The oxidation of propan-1-ol can form propanal and propanoic acid - AQA - A-Level Chemistry - Question 3 - 2018 - Paper 2

1. Maintain the temperature of the distillation apparatus below the boiling point of propan-1-ol (97 °C) to prevent its vaporization, ensuring predominantly propanal condenses.

2. Use a efficient distillation apparatus such as a fractionating column to enhance the separation of propanal from higher boiling components.

To confirm the absence of propanoic acid, add a small amount of the distilled sample to a test tube followed by a few drops of sodium bicarbonate solution. If effervescence (bubbling) occurs, it indicates the presence of an acid such as propanoic acid, as it will react to produce carbon dioxide. If no bubbles are observed, it suggests that the sample is free from propanoic acid.

First, calculate the amount of heat absorbed by the water:

$q = mc\Delta T$

where:

• $m = 150 \text{ g}$ (mass of water)
• $c = 4.18 \text{ J g}^{-1} \text{ K}^{-1}$ (specific heat capacity of water)
• $\Delta T = 40.2 \text{ °C} - 25.1 \text{ °C} = 15.1 \text{ °C}$

Thus,

$q = 150 \times 4.18 \times 15.1 = 9477 \text{ J} = 9.477 \text{ kJ}$

Now, calculate the number of moles of ethanol burned:

$n = \frac{457 \text{ mg}}{1000} \times \frac{1 \text{ g}}{1 \text{ mg}} \times \frac{1 \text{ mol}}{46.07 \text{ g}} = 0.00990 \text{ mol}$

The enthalpy change per mole of ethanol:

$\text{Enthalpy of combustion} = -\frac{q}{n} = -\frac{9.477 \text{ kJ}}{0.00990 \text{ mol}} = -955.5 \text{ kJ mol}^{-1}$

Rounding to three significant figures gives:

$-956 \text{ kJ mol}^{-1}$

Name of mechanism: ( E1 ) or ( E2 ) elimination.

Outline: The dehydration of pentan-2-ol involves two main steps.

1. Protonation of the alcohol to form a better leaving group (water).
2. Loss of a β-hydrogen and the leaving group (water) to form the alkene (pent-1-ene). The mechanism results in the formation of the double bond between the first and second carbons.

The less polar stereoisomer formed is likely to be the trans isomer.

Explanation: Trans stereoisomers have less steric hindrance and are typically less polar than their cis counterparts due to the arrangement of substituent groups across the double bond, which allows for a more linear structure and a reduction in dipole moments.