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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

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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 p... 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.

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Answer

The separation of propanal from the reaction mixture via distillation is attributed to the differences in boiling points, which are influenced by intermolecular forces. Propan-1-ol and propanoic acid exhibit stronger hydrogen bonding due to their hydroxyl groups, resulting in higher boiling points (97 °C and 141 °C respectively). In contrast, propanal has a lower boiling point of 49 °C, indicating weaker dipole-dipole interactions. During distillation, the reaction mixture is heated, and when the temperature reaches 49 °C, propanal vaporizes while the other compounds remain in the liquid state, facilitating its collection as a distillate.

Step 2

Give two ways of maximising the yield of propanal obtained by distillation of the reaction mixture.

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Answer

  1. Maintain the temperature of the reaction mixture just below the boiling point of propanal (below 49 °C) to prevent the loss of propanal through evaporation.

  2. Use a more efficient distillation apparatus, such as a fractionating column, to allow for better separation of the components based on their boiling points.

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.

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Answer

To confirm the absence of propanoic acid, add a small amount of sodium bicarbonate (NaHCO₃) to the test tube containing the distilled propanal. If effervescence (bubbling) occurs, it indicates the presence of an acidic compound (like propanoic acid). In absence of reaction, it suggests that propanoic acid is not present.

Step 4

Calculate a value, in kJ mol⁻¹, for the enthalpy of combustion of ethanol in this experiment.

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Answer

First, calculate the temperature change of water:

ΔT=40.2°C25.1°C=15.1°C\Delta T = 40.2 °C - 25.1 °C = 15.1 °C

Next, calculate the heat absorbed by water using the formula:

q=mass×specific_heat_capacity×ΔTq = mass \times specific\_heat\_capacity \times \Delta T
q=150g×4.18JK1g1×15.1K=947.7Jq = 150 g \times 4.18 J K^{-1} g^{-1} \times 15.1 K = 947.7 J

Convert to kJ:
q=0.9477kJq = 0.9477 kJ

The number of moles of ethanol combusted:

Molar mass of ethanol (C2H5OH) = 46.07 g mol1\text{Molar mass of ethanol (C}_2\text{H}_5\text{OH) = 46.07 g mol}^{-1}
moles=457mg46.07gmol1×1000mg/g=0.0099mol\text{moles} = \frac{457 mg}{46.07 g mol^{-1} \times 1000 mg/g} = 0.0099 mol

Now, calculate the enthalpy of combustion per mole:

ΔHc=0.9477kJ0.0099mol=95.7kJmol1\Delta H_c = \frac{0.9477 kJ}{0.0099 mol} = 95.7 kJ mol^{-1}

Thus, the enthalpy of combustion of ethanol is approximately 95.7 kJ mol⁻¹.

Step 5

Name and outline a mechanism for the reaction producing pent-1-ene.

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Answer

Name of mechanism: Elimination

Mechanism: The reaction involves the elimination of water from pentan-2-ol to form pent-1-ene. The protonation of the hydroxyl group occurs first, forming a better leaving group. Then, a base abstracts a proton from a β-carbon, resulting in the formation of a double bond between the α-carbon and β-carbon as water is eliminated, leading to the formation of pent-1-ene.

Step 6

Name the less polar stereoisomer formed.

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Answer

Name: E-pent-2-ene.

Explanation: This stereoisomer is less polar because it has larger groups positioned across from each other in the double bond, reducing the overall dipole moment compared to the Z-isomer where similar groups are on the same side.

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