Organic Chemistry: Concepts and Applications - Headley Allan D. 2020

Bonding and Structure of Organic Compounds
1.8 Intermolecular Molecular Interactions — Concept Summary and Applications

The types of intermolecular interactions discussed in the previous section, especially hydrogen bonds, drastically alter the expected physical properties of some compounds. The physical properties of any compound are measurable and observable characteristics of that compound. Sets of characteristics are unique for a specific compound, and these characteristic properties are often used to identify unknown compounds. The boiling and melting points of compounds are measurable properties and that a liquid is colorless is an observable property. Knowledge of the physical properties of compounds can give an indication of certain structural features of the molecule. For example, ethanol has a boiling point of 78.5 °C, whereas dimethylether has a boiling point of −23.6 °C, even though they both have the same molecular formula and molecular weight and the same type of atoms.

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Figure 1.34 Graphical illustration of the attraction between two nonpolar molecules (a) and illustration of these attractions of pentane (b).

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The boiling point of a liquid is the temperature at which the vapor phase of the liquid is the same as that of the atmospheric pressure. At that temperature, the molecules move freely from the liquid phase to the vapor phase. Since the two compounds mentioned above have the same molecular mass, it is expected that the same amount of energy would be required to vaporize both, and hence they should have the same boiling point. The boiling point of a liquid has a direct relationship with possible intermolecular attractions and the types of attractive forces that exist. If the intermolecular attractions for one of these compounds are very strong, the individual molecules are not as free to escape from the liquid phase to the vapor phase, and thus more energy must be supplied to overcome these intermolecular attractions before boiling of that liquid can take place. This increase in energy input results in higher boiling points for liquids that have strong intermolecular attractions. The type of intermolecular attractions that are encountered in ethanol is hydrogen bonding, and this type of interaction is not possible for dimethylether since all the hydrogens are bonded to carbons and the partial positive charge is not as great, compared to those of ethanol.

Problem 1.23

Butaneamine (CH3CH2CH2NH2) has approximately the same molecular weight as propanol (CH3CH2CH2OH), yet the boiling point of butaneamine is 48 °C and that of propanol is 97 °C. Explain this difference in boiling points.

London forces, even though much weaker than hydrogen bonding and dipole—dipole attractions, also affect the expected boiling points of liquids. For example, the boiling points of most alkanes are usually very low, compared to other organic compounds of similar molecular weights that contain heteroatoms. In addition, the boiling points of highly branched alkanes are typically lower than alkanes of similar molecular weights that are not highly branched as shown in Figure 1.35.

Since hydrocarbons are nonpolar molecules, the type of intermolecular attractions that are expected are London forces. For branched hydrocarbons, there is less surface area and hence the attractive forces that are created will be less than those of straight chain hydrocarbons that have greater surface areas as shown in Figure 1.34. The boiling point of pentane is 36.0 °C, whereas the boiling point of 2,2-dimethylpropane is 4.4 °C, a compound that has the same molecular formula, but a lower boiling point. Intermolecular attractions can be used to explain this observation. For branched alkanes, the electrons are close to each other, compared to a straight chain alkane, where the electrons are more dispersed throughout the molecule. As a result, the surface area of a branched alkane is less than that of a straight chain alkane. Thus, the electrons of a branched alkane are less polarizable; whereas for the less branched alkanes, the electrons are more polarizable. Due to the increased Van der Waals attractions for the more polarizable molecules, straight chain alkanes have higher boiling points than highly branched alkanes. 2,2-Dimethylpropane has less surface area and more compact and hence a lower boiling point than pentane, which has a greater surface area and is less compact. In general, isomers/molecules with larger surface areas have higher boiling points than those with less surface area. Since alkanes are nonpolar molecules, the attractions between the different molecules are minimal, resulting in low boiling points in general, compared to more polar molecules.

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Figure 1.35 Boiling points of different alkanes with the same molecular weight and same kinds of atoms (carbon and hydrogen atoms).

Problem 1.24

Arrange the following compounds in the order of boiling points, i.e. the lowest boiling liquid first:

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The solubility of compounds in different solvents is also dictated to a large extent by the intermolecular attractions that are created between the solute and solvent molecules. As a result, different compounds have different solubilities in water, which is a polar solvent. Water has two hydrogens bonded to the very electronegative oxygen. Solvents that have such hydrogens carry a partial positive charge on the hydrogens and these solvents are called polar-protic solvents. Methanol (CH3OH) and ethanol (CH3CH2OH) are examples of polar-protic solvents. Based on potential intermolecular interactions, it is possible to predict the relative solubilities of different solutes in water. Glucose (structure shown below) is very soluble in water, whereas cyclohexane (structure shown below), which has a similar appearance, is not soluble in water. In a solution of glucose and water, intermolecular attractions via hydrogen bonds are possible, whereas for cyclohexane, all the hydrogens are bonded to carbons and a similar type of solvent solute intermolecular attraction as that for glucose—water does not exist.

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Problem 1.25

Draw structures to demonstrate the hydrogen bonds that are possible with water and glucose (structure shown above).