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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 p... show full transcript
Step 1
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
Answer
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.
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
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
Answer
First, calculate the temperature change of water:
Next, calculate the heat absorbed by water using the formula:
Convert to kJ:
The number of moles of ethanol combusted:
Now, calculate the enthalpy of combustion per mole:
Thus, the enthalpy of combustion of ethanol is approximately 95.7 kJ mol⁻¹.
Step 5
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
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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