the enthalpy of combustion of ethanol in this experiment.
Answer
To calculate the enthalpy of combustion, use the formula:
where:
- = mass of water (150 g)
- = specific heat capacity of water (4.18 J K⁻¹ g⁻¹)
- = change in temperature (40.2 °C - 25.1 °C = 15.1 °C)
Substituting the values:
Converting this to kJ:
Now, to find the enthalpy change per mole of ethanol burned:
- Calculate the number of moles of ethanol from its mass:
- Molar mass of ethanol (C₂H₅OH) = 46 g/mol;
- Moles of ethanol burned = .
- The enthalpy change is therefore:
Thus, reporting with appropriate significant figures, we find:
Step 5
03.5 Name and outline a mechanism for the reaction producing pent-1-ene.
Answer
Name of mechanism: Elimination (E1)
Mechanism: In the dehydration of pentan-2-ol, the hydroxyl group (-OH) is protonated by sulfuric acid, forming a better leaving group (water). This leads to the formation of a carbocation intermediate. Subsequently, elimination of a proton occurs, resulting in the formation of the double bond, yielding pent-1-ene. The key steps are:
- Protonation of the alcohol group.
- Loss of water leading to carbocation formation.
- Elimination of a proton to form the alkene.
Step 6
03.6 Name the less polar stereoisomer formed.
Answer
Name: E-pent-2-ene.
Explanation: The less polar stereoisomer arises from the relative orientation of substituents across the double bond. In the E-isomer, substituents are on opposite sides, resulting in less steric hindrance and, consequently, a lower dipole moment compared to the Z-isomer where the substituents are on the same side.