Assigning C3H8O Isomers from Boiling Point and Solubility Data
A student is given three unlabeled molecular models, A, B, and C, all with molecular formula C3H8O. The models represent propan-1-ol, propan-2-ol, and methoxyethane. Measured boiling points are A = 97 °C, B = 82 °C, and C = 11 °C. A and B mix completely with water; C dissolves only slightly. None of the liquids conducts electricity. The student claims that C must have the highest vapor pressure because it has the lowest boiling point and that A and B differ only by branching, so their boiling points should be nearly identical. The teacher asks the student to assign a structure to each model, explain the boiling-point pattern using intermolecular forces, and evaluate the vapor-pressure claim. The student must also predict which liquid would evaporate fastest from an open watch glass at 20 °C and justify the prediction with molecular-level reasoning.
1. Assign each model to a structure and explain the boiling-point order A > B > C using intermolecular forces. Include why A and B are both water-miscible but C is not.
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Boiling point reflects the strength of intermolecular forces, not molar mass, because all three are C3H8O. A = propan-1-ol, B = propan-2-ol, C = methoxyethane. A and B both have an O–H group, so they form hydrogen bonds; C has no O–H and cannot donate hydrogen bonds, so it has only dipole-dipole attractions and London dispersion forces. That explains C’s much lower boiling point (11 °C). Between A and B, both are alcohols, but propan-1-ol is more linear and has greater surface contact, so its London dispersion forces are stronger; propan-2-ol is more branched and has weaker dispersion, giving the lower boiling point (82 °C vs 97 °C). A and B are water-miscible because their O–H groups donate and accept hydrogen bonds with water; C is only slightly soluble because it can accept but not donate hydrogen bonds.
2. Evaluate the student’s vapor-pressure claim and predict which liquid evaporates fastest from an open watch glass at 20 °C. Justify using molecular-level reasoning.
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The student is right that C has the highest vapor pressure, but the reasoning should be molecular: C has the weakest intermolecular forces because it lacks O–H hydrogen bonding, so more molecules escape into the vapor phase at a given temperature. Its low boiling point is a consequence of that weak attraction, not the cause. At 20 °C, C will evaporate fastest from an open watch glass because it has the highest vapor pressure; A and B have much lower vapor pressures due to hydrogen bonding. A and B are not nearly identical: B (propan-2-ol) has a higher vapor pressure and evaporates somewhat faster than A because its branched structure reduces dispersion forces. Equal surface area and equal volume would be needed for a fair evaporation comparison.